How to invoke constants badly

Question

In a nice and witty lecture titled "how to write mathematics badly" (available on YouTube at https://www.youtube.com/watch?v=ECQyFzzBHlo&t=23s), Jean-Pierre Serre describes various ways in which a paper can be poorly/confusingly/inaccurately written.

Around min 34:00 in the previous link, he criticizes the use of the word "constant", in particular in inequalities. The example he provides is of the type:

$$\|Af\|\le C\|f\|$$ for some constant $C$

where $A$ is a complicated operator depending on many parameters. In this case, he says, usually the only thing that the writer means is that $C$ does not depend on "some of the data" of the problem. He adds that this attitude "caused lots of mistakes".

What are examples of these mistakes? Has any significant piece of mathematics been rewritten or erased altogether because of some problem with proofs invoking "constants" too nonchalantly?

I can't cite any mistakes caused by this practice, but I have no doubt there have been many. When I first ran into this when studying PDEs, it made me very uncomfortable, so I always wrote explicitly what parameters $C$ did depend on. So my $C$ always looked something like $C(n,a,b,P, Q, \beta)$. If I could write an explicit formula for $C$, I often did, since the dependence of the parameter sometimes mattered, too. — Deane Yang, Dec 21 '20 at 17:37
It is the kind of mistake students make every day, and I have certainly seen it in submitted manuscripts, which were insufficiently memorable to recall specifics. I do not see the point of creating a list here. — Michael Renardy, Dec 21 '20 at 18:18
Indeed I was not sure about creating a list, you can remove the tag if you want. However, I think Serre was referring to published papers. — Alessandro Della Corte, Dec 21 '20 at 18:20
I, a nephew of Jean-Pierre Serre, am certainly guilty of invoquing such constants in my research activity. But the inequality I am proud of (Compensated Integrability) has an explicit and accurate constant. — Denis Serre, Dec 21 '20 at 19:26
In the rest of mathematics, this problem is solved by introducing quantifiers at the right place in the statement (usually written out in words). I've never fully understood why this is not [always] used when doing estimates... — R. van Dobben de Bruyn, Dec 21 '20 at 19:28
On a related note, the "little-oh" notation $o(1)$ has always bothered me, and for a while I tried using the notation $o(n^0)$ instead, but I couldn't persuade anyone else to adopt it. — Timothy Chow, Dec 21 '20 at 23:13
@R.vanDobbendeBruyn In any field of math, if quantifiers were introduced separately for each statement, the exposition would become very cluttered (esp. if there are a lot of nonce variables that occur in just one line). Outside of "hard" analysis, though, there are often convenient definitions that hide a lot of the quantifiers (e.g., continuity hides epsilon and delta quantifiers, a universal property in a category hides universal and existential quantifiers, etc.). But in the "fuzzy" world of estimates we often don't have this luxury (unless one uses something like nonstandard analysis). — Terry Tao, Dec 22 '20 at 01:31
@TimothyChow My guess what is bothering you about $o(1)$ (and possibly also $O(1)$) is that it looks like a function applied to $1$, but it really is a rather different use of the (highly overloaded) parenthesis symbol. Personally I like to use $o_{n \to \infty}(1)$ instead of $o(1)$ in order to explicitly specify the asymptotic limit involved here (and there are variants such as $o_{n \to \infty;k}(1)$ when one wants to permit the decay rate to depend on additional parameters such as $k$). — Terry Tao, Dec 22 '20 at 01:37
@TerryTao : Actually, I think that what bothers me is that it strikes me as an example of misguided "simplification." In the expression $o(n)$ or $O(n)$, the $n$ is there not just to indicate the rate of growth, but to indicate what variable is going to infinity. So $o(n^0)$ or $O(n^0)$ is the logical extension of that notation, but then people reflexively "simplify" the expression, perhaps without consciously realizing that they're not really simplifying, but actually destroying information. — Timothy Chow, Dec 22 '20 at 02:12
A very similar and common sloppiness occurs with "for sufficiently small $\varepsilon>0$" when the required smallness actually depends on other parameters lurking in the background. — Brendan McKay, Dec 22 '20 at 02:43
The opposite question is how one conveys to the reader that a constant doesn't depend on anything; i.e. that it could be replaced by an explicit number. I've see "universal constant" and "numerical constant" but I don't know if everyone would get the message. — Brendan McKay, Dec 22 '20 at 03:51
@BrendanMcKay A standard phrasing in analytic number theory is "The implied constant is absolute". (There is also the further distinction of constants into "effective" and "ineffective"; what you refer to as a "numerical constant" would likely be called an "effective absolute constant" (or "effectively computable absolute constant") in analytic number theory.) — Terry Tao, Dec 22 '20 at 04:49
@DeaneYang I prefer to state on which parameters a constant does not depend. I see no point in tracking the dependence on stuff that's not relevant to my problem. An incarnation of this paradigm is to state how a constant depends on certain things, like it depends on an elliptic operator only via its coefficient bounds (i.e., it does not depend on the remaining data of the operator). Of course it is then the responsibility of the user to check that one really only varies parameters that were excluded from the dependence list. — Sebastian Bechtel, Dec 22 '20 at 10:33
@TimothyChow Having the asymptotic variable appear inside $o()$ can sometimes help identify the asymptotic limit but is not a reliable solution for doing so. For instance the statement $t = o(t^2)$ without context does not identify whether the asymptotic limit is $t \to \infty$ (where the statement is true) or $t \to 0$ (false). Similarly, the statement $n^k = o((n+1)^k)$ for natural numbers $n,k$ without additional context does not identify whether the asymptotic limit is $k \to \infty$ with $n$ fixed (true) or $n \to \infty$ with $k$ fixed (false). Subscripting is my recommended solution. — Terry Tao, Dec 22 '20 at 17:36
@TerryTao I guess the writing style of mathematicians working in "hard" analysis estimates is more or less optimized for their aims - after all, while mathematics is an exact science, writing it and maybe even reading it is somewhere in the middle between an empirical science and an art. — Alessandro Della Corte, Dec 22 '20 at 17:41
@TimothyChow, re, isn't the information you're trying to recover precisely that provided by @‍TerryTao's suggestion $o_{n \to \infty}(1)$? — LSpice, Dec 22 '20 at 19:54
@LSpice : Terry Tao's suggestion includes more information because it gives the limiting value of $n$. But I doubt that many people will adopt it except in circumstances where it is really important to emphasize what the limit is. People are fundamentally lazy, after all. Who is going to write, "Comparison sorting takes $O_{n\to\infty}(n\log n)$ comparisons"? I was hoping that maybe people could be induced to switch from $o(1)$ to $o(n^0)$ since it's not all that much more work. But either because old habits die hard or because it's easier to write $1$ than $n^0$, I have zero converts so far. — Timothy Chow, Dec 22 '20 at 23:27
(By the way, when I say zero converts, that includes myself!) — Timothy Chow, Dec 22 '20 at 23:28
@TimothyChow But what if you have many variables and only one is going to infinity? Like, let's say you want to note $kn = o(kn^2)$, where $k$ is some fixed thing and $n \to \infty$. I really don't think it's wise to adopt the convention that "whatever is in the o/O is the parameter growing or shrinking". — mathworker21, Dec 25 '20 at 11:34
@mathworker21 : I'm not claiming to solve all problems with my suggestion. I'm claiming only that the notation $o(n^0)$ is no worse than, and probably better than, the notation $o(1)$. Do you disagree, and believe that the notation $o(1)$ is strictly better than the notation $o(n^0)$? (Assume that no other changes are made to the standard notation.) — Timothy Chow, Dec 27 '20 at 04:09
@TimothyChow "Do you disagree, and believe that the notation $o(1)$ is strictly better than the notation $o(n^0)$?" Yes, of course. The notation $o(1)$ is easier on the eyes and the notation $o(n^0)$ reinforces the false conception that variables inside the parentheses are the relevant parameters going to infinity or $0$. — mathworker21, Dec 27 '20 at 04:17
@mathworker21 : So $o(n)$ is also bad, I guess, because it reinforces the same false conception? — Timothy Chow, Dec 27 '20 at 12:51
@TimothyChow No. I said $o(n^0)$ is reinforcing the false conception because it's clear the author went out of their way to emphasize that $n$ is the variable going to infinity or $0$. If $o(n^0)$ genuinely popped up naturally, that would be fine. — mathworker21, Dec 27 '20 at 14:29
@mathworker21 : The fact that we use $o(n)$ means that we trust the reader to interpret the notation correctly. If we trust readers that far, then why not trust them with $o(n^0)$? It seems no more unnatural to me than $o(1)$. The trouble with $o(1)$ is that we have intentionally destroyed information irretrievably for no good reason. Note, by the way, that I am not adopting the convention that whatever is in the o/O is the parameter growing or shrinking. I am making only one change to the standard notation. — Timothy Chow, Dec 27 '20 at 14:39
If there are many variables and only one is going to infinity, then I will happily adopt whatever notation you suggest. — Timothy Chow, Dec 27 '20 at 14:45
@TimothyChow But the reason you're making the change is to reinforce the idea that whatever is in the o/O is the parameter growing or shrinking. My issue is not only when there are multiple variables. Suppose we have some process changing over time $t \to \infty$. As $t \to \infty$, our position $x(t)$ changes in some way. And we have two functions $f(x)$ and $g(x)$ that are functions of position $x$. If we want to say $f(x) = o(g(x))$, we'll be referring to $t \to \infty$ and not any limit of $x$. — mathworker21, Dec 27 '20 at 16:04
My point is, we should always just rely on context and the reader to understand. The $o(1)$ should be clear from context. If there is a need to emphasize to the reader that $n$ is the relevant parameter shrinking/growing, then one should write $o_{n \to \infty}(1)$. I just see no point to $o(n^0)$. — mathworker21, Dec 27 '20 at 16:06

score 41 · Answer 1 · answered Dec 22 '20 at 07:23

It came to my mind what's perhaps the oldest example of this kind of mistake, so I add an answer to my own question: in 1821 Cauchy 'proved' that convergent sums of continuous functions are continuous, and later on Abel found counterexamples (see [1] for historical details). Of course Cauchy implicitly assumed uniform convergence, which means that he treated his $\delta$ as "more constant" than it was...

[1]: Sørensen, H. K. (2005). Exceptions and counterexamples: Understanding Abel's comment on Cauchy's Theorem. Historia Mathematica, 32(4), 453-480.

The nLab article on the Cauchy sum theorem gives a thorough discussion of this theorem, including various suggestions for how to argue that Cauchy's argument was correct after all. But someone is confused here— if not Cauchy, then his critics. — Timothy Chow, Dec 22 '20 at 17:24

score 16 · Accepted Answer · edited Dec 24 '20 at 14:08

16

Edit: The original answer below refers to Nelson's attempt from 2011. Upon a cursory look at the afterword by Sam Buss and Terence Tao to Nelson's paper placed in arxiv in 2015 (after his death), it seems he later attempted to address the error referred to in the original answer below; it would be interesting to know what the experts think on how successful his efforts were or potentially can be.

Original Answer: Edward Nelson's recent project on finding inconsistency of arithmetic (which was the subject of a MathOverflow Question) might be pertinent. The error, discovered by Terence Tao, seems to be the dependence of a constant on the underlying theory that Nelson did not account for.

edited Dec 24 '20 at 14:08

John Coleman

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answered Dec 22 '20 at 18:03

pinaki

5,064

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This is fascinating. The afterword by Buss and Tao indicates that the final paper (while not establishing what it set out to do) contains many interesting ideas worthy of further exploration. It was also incomplete at the time of Nelson's death, containing only 6 of a planned 10 parts. Does anyone know if there has been any work done since 2015 in separating the wheat from the chaff in Nelson's paper? – John Coleman Dec 24 '20 at 14:15

score 15 · Answer 3 · answered Dec 22 '20 at 02:20

15

A good example (of a somewhat different kind though) was given by Adian in the introduction to his book "The Burnside problem and identities in groups" (1975, English edition 1979), where he refutes the proof of the main result from his rival's book by stating that

``the conditions $$ \begin{aligned} &u_4 = u_1 +r_{25} \; \text{(p.145, line 10 from below)} \\ &r_{25} \ge u_{37} +54/e \; \text{(p.283, line 4 from below)} \\ &u_{37} >14\alpha +214/e, \; \text{where}\; \alpha=\varepsilon_{30}+u_{13} +6u_4 \; \text{(p.221, lines 11 and 12 from below)} \end{aligned} $$ give an obvious contradiction $u_4>r_{25}>u_{37}>u_4$.''

answered Dec 22 '20 at 02:20

R W

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Well if we want to be pedantic, $u_1<0$ would solve everything here, and so far I don't see that excluded. – Torsten Schoeneberg Dec 22 '20 at 18:37
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Isn't the implied inequality $u_4 > r_{25} > u_{37} > 84\cdot u_4$? Am I being dense? (Assuming all constants are positive) – o r Dec 22 '20 at 19:35
@or, re, of course $84u_4 > u_4$ if everything is positive. – LSpice Dec 22 '20 at 19:48
This is an exact quote – R W Dec 22 '20 at 21:54
1

@LSpice ok thanks I was being dense, I don't even know what I mixed up here ... – o r Dec 22 '20 at 23:21

score 8 · Answer 4 · edited Dec 22 '20 at 19:50

There are periodically false proofs of the rapid decay (RD) Property (see Chatterji and Saloff-Coste - Introduction to the property of rapid decay) for cocompact lattices in semisimple Lie groups (Valette's conjecture). This is a functional-analytic property of these groups. As far as I understand, wrong proofs typically make use of some quantitative decreasing of coefficients, and this involves a "constant". The issue (sorry if I'm a bit imprecise) being that this "constant" actually depends on the dimension of something related to the induced representation restricted to a maximal compact subgroup.

How to invoke constants badly

4 Answers4