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$\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}$This question is prompted by a recent highly-upvoted question, Conceptual reason why the sign of a permutation is well-defined? The responses made me realize that my intuition differs from that of many other mathematicians, in a way that I had previously been unaware of.

As a bit of personal background, I recall being taught in middle school that (in not so many words) that there are "24 symmetries of a cube," since you can rotate any face to the bottom (a factor of 6) and then rotate that face in place (a factor of 4). Similarly, for each of the Platonic solids, we can count $4\times 3 = 12$ "symmetries" of a tetrahedron, $8\times 3 = 24$ "symmetries" of an octahedron, and so forth.

One of the various equivalent formulations of the aforementioned MO question is, "What is the conceptual explanation for the existence of the alternating group?" In part because of my background, the answer that came to my mind was, "Because it's the group of symmetries of a simplex." However, people rightly pointed out that to conform to standard usage, this answer should really be phrased, "Because it's the group of orientation-preserving symmetries of a simplex." But once the statement is phrased this way, it suggests that maybe the existence of the alternating group isn't such a "basic" fact after all; maybe it's only the symmetric group whose existence can be taken as basic, and we have to "explain" the existence of a subgroup of index 2 using some kind of algebraic argument.

There's no question that Poonen's version of Cartier's argument is slick and beautiful. Nevertheless, something doesn't quite sit right with me if we call this an "explanation" of the existence of the alternating group. It still seems to me that orientation-preserving rigid motions in Euclidean space are mathematically fundamental, because they are rooted in our physical intuition. Admittedly, in modern mathematics, formally capturing our physical intuition in a direct manner is a cumbersome process; we have to construct the real numbers, and then talk about continuous transformations that preserve the metric. But for me, the complexity of this formal definition does not imply that the underlying concept is complex; rather, it says more about the difficulties involved in formalizing our geometric intuition. It is not hard for me to imagine an alternative universe in which we begin with $\SO(n)$, and think of $\O(n)$ as an extension of $\SO(n)$, much as Spin came before Pin.

But let me now finally come to my question. For those who take the symmetric group as given, and feel the need for an algebraic explanation of a subgroup of index 2, do you feel the same way about other finite subgroups of $\SO(n)$? For example, do you think the existence of the hyperoctahedral group of order $2^n n!$ is "obvious," but the subgroup of index 2 is in need of an algebraic "explanation"? If so, is there an algebraic proof that you feel explains all these "miraculous" subgroups of index 2 in a uniform way? Or is the existence of orientability itself a miraculous fact?

EDIT: Let me make a few additional comments to nudge the discussion in a more technical and less opinion-based direction.

Suppose I don't buy the "argument from geometric intuition" and I seek a more algebraic (or combinatorial) explanation of the existence of the alternating group. It still seems to me that an argument based on determinants, or the distinction between $\SO(n)$ and $\O(n)$, would be a more "conceptual" route than an argument based on actions on the complete graph. For example, suppose I ask the corresponding question for the hyperoctahedral group ("why is there this subgroup of index 2?"). If we've already decided that the relationship between $\SO(n)$ and $\O(n)$ is the key point, then the same explanation works for both questions. Whereas, the graph-theoretic approach doesn't seem to have the same ability to give a unified explanation for both cases.

Am I mistaken about this? Does Cartier's proof generalize to cover all situations that I might try to explain by appealing to the distinction between $\SO(n)$ and $\O(n)$? Does it perhaps generalize even further to explain index-2 subgroups that don't seem to be explicable in terms of $\SO(n)$?

Timothy Chow
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    As someone who also voted to close the original question, I think that this question is fantastic, but surely, if any question is opinion-based, "do you feel the same way about other finite subgroups of $\operatorname{SO}(n)$?" is. – LSpice Mar 14 '22 at 03:03
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    @LSpice The main question is the one in bold. The preceding questions are there to acknowledge that some people (including myself!) may not agree with some of the tacit presuppositions behind the question. It is true that any question that asks for an "explanation" is going to be somewhat subjective and opinion-based, but not (in my opinion!) in a way that disqualifies it from being suitable for MO. If we were to adopt such a strict interpretation of "opinion-based" then (for example) any question asking for intuition or motivation would be off-topic. – Timothy Chow Mar 14 '22 at 03:24
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    Regarding intuition, you need only look in a mirror, or look at your left and right hands, or look at any of many other examples of reflective symmetry, to perceive orientation reversing symmetry in a completely natural and intuitive manner. – Lee Mosher Mar 14 '22 at 03:34
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    I also recommend watching the "redrum" scene in The Shining. – Lee Mosher Mar 14 '22 at 03:35
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    Are you sure that "being taught in middle school" isn't a euphemism for "I was fiddling with a Rubik's cube below the desk ... creak ... creak ... creak ..."? – Michael Engelhardt Mar 14 '22 at 03:41
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    @MichaelEngelhardt Ha! Good one. But I distinctly remember a teacher explaining it to me. The only poetic license I used was the term "middle school": I was educated in British schools, and I think I was in Primary 6 or Form 1 or Form 2 when I learned this. – Timothy Chow Mar 14 '22 at 03:56
  • Constructing the rotational symmetry group of a tetrahedron does not require constructing the reals, since $A_3^+ \subset SO(3,\mathbb{Q})$. – Daniel Sebald Mar 14 '22 at 04:46
  • @DanielSebald: each of the finite subgroups of $SO(n)$ is in a finite extension of the rationals. – Ben McKay Mar 14 '22 at 10:01
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    @DanielSebald In a sense you are right, but I want to describe a physical rotation in a way that automatically excludes reflections. The most direct way to do this is to insist that each point of the tetrahedron trace out a continuous path from $t=0$ to $t=1$, with all distances being preserved at any intermediate time $t$. For that I think I need the reals. If instead we try to say that there's an isometry between the starting and ending states, without saying anything about what happens in between, then it's hard to prevent reflections from sneaking in the door. – Timothy Chow Mar 14 '22 at 11:21
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    Isn't the 1-dimensional case the source of all of this? The general linear group $GL_1(\mathbb{R})$ consists of all non-zero real numbers and has two components. If you believe in the existence and some properties of the determinant then the same follows for $GL_n(\mathbb{R})$ for all $n$. – IJL Mar 14 '22 at 11:31
  • there is a textbook "Course in Algebra" by E.Vinberg, with a short and convincing treatment of determinants and orientation. https://www.google.com/search?q=e.vinberg+algebra&oq=e.vinberg+algebra&aqs=chrome..69i57.8647j0j7&client=ms-android-motorola-gfw&sourceid=chrome-mobile&ie=UTF-8&stick=H4sIAAAAAAAAAONgVuLSz9U3SDZJTqmMf8Royi3w8sc9YSmdSWtOXmNU4-IKzsgvd80rySypFJLgYoOy-KR4uJC08exi4sgvSy0qy0wtX8QqmKpXlpmXlFqUrpCYk56aVJQIAG6CzGhlAAAA&ictx=1&ved=2ahUKEwjWh7Cy_MX2AhVNh1wKHRwLDpsQyNoBKAB6BAgDEAg – Dima Pasechnik Mar 14 '22 at 15:59
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    @IJL: determinants are defined involving the sign $S_n \to {\pm 1}$, which is what the other post was about. The OP wants to define signs/determinants/etcetera without ever constructing algebraically a homomorphism $\operatorname{GL}_n(\mathbf R) \to \mathbf R^\times$, but instead finding an a priori reason that $O(n)$ has two components. – R. van Dobben de Bruyn Mar 14 '22 at 22:58
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    I would say the issue has to do with dimension. In dimension 2 and maybe 3, it's intuitively obvious that one can't obtain a reflection by continuous rotations. The question is whether your intuition allows you to transfer that from those dimensions to higher dimensions that can't be visualized. For some people it might, for others it might not. An algebraic proof should work for all (at least once they learn the relevant algebraic notation). – Will Sawin Mar 14 '22 at 23:20
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    It’s interesting that the Symplectic Groups don’t have an index 2 subgroup. – Noah Snyder Mar 15 '22 at 03:42
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    I was always wondering about the special properties of the number $2$. I believe that its existence is the unique collaboration between God and the Devil. – Roland Bacher Mar 15 '22 at 11:22
  • @WillSawin One might be able to argue for the extrapolation to higher dimensions by developing the connection between the sign and the first stable homotopy group of spheres, and then appeal to the Freudenthal suspension theorem, which gives a good upper bound on how many suspensions need to be taken before the groups stabilize. There would be quite a bit to flesh out, of course, and many would object that Freudenthal has no place in a discussion of intuition :) – Tim Campion Mar 15 '22 at 16:06
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    @TimCampion A different topological approach is the action of $O(n)$ on $S^{n-1}$ with stabilizer $O(n-1)$. The fibration exact sequence should give an isomorphism between the component groups of $O(n)$ and $O(n-1)$ as soon as $S^{n-1}$ is simply connected, which reduces you to the $n=2$ case which can be proven by manipulating a physical object. I think this will be easier / more intuitive than stable homotopy but am not completely sure. – Will Sawin Mar 15 '22 at 16:09
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    I think the question is a "baby instance" of a fairly general phenomenon. Setting up the theory behind talking about geometry or topology effectively, this was a major human undertaking that involved thinking about how we decide to structure our thought, and what tools we use to "encase" them in language. Geometry and topology involve a different mode of thinking than the purely verbal. The proofs reflect this -- there is a big formal gap between Lee Mosher's "look in a mirror" intuition and a short proof, written in a printed language. – Ryan Budney Mar 15 '22 at 16:36
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    Once you have the determinant homomorphism, it's clear that $GL_n(\mathbb{R})$ has two components (corresponding to positive and negative determinant). Orientation is just our name for those two components, so one then only needs to note that some elements of the symmetric group have negative determinant. So it seems to me that what's really needed is a clean/intuitive (according to taste) construction of the determinant, along the lines of Bjorn Poonen's construction of the sign. – HJRW Mar 15 '22 at 16:50
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    @TimothyChow Would a proof along these lines satisfy: Prove the complex spectral theorem without reference to determinants/orientations, and so we have eigenvalues which are either $\pm 1$ or come in conjugate pairs. Given a continuous curve $\gamma: [0,1] \to \mathcal{O}(n)$, prove that the eigenvalues can be continuously parameterized. Conclude that the product of the eigenvalues is a continuous map from $[0,1]$ to the discrete space ${0,1}$. I am not absolutely sure (especially about continuous variation of the eigenvalues), but I think this could work. – Steven Gubkin Mar 15 '22 at 16:53
  • @RolandBacher, re, there's the witticism "$2$ is the oddest prime." (But we might also reflect that perhaps it seems so odd only because it's the only number small enough for us to see all the situations where weirdness emerges. If I worked with $\mathsf G_2$ as much as with $\mathsf B_n$, $\mathsf C_n$, and $\mathsf D_n$, then I am sure I would find $3$ exceedingly strange, too.) – LSpice Mar 15 '22 at 20:12
  • @TimothyChow I think that the answer here gives a completely satisfying geometrical argument. Please let me know if you agree: https://mathoverflow.net/a/417730/1106 – Steven Gubkin Mar 16 '22 at 14:47
  • I know this is not the point of your question at all, but did you know you can count rotational symmetries by just doubling the number of edges? $2\times 6 = 12, 2 \times 12 = 24, 2 \times 30 = 60$ – Sean Eberhard Mar 18 '22 at 07:51

4 Answers4

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To me, it is not that hard to imagine an alternate universe where the fact that $|\pi_0(GL_n(\mathbb{R}))| = |\pi_0(O_n(\mathbb{R}))| = 2$ is an unstable fact that holds for small $n$, but not in very high dimensions. In such a universe, determinant would only be defined for small square matrices, and $\bigwedge^n \mathbb{R}^n$ would be $0$-dimensional for large $n$.

By way of analogy, suppose we lived in a two dimensional universe. It would be completely intuitive to us that $\pi_1(SL_2)$ was $\mathbb{Z}$. Elementary school textbooks would say something like "every person, throughout their life, has made a certain number of full turns to the left, and a certain number of full turns to the right, and the difference between these numbers is called the 'winding number'." It would then be extremely surprising when we studied three dimensional geometry (that arcane, counterintuitive subject!) and learned that winding number is only defined modulo $2$.

David E Speyer
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    By the way, I strongly agree with your comment that the previous thread taught me more about mathematicians different intuitions than about mathematics. To me, the fact that $\sigma \left( \prod_{i<j} (x_i-x_j)\right) = \pm \prod_{i<j} (x_i-x_j)$ for any permutation $\sigma$ is obvious, and the fact that this defines a group character is likewise. Any statement that involves the word "torsor" requires some thought. – David E Speyer Mar 14 '22 at 13:44
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    I'm not sure that anyone in the previous MO question said the polynomial proof relied on "not obvious" facts- just that it felt contrived in some way. – Sam Hopkins Mar 14 '22 at 14:30
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    I never liked the polynomial proof for the existence of the sign map because whenever I have taught algebra we do groups before rings and so I end up introducing automorphisms of polynomial rings in order to show S_n acts by automorphisms of that ring. So I prefer for that reason some variant of the proof not using polynomials – Benjamin Steinberg Mar 14 '22 at 18:33
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    @DavidESpeyer: One could replace the t-word with saying that the action is simply transitive, which might be less offensive to some! – Bjorn Poonen Mar 14 '22 at 20:45
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    @DavidESpeyer: Part of what I was trying to do was to avoid the implicit choice of ordering of the $n$-element set in the polynomial construction, by making all possible choices. I guess there is also a version of the argument that is kind of between the torsor argument and the polynomial argument, namely to consider the action of $S_n$ on the set of both square roots of the symmetric polynomial $\prod_{{i,j}} (x_i-x_j)^2$. This slightly simplifies the proof in the polynomial argument that one gets a homomorphism (and avoids having to order the $n$-element set). – Bjorn Poonen Mar 14 '22 at 20:50
  • XKCD considers my scenario https://xkcd.com/2882/ . – David E Speyer Jan 30 '24 at 12:19
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One interesting phenomenon is that there’s a good notion of symmetric tensor category $\mathrm{Rep}(O_t)$ when $t$ is not an integer, but no good notion of $\mathrm{Rep}(SO_t)$ for generic $t$. (Equivalently, $\mathrm{Rep}(O_t)$ doesn’t have a determinant object.). So that’s a clear sense in which something special and surprising happens to allow SO(n) to exist.

Noah Snyder
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In the spirit of the answer by David Speyer, one could also point out that perhaps the important role of 2 (or bilateral symmetry) in our study of symmetry can be seen as an artefact of living in a world where the "special valuation at $\infty$" dominates our intuition. (Since $\mathbb{Z}/\langle 2\rangle$ occurs as the maximal finite subgroup of $\mathbb{R}^{*}$.)

If we lived in a $p$-adic universe for a specific $p\neq\infty$, then the maximal finite subgroup of the group $\mathbb{Z}_p^{*}$ (the units in $p$-adic integers) may have played a bigger role in our intuition.

The above does not quite work the way I imagined!

The group generated by rotations (defined as the product of two reflections) always (over a field of characteristic different from 2) generates a subgroup of index 2 in the group generated by reflections. My answer to the earlier question mentioned above gives an elementary proof.

So, in this context, the homomorphism to $\mathbb{Z}/\langle 2\rangle$ does appear to play a special role.

Another way to see is by appealing to the Cartan-Dieudonné theorem and determinants.

Kapil
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(Just a long comment that doesn't directly answer the bold question, but does [I hope] address an implicit question.) What I've found fascinating in thinking about your answer from that other thread are the differences between intuition and proof.

Some of the benefits of intuition are:

  • It gives us a sense of what is right and wrong, without the need for long complicated proofs. This can significantly shorten the time needed for learning, or the time needed when looking for new ideas, and proofs.
  • Similarly, intuition can also tell us when something surprising is happening. It helps us know when to double-check that a proof didn't take a wrong turn.
  • It can help us develop a simpler model of the world, and a way to explain known facts to others.

Some of the weaknesses of intuition are:

  • It often doesn't give us the greater information included in a full proof.
  • It might rely on a single model of a situation, rather than multiple models.
  • Intuition might be "just wrong". (Though, those intuitions guided by proofs are often more robust.)
  • Even when intuition is right, it is not always possible to convince others that it is helpful/useful/right without a proof.

We've all gone through the process of refining our intuitions. I thought David's example of the 2D beings learning about winding numbers in higher dimensions was perfectly apropos, because I can think of similar "aha!" moments where my intuition had to change. This other question about counterexamples in algebra contains quite a few of them.

Now, the fact that our intuitions differ is a good thing! It means we look at problems in different ways, which can help guide us to new, surprising solutions. Thus, just because I may not (currently) find "orientation-preserving" as intuitive as you, that's not a flaw in either of us. Moreover, the existence of algebraic proofs (whether natural or uniform or not) in this case tell me that your intuition (whatever it is) is a good one!

In the case of the ultrafinitists who question the consistency of PA, we'll really only ever know if their intuition is correct if they eventually find a provable flaw in the fabric of mathematics. My intuition says they won't find one. They, at least, have a hope of proving their intuition is correct.

LSpice
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Pace Nielsen
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  • I suppose the question is whether a mathematical proof which starts with the intuitive idea should be replaced with a "simpler" or more "direct" proof that seems to be primarily computational. Is it possible that the latter proof is then too narrow to take care of (for example) all cases where index 2 subgroups arise. – Kapil Mar 18 '22 at 10:28