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I would like to ask if anyone could share any specific experiences of discovering nonequivalent definitions in their field of mathematical research.

By that I mean discovering that in different places in the literature, the same name is used for two different mathematical objects. This can happen when the mathematical literature grows quickly and becomes chaotic and of course this could be a source of serious errors. I have heard of such complaints by colleagues, mostly with respect to definitions of various spaces and operators, but I do not recall the specific (and very specialised) examples.

My reason for asking is that I'm currently experimenting with verification using proof assistants and I'd like to test some cases that might be a source of future errors.

EDIT: Obviously, I would be interested to see as many instances of this issue as possible so I would like to ask for more answers. Also, it would be even more helpful if you could provide references.

Ben McKay
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Angeliki Koutsoukou Argyraki
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    A classic example is "compact", which can mean either "Hausdorff and every open cover has a finite subcover", or just "every open cover has a finite subcover". – Arturo Magidin Nov 22 '17 at 19:12
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    closed manifold has different meaning than closed set – Piyush Grover Nov 22 '17 at 19:14
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    Closely related, may be a duplicate: https://mathoverflow.net/questions/7389/what-are-the-most-overloaded-words-in-mathematics – Qiaochu Yuan Nov 22 '17 at 19:17
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    I was surprised to recently find the term "semi-local" being used for a commutative ring to mean having exactly one maximal ideal and "local" to mean a "semi-local" ring that is also Noetherian. This struck me as very odd, but perhaps has historical roots? – M.G. Nov 22 '17 at 19:20
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    @QiaochuYuan I don't mean this in the sense of the question in the link that you provide, what you mention has to do with common adjectives used to describe different mathematical objects in a different sense, I mean this in the sense of Arturo Magidin 's answer above i.e. calling a space X if some condition A holds, and also calling a space X if some condition B different than A holds. – Angeliki Koutsoukou Argyraki Nov 22 '17 at 19:31
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    @ArturoMagidin This leads to the classic dialogue "The space $X$ is quasicompact, so..." "What does quasicompact mean?" "It means compact." – Will Sawin Nov 22 '17 at 19:35
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    @Guest: okay, that's why I didn't want to say "is a duplicate." That's an important clarification over "the same name is used for two different mathematical objects." An example that I've bumped into a lot is "semisimple": some people use it to mean a ring with trivial Jacobson radical (what I would call semiprimitive), and some use it to mean a ring whose representation theory is semisimple, or equivalently a ring with trivial Jacobson radical which is also artinian. – Qiaochu Yuan Nov 22 '17 at 19:37
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    There are several different non-equivalent definitions of $p$-adic multiple zeta values, all of which have appeared in the literature with the same notation $\zeta_p(s_1,\ldots,s_k)$. – Julian Rosen Nov 22 '17 at 20:01
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    In my experience at least, the only people who think "compact" includes Hausdorff are algebraic geometers. I've never heard a topologist use "compact" to mean anything other than "every open cover has a finite subcover". – John Pardon Nov 23 '17 at 03:29
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    One can often avoid these ambiguities: for example, by never speaking of "compactness", but only of "quasicompactness" or "quasicompact and Hausdorff", never of "positive", but only of "non-negative" or "strictly positive", etc. Fortunately, nobody does that - it makes for rather inelegant language. – Lukas Lewark Nov 23 '17 at 05:46
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    @JohnPardon: there is also a cultural thing here. In France, it is standard to include "Hausdorff" in the definition of a compact space. – Taladris Nov 23 '17 at 14:49
  • Do you mean definitions that contradict? Or definitions that are simply different? One of the earliest examples in my mathematical studies I remember is the constant $e$. Historically, $e=\lim_{n\to\infty} (1+\frac{1}{n})^n$. First and, if you ask me, only true definition. Sometimes you see definitions like $\frac{d}{dx}e^x=e^x$ but to me this is a result stemming from prior known properties. These may be equivalent in the sense once can be derived from the other - proven theorems - but no mathematical concept can have more than one definition. At all. Whatsoever. Thats just absurd. – CogitoErgoCogitoSum Nov 23 '17 at 22:26
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    @CogitoErgoCogitoSum: Then you can think of this question as asking for distinct (inequivalent) concepts that are given the same name. – Nate Eldredge Nov 28 '17 at 00:25
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    I seem to remember reading in an essay by Kohlhase that some MKM project ran into surprising subtleties coming from the fact that different function tables use different branch cuts to define inverse trigonometric functions, and that each is serenely confident that its own is the universal convention. – LSpice Nov 28 '17 at 23:19
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    @Taladris, maybe in france, everyone's an algebraic geometer – Vivek Shende Nov 29 '17 at 00:54
  • @VivekShende: not everyone. But it is possible that some important algebraic geometers influenced the community. – Taladris Nov 29 '17 at 01:25
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    @JohnPardon I have a book on topological vector spaces which assumes compact means Hausdorff. Interestingly, the author is French, which supports other commenters assertions that such a convention is common in France. – silvascientist Nov 30 '17 at 21:43
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    @Taladris: in my opinion, quasicompact is French for compact, whereas compact is French for compact and Hausdorff. This makes it a simple matter of translation, rather than a cultural issue. – R. van Dobben de Bruyn Oct 16 '18 at 18:14
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    In algebraic geometry (the way I think about it), a scheme is quasicompact if its underlying topological space is compact. The notion is ubiquitous, but the word compact would give the wrong intuition. Although one usually does not call proper varieties compact, we do speak of compactification rather than properification. – R. van Dobben de Bruyn Oct 16 '18 at 18:18
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    Also many people, beyond algebraic geometry and beyond France, assume that "locally compact group" implicitly implies means Hausdorff. Including by English-speaking people, who don't assume Hausdorff in compact space. – YCor Jan 02 '19 at 09:57
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    I think it would be more interesting to ask a question "Definitions that slowed down the development of mathematics". I would nominate the definition of topology on a dual topological vector space as the topology of uniform convergence on bounded sets. – Sergei Akbarov May 25 '19 at 09:33
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    @CogitoErgoCogitoSum On the other hand, one can define the exponential function $\exp(x)$ as common value $\lim_{n\to\infty}(1+x/n)^n=\sum_{k=0}^\infty x^k/k!$ and prove its properties, without using nor mentioning the constant $e$. So it is not that crazy defining the constant $e$ simultaneously as $e=\exp(1)= \lim_{n\to\infty}(1+1/n)^n=\sum_{k=0}^\infty1/k!$. After all, what is of fundamental importance is the exponential function, rather than the constant. – Pietro Majer Jun 12 '19 at 10:34
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    I think words that end up having a precise modern definition, but also have a vernacular meaning in mathematical writing, might be a more serious problem. Ex1: domain as the "source" of a set-theoretic map, or an open connected subset of $\mathbb C$; Ex2: function as just a set-theoretic map, or more specifically one from set of numbers to set of numbers, or in expressions like "some formula/expression is a function of x". – liuyao Jul 06 '19 at 05:06
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    @JohnPardon In the olden days topologists used "compact" to mean that every sequence had a convergent subsequence, and compact spaces as we now know them were called "bicompact" (e.g. in Tychonoff's paper with his eponymous theorem). Of course, it was eventually realized that "bicompactness" was the more important notion, so the terminology changed. – Robert Furber Dec 21 '19 at 22:39
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    Let $F(x,y)=x^2+y^2$. What is $F(r,\theta)$? To a mathematician, it is $r^2+\theta^2$. To a physicist, it is $r^2$. – Michael Renardy Sep 09 '23 at 21:20

71 Answers71

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Perhaps the mother of all examples is "natural number". You can start an internet flame war by asking whether zero is a natural number.

Gerry Myerson
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    Of course, $0$ is not a natural number, otherwise $\mathbb{N}_0$ would be a waste of a perfectly good notation :P – M.G. Nov 22 '17 at 21:13
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    I failed to get full score on a final because I assumed 0 was a natural number... – Per Alexandersson Nov 22 '17 at 21:14
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    Perhaps not the mother of all examples, but certainly the mother of lots of other examples. Excluding 0 from the natural numbers is closely related to excluding trivial cases from many other definitions, e.g. "an X is a nonempty set together with ...", which often leads to other examples of this sort when not everyone excludes it. – Mike Shulman Nov 22 '17 at 22:14
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    @July, so is $\mathbb{N}^+$ not a waste of notation? :P – Peter Taylor Nov 23 '17 at 11:20
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    @PeterTaylor That would most likely be understood as "the additive semigroup of natural numbers". Although $\mathbb{N}^*$ is likely to be evaluated to ${1}$ and not to the multiplicative monoid of all natural numbers. – Andrei Smolensky Nov 23 '17 at 11:53
  • @PeterTaylor: $+$ is an algebraic operation, so unless you mean the set of natural numbers equipped with the operation addition, this is a bad notation :P Like the negative numbers or the empty set, $0$ is a "made-up" object, say for practical purposes, hence not really natural. – M.G. Nov 23 '17 at 11:54
  • @MikeShulman: case in point: https://mathoverflow.net/questions/286732/nonequivalent-definitions-in-mathematics/286758?noredirect=1#comment709029_286758 – M.G. Nov 23 '17 at 13:42
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    @July In France it is actually common to write $\mathbb N$ for ${0,1,2,\ldots}$ ! – Adrien Hardy Nov 23 '17 at 14:49
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    I always told my students that the natural numbers were what you used to tell the number of grains of sand in a bowl. – Lubin Nov 23 '17 at 15:28
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    I didn't say this because I didn't want to start the flame war, but I think a better response to @July is "clearly $\pi$ is then not a real number, since otherwise $\mathbb{R}_\pi$ would be a waste of a perfectly good notation". (-: Category theorists almost universally include 0 as a natural number, whatever their nationality; what could be more "natural" than the cardinality of the initial object in the category of sets? – Mike Shulman Nov 23 '17 at 16:00
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    @MikeShulman: replacing $\mathbb{N}$ by $\mathbb{R}$ and $0$ by $\pi$ in my comment is not a good response because it suggests a false comparison and is missing the point of my comment, which was a joking way to point out a most simple notation that makes both sides of the debate happy. Your arguments support why one should consider the set $\mathbb{N}_0$ instead of $\mathbb{N}$ most of the time (and I don't object to that in any way), but not why $0$ should be called natural in the sense $1,2,3...$ are called natural. There is an unfortunate overload of meanings of "natural" here. Cont. – M.G. Nov 23 '17 at 16:27
  • cont. $0$ is a logical construction just like $\emptyset$ to indicate absence, which itself is a tad more of an abstract notion than counting your apples. – M.G. Nov 23 '17 at 16:30
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    In other words, while introducing $0$ is a natural thing to do from mathematical POV, $0$ is not a natural object in the same way $1,2,3...$ are, they are much more basic. – M.G. Nov 23 '17 at 16:40
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    @July this will be my last comment on the subject. The notation $\mathbb{N}_0$ does not make me happy at all; it is far too cumbersome for such a basic object. If the set including zero is a more important mathematical object, then it deserves the simpler notation. The word "natural", when used in mathematics, should refer to naturalness in mathematics; there's no reason a historical accident of terminology should guide the choice of appropriate notation; and I dispute that there's anything less natural about "the number of elephants in this room" than "the number of fingers on my hand". – Mike Shulman Nov 23 '17 at 17:44
  • @MikeShulman: too bad, I liked addressing the elephants in your room :-) I disagree that $\mathbb{N}_0$ is more important than $\mathbb{N}$: both occur often enough throughout mathematics to deserve a right to equal treatment. I fail to see how $\mathbb{N}_0$ is such a cumbersome notation, but I guess it is truly subjective. The name "natural" may be of historical origin, but I wouldn't call it accidental: certainly, there are also philosophical and meta-mathematical considerations. "0 number of elephants" is just a colloquial/linguistic convenience. I hope you clean up after the elephants :-) – M.G. Nov 23 '17 at 18:18
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    @July: "Linguistic convenience" is precisely the sense in which I would agree its accidental. Natural language optimizes to distinguish "zero / one / more than one", and its unfortunate that the distinction in natural language leads people to think the distinction is more profound! At least people seem to universally accept one as a number these days. –  Nov 23 '17 at 18:42
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    @July: Incidentally, when I see $\mathbb{N}_0$, my first instinct is to read that as "the natural numbers, excluding zero", since I've seen the $0$ subscript used that way in other cases. –  Nov 23 '17 at 18:48
  • @Hurkyl: perhaps not profound, but there is at least one important difference between $0$ and $1,2,3...$ that is not of purely linguistic nature: while the former indicates absence or non-existence, all members of the latter indicate the opposite. In that regard, $0$ is much closer to the negative integers (no pun intended). PS: This is not a typical first instinct since an integer in a lower subscript to a set of integers often stands for the starting integer, i.e. cutting off the said sequence of integers, this is not uncommon. – M.G. Nov 23 '17 at 18:55
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    @July: I assert that's just the "zero / not zero" distinction from natural language speaking. There is actually important mathematical example where using 0 to quantify something is very pointedly not a case of absence or nonexistence -- the notion of a set having measure zero is very different from the notion of a set not having measure! I feel like the general case is similar; that 0 is a clear and accurate quantification of something, and its just natural language that leads people to reject the idea that the thing can be quantified. –  Nov 23 '17 at 21:01
  • @Hurkyl: this distinction is not just the natural language speaking, it is exactly what it means when you count things, it has physical presentation. To be more precise, since $0$ does not exist to me (naturally), such distinction does not either (but, yes, it does exist on logical and linguistic level). I am not convinced by your measure and quantification argument: certainly, negative numbers, rationals, reals, complex numbers, p-adics also quantify things, but that doesn't make them part of the natural numbers. They are simply more sophisticated concepts of quantification, and so is $0$. – M.G. Nov 23 '17 at 21:33
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    Another argument in favour of the notation $\mathbb{N}_0$ (regardless of which side of the naming debate you are on) is that it emphasizes the presence of $0$: it reminds you to check the degenerate cases (as per Mike Shulman's comment) like recently $\mathbb{R}^0$ and probably many others. – M.G. Nov 24 '17 at 13:06
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    Richard Stanley likes to use $\mathbb{N}$ for the Nonnegative integers and $\mathbb{P}$ for the Positive integers. I have also seen people try the notation $\mathbb{Z}^+$. Of course, one can then ask whether "positive" includes zero... – Timothy Chow Nov 24 '17 at 15:10
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    @TimothyChow: if I see $\mathbb{P}$ in the context of integers, it would probaby be obvious to exclude "projective" and "probability", but my next reaction would be "primes" (personally I don't use it to denote the primes, but many others do). – M.G. Nov 24 '17 at 16:01
  • @Hurkyl: even if you are not happy with my "absence-presence" analogy, there are other purely mathematical properties that distinguish $0$ from the natural numbers. $0$ cannot be written as a product of powers of primes (the irony is not lost on me that one has to use $0$ exponents for $1$), and having the minimal element of $\mathbb{N}$ also being the additive generator of $\mathbb{N}$ is kinda nice, don't you think? – M.G. Nov 24 '17 at 16:08
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    I like to write $\Bbb{Z}{>0}$ and $\Bbb{Z}{\ge0}$. Upside: impossible to misinterpret, doesn't annoy people by picking favourites between the two. Downside: a little clunky. – Gareth McCaughan Nov 24 '17 at 23:16
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    It turns out that one can start an internet flamewar merely by mentioning that one can start an internet flamewar by asking whether zero is a natural number. – David Richerby Nov 25 '17 at 12:28
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    I wonder if whether one's inclination to include $0\in\Bbb{N}$ is related to how much experience one has computer programming, where indices universally begin with $0$ instead of $1$. After one has been burned by enough IndexErrors and segfaults, one naturally begins to count with $0$ instead of $1$. – Neal Nov 25 '17 at 15:50
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    I resolve this "mother" problem denoting by $\mathbb N$ the set ${1,2,\dots}$ and by $\omega$ the set ${0,1,2,\dots}$ and this well agrees both with Number Theory and Set Theory: elements of $\mathbb N$ are natural numbers and elements of $\omega$ are finite ordinals. – Taras Banakh Nov 25 '17 at 23:29
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    @Taras, I thought $\omega$ stood for the $\omega$hole numbers. – Gerry Myerson Nov 25 '17 at 23:45
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    @GerryMyerson Exactly, $\omega$ denote $\omega$hole numbers and $\mathbb N$ -- $\mathbb N$ot whole numbers:) – Taras Banakh Nov 26 '17 at 00:08
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    I've heard people suggest using "whole numbers" and "natural numbers" to distinguish the two sets. The funny thing is, sometimes these suggests are for the whole numbers to include zero, and other times its for the natural numbers to include zero. :( –  Nov 26 '17 at 03:50
  • @Neal: I think that the idea to use half-open intervals of integers is the real insight here. But for some reason, working with intervals $[a, b)$ feels much more natural than intervals $(a, b]$, and thus 0-up counting for the standard $n$-long interval $[0,n)$. –  Nov 26 '17 at 03:52
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    I really think mathematicians should gather to vote on this in the near future so we can settle this one and for all. (Also, I think they should give set and category theorists the wrong address for the meeting.) – k.stm Nov 26 '17 at 08:12
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    I am OK with folks who support the idea that zero is not a natural number. All I ask is they be consistent in their notation and write $\mathbb{N}={I, II, III, IV, V, VI, VII, VIII, IX, X,...}$ – Abdelmalek Abdesselam Nov 26 '17 at 21:14
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    @GarethMcCaughan: Actually that notation annoys me, because clearly it's much better to write $\mathbb{Z}^{>0}$ and $\mathbb{Z}^{\geq 0}$ (superscripts instead of subscripts). Only a heretic would suggest otherwise ;-) – psmears Dec 01 '17 at 11:20
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    @psmears But superscripts are annoying when one needs to make cartesian powers. I personally find ${>0}$, $^{>0}$, ${\ge0}$, and $^{\ge0}$ equally ugly as sub- and superscripts, and I rather had $\mathbb{R}+$ $\mathbb{Q}+$ $\mathbb{Z}+$ and, yes, $\mathbb{N}+$. – Pietro Majer Jan 22 '18 at 22:26
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    @PietroMajer: I think you may have missed the joke... – psmears Jan 22 '18 at 22:51
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    @DavidRicherby Now I see a natural-indexed tower of internet flamewars $(IFW_n)_{n\in ?}$. – Matemáticos Chibchas Dec 18 '18 at 12:13
  • @TarasBanakh This is also how I use $\omega$ and $\mathbb N$. I just make sure to write $\mathbb N={1,2,3,...}$ the first time I use the symbol in a paper. If it's good enough for you, it's good enough for me! – Forever Mozart Jan 02 '19 at 07:08
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    In US-conventions, $\mathbf{N}={3.281$ feet, $6.562$ feet, $\dots}$. – YCor Jan 02 '19 at 09:59
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    @Neal, unless they're Lua programmers …. – LSpice May 25 '19 at 16:41
  • Whatever the views expressed here, the passage of time will decide the notation for the natural numbers and the positive integers. Earlier, $\mathbf{N}$ more commonly meant the positive integers, while present usage is tending towards $\Bbb N$ for the natural numbers. They are distinctive enough to have both, but unfortunately this is not accepted. – John Bentin Dec 05 '19 at 09:25
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    the natural numbers $(N, \mathrm{pt}, \mathrm{succ})$ as defined by the second-order Peano axioms or by a universal property in the category of sets could refer to either. It is really the semigroup structure resulting from the recursive/inductive definition of addition that determines whether $\mathrm{pt}$ is zero or one. – Madeleine Birchfield Jul 19 '22 at 15:42
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    My rule of thumb is: if the additive structure of the naturals is more relevant for one's application than the multiplicative structure, start at 0; if the multiplicative structure is more important, start at 1; and if neither is particularly important, leave it up to personal taste. – Terry Tao Aug 10 '23 at 04:29
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    @TerryTao And if both are important, start at $\frac{1}{2}$! – Carl-Fredrik Nyberg Brodda Sep 12 '23 at 04:23
  • When I'm feeling wild and crazy, I like to think of $0$ as at the very top of the divisibility lattice for $\mathbb{Z}{\geq 0}$, acting as an infinity of sorts, and $0$ as the zeroth prime. Also, $\aleph_0$ is the first infinite cardinal, while $\omega$ is the first infinite ordinal, and they are different, both, in sense, than $\mathbb{Z}{\geq 0}$. – Jesse Elliott Sep 13 '23 at 08:00
  • @Jesse, zero/divisibility/infinity see https://math.stackexchange.com/q/1014838 – Gerry Myerson Sep 13 '23 at 08:05
  • @GerryMyerson Of course, my comment was somewhat tongue-in-cheek, in response to Terry Tao's suggestion that $0$ is not multiplicatively relevant. If $0$ is like infinity, then the $0$ is like the prime at infinity, which makes sense in terms of Ostrowski's theorem, for example. In ring-theoretic terms, $(0)$ is prime in any integral domain, which is why one might consider $0$ to be a prime integer. – Jesse Elliott Sep 13 '23 at 20:09
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    @Jesse, I've seen $-1$ called a prime, so as to extend unique factorization to the negative integers. – Gerry Myerson Sep 13 '23 at 22:06
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Linear functions:

In high-school algebra (sometimes called "pre-calculus"), we are taught that linear functions are those of the form $y=mx+b$, because they are graphed by a straight line in the plane.

Then we study linear algebra in university, and realize that for a function $f$ to be linear it must satisfy $f(0)=0$, as a special case $a=0$ of the linearity requirement that $f(ax)=af(x)$, and the functions that were called "linear" in high school are not really linear (unless $b=0$) but affine.

Ari Brodsky
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    I teach my algebra students the term affine, but let them know that in most cases they are called "linear" despite not actually being linear. – Alex Jones Nov 23 '17 at 08:40
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    Computer science also uses "linear" to mean "cost grows directly proportional to problem size, ignoring a constant fixed cost", which is affine. – Eric Lippert Nov 24 '17 at 15:37
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    @EricLippert: They're only interested in the asymptotic behaviour when $x$ gets to infinity. That's why $b$ is ignored. The number and speed of CPU is arbitrary, so $a$ is set to $1$. – Eric Duminil Nov 24 '17 at 16:35
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    @EricLippert, similarly "exponential" in CS often refers to exp(polynomial), where polynomial doesn't have to be linear. – Michael Nov 26 '17 at 20:54
  • @EricLippert: Although in CS there is never really any confusion about this. If we want to specify, we might say "linear growth rate." Also, sometimes we do talk about linear functions (lookup, e.g., "matrix rigidity"), and sometimes we talk about affine functions, and have to be careful about the distinction between the two. – Joshua Grochow Nov 28 '17 at 05:14
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    @EricLippert Computer scientists would also call $n + \log{n}$ 'linear', even though it's not even affine. – Adam P. Goucher Dec 20 '17 at 17:08
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    An ODE $\dot u=Au+b$ is still called a "linear equation", and $\dot u=Au$ "linear homogeneous" – Pietro Majer Jan 22 '18 at 22:16
  • Something I have problems with when teaching Calculus II is this: if I call "tangent plane" the one physically tangent to the surface $S$ at $P$, how do I call the space of vectors tangent to $S$ to $P$? I call it "vector space tangent to $S$ at $P$", and I spend time to tell why the two objects are distinct, but closely related, but I'm not especially happy about this choice. – Nicola Arcozzi Jun 15 '19 at 10:01
  • Possible culprit Bourbaki, in La Tribu Nº 6 (1941): “adopted: *Linear function* (for “linear and homogeneous”); *Affine function* (for “linear (non homogeneous)”).” Borel (2001, p. xi) emphasizes: $()$ and $()$ are for us “the general and special linear groups, but this is not Lie’s terminology. For him, they are the general and special homogeneous linear groups. He reserves the term general linear group for our affine group $\textrm{Aff}(^)$”. – Francois Ziegler Aug 14 '19 at 17:24
  • I was once told that in Romanian afina means blueberry, and undergraduates always wonder why people talk about blueberry spaces and blueberry mappings. – Michael Renardy Sep 10 '23 at 03:23
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Not a word but a piece of notation: Sometimes I have seen $\subset$ used to mean "is a proper subset of" while other times I have seen it used to mean "is a subset of".

Timothy Chow
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    In my university this is quite an interesting example: my first course there was an introductory notation and structures course, introducing $\subset$ as proper subset. Then literally every other teacher proceeded to use $\subset$ as a generic subset. – tomsmeding Nov 23 '17 at 08:12
  • I heard that algebraists tend to use it for a proper subset, and analysts for a generic subset. – Fedor Petrov Nov 23 '17 at 10:14
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    That is why I always use $\subseteq$ and $\subsetneqq$... – Dirk Nov 23 '17 at 10:31
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    @DirkLiebhold As an alternative to the latter there are also the (at least to me) aesthetically more pleasing $\varsubsetneq$ and $\subsetneq$ – Jules Lamers Nov 23 '17 at 12:48
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    @JulesLamers one careless stroke in handwriting or rubbed off fiber of very worn out paper, and your notation confuses the reader. – Ruslan Nov 24 '17 at 10:13
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    Interestingly, I have never seen the symbol $<$ used to mean "less than or equal to" nor have I seen anyone feel compelled to use the analogue of Dirk Liebhold's solution (by inventing a symbol for "less than but not equal to"). I wonder why the two notations have evolved differently even though they seem extremely closely related. Maybe "proper subset" just doesn't come up that often. – Timothy Chow Nov 24 '17 at 15:05
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    Even more confusingly: In at least one book and logic and set theory (Fitting, Smullyan - Set theory and the continuums hypothesis), I have seen $\supset$ used to denote implications while at the same time using $\subset$ for inclusion of sets. That one took me way too long to figure out. – Johannes Hahn Nov 24 '17 at 15:11
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    @TimothyChow I was quite distressed to learn that $M\prec N$ was standard notation for "$M$ is an elementary substructure of $N$", used interchangeably with $M\preceq N$ in my model theory course. – Mario Carneiro Dec 04 '17 at 19:53
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    In the introduction of some book (I don't remember which) I read the following convention: $\subset$ and $\subsetneq$ are synonymous but use $\subset$ if the equality case does not occur for trivial reasons; use $\subsetneq$ if equality needs to be explicitly ruled out or if the fact that the inclusion is proper is the main point. Examples: "$\mathbb{Z}\subset\mathbb{Q}$" and "let $\mathfrak{m}\subset A$ be a maximal ideal" and "let $A\subseteq B$ be a subset..." and "let $A\subsetneq B$ be a proper subset...". Note the difference between "...hence $A\subset B$" and "...hence $A\subsetneq B$" – Tashi Walde Dec 20 '17 at 23:38
  • @TimothyChow In my papers on modal logic, I use $<$ as a generic symbol for arbitrary transitive relations, with $x\le y$ a shorthand for $x<y\lor x=y$, and $x\lnsim$ for $x<y\land y\nless x$. (I would use $x\lneq y$ for $x<y\land x\ne y$, but it does not occur naturally in this context.) It may well happen that $<$ is already reflexive, in which case ${<}={\le}$. – Emil Jeřábek Jan 23 '18 at 12:17
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    @TimothyChow Actually I've seen many people in group theory writing $H<G$ to mean "$H$ is a subgroup of $G$", other using the more correct notation $H\le G$. I guess the only reason to use the incoherent first one is (a) laziness in latex writing (b) being use to read people using (a) (so I've also seen this on blackboard). Also the setting in which this is used often makes the case $H=G$ uninteresting (vacuously true), whence the sloppy notation. – YCor Jan 02 '19 at 10:04
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    @Johannes Worse, in my opinion, is that $A \subset B$ iff $\forall x . x \in A \supset x \in B$ (it switches direction). – user76284 Apr 21 '19 at 15:26
  • Why did TeX (or Kunth) program \subset to be "subset"? – Colin Tan Dec 15 '22 at 14:06
  • @JohannesHahn Russell & Whitehead's Principia Mathematica, and a lot of classical texts on logic from that time (including Wittgenstein's Tractatus) all use $\supset$ for implication. – Carl-Fredrik Nyberg Brodda Sep 12 '23 at 04:29
  • @ColinTan I'm not sure Knuth would be too happy to read that particular typo... :-) – Carl-Fredrik Nyberg Brodda Sep 12 '23 at 04:29
72

Everyone knows what a curve is, until he has studied enough mathematics to become confused through the countless number of possible exceptions. — F. Klein

Francois Ziegler
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    Of course, sometimes a curve is a map from $\mathbb{R}$ into a manifold, and other times it is a Riemann surface, and still other times it is a reduced 1-dimensional separated scheme of finite type over a field. – Robert Furber Nov 23 '17 at 14:37
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    Heck, people keep telling me $\mathbb{C}$ is a line. – Eric Towers Nov 24 '17 at 06:32
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    I remember being confused about how some algebraic geometers and number theorists were using the word "curve" until I realized that I was tacitly (and incorrectly) assuming that a "curve" was what, in their language, was a "set of $F$-rational points" of the curve. It takes a while to unlearn the idea that a curve is a set of points. – Timothy Chow Nov 24 '17 at 15:21
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    @TimothyChow, which is funny, because it is undoing the complicated process (foreign to Euclid, for example) of first learning that a curve is a set of points …. – LSpice May 25 '19 at 16:55
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    @LSpice : Partially undoing it at least. It was also foreign to Euclid to define a curve using algebraic equations and/or morphisms. – Timothy Chow May 26 '19 at 15:58
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    @TimothyChow, good point; but (at the risk of de-railing a years-old comment section with this) I feel that, with the idea of tracing out a curve as the locus (of points, I can't help saying) satisfying a certain condition being so familiar to him, Euclid would have been much more comfortable with the algebraic-geometry than with the point-set approach to curves. – LSpice May 26 '19 at 20:00
68

Another kind of answer. There is

increasing; strictly increasing

and there is

nondecreasing; increasing

Gerald Edgar
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Polygons!

Is a polygon a sequence of vertices together with the edges that connect consecutive vertices? If so, can two distinct vertices be the same point? How about two consecutive vertices? Or a pair of vertices two indices apart: can we go from A, to B, and then directly back to A?

Or if not, can the edges of a polygon intersect each other, aside from the necessary intersections between consecutive edges at vertices?

Can we say that points are inside or outside the polygon? If so, and the polygon is self intersecting, do the points within the intersection cancel, or stay, or get counted with multiplicity?

Or is a polygon an area whose boundary consists of line segments? If so, can its boundary be disconnected (can It have holes)? Can it pinch to a single point like a figure eight?

Are the vertices ordered? Or at least ordered cyclically? Are the edges directed? If so, do the directions of consecutive edges need to match up?

For that matter, do the edges need to connect up at all, or can our polygon have a starting point and an ending point with no edge connecting them?

And that's not even getting into polyhedra...

Thamus Panmegas
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    Does every polygon in a space necessary lie in a plane? – Andrei Smolensky Nov 23 '17 at 11:57
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    I participated in a consortium that has the task of defining rules for a map format for vehicle routing and one of problems was to define, what kind of planar polygons are allowed; trying to work out mathematically sound mandatory rules was a major pain because different participants had different ideas of what is a simple polygon (should self-touching be allowed and if yes, what kind of restrictions should be imposed on edge overlaps) – Manfred Weis Nov 26 '17 at 11:24
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    Does a polygon have finitely many vertices? – Christopher King Jan 22 '18 at 22:37
  • @AndreiSmolensky what kind of plane? – Christopher King Jan 22 '18 at 22:38
  • For polyhedra, we have Lakatos' whole book of course. – Balazs Sep 19 '23 at 09:37
54

Positive definite matrix (and related terms). Most authors require these to be hermitian (or symmetric in the real case), but not all.

EDIT: also, "positive" can be ambiguous even for numbers: most authors, especially in English, consider $0$ to be neither positive nor negative, but according to Bourbaki it is both, and many authors (especially French ones) follow Bourbaki's convention, using "strictly positive" or "strictly negative" for what most authors would call "positive" or "negative".

Robert Israel
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    Many prefer the distinction between "positive" vs. "non-negative", which clears possible confusion. "Strictly positive" is more of emphasizing the positiveness so as to avoid any potential source of confusion. Using both "positive" and "strictly positive" is kinda weird. – M.G. Nov 22 '17 at 22:25
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    Functional analysis generally uses "positive" to mean "nonnegative" because the notion of strictly positive does not generalize well to infinite dimensional spaces (the interior of the positive cone of an ordered Banach space can be empty). Functional analysts also call "nonexpansive" mappings "contractions". It makes sense that the more frequently used concept gets the shorter name. – Robert Furber Nov 23 '17 at 13:35
  • @RobertFurber The notion of a strictly positive element in an ordered linear space I'm familiar with is an element that has a strictly positive value under every nonzero positive linear functional. That works even for ordered Banach spaces in which the positive orthant has no interior points. – Michael Greinecker Nov 23 '17 at 22:42
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    @July and "not non-negative" is not equivalent to "negative." You get other issues. – Clement C. Nov 24 '17 at 03:13
  • @ClementC., I have never seen anyone using the term "not non-negative". – M.G. Nov 24 '17 at 11:59
  • @MichaelGreinecker That definition works better, but can still be an empty set. For instance, in the case of the C$^*$-algebra $c_0(X)$ for an uncountable set $X$. – Robert Furber Nov 24 '17 at 14:45
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    @July "by contradiction, suppose f is not non-negative [...]" – Clement C. Nov 24 '17 at 15:05
  • @ClementC., ok, but certainly there must be a better wording than that, depending on the context. – M.G. Nov 24 '17 at 15:48
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    @July I think there are far fewer issues with "positive, strictly positive" than with "non-negative, positive". After all we have "contains, strictly contains", and nobody considers that kinda weird. Trying to weaken a notion by negating its opposite is a bad idea in general, because it depends on just how the opposite is taken, and may not give the right notion anyway. See https://math.stackexchange.com/a/115951/18880 – Marc van Leeuwen Dec 01 '17 at 09:06
  • @MarcvanLeeuwen: it seems a common issue in this thread is that language does not work like mathematics, and it is a really a bad idea for mathematicians to try and change meanings as they see fit. "Contains" and "bigger than" are different relations: every set naturally contains itself, i.e. the default state of "contains" is nonstrict, while "any number is bigger than itself" is absurd, i.e. the default state of "bigger than" is strict. "Positive" really means "bigger than $0$" and not "bigger than or equal 0" (in many languages, not just in English). – M.G. Dec 01 '17 at 13:45
  • Same thing applies to "increasing" and as to which numbers occur in natura. – M.G. Dec 01 '17 at 13:50
51

Tensor: for some people, a tensor is an element of a tensor product of vector spaces. For others it’s a section of a tensor product of tangent and cotangent bundles on a manifold. Members of the first group would call that latter notion a tensor field.

More interestingly:

Tensor rank: a tensor of rank $r$ is either a sum of $r$ simple tensors (outer products) or an element of a tensor product of $r$ vector spaces (an “$r$-dimensional array of numbers”). For example, a matrix is a rank $2$ tensor in the latter sense.

I speculate that at least 25% of Math Stack Exchange questions about tensors are about confusion over one or both of the above.

Zach Teitler
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    I was half expecting the physicist's "definition" here: "A tensor is an object that transforms like a tensor." (Student: "?!" Professor: "It's basically anything with indices on it"). ;) – balu Nov 23 '17 at 20:27
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    @balu Followed up by "oh, so like the Christoffel symbols?" "No." – A. Thomas Yerger Nov 25 '17 at 08:47
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    So the zero matrix would have rank $2$. Interesting :P – Qfwfq Apr 28 '18 at 13:19
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    Maybe it would be better to say "tensor of degree $r$" to mean an element of $V^{\otimes r}$, which is consistent with the fact that the tensor algebra is a graded algebra so its elements have a degree. – Qfwfq Apr 28 '18 at 13:21
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    Let's not forget the machine learning definition of tensor: a multidimensional array (no further properties required). – littleO Apr 28 '18 at 16:46
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    Degree would make sense. More common are order (a matrix is an order $2$ tensor); "way", as in $r$-way tensor (a matrix is a $2$-way tensor); or simply $r$-tensor (a matrix is a $2$-tensor). I personally don't like "way". – Zach Teitler Aug 22 '19 at 17:29
51

"Function", prior to about 1910 always meant the $y$ in $y=f(x)$ (Look up any definition form that period). Since roughly 1920 it's officially the $f$. Physicists, engineers and many applied mathematicians still mean $y$ when they talk about functions today. But, since the hijacking of the term, they lack adequate terminology to make the distinction between $y$ and $f$. (Probably they are not even aware of the distinction, due to the sloppy use of the word function in most calculus textbooks combined with the common notation $y=y(x)$.)

More on the history of this strange development can be found here and here.

Michael Bächtold
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    People who regard $y$ in $y = f(x)$ as a function actually implicitly mean a “function in / of $x$”. However, noone speaks of $f$ as a “function in / of $x$” (at least I haven’t heard that). Thereby, both usages may live together in peace as long as one exercises caution. – k.stm Nov 26 '17 at 08:26
  • @k.stm: I agree in principle, but have personally found it almost impossible to exercise this caution in practice. For example, if one wishes to consistently and explicitly make the distinction while teaching calculus, one ends up constantly saying things like: "the function of $x$, $x^2$...". This was not necessary during the 200 years prior to 1910, as $f$ was not yet called a function then. – Michael Bächtold Nov 26 '17 at 16:14
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    one could "dependent variable" as in books on ODEs. – Abdelmalek Abdesselam Nov 26 '17 at 21:06
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    @k.stm "noone speaks of $f$ as a "function of $x$"". Well, every calculus book writes $\frac{\partial f}{\partial x}$ when $f:\mathbb{R}^2\to \mathbb{R}$, which is basically another way of doing the same. The origin of this seems to be the article of Jacobi. – Michael Bächtold Nov 29 '17 at 14:36
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The Fourier transform is defined in at least 3 different ways depending on which subject and school one comes from: $$ \hat{f}(\xi)=\int_{\mathbb{R}^n} f(x)e^{-2\pi ix\cdot\xi}dx $$ or $$ \hat{f}(\xi) = \int_{\mathbb{R}^n} f(x)e^{-ix\cdot\xi}dx $$ or $$ \hat{f}(\xi) = \frac{1}{(2\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^n} f(x)e^{-ix\cdot\xi}dx $$ The second definition is no longer an unitary operator on $L^2(\mathbb{R}^n)$ and loses the usual symmetry with the inverse, both of which are however rectified again by the third definition.

M.G.
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    Also, some definitions of the Fourier transform will change the complex exponential's sign. – J.G. Dec 03 '17 at 22:52
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    For the second definition you may use $L^2$ space with the measure $(2\pi)^{-d} d\xi$. This is sometimes used in microlocal analysis to get rid of all the annoying powers of $2\pi$. – mcd Apr 28 '18 at 11:12
  • I use the second in teaching (engineers) and the first in research. Including a factor in the definition to make the $L^2$ theory fluid is not good in a course, I think. And the $L^1$ theory will have the extra factor. – Nicola Arcozzi Jun 15 '19 at 10:06
  • @NicolaArcozzi, re, what does it mean to make the $L^2$ theory fluid? – LSpice Sep 12 '23 at 02:54
47

How about "algebra"? Usually an algebra over a field is assumed to be associative by default, but sometimes it is not.

Not to mention the various category-theory uses of "algebra" (over a monad, over an operad, for a Lawvere theory, for an endofunctor...).

Stefan Perko
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    The category-theory sense of "algebra" is related to the one in Universal Algebra: An algebraic signature is a set of function symbols, each with a non-negative arity. An algebra (over a signature) is a any non-empty set together with n-ary functions for the n-ary function symbols. (These algebras are occasionally called universal algebras or algebraic structures.) –  Nov 23 '17 at 13:48
  • @HansAdler Still, I think the average mathematician might be a least a bit confused if I claim that compact Hausdorff spaces are examples of "algebras". Also sometimes one has to be very precise, e.g. to not confuse algebras over a monad and algebras for the corresponding endofunctor. – Stefan Perko Nov 24 '17 at 16:18
  • Not sure how you read it, but I just wanted to expand on your second paragraph by mentioning a definition of 'algebra' that most mathematicians find easier to relate to than category-theoretic notions of algebra, but which is still vastly more general than any of the classical notions of an algebra. - Btw, I wasn't aware that Hausdorff spaces are examples of the category theoretic notion of algebra. Is that by cheating (working in a category of Hausdorff spaces and with an empty signature)? –  Nov 28 '17 at 15:42
  • @HansAdler Ah, I see. Yes, thank you. Feel free to edit my answer as well. - Well, I guess this the problem with the word "algebra". Noone knows what it means. I'm kidding. I was referring to the fact that compact Hausdorff spaces are monadic over $\mathsf{Set}$. This is "Handbook of categorical algebra 2" Proposition 4.6.6. – Stefan Perko Nov 28 '17 at 20:16
  • @HansAdler I guess the choice of the word "algebra" in universal algebra is the main reason this is not widely known. (Calling the field universal algebra is not a problem, but calling its structures "algebras" is dire) – YCor Jan 02 '19 at 10:11
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To many authors a "category" is necessarily locally small, but to others it need not be.

Mike Shulman
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    Mac Lane understands small to mean “within some fixed Grothendieck universe $U$” – is this nowadays completely out of fashion or do usages of this term here, too? – k.stm Nov 26 '17 at 08:33
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    Sometimes "small" means that; other times it means "a set" (as opposed to a proper class). But I wouldn't consider that two different meanings for the same word, rather two different ways to formalize the same meaning. (-: – Mike Shulman Nov 26 '17 at 12:18
41

What is a "set"? ZFC has one answer; ETCS has another; Bishop had another; HoTT has yet another.

Mike Shulman
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  • Are they really part of the same field? – Michael Greinecker Nov 22 '17 at 22:10
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    @MichaelGreinecker Yes and no. No in that people who study ZF-sets for their own sake are a distinct community from people who study ETCS-sets and people who study HoTT-sets. But yes in that all of them have been proposed as a precise meaning of the word "set" as it is used foundationally in the rest of mathematics. – Mike Shulman Nov 22 '17 at 22:18
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    Maybe the biggest conceptual difference is between ZFC on the one hand, and ETCS and HoTT on the other (I don't know where Bishop lies). In ZFC, a "set" is (conceptually) a subset of the universal class; set membership is a predicate and one set can be a subset of another. In HoTT, set membership is not a predicate and an element of a set cannot also belong to another set. But there is a separate concept, "subset of the set $S$", where membership is a predicate (over $S$), and one subset-of-$S$ can be a sub-(subset-of-$S$) of another subset-of-$S$. – Tanner Swett Nov 22 '17 at 23:04
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    The reason for this is that HoTT (and I think also ETCS, but I'm not sure) are type theories, in which "sets" are a particular kind of type, and all types are disjoint from each other. The word "set" does not refer to a subset of a type. ZFC, on the other hand, only has one "type", which is the universal class; and the word "set" does refer to a subset of this "type". – Tanner Swett Nov 22 '17 at 23:07
  • @TannerSwett ETCS is not technically a type theory, but it shares many features of type theories. – Mike Shulman Nov 23 '17 at 00:04
  • Don't forget ZF + not axiom of choice. – Christopher King Jan 22 '18 at 22:39
40

Some authors formulate separability axioms T3/T4 as normality/regularity plus T1. Some other authors do not require T1, on the contrast, they define normality/regularity as T3/T4 plus T1.

Fedor Petrov
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  • Really? I've seen "regular" used for what others called "regular Hausdorff", but I don't recall anyone using $T_3$ without also meaning $T_2$. Do you have any references? – tomasz Nov 27 '17 at 17:18
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    I quote wikipedia: "Although the definitions presented here for "regular" and "T3" are not uncommon, there is significant variation in the literature: some authors switch the definitions of "regular" and "T3" as they are used here, or use both terms interchangeably." They do not provide a reference in this place, however. – Fedor Petrov Nov 27 '17 at 19:55
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    Some authors (e.g., Cullen 1968, pp. 113 and 118) interchange axiom T_3 and regularity, and axiom T_4 and normality (from http://mathworld.wolfram.com/SeparationAxioms.html) – Fedor Petrov Nov 28 '17 at 19:35
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Simply connected : for some authors, such a space is necessarily path-connected, for others, not.

Duchamp Gérard H. E.
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    For that matter, is the empty space connected? – Mike Shulman Nov 22 '17 at 22:16
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    The empty space should not be seen as connected, for the same reason as units in a ring are not primes: you want the decomposition into connected components to be unique up to permutation. I'm sure others will disagree, though, so it's another good example :) – Lukas Lewark Nov 23 '17 at 05:41
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    @LukasLewark I agree ... connected components are seen as "atoms" and therefore empty space should not be considered connected although this does not match with the formal definition. – Duchamp Gérard H. E. Nov 23 '17 at 10:50
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    This mini-debate is exactly the simply-connected+(conntected or not) debate one dimension down! – Jeff Strom Nov 23 '17 at 14:49
  • @LukasLewark If you view the decomposition into connected components as a partition (i.e., a set of nonempty equivalence classes), then it is truly unique, not unique up to permutations. – Goldstern Nov 24 '17 at 08:18
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    The empty space is connected, but it is not $0$-connected. (Of course it is not even $(-1)$-connected, that is the point.) – Carsten S Nov 24 '17 at 11:54
  • This has been discussed on MO before. – Jarek Kuben Apr 28 '18 at 11:38
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    The question about connectedness of the empty set is the question of whether the set of components of the empty set is $\emptyset$ or ${\emptyset}$. (Fortunately, according to my Discrete Mathematics students, there is no difference between these two sets.) – LSpice May 25 '19 at 16:45
34

"Topos" sometimes means "elementary topos" and sometimes "Grothendieck topos".

Mike Shulman
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    There is a correlation between the intended meaning and the pluralisation. Algebraic geometers usually study Grothendieck topoi, and logicians typically study elementary toposes. – R. van Dobben de Bruyn Oct 16 '18 at 18:04
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A tree can be a very different thing in different parts of mathematics. It might be a certain kind of acyclic graph; or a partial order such that the predecessors of every node are linearly ordered; or a partial order where the predecessors are well-ordered; or either of these, except with successors instead of predecessors. In some parts of mathematics, trees are presumed finite, while in others, they are nontrivial only when infinite.

Joel David Hamkins
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    For that matter, so can a "graph". – Mike Shulman Nov 22 '17 at 21:59
  • ... and a field – Fedor Petrov Nov 22 '17 at 22:00
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    @FedorPetrov If you mean "commutative division ring" versus "function on a space", then I don't think that's an example for this question, as noted in the comments comparing it to the other question about overloaded words. But the various meanings of "graph" and "tree" are very closely related, yet distinct. – Mike Shulman Nov 22 '17 at 22:04
  • (I'm not, of course, talking about "the graph of a function" in the precalculus sense.) – Mike Shulman Nov 22 '17 at 22:05
  • @MikeShulman I thought that you mean the graph of a function (it is used not only in precalculus, cf. https://en.wikipedia.org/wiki/Closed_graph_theorem ) – Fedor Petrov Nov 22 '17 at 22:09
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    I interpreted Mike's remark to refer to the fact that graphs have different conceptions even in graph theory, depending on whether self-edges or parallel edges are allowed. – Joel David Hamkins Nov 22 '17 at 22:11
  • Joel guessed correctly what I meant. Sorry for the confusion. – Mike Shulman Nov 22 '17 at 22:12
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    If we were to go along this road, I'd suggest to look for all the various definitions of "normal" in different branches of mathematics ... – Hagen von Eitzen Nov 23 '17 at 12:19
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I would say the word "kernel" is probably among the most overloaded terms in mathematics. You've got kernels of linear operators, convolutional kernels, distribution kernels, Markov kernels, and Reproducing Hilbert Space kernels. All of these notions are related but are, strictly speaking, distinct objects. Thus, the kernel of a linear operator is a subspace, while an RKHS kernel is a positive definite map from some set times itself to the reals.

Aryeh Kontorovich
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27

Topology: Some authors use the term neighborhood while other use open neighborhood instead.

Markus Scheuer
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    This is worse than you indicate. "Neighborhood" is defined in two different ways. J.L.Kelley (General Topology) defines neighborhood of $x$ as an open set containing $x$, then asserts that analysts and Bourbaki use "neighborhood" to be any set containing an open set containing $x$. (References are in my math.SE answer here.) – Eric Towers Nov 24 '17 at 06:56
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    Most of the time people use neighbourhoods where it does not matter if it is open or not and the end result is the same. Problems begin if one insists on taking, say, complements or unions of neighbourhoods, in which case it is better to simply speak of open sets instead, or introduces a connectedness hypothesis in the definition of neighbourhood... – M.G. Nov 24 '17 at 13:15
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    @July It matters very much when one wants to work with neighborhood filter. – Michael Greinecker Nov 27 '17 at 08:40
  • @MichaelGreinecker: does the problem disappear if one speaks of open sets containing the point? – M.G. Nov 27 '17 at 12:46
  • My point is that one should use the term "neighbourhood" when it's irrelevant if it is open or not as long as it contains an open set containing the point in question. If openness of the neighbourhood is important, then one should simply say "open set containing the point $x$", although admittedly "open neighbourhood of $x$" sounds better. No reason not to say "open" explicitly when relevant. On the other hand, one also uses compact neighbourhoods or connected neighbourhoods, which is certainly easier on the tongue than "compact set containing an open set containing the point". – M.G. Nov 27 '17 at 12:51
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    @July The point is that filters are closed under supersets, so this pretty much requires the "French" definition. Or talking about the filter generated by open neighborhoods, but I think filters are a big part of what terminology is favored. – Michael Greinecker Nov 27 '17 at 12:57
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    I think the language used in the theory of locally compact spaces (where we talk about compact neighbourhoods) and the theory of locally convex vector spaces (where we make statements like "the polar of an equicontinuous set is a neighbourhood of 0") would become very awkward if neighbourhood could only be used to mean "open neighbourhood". I can't think of an opposing situation in which it is beneficial to have "neighbourhood" mean "open neighbourhood". – Robert Furber Nov 27 '17 at 23:01
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Perhaps a prominent example is the definition of a smooth manifold. Some authors require the underlying locally Euclidean Hausdorff space to be 2nd countable, while others require it to be only paracompact. The latter is more general since it allows for uncountable number of connected components, although I'm not sure if that makes such a big difference in the everyday life of a differential geometer. But usually there is no room for confusion since most authors state at the beginning which convention they use.

EDIT: It turns out it actually does make a pretty big difference, see Stefan Waldmann's comment below.

M.G.
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    It actually does make a huge difference: any set can be a zero-dimensional manifold if you only require paracompactness. In particular, nice theorems like: a bijective immersion is a diffeomorphism will fail in this context while they hold in the 2nd countable context. A simple consequence/example: a transitive Lie group action of $G$ on $M$ will not yield a homogeneous space structure $G/G_p$ for $M$ if you are allowed to topologize $G$ discretely... Very nasty. – Stefan Waldmann Nov 23 '17 at 06:31
  • @StefanWaldmann: this is great, thanks for the enlightening comment! But what do you mean by "any set can be a zero-dimensional manifold"? – M.G. Nov 23 '17 at 11:50
  • @July Take the (well, OK, non-empty if you like) set $M$ with the discrete topology. This is a zero-dimensional paracompact Hausdorff manifold with a lot of connected components. – Stefan Waldmann Nov 23 '17 at 12:05
  • @StefanWaldmann: I am probably missing something, but how is this locally Euclidean? Any singleton is open, but any chart into $\mathbb{R}^n$ maps the singleton into a single point, which is not open in the Euclidean topology, hence the chart is not an open map... – M.G. Nov 23 '17 at 12:15
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    @July Consider $\mathbb{R}^0$. – Robert Furber Nov 23 '17 at 13:36
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    @RobertFurber: right, thanks, I forgot that counts as Euclidean, I feel so silly now. So, ironically, my other comment about excluding $0$ from the naturals came back to hunt me :D – M.G. Nov 23 '17 at 13:40
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    When I first learned of n-manifolds in the early 1960s, they were required only to be locally homeomorphic to n-dimensional euclidean space. Then people started adding additional conditions like Hausdorff, second-countable or paracompact — to the very definition of the term. Then there are also the topological, piecewise-linear, C^r (1 ≤ r ≤ ∞), and real-analytic (C^), not to mention complex manifolds. So many definitions make it a very good idea for anyone using the term "manifold" to state exactly what the assumptions are. – Daniel Asimov Aug 10 '23 at 20:01
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    The so-called Prüfer manifold is a real-analytic surface that is not second countable. It is extremely elegant, and in fact its complexification is a 2-dimensional complex manifold that is also not second countable — which thanks to Radó's theorem every Riemann surface must in fact be. I would hate to exclude the Prüfer manifold on the grounds that it is "not a real manifold". – Daniel Asimov Aug 10 '23 at 20:17
  • @DanielAsimov: nice example! Thanks! – M.G. Aug 10 '23 at 21:21
25

$(a,b)$

Is that a coordinate pair representing a point in the plane? or,

The open interval from $a$ to $b$? or

The greatest (highest) common factor (divisor) of $a$ and $b$? or

The ideal generated by $a$ and $b$? or

The inner product of $a$ and $b$? or...

Zach Teitler
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Amir Asghari
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    This is surely a case of notation, rather than a definition – Yemon Choi Nov 26 '17 at 20:30
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    This was my spontaneous answer to the part of the question saying that "different places in the literature the same name is used for two different mathematical objects." Plus, I am thinking of a notation as a written form of a name. – Amir Asghari Nov 26 '17 at 21:31
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    This is why I always use $\langle a,b\rangle$ for points in the plane. I've even used $\langle a,b,c,...,h,...\rangle$ to indicate points in the hilbert cube! – Forever Mozart Jan 02 '19 at 07:18
  • @ForeverMozart, re, what is the 27th component of that point in the Hilbert cube? – LSpice Sep 12 '23 at 02:52
25

The definition of a Turing Machine is a great example, where multiplied all together there are at least hundreds of possible definitions.

  • Is the tape doubly infinite or singly infinite? (If singly infinite, what happens when you move left at the left at the end of the tape?)

  • Does the machine have special "accept" and "reject" states, just a "halt" state, or neither of these?

  • Is the input string written on a separate read-only tape? Is the output string written on a separate write-only tape?

  • Does the number of states include the accept and reject states?

  • Does the space usage include input and output? Does the running time include the step when the machine halts?

  • Is the input alphabet $\{0,1\}$, or an arbitrary set? May it contain only 1 symbol or even 0 symbols?

  • Is the tape alphabet just adding a blank symbol, or may it contain other convenient working symbols?

  • Are multiple tapes allowed? Is staying still instead of moving left or right allowed?

One usually spends a week in a theory of computation course fleshing out some of the equivalences between various definitions. But I don't know that anyone has ever done this in machine-checked or at least formally exhaustive detail, avoiding all pitfalls and edge cases. Rather, researchers rely heavily on the informal Church-Turing thesis and assume some kind of programming language or machine representation which suits their needs and is equivalent to all these definitions of Turing Machine at once.

Yet they are not at all equivalent for many purposes. (1) If worried about low-level running time (not just up-to-a-polynomial), doubly-infinite vs singly-infinite, how many symbols are allowed, multiple vs. single tapes, and whether accept/reject is part of the finite state machine all have an impact. (2) If worried about sub-linear complexity classes, we know that it is crucial that the input string be read-only. (3) Even if only worried about defining polynomial time computability, at least two symbols are required in the input alphabet. (4) To define the Busy Beaver function, as well as many other non-computable functions, requires a specific representation to be fixed. This one is a big "gotcha" that one always runs into when asking students (as an exercise) to compute the Busy Beaver function of $1$ or $2$.

Caleb Stanford
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Two examples from point-set topology:

  • a topological space $(X,\tau)$ is locally compact if "any point has a compact neighborhood" or if "any point has a local basis of compact neighborhoods". The two are equivalent if the space is Hausdorff, but not in general!
  • Even more subtle: a topological space $(X,\tau)$ is locally connected at $x$ if "$x$ has a local basis of connected neighborhoods" or if "$x$ has a local basis of open connected neighborhoods". The two are equivalent if we ask them for all points $x\in X$, but not in general!
Karl Kronenfeld
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Mizar
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    The second point is actually frustrating in practice. Most authors use connected im-kleinen for your first definition of locally connected, but I've also seen weakly locally connected. – Forever Mozart Jan 02 '19 at 07:23
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Antisymmetrization and symmetrization of tensors. Should we divide it by $(n!)$ ? This affects the relations between tensor and (anti-)symmetric algebra, the theory over $\mathbb Q$ and $\mathbb Z$ and is generally a mess. A special case: is a quadratic form over $\mathbb Z$ represented by a polynomial with integer coefficients or by a symmetric matrix with integer elements?

Anton Fetisov
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    It also affects whether you can symmetrise or antisymmetrise in positive characteristic …. – LSpice May 25 '19 at 16:49
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For some strange reason, algebraic order theory choose "lattice theory" as its name, and "lattice" as its central object of study. But of course, for everybody else "lattice" still means what it always meant, and what it also means in plain English.

Thomas Klimpel
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    Even for the "classical" lattices, there are many different definitions depending on the book or paper you look at, and some of them only coincide for the special case of a full, free $\mathbb{Z}$-module in a $\mathbb{Q}$-vector-space... – Dirk Nov 23 '17 at 10:36
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    The term lattice is also used in functional analysis, probability theory, computer science, economics,... Everybody else might be a minority nowadays. – Michael Greinecker Nov 23 '17 at 13:10
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    This would be a better answer if you spelled out what lattice "always meant". In any case, this seems to me to be more an example of a common word being overloaded. Certainly the name "lattice" for the algebraic structure seems very reasonable once you see a picture of a Hasse diagram. – Alex Kruckman Nov 29 '17 at 16:27
  • @AlexKruckman I thought of the English word lattice in the sense of crystal lattice. Here is an animated picture of a Hasse diagram of the free modular lattice on 3 generators. Feels different than a crystal lattice to me. OK, lattice is an English word which could also mean something like framework. Maybe that animated image really feels somewhat like a framework. But the Hasse diagram is only one possible way of thinking about ordered structures, so even this is not a good excuse for that name. – Thomas Klimpel Nov 29 '17 at 18:25
  • Even in algebra the term "lattice" may have different meanings: it could be endowed with an (integer? positive definite?) quadratic form or not, it could even be a discrete subgroup of a Lie group... To eliminate the confusion one may call order lattice the main object of study of algebraic order theory, orthogonal lattice a free $\mathbb{Z}$-module equipped with a quadratic form, Euclidean lattice if such form is positive definite, maybe integer/integral orthogonal lattice if the quadratic form is $\mathbb{Z}$-valued. – Qfwfq Apr 28 '18 at 13:38
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    (cont'd) Also, one could speak about a Lie(-theoretic) lattice, or lattice subgroup, when referring to the discrete subgroup of Lie group case. – Qfwfq Apr 28 '18 at 13:38
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    This one has confused the Library of Congress. For example, Bremner's book on lattice basis reduction has an LC call number of QA171.5 .B74 2012 which lumps it together with Birkhoff's book on lattice theory. Bremner's book should really be QA 9.58 something. – Timothy Chow Sep 01 '19 at 16:54
  • Lattice is used in other areas, with other meanings. For instance holomorphic sections of a complex vector bundle are a lattice in the sheaf of meromorphic sections. I think there are also several definitions of lattice within number theory alone. – plm Aug 08 '23 at 19:21
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The term 'Macdonald polynomial' might refer to:

  • The symmetric function $P_\lambda(x;q,t)$ with coefficients being rational in $q,t$.
  • The symmetric function $J_\lambda(x;q,t)$, being $P_\lambda(x;q,t)$ multiplied by a normalization factor.
  • The one of the non-symmetric polynomial $E_\alpha(x;q,t)$ (Haglund vs. Marshall normalization/convention).
  • The non-symmetric polynomial $E_\lambda(x;q,t)$ indexed by weights in a root lattice, and there is a formula over alcove walks.

  • The modified Macdonald polynomial $\tilde{H}_\lambda(x;q,t)$, which is the most 'combinatorial' version, which is the bigraded Frobenius character of a certain $S_n$-modules indexed by $\lambda$. Combinatorial formula due to Haglund.

  • There are also non-homogeneous Macdonald polynomials...

This specializes to confusion regarding Hall-Littlewood polynomials (the 'standard' $P_\lambda$, the 'dual' $Q'_\lambda$ or the modified, $\tilde{H}_\lambda$).

Per Alexandersson
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An extension of a group $A$ by a group $B$ can be either a group $G$ with a normal subgroup isomorphic to $B$ with $G/B$ isomorphic to $A$ or a group $G$ with a normal subgroup isomorphic to $A$ with $G/A$ isomorphic to $B$. (See, for example, https://en.wikipedia.org/wiki/Group_extension especially section 1.2.3.)

Andreas Blass
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In a real vector space $E$, some people use cone to mean a subset $C \subseteq E$ closed under multiplication by positive reals. If it is additionally closed under addition, they call it a convex cone, and if $C \cap -C = \{0\}$, they say it is a proper cone, or that $C$ is proper.

A second group of people has no name for the first thing, but calls the second thing just a cone, also using proper cone for the third thing.

A third group of people also never use the first thing, but call the second thing a wedge and the third thing a cone. So instead of saying "we prove that this cone is proper", they say "we prove that this wedge is a cone".

Since the preorder defined by a convex cone is a partial order iff that cone is proper, I am inclined to put myself in the third group of people and use the short word cone for the most useful notion.

Robert Furber
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  • I would have said multiplication by nonnegative reals. – Daniel Asimov Aug 10 '23 at 20:07
  • @DanielAsimov See my comment to another answer to this same question. "Positive reals" means reals $\geq 0$ in functional analysis, those $> 0$ are "strictly positive reals". – Robert Furber Nov 08 '23 at 11:04
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    Wait until the fact 0 is sometimes considered a positive integer in France is brought up. But more seriously, this is not worth arguing over. (I have deleted some unproductive comments, but left the first two to highlight the ambiguity and the clarification) – David Roberts Nov 10 '23 at 02:18
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Speaking of functional analysis: While algebraists seem (at least as far as I can tell) to finally accept that things called ring and in particular things called algebra really should have units, people in functional analysis seem adamantly opposed to this, especially those who work with $C^\ast$-"algebras". I can sort of see why they do it, but I don't find their reasons particularly compelling; they rarely seem to talk about actual honest to god $C^\ast$-algebras-without-unit. Most of the time an "algebra" is really better viewed as an ideal in some proper algebra. All the proofs I have seen work the same way: First step is always to go to the unital case by embedding the "algebra" in question into an unital algebra.

As an algebraist myself, talking to a friend of mine who lives in $C^\ast$-land sometimes is a hassle because of this.

Johannes Hahn
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    What? We talk about nonunital C*-algebras all the time. "All the proofs I have seen" --- well, then you're not very familiar with the subject. – Nik Weaver Nov 22 '17 at 21:51
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    Even stripped of any opinion as to what the "right" thing to do is, I think this is still a good example of what the question asks for. – Mike Shulman Nov 22 '17 at 22:00
  • @NikWeaver You're right. It is definitely not my field. However, I have that friend who sometimes talks with me about it so I can at least attest to the fact that some of the proof that C*-people insist are fundamentally non-unital really aren't and smuggle in a unital algebra through the backdoor. – Johannes Hahn Nov 23 '17 at 00:00
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    Adding a unit is certainly a technique that gets used, but there's a famous quote by Gert Pedersen about how "in the old days" we used to always start by adding a unit, but now we start by tensoring with the compacts and getting rid of the unit. It's an overstatement, but tensoring with the compacts is certainly also a common technique. – Nik Weaver Nov 23 '17 at 01:27
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    Also there is a reason why "operator algebraist" are not as attached to units as "ordinary algebraist" are: while non unital rings can be quite unusual (for example, the multiplication can be $0$ every where) $C^*$-algebras always have approximate units. – Simon Henry Nov 23 '17 at 10:40
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    @NikWeaver But that is also a prime example of my claim that there really is a unit somewhere and algebras-without-unit are really just ideals as something bigger. $\mathcal{K}$ is an ideal of $B(H)$. – Johannes Hahn Nov 23 '17 at 11:02
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    Upvoted because this is a really important case of conflicting definitions, but I wish you’d mentioned the related question of whether ring homomorphism means unital and left out the polemics about which was right. – Noah Snyder Nov 23 '17 at 15:42
  • @JohannesHahn: any nonunital C-algebra is an ideal in its unitization, so I'm not sure what your point is ... they aren't "really" C-algebras, they're "really" ideals? – Nik Weaver Nov 23 '17 at 16:57
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    My main objection is to "they rarely seem to talk about actual honest to god C∗-algebras-without-unit", which is totally false. – Nik Weaver Nov 23 '17 at 17:01
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    Yeah, maybe I didn't make myself very clear there. I know that examples like $C_0(X)$ get talked about all the time. My point is that it seems like they never get talked about in a way that makes them stand on their own so to speak. The proofs, the concepts, everything seems very unital all the time. To me (and again, I'm not an expert, I'm just saying what has been my experience thus far) it seems that the only reason for non-unital algebras is historical and if one were to invent the whole theory today from scratch, ... – Johannes Hahn Nov 23 '17 at 17:41
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    ... one would phrase and prove everything either about unital algebras or pairs (unital-algebra,ideal). Then certain proofs would even get clearer in my opinion. Why do certain proofs use the unitalisation, others approximate units and others yet again embed into a multiplier-algebra or some such? Because the same algebra-without-unit is really three different objects: $(A^+,A)$, $(\tilde{A},A)$ (or however the functor going from algebras-with-approximate-units to proper algebras is called) and $(M(A),A)$, ... – Johannes Hahn Nov 23 '17 at 17:45
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    ... which describe three different geometric pictures, three different versions of gelfand duality corresponding to three different notions of morphisms etc. I feel that conflating all these as one single algebra-without-unit hides what is "really going on". Focusing on unital algebras and talking about their ideals would clear that up. – Johannes Hahn Nov 23 '17 at 17:47
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    The same happens with representations of C*-algebras. As far as I have observed it (which may not be representative, I know) all theory is essentially about "non-degenerate" representations, which are really the same as representations of the appropriate unital algebra lurking in the background. If one were to invent that notion anew, I don't think anyone would phrase it in terms of non-denegeracy, but simply in terms of representations of unital algebras. – Johannes Hahn Nov 23 '17 at 17:52
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    Well, these are more nuanced and more interesting comments. But would you also tell topologists that locally compact Hausdorff spaces should not be considered as "proper" topological spaces, but rather as open subsets of compact Hausdorff spaces? – Nik Weaver Nov 23 '17 at 18:38
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    Also, you didn't mention K-theory, which is a good example of an area where you really want nonunital C-algebras. You nonunitally* embed $A$ in $M_2(A)$ in $M_3(A)$ etc. as upper-left-corners, then you take the nonunital direct limit to get the nonunital algebra $A \otimes K(H)$, and it is the projections living in here that yield the $K_0$ group of $A$. – Nik Weaver Nov 23 '17 at 18:40
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    @NikWeaver In some cases, yes. If one works with LCH spaces+proper maps for example, I would always ask what each definition, theorem, ... means for the one-point compactification. That's just part of what proper maps are. For LCH+continuous maps my gut feeling is different, because $\beta X$ is usually so much uglier than the spaces they came from (whereas unital algebras are so much nicer). Additionally: Some topologists always think of "manifold" when LCH spaces are discussed. Here is where my knowledge of C*-algebras ends. I do not know how well the concept of manifolds translates. – Johannes Hahn Nov 23 '17 at 19:20
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    ... Maybe if I obtain that knowledge, I will finally be convinced of the usefulness of forgetting units.

    Until then: K-theory is another one of those cases: The idempotent construction always felt unnatural and ad hoc to me, the definition as "Grothendieck group of the fin.gen. projectives" seems more conceptual to me and works best with unital algebras. Maybe that's also algebraist-bias, but I never looked back once I learned the Grothendieck-group definition. Everything about K_0 just makes so much more sense in the language of projective modules.

     – Johannes Hahn Nov 23 '17 at 19:23
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    @JohannesHahn It's not wrong to view a non-unital C*-algebra as an ideal in a unital one, just as it's not wrong to view every LCH space as an open set in a CH space. But this perspective falls over when it makes contact with examples. Should you be required to embed an open subset of Euclidean space into a compact space whenever you try to do calculus? (And which one? The one-point compactification? Its closure?) And what compactification should you use for the total space of a vector bundle over a torus? These aren't just analogies - this is literally the commutative case! – Paul Siegel Nov 25 '17 at 15:55
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    The perspective also doesn't survive contact with applications. In applications of C-algebras to foliations the leaves are usually non-compact. In applications to positive scalar curvature or the Novikov conjecture one often works $\pi_1(M)$-equivariantly on the universal cover, which is non-compact unless $\pi_1(M)$ is finite. In applications to representation theory one often attaches C-algebras to a group by completing algebras of functions on spaces on which the group acts, and in the case of infinite groups the spaces are usually non-compact and the algebras non-unital. – Paul Siegel Nov 25 '17 at 17:02
  • (Again, in all of these cases you can certainly embed the relevant C-algebras into unital C-algebras, but carrying around the unital baggage doesn't necessarily clarify your thinking about the object of interest, which is the non-unital thing. In fact, constructing a well-chosen unitalization is sometimes the goal of the theory, rather than the first step!) – Paul Siegel Nov 25 '17 at 17:04
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    @SimonHenry's comment looks like it might give a clear route out of this conundrum. Why would an algebraist be surprised that for operator algebras you might want an approximate unit instead of an exact one? Some natural follow-up questions though are "Is there a good reason why the multiplication is not approximate?" and "Do homomorphisms preserve approximate units?" Anyway, this is all a special case of the "natural numbers" answer because the reason to want units is that way you can multiply an empty set of elements. – Noah Snyder Nov 25 '17 at 17:50
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    Not all morphism preserves approximate units but those who do have a name : "non-degenerate morphisms", which in my opinion somehow say what is the general view about them: they are the interesting morphisms. (But of course there are situation where looking at degenerate morphisms is interesting, for example corner inclusion are generally not non-degenerate), And between unital algebra, non-degenerate = preserves the unit. – Simon Henry Nov 25 '17 at 18:22
  • I don't understand this discussion : $c_0$ is a simple example of a non-unital $C^\ast$-Algebra. See https://math.stackexchange.com/questions/3036068/examples-of-non-unital-commutative-c-algebras . – jjcale Sep 10 '23 at 07:45
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The limit of a function may be a deleted limit or a non-deleted limit.

A number $L$ is the deleted limit of $f$ as $x$ approaches $p$ if, for every $\epsilon>0$, there exists a $\delta>0$ such that $0<|x−p|<\delta$ and $x$ is in the domain of $f$ implies $|f(x)−L|<\epsilon$.

The non-deleted limit is exactly the same except that instead of $0<|x-p|<\delta$ you have $|x−p|<\delta$.

For example, if

$$f(x)=\begin{cases} 1,&x=0,\\0,&x\ne 0,\end{cases}$$

then $\lim_{x\to 0}f(x)=0$ if deleted limits are used, but $\lim_{x\to 0}f(x)$ does not exist if non-deleted limits are used.

LSpice
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JRN
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    I have never seen a non-deleted limit in practice. Is there a reference with such definition? – Fedor Petrov Nov 23 '17 at 10:19
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    @FedorPetrov, Yes, "A First Course in Calculus" by Serge Lang, 5th edition (Springer, 1986). – JRN Nov 23 '17 at 12:57
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    B. Gelbaum has an analogous definition for the "non-deleted" version of $\liminf$ and $\limsup$ in Problems in Analysis – Pietro Majer Jan 22 '18 at 22:12
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Variety has a number of slightly different definitions. Apparently some authors use reduced of finite type over a field, whereas I would want separated to rule out the line with two origins. Most people use scheme-theoretic language to solve this problem.

Conway-Sloane jokes the discriminant of a quadratic form is a function of the form and the author. They are not wrong. That same subject has two distinct definitions of integral, which are basically the same (multiply by $2$) but which are a pain to keep track of. The Hasse-Witt invariant has two different definitions as well.

Watson Ladd
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15

I believe the number of proper definitions of the term "quantum group" is some element of $\mathbb R \setminus \{1\}$. Even the simplest example, the quantized enveloping algebra of $\mathfrak{sl}_2(\mathbb C)$, has at least two definitions (more if you count the coproduct as part of the definition).

If you are careful with the details, as for example in Jantzen's book "Introduction to Quantum Groups", you either get bogged down in the details of left-right compatibilites and twisted coproducts or, more in the spirit of the subject, you change the definition midway.

Pablo Zadunaisky
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Are Hermite polynomials $H_n(x)$ orthogonal w.r.t. the weight $e^{-x^2}$ or $e^{-x^2/2}$?

Fedor Petrov
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    The probability measure $\text{constant} \times e^{-x^2/2}, dx$ has variance $1.$ That's why it's done that way in probability theory. – Michael Hardy Sep 12 '23 at 02:19
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What some call the adjugate matrix is often called the adjoint (or sometimes classical adjoint) matrix, while adjoint matrix commonly also means the transpose or Hermitian transpose. I doubt this double use ever causes confusion to anyone but students, but it is an example of the sort the questioner seeks.

A more serious example is that regular elements of (algebraic) Lie groups are defined differently by different authors. Correspondingly one encounters various definitions of what is a Cartan subalgebra. Here the issue tends to be the scope of definition; usually two different definitions are equivalent in the semisimple setting, or over characteristic zero, but not in more general settings.

Dan Fox
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    What do you have in mind for the different definitions of 'regular'? I am used to seeing it used sometimes to mean 'regular' in the sense that the centraliser is of minimal dimension, and sometimes to mean 'regular' in that sense + semisimple; is that what you mean by the definitions being equivalent "in the semisimple setting"? – LSpice May 25 '19 at 17:09
14

One of the most classical examples is the word «algebra», which denotes not only a branch of mathematics, but also the following mathematical objects:

In linear algebra an algebra is a vector space, equipped with a bilinear operator (called product).

In set theory an algebra is a collection of sets closed under finite union, finite intersection and complement.

In universal algebra an algebra is a set, equipped with collection of finitary operations.

Note, that any algebra in linear algebraic sense is also an algebra in universal algebraic set, but not vice versa.

A quite similar thing happens with varieties:

In algebraic geometry a variety is the set of solutions of a system of algebraic equations.

In universal algebra a variety is a class of all algebras (in universal algebraic sense) with a given signature, satisfying a given set of identities.

Moreover, both those «varieties» are translated to Russian as «многообразия» - the same word, that is used for manifolds (topological spaces, such that each point of them has a neighbourhood, that is homeomorphic to $\mathbb{R}^n$ for some fixed $n$)

And if I have reached the theme of ambiguous translations, I think, that I should mention that «perfect groups» (groups equal to their derived subgroup), «complete groups» (centerless groups isomorphic to their automorphism group) and «immaculate groups» (finite groups, whose order is equal to the sum of orders of their proper normal subgroups) are all translated to Russian as «совершенные группы».

Also, the following examples deserve to be mentioned:

Two abstract groups are called commensurate (or commensurable), if they have isomorphic subgroups of finite index. Two subgroups of a group are called commensurate (or commensurable) if their intersection has finite index in both of them. Note, that two subgroups of a group may be commensurate as abstract groups, but not commensurate as subgroups.

Artinian groups are groups, that satisfy the minimum condition on subgroups. Artin groups are groups, that have a presentation of specific form. Both of them are «Артиновы группы» in Russian.

A right (left) ideal of a ring is a subring, that is closed under right (left) multiplication on arbitrary element of the ring. A right (left) ideal of a semigroup is a subsemigroup, that is closed under right (left) multiplication on arbitrary element of the semigroup. Note, that an ideal of the multiplicative semigroup of an associative ring is not always an ideal of that ring (because it does not need to be closed under addition).

Cubic graphs are usually defined as finite simple 3-regular graphs. However, the Hamming graph $H(3, 2)$ is also referred by some authors as «The Cubic Graph». Well, it is indeed finite, simple and 3-regular, but not the only one with this property.

The definition of simple graph I am used to is "graph without loops and multiple edges", however I know, that some people define simple graph as "graph without multiple edges" (loops are allowed).

In different sources $D_{2n}$ means either $C_{2n} \rtimes C_2$ or $C_n \rtimes C_2$.

The word "automaton" is sometimes used as a synonym for "acceptor" and sometimes for "transducer"

Sometimes $C_n^k$ means binomial coefficient $\frac{n}{(n - k)!k!}$ and sometimes direct product of $k$ isomorphic copies of a cyclic group of order $n$.

Chain Markov
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An $n$-category used to mean a strict $n$-category (a category enriched in the cartesian closed category $(n-1)$-$\mathsf{Cat}$), and now is often used to mean a weak $n$-category (which itself is defined in multiple ways).

I'd like to mention also that many of the examples in the answers and comments are examples of what the nLab calls red herrings, which need be neither red nor herrings, or sometimes all herrings are red herrings. For example, a multi-valued function is not actually a function, but all functions are multi-valued functions. A manifold with boundary is not actually a manifold, but all manifolds fit the definition of manifold with boundary, hence the locution "manifold without boundary". The category of non-associative algebras includes associative algebras.

Todd Trimble
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A distribution can either be related to probability or to generalized functions

Manfred Weis
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In functional analysis, an embedding of one topological vector space $X$ into another $Y$ is simply a continuous linear injection $T$ from $X$ to $Y$. In particular it typically is not an embedding in the usual sense of general topology, since $T$ need not be a homeomorphism onto its image.

Nate Eldredge
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I can't believe this hasn't mentioned already.

What is a graph? Everybody agrees that a graph has vertices and edges (which are usually not oriented). Depending on whom you ask, it might not be allowed to have multiple parallel edges between two vertices. Often a graph is not allowed to have loops at a vertex. Also, is a graph necessarily finite?

And I am not even talking about all the non combinatorial definitions of a graph, e.g. $1$-dimensional CW-complex.

Tashi Walde
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Some people take division ring, division algebra, and skew field to be the same thing; namely a (possibly noncommutative) ring with $1\neq 0$ such that all nonzero elements are units. Some authors however add one or two extra qualifiers to some of these definitions. [Such as forcing noncommutativity, or being finite dimensional over the center, or...]

Pace Nielsen
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The distinction between compact and quasi-compact spaces is another one. If a topological space has the property that any open cover contains a finite subcover, then to some people this space is compact, to others it is quasi-compact. In the latter case, a compact space would then be a quasi-compact space that is also Hausdorff.

Nephry
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The notion of a category:

It sometimes happens that some people define a category as a locally small category without stating it, or probably without knowing that the general definition is where a category consists of a class of morphisms, instead of that morphisms between any two of its objects has to be a set. Thus, if you see the term "category" in the literature/ in lecture notes, the definition of a category might denote a category which is locally small.

The notion of a topological category is another example from this area https://en.wikipedia.org/wiki/Topological_category.

Furthermore, it often happens that there are definitions of something which are equivalent for cartain classes of objects, but not in general:

In $C^*$-algebras: There are notions like "stably finite" (and other notions like this), which can be non-equivalent, see here https://math.stackexchange.com/questions/2073741/stably-finite-c-ast-algebras for instance, which are defined in many ways and which are equivalent for a certain classes of $C^*$-algebras (simple $C^*$-algebras in this case).

The notion 'quasidiagonality' of $C^*$-algebras is another example.

Sabrina Gemsa
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Several distributions from probability theory share names:

The log-gamma distribution. Similarly to the log-normal distribution where we say $X$ is log-normal if $Y = \log X$ is normally (Gaussian) distributed, we can say that $A$ is log-gamma if $B = \log A$ is gamma distributed.

But I have also seen $A$ being called log-gamma if $B = \exp A$ is gamma distributed.

The Weibull distribution may refer to the heavy-tailed distribution function or one of the three max-stable distributions.

Another example is given by Mittag-Leffler distributions, which are distributions on $\mathbb{R}_+$. For $\alpha \in (0,1]$, let $E_\alpha (z) := \sum_{n \geq 0} \Gamma (1+n\alpha)^{-1} z^n$ be the Mittag-Leffler function of index $\alpha$. Then a (normalized) Mittag-Leffler distribution may refer to:

  • a distribution whose CDF is defined with the Mittag-Leffler function:

$$\mathbb{P} (X_\alpha \geq t) = 1-E_\alpha (-t^ \alpha) \quad \forall t \geq 0;$$

these are heavy-tailed distributions, with Laplace transform $(1+\lambda^ \alpha)$, and used mostly in statistics, economics, etc.

  • a distribution whose MGF is defined with the Mittag-Leffler function:

$$\mathbb{E} (e^{zX_\alpha}) = E_\alpha (z) \quad \forall z \in \mathbb{C};$$

their densities decay quickly at infinity, and they appear as the limit of local times of Markov processes.

D. Thomine
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Therkel
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    Another example is the Geometric distribution. In some books it is the trial of first success and in others it is the number of failures before the first success. Not a big deal perhaps, but I do find the ambiguity annoying at times. Different supports, different means, different MGF, but the same name. – John Coleman Nov 24 '17 at 04:24
  • @D.Thomine I have never stumbled upon the Mittag-Leffler distribution so I do not know about it. Since this is a community wiki, perhaps you can add it to the answer? – Therkel Nov 25 '17 at 20:37
  • @Therkel : good idea. That's done. – D. Thomine Nov 26 '17 at 10:34
8

Non-cooperative game theory has the odd property that essentially all authors have their own notions of what a game in extensive form is and thus prove results about principally different mathematical objects.The used notions tend to be equivalent under some reasonable isomorphism, but (almost) nobody ever bothers to make these isomorphisms explicit.

Michael Greinecker
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For a commutative ring with unity $R$, a primitive polynomial in $R[x]$ is a polynomial whose coefficients generate the ideal $(1)$. If $R$ is a UFD, a primitive polynomial in $R[x]$ is a polynomial whose coefficients have the greatest common divisor $1$. These defintions coincide iff $R$ is a PID, hence for higher-dimensional UFD's (like $k[x,y]$) we get two different definitions of the same notion.

A perfect ring is a ring, such that every left module admits a project cover. In positive characteristics, a perfect ring is a ring on which the $p$-th power map is an isomorphism.

Also note the slightly different definitions of a projective morphism in EGA and Hartshorne. To be fair Hartshorne adresses the difference after his definition.

Another (not perfectly fitting) example would be a noetherian scheme and a scheme whose underlying topological space is noetherian.

MooS
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  • "A perfect ring is a ring, such that every left module admits a project cover." Who says that? Never heard this one. – darij grinberg Nov 24 '17 at 23:35
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    @darij: For example T. Y. Lam in A First Course in Noncommutative Rings, Springer GTM 131 (2001). (Well, of course he says "left perfect".) – Fred Rohrer Nov 25 '17 at 11:57
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  1. The category of models of a (finitary or infinitary) first-order theory
  2. A model category which is an abstract setting for doing homotopy theory

The first notion is expounded for example in the Adamek and Rosicky book "Locally Presentable and Accessible categories", chapter 5. As for the second notion, a possible starting point is https://ncatlab.org/nlab/show/model+category.

Philippe Gaucher
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7

Is a parallelogram also a trapezoid?

Fedor Petrov
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    Does the disagreement about this question correlate with the level of the mathematics? I don't recall ever seeing a negative answer to this question (or analogous ones, like square vs. rectangle, circle vs. ellipse, equilateral vs. isosceles) in research-level mathematics, but I can easily imagine a negative answer in high-school textbooks. – Andreas Blass Nov 23 '17 at 16:46
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    Trapezoid or trapezium, depending on US or UK. – Gerald Edgar Nov 23 '17 at 17:16
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    For square vs. rectangle and other your examples I have also never seen negative answer, but for parallelogram vs. trapezoid/trapezium it is usual at least in Russian tradition. It has the following sense for me: parallelogram does not behave like (other) isosceles trapezoids, which, for instance, are always inscribed in a circle. The only parallelogram which is a limit of isosceles trapezoids is rectangular. I would like rectangulars being called isosceles trapezoids, but not other parallelograms. – Fedor Petrov Nov 24 '17 at 08:19
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    @FedorPetrov Your comment (which I saw only now) suggests to me that perhaps the problem lies not in the definition of "trapezoid" but in the definition of "isosceles trapezoid". That is, a parallelogram (other than a rectangle) might reasonably be called a trapezoid, but we might (for the reasons in your comment) decide not to call it isosceles. – Andreas Blass Jan 02 '19 at 15:18
  • See https://matheducators.stackexchange.com/questions/13700/in-what-curricula-are-rectangles-defined-so-as-to-exclude-squares/13766#13766 for an interesting historical take. – kjetil b halvorsen Jan 15 '20 at 01:25
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The notion holomorph in finite group theory

Robert L. Griess writes in [1], Definition 2.21:

A holomorph of a group $G$ is a group $E$ containing $G$ as a normal subgroup such that $C_E(G) = C(G)$ and the action of $E$ by conjugation on $G$ induces $\mbox{Aut}(G)$.

Later in [1] he explains:

Our term holomorph replaces an older and rare usage which means the semidirect product of a group with its automorphism group.

The term holomorph in the new sense is also used in [2]. The term holomorph in the old sense is used in

https://groupprops.subwiki.org/wiki/Holomorph_of_a_group ,

https://en.wikipedia.org/wiki/Holomorph_(mathematics) .

$ $

[1] Robert L. Griess, Jr., Twelve Sporadic groups, Springer Verlag 1998

[2] A. A. Ivanov, The Monster Group and Majorana Involutions, Cambridge University Press 2009

Martin Seysen
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  • So the difference to the old definition is that $E\to Aut(G)$ does not need to be split? Am I understanding this correctly? – Johannes Hahn Dec 11 '17 at 17:31
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    An nontrivial inner automorphism $e$ of $G$ occurs twice in a holomorph $E$ in the old sense: once as an element $e_1$ of $\mbox{Aut}(G)$ and once as an element $e_2$ of $G$. So $e_3 = e_1 e_2^{-1}$ is in $ E \setminus G$ and also in the centralizer $C_E(G)$ of $G$.

    There is no such $e_3$ in a holomorph in the sense of R. Griess.

     – Martin Seysen Dec 11 '17 at 21:45
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Injective: Semigroups can be completely (right/left) injective, while a Banach algebra is injective if the multiplication induces a continuous linear map of the injective tensor product $X\check{\otimes}X$ into $X$ (Varopoulos). I discovered the former when a reviewer insisted on a hyphen in the title of a paper I had submitted: "Injective seimgroup algebras".

J.J. Green
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    And also : injective object in a category with respect to a class of maps, injective set map in the sense of one-to-one set map, or the notion of injective object in homological algebra. – Philippe Gaucher Nov 23 '17 at 09:50
6

An affine stratification can be a stratification into copies of affine space ($\mathbb{A}^{n_i}$) or into affine schemes ($\operatorname{Spec} R_i$).

Z. M
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Daniel McLaury
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6

Differential geometry seems to have many cases of conflicting conventions:

1) Is the Laplacian a non-negative operator? Or a non-positive one?

2) Is the mean curvature the... mean... of the principal curvatures? Or their sum?

3) Is the directional derivative linear in the magnitude of the direction? Or invariant to it?

Etienne Vouga
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  • For (1), if you ever talk to probabilists, you will have another question of whether the Laplacian includes a factor of $1/2$. This is somewhat related to Fedor Petrov's note about Hermite polynomials. – Nate Eldredge Nov 28 '17 at 00:38
6

I'm surprised nobody has mentioned "range" which can by synonymous with "co-domain" or "image" depending on personal preference.

user123824
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    I might be wrong, but I've never seen "range" used to refer to codomain in a modern source. – Noah Schweber Apr 28 '18 at 03:22
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    @NoahSchweber: I'm five years late, but I have seen "range" being used for codomain in at least two well-known modern books: Multivariable Mathematics (2005) by Ted Shifrin on page 24, and The Princeton Companion to Mathematics (2008) on page 11. As a result, I prefer to avoid using the word "range" entirely, instead opting for "image" and "codomain". – Joe Jul 26 '23 at 14:04
6

A partition of a set.

A partition of an integer.

A partition of an interval (in defining Riemann and Riemann–Stieltjes integrals.

A partition of unity.

A partition of a sum of squares in multiple regression, multi-way analysis of variance, experiments with both random effects and fixed effects, the partition of residual sums of squares into pure-error and lack-of-fit sums of squares, etc.

A partition of a matrix.

Quotition versus partition.

  • $6\div2=3$ because $3$ is how many $2$s must be added to get $6$ (quotition).
  • $6\div2=3$ because when $6$ is a sum of $2$ equal terms, then each term is $3$ (partition).
Michael Hardy
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In graph theory, there are at least two different meanings of the word "hereditary".

Some definitions first. Let G be a graph. If graph H is obtained by deleting 0 or more vertices from G, then H is an induced subgraph of G. If graph K is obtained by deleting 0 or more vertices, and deleting 0 or more edges from G, then H is a subgraph of G. So every induced subgraph is a subgraph, but not vice versa.

Let X be a set of graphs. If, for any G in X, any subgraph H of G is also in X, then X is hereditary. If, for any G in X, any induced subgraph H of G is also in X, then X is induced hereditary. So every hereditary set is also induced hereditary.

Now, some authors use the term "hereditary" to refer to induced hereditary sets.

AFarr
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5

Everybody agrees that an isometry is a distance-preserving map. In the context of functional analysis and in particular operator algebras, this is indeed the definition. But in geometry, an isometry is usually required to be bijective, leading e.g. to the isometry group of a metric space.

If there is a danger of confusion, one can still say isometric embedding or isometric isomorphism for disambiguation, whenever no specific term such as unitary is available. (Incidentally, unitary is also an overloaded term meaning either "unitary operator" or "having a unit element".)

Tobias Fritz
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4

I believe that the notion of Euclidean domain provides another example of this situation.

Some authors define Euclidean domain as an integral domain $D$ endowed with a function $d \colon D\setminus \{0\} \to \mathbb{Z}^{+}$ such that

(i) If $a,b \in D \setminus \{0\}$ and $a \mid b$, then $d(a) \leq d(b)$.

(ii) If $a, b \in D\setminus \{0\}$, then there exist $q, r\in D$ such that $a=bq+r$ where $r=0$ or $d(r)<d(b)$.

(cf. Herstein's Topics in Algebra, Stewart's Algebraic number theory and Fermat's Last Theorem, etc.). Since this is the definition I was more accustomed to, I must confess I felt a wee bit uneasy when I learnt that, in their book on problems in algebraic number theory, R. Murty and J. Esmonde had chosen to define this concept as follows:

If $R$ is [an integral] domain with a map $\phi \colon R \to \mathbb{N}$, and given $a, b\in R$, there exist $q, r \in R$ such that $a=bq+r$ with $r=0$ or $\phi(r)<\phi(b)$, we call $R$ an Euclidean domain.

Later on, I would come to the conclusion that they'd decided to disregard the first condition in the definition à la Herstein because in their examples, etc., they were to analize "euclidianness" with respect to the norm in the extension which determines an application that is even completely multiplicative.

José Hdz. Stgo.
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    I seem to recall once seeing a generalization of Euclidean domains, where $d$ or $\phi$ or whatever took values not necessarily in $\mathbb{N}$ but in any well-ordered set. But I don't think they used a different name; I think they still just called them "Euclidean domains"! (I mean, to my mind, this should probably be the standard definition, but it isn't, so, clarity would be a good thing...) – Harry Altman Dec 01 '17 at 01:51
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    The situation is better than you expected: a Euclidean domain without condition (i) on the Euclidean function always has another Euclidean function satisfying condition (i). See Theorem 2.1 in https://kconrad.math.uconn.edu/blurbs/ringtheory/euclideanrk.pdf. – KConrad Oct 30 '22 at 20:35
4

Singular support of a sheaf seems to be a subset of the cotangent bundle, whereas the singular support of a distribution is a subset of the base space. The former is more like the wavefront set, as far as I can intuit.

People try to avoid the confusion by denoting the former by S.S., and the latter by sing supp, but in my mind they read the same.

liuyao
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4

For some people, the dihedral group $D_n$ has order $n$ and for others, its order is $2n$. See the discussion on MathStackExchange. I'll look it up when I get a chance.

Michael Hardy
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Stephen Meskin
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    Have you had a chance to look it up? – Gerry Myerson May 24 '19 at 23:38
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    @GerryMyerson, probably https://math.stackexchange.com/questions/2560348/terminology-for-dihedral-groups . (Of course I think almost no-one writes $Dn$, as opposed to $D_n$, so that is probably a typo.) Brian Conrad's (and DavidWheeler's) argument was that one can use $\operatorname D_{2n}$ for the symmetries of an $n$-gon exactly when one uses $\operatorname S_{n!}$ for the symmetric group on $n$ letters. – LSpice May 25 '19 at 17:17
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    @LSpice there is a reason group theorists think it's okay to write dihedral groups as $D_{2n}$ and symmetric groups as $S_n$ instead of $S_{n!}$. I once asked Richard Foote about this ("Would you call the permutations of 5 letters $S_{120}$?") and here was his reason: in practice, if an abstract symmetric group arises you can nearly always find a set of $n$ things that it permutes, but abstract dihedral groups usually arise with no regular $n$-gon involved in some way, so group theorists prefer to label those groups by their size, just as we do for the quaternion group $Q_8$ (and $Q_{2^n}$). – KConrad Oct 30 '22 at 20:42
4

I think it's fair to mention that in mathematical physics many terms used in pure mathematics with some definition are applied to describe cognate objects with another definition, in a potentially irking way. :)

  1. A Lie algebra may be called "Lie group".
  2. The renormalization group is usually only a semigroup.
  3. Distributions may be called "functions", like "the Dirac delta function"
  4. If you go into gauge theory you have many terms with different meanings from mathematics. You can check the dictionary in the wiki page
  5. Talking of "gauge", there is also the gauge of the "gauge or Henstock-Kurweil integral", a generalization of the Riemann integral.
  6. A vector field on Minkowski spacetime will be called a "scalar field", if it transforms under Lorentz transformations in the trivial representation (on $\mathbb R^4$, but it can be generalized), instead of the defining representation. We can define these terms on space or galilean spacetime with respect to orthogonal or galilean transformations.
  7. One example i came across is in stochastic PDEs: "stochastic quantization" may refer to adding a fictitious time dimension à la Parisi-Wu and averaging over a stochastic process to get a quantization of a classical PDE (say of the equations of motion of euclidian 3D $\phi^4$-theory), but "stochastic quantization" may be just à la Nelson, without additional dimension, with the equilibrium measure of a Markov process corresponding to the ground state of the associated Schrödinger equation. It may be confusing, especially when considering yet other dynamics related to the above by Wick-rotation.

Generally when there are different mathematical models for a single physical theory we may find a given mathematical term "interpreted" differently, applied to objects of a different mathematical nature. Examples of more physical than mathematical terms come to mind easily: "state", "position", "momentum", "energy"; in classical physics those are real numbers or functions on state space which is a manifold (finite dimensional for point physics or infinite-dimensional for continuum physics), while in quantum physics those are self-adjoint operators on state space -though actually they are more general than that, as a single self-adjoint operator only corresponds to position/momentum along a single dimension, and if we want the full position we must consider several such operators, and this is usually not described in textbooks. It is more difficult to find purely mathematical terms with different interpretations. Mathematics tends to use physically-inspired terms: for instance a "sheaf" or a "bundle", "separated", "compact", "rigid",... "Rigid" is actually an example of a term with different definitions: in topology "rigid" refers to classes of spaces which are topologically determined by their homotopy type, whenever in that class 2 spaces have the same homotopy type they are homeomorphic (or diffeomorphic, or even isomorphic with respect to generally finer structures), like in Mostow's rigidity theorem. But "rigid" is also used in "rigid analytic space". But i'm leaving mathematical physics with these examples, and this term has probably been given in another answer.

Coming back to different physical models of the same phenomenon: in general relativity you have black holes, which are solutions with a light-trapping bounded region of space (or the bounded region itself). But there are also several theories of quantum gravity, with black holes as they describe the same natural objects. In string theory we have a Hilbert space of states and a black hole would be certain states or "ensembles" of states (a set with a locally finite measure). We may even add dualities to this picture, and then black holes may correspond to free states on a related space, or thermal (equilibrium) states. My description is vague but it is only meant to give a sense of how words can become overloaded: the point is that practitioners would find it more practical to use the same word and implicitly sort out possible ambiguities, than to be super-rigorous. This tradeoff happens in mathematics, but mathematics users tend to nitpick more on possible ambiguities.

plm
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    And, as far as I can tell, a Lie group = Lie algebra is regarded not as an algebra, but as a collection of structure constants …. At least that's an equivalent, if confusing, way of thinking about (based) algebras! – LSpice Aug 09 '23 at 19:24
3

In the field of dynamical systems, you would see various definitions for a stable system. There are even different definitions for specific names, e.g. Lyapunov stability in the literature.

I have seen that some call the limit cycle a stable manifold, while others consider it an unstable one.

polfosol
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3

The wide range of choices in the definition of an automorphic form is particularly annoying. Depending on the purposes, it could be a meromorphic function fully invariant by a certain discrete group of transformation, a holomorphic function almost-invariant, a differential form, a subrepresentation of an $L^2$ space, a classical or an adelic object, a solution to a partial differential equation, etc.

These are sometimes related, sometimes definitely different, and the lack of vocabulary consistency in such an active field requires some care when dealing with object (leading authors to usually restate a precise definition of the objects they are dealing with, or at least quoting the literature associated with the paradigm they have chosen.

Desiderius Severus
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3

Graphs.

Some define them without loops or multiple edges. Some with both. They can also be directed or undirected too. Some define quivers to be directed graphs with multiple edges and loops which is a atart on the naming confusion.

Mozibur Ullah
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The word conjugate has several different meanings. Wikipedia lists 16 mathematical meanings, some of which are related, but many of which are very different.

Ira Gessel
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    Is this thread supposed to be about particular words that have disparate definitions or about particular concepts with actually conflicting definitions? – Michael Hardy Sep 12 '23 at 02:14
  • ${}\qquad\uparrow\qquad\ldots,$and now I see that I took it as the former in my posted answer to this question concerning the word partition, and as the latter in may posted answer about linear regression. – Michael Hardy Sep 12 '23 at 18:06
2

The notion of spectrum in operator theory. Some people assume that if $\lambda$ has a property that $T-\lambda I$ is injective with dense range which is not the whole space and with continuous inverse then $\lambda$ is not an element in spectrum since $(T-\lambda I)^{-1}$ can be extended continuously other authors consider such $\lambda$ as an element of spectrum.

truebaran
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1

In set theory a forcing notion can be a pre order with a largest element or lowest element, depending on the style of the author.

Martin Sleziak
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Port
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1

"Linear regression"

Statisticians know better than to think that the term "linear regression" is so called because one is fitting a straight line. Often in linear regression one fits a parabola or a sinusoid or some other sort of functions via ordinary least squares, and the word "linear" is just as apt as when one fits a straight line, and is more appropriate than it would be if "linear regression" meant fitting a straight line, since, if "linear" referred to the function one is fitting rather than to how it is done, then the word "affine" would be more appropriate.

So where is the "other" definition --- the one according to which it does mean fitting a straight line? One example is when an astronomer taught a course in a statistics department on applications of statistics to astronomy, and reserved the word "linear" for fitting a straight line. Respectable scientists from outside of statistics do that.

Consider the model $$ Y_i = a + bx_i + cx_i^2 + \text{error}_i, $$ and the model $$ Y_i = a + b \cos x_i + c\sin x_i + \text{error}_i, $$ where the ordinary least-squares estimates are $\hat a, \hat b, \hat c.$

For either of those two models, the mapping $$ \left[ \begin{array}{c} Y_1 \\ \vdots \\ Y_n \end{array} \right] \mapsto \left[ \begin{array}{c} \hat a \\ \hat b \\ \hat c \end{array} \right] $$ (with the $x$s fixed) is linear.

Contrast this with the model \begin{align} & \operatorname{logit} \Pr(Y_i=1) = \operatorname{logit}( 1- \Pr(Y_i = 0) ) = a + bx_i, \\[8pt] & \text{where } \operatorname{logit} p = \log \frac p {1-p} \text{ for } 0<p<1. \end{align} (long "o", soft "g")

Here the usual estimates $\hat a, \hat b$ are not found by least squares, but rather \begin{align} \left(\hat a, \hat b \right) = {} & \operatorname*{argmax}_{a,b} \Pr( Y_1=y_1\ \& \cdots \&\ Y_n=y_n\mid a,b) \\[8pt] = {} & \operatorname*{argmax}_{a,b} \prod_{i\,:\,y_i\,=\,1} \operatorname{logit}^{-1}(a+bx_i) \\ & \qquad \times \prod_{i\,:\,y_i\,=\,0} \left( 1-\operatorname{logit}^{-1} (a+bx_i) \right). \end{align} This is one of many instances of nonlinear regression.

Michael Hardy
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1

sample

Thoroughly stir the population of jabberwockies and pick one at random, then stir again and pick another independently of the first, and so on until you have $50$ jabberwockies.

Do you have $50$ samples, or one sample consisting of $50$ independent observations?

It seems that in signal processing and data science and various other disciplines, this is called $50$ samples, and in statistics it is one sample.

Notice that the term "sample size" is commonplace: the size of this one sample is $50.$ And the two-sample t-test is called the two-sample t-test. A two-sample t-test may use one sample of $50$ northern jabberwockies and another sample of $40$ southern jabberwockies.

Michael Hardy
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1

Does entropy decrease or increase? What mathematicians call "entropy" is actually what physicists would call "free energy." Free energy includes a term proportional to entropy, but with a minus sign.

Michael Renardy
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0

When is a function concave? When is it convex? Do you determine this by looking at the graph "from above", or "from below"?

Rodrigo A. Pérez
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  • I can never remember which way this goes without looking it up, but are there really varying conventions in upper-level mathematics (as opposed to calculus textbooks)? I thought one spoke exclusively about convex functions in, say, measure theory. – LSpice Nov 29 '17 at 18:59
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    @LSpice Instead of looking it up, you could just look up (i.e., look at the graph from below) and you'd have the standard convention. – Andreas Blass Dec 01 '17 at 23:55
  • @AndreasBlass, hah, very nice! Indeed, once I know the answer I can justify it ex post facto, but unfortunately that doesn't help me personally remember, as @‍RodrigoA.Pérez puts it, whether to look "from above" or "from below". Maybe thinking of it as looking up will help. – LSpice Dec 02 '17 at 01:39
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    The definition of "convex function" is universally agreed and standard, luckily – Pietro Majer Jun 12 '19 at 12:11
0

Let $X_1,X_2,X_3,\ldots$ be independent random variables.

Sometimes it is said that a stopping time for this random process is a random variable $T$ for which the truth value of $T=n$ (for $n=1,2,3,\ldots$) is determined by the values of $X_1,\ldots,X_n.$

And sometimes it is said that a stopping time for this random process is a random variable $T$ for which the each of the events $\big[T=n\big]$ for $n=1,2,3,\ldots$ is independent of the sequence $X_{n+1}, X_{n+2}, X_{n+3},\ldots\,.$

The second definition is more inclusive than the first.

Michael Hardy
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-1

I think the concept of simple Lie groups has some non equivalent definitions.

Ali Taghavi
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