Timeline for Conceptual reason why the sign of a permutation is well-defined?
Current License: CC BY-SA 4.0
105 events
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| Jan 3 at 1:41 | answer | added | John Wiltshire-Gordon | timeline score: 1 | |
| Nov 28, 2023 at 6:17 | answer | added | Peter Mueller | timeline score: 4 | |
| Sep 28, 2023 at 7:10 | comment | added | Lennart Meier | I couldn't find if this is mentioned yet: you can define the sign of a permutation of $\{0,...n\}$ to be the degree of the map $S^n \to S^n$ it induces. (There are certainly ways to define the degree that already involve determinants, e.g. via oriented point counts; using algebro-topological definitions, this can be circumvented.) | |
| Sep 27, 2023 at 21:55 | comment | added | Ian Agol | Another silly way to define the determinant: compute the inverse of a square matrix with general variable entries, so that the entries of the inverse are rational functions of the matrix entries. Then there is a common denominator which is the determinant (up to scaling). Probably not very helpful, but maybe makes it seem inevitable. | |
| Sep 27, 2023 at 17:06 | comment | added | Ian Agol | One could define the determinant via the Whitehead group. See Example 1.2. maths.ed.ac.uk/~v1ranick/papers/milnorwh.pdf | |
| Sep 27, 2023 at 5:56 | answer | added | Ian Agol | timeline score: 7 | |
| Sep 20, 2023 at 13:20 | answer | added | ACL | timeline score: 4 | |
| Aug 14, 2023 at 17:04 | review | Close votes | |||
| Aug 14, 2023 at 20:02 | |||||
| Aug 14, 2023 at 16:46 | comment | added | Mikhail Katz | I’m voting to close this question because enough conceptual reasons already. | |
| Aug 14, 2023 at 1:23 | answer | added | orangeskid | timeline score: 0 | |
| Jun 19, 2023 at 11:18 | history | edited | Martin Sleziak |
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| Jun 11, 2023 at 19:09 | comment | added | Raphael J.F. Berger | @TimCampion I had the recursive definition via the determinants of the minors in mind, actually. Then one needs of course to develop the properties of these determinants. | |
| Jun 11, 2023 at 14:00 | comment | added | Tim Campion | @RaphaelJ.F.Berger This reasoning is circular because the sign of a permutation appears in the usual formula for the determinant. | |
| Jun 11, 2023 at 5:05 | comment | added | Raphael J.F. Berger | I am writing mostly since your question caused me doubts in my own understanding and maybe this is also stupid (on top from a layman): I find it directly plausible that a permutation can by uniquely represented by its permutation matrix (once every element in X has been assigned to a integer from \{1,..,n\}). The sign is defined as the determinant of this matrix. Determinants don't change upon reordering the axis, and due to the nature of the matrix it's either 1 or -1. Where would be the magic bit in this approach in your eyes? | |
| Jun 6, 2022 at 14:38 | answer | added | Andrea Marino | timeline score: 4 | |
| Apr 1, 2022 at 13:42 | answer | added | 5th decile | timeline score: 5 | |
| Mar 18, 2022 at 22:14 | history | protected | Yemon Choi | ||
| Mar 18, 2022 at 21:59 | answer | added | Venkata Karthik Bandaru | timeline score: 0 | |
| Mar 16, 2022 at 18:18 | answer | added | Syzygies | timeline score: 2 | |
| Mar 16, 2022 at 14:51 | answer | added | Omar Antolín-Camarena | timeline score: 14 | |
| Mar 16, 2022 at 13:54 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| Mar 15, 2022 at 3:01 | review | Close votes | |||
| Mar 16, 2022 at 3:04 | |||||
| Mar 13, 2022 at 23:01 | answer | added | Eugene Stern | timeline score: 3 | |
| Mar 13, 2022 at 15:01 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| Mar 11, 2022 at 18:12 | comment | added | LSpice | @SridharRamesh, re, I see your point. Would you buy it if I claimed that $2^n{\bigwedge}^n\mathbb Z^n$ equalled ${\bigwedge}^n(2\mathbb Z)^n$, showed that the latter was non-$0$ by mapping onto $2^n\mathbb Z/2^{n + 1}\mathbb Z$, and then concluded that $2{\bigwedge}^n\mathbb Z^n$ was non-$0$? Or maybe showing that the obvious map $\leftarrrow$ in my claim is an isomorphism would involve a circularity. | |
| Mar 11, 2022 at 18:02 | answer | added | Olof Sisask | timeline score: 9 | |
| Mar 11, 2022 at 8:32 | answer | added | Wilberd van der Kallen | timeline score: 53 | |
| Mar 10, 2022 at 21:01 | comment | added | Sridhar Ramesh | Ah, we all make mistakes. I just realized I said "counit" where I should have said "unit". | |
| Mar 10, 2022 at 20:53 | comment | added | David Roberts♦ | @SridharRamesh because abelianisation is left adjoint to the inclusion, but you are right I completely butchered the functorial setup I was thinking of. What I really want is to have is the sign map as some natural transformation that incorporates not just bijections, so capturing the symmetric groups, but also injections between diff sets, hence capturing the inclusions between diff rank symmetric groups. | |
| Mar 10, 2022 at 20:10 | answer | added | მამუკა ჯიბლაძე | timeline score: 6 | |
| Mar 10, 2022 at 19:03 | answer | added | Benjamin Steinberg | timeline score: 4 | |
| Mar 10, 2022 at 17:40 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Mar 10, 2022 at 15:46 | comment | added | Sridhar Ramesh | Why do you feel a left adjoint exists? The counit would have to be an injection from an arbitrary finite set into a 2-element set. But most finite sets don't have injections into a 2-element set. Another counterargument: A left adjoint would have to preserve initial objects, but the category of finite sets and injections has an initial object (the empty set has a unique injection into every set), while the category of 2-element sets and injections has no initial object (any 2-element set has two different injections to any other 2-element set). | |
| Mar 10, 2022 at 10:16 | comment | added | David Roberts♦ | I'm wondering about a species-like description. Consider the category of finite sets and injections, and the subcategory of 2-element sets. Shouldn't the inclusion have a left adjoint where the counit is, more or less, the sign map? The functoriality and naturality may force some interesting behaviour. In particular, what is the 2-element set output by the left adjoint evaluated on some finite set? | |
| Mar 10, 2022 at 6:04 | answer | added | Pace Nielsen | timeline score: 15 | |
| Mar 10, 2022 at 3:22 | answer | added | manzana | timeline score: 11 | |
| Mar 10, 2022 at 1:33 | vote | accept | Tim Campion | ||
| Mar 9, 2022 at 23:52 | comment | added | Sridhar Ramesh | That is, does $2\wedge^n \mathbb{Z}^n$ have some nice universal property in itself that makes this go through, that is automatically known without needing to establish $E \neq -E$ by some separate permutation parity argument? I'm not aware of that. | |
| Mar 9, 2022 at 23:48 | answer | added | Carl-Fredrik Nyberg Brodda | timeline score: 13 | |
| Mar 9, 2022 at 23:46 | comment | added | Sridhar Ramesh | I know how to define a linear map on all of $\wedge^n \mathbb{Z}^n$ using its universal property, but how are you proposing to define this map on only $2 \wedge^n \mathbb{Z}^n$? I'm worried about the non-uniqueness of representations of an element of $2 \wedge^n \mathbb{Z}^n$ by an $n$-tuple of elements of $\mathbb{Z}^n$. I would need to be assured that any two such representations yield the same matrix determinant or mod 4 permanent. The only way I know to assure this is defining via universal property a determinant on all of $\wedge^n \mathbb{Z}^n$, but for that the omitted signs DO matter. | |
| Mar 9, 2022 at 22:58 | answer | added | Antoine Labelle | timeline score: 3 | |
| Mar 9, 2022 at 22:50 | comment | added | LSpice | @SridharRamesh, you're right, I meant $2\mathbb Z/4\mathbb Z$. That determinant map I was claiming only on $2{\bigwedge}^n\mathbb Z^n$, not on all of ${\bigwedge}^n\mathbb Z^n$. It is defined by the usual formula, just with signs omitted. (Since every term is divisible by $2$, the signs don't matter modulo $4$.) | |
| Mar 9, 2022 at 22:36 | comment | added | Tim Campion | @LSpice re My inclination is to let it breathe for a little longer -- the question is still less than two days old. I think it's not a bad thing for answers from the comments to be "duplicated" as actual answers, and the level of duplication from one answer to another is acceptable so far at least to me (it helps that the eagle-eyed such as yourself have been pointing out in the comments when this occurs). Anyway, I will likely "accept" Bjorn Poonen's answer in a little while, maybe lending the question some air of "finality". | |
| Mar 9, 2022 at 22:32 | comment | added | Sridhar Ramesh | Re: @LSpice's comment at mathoverflow.net/questions/417690/…. How do you define the "sign-free, non-trivial determinant map to $2^n \mathbb{Z} / 2^{n + 1} \mathbb{Z}$"? | |
| Mar 9, 2022 at 21:25 | answer | added | Karol Szumiło | timeline score: 21 | |
| Mar 9, 2022 at 19:41 | comment | added | LSpice | This question seems to have attracted a lot of attention … in some sense, too much, since the new answers are often duplicating answers that have already appeared in comments or other answers. Is there some appropriate step to take when a question reaches that stage? (Or perhaps we're not agreed that the question has reached that stage.) | |
| Mar 9, 2022 at 17:32 | comment | added | JP McCarthy | Relevant: math.stackexchange.com/a/94348/19352 | |
| Mar 9, 2022 at 14:06 | answer | added | André Henriques | timeline score: 0 | |
| Mar 9, 2022 at 13:21 | answer | added | Steven Gubkin | timeline score: 4 | |
| Mar 9, 2022 at 9:55 | answer | added | user171227 | timeline score: 3 | |
| Mar 9, 2022 at 9:47 | answer | added | Roland Bacher | timeline score: 24 | |
| Mar 9, 2022 at 4:26 | answer | added | Andy Jiang | timeline score: 10 | |
| Mar 9, 2022 at 4:15 | answer | added | Bjorn Poonen | timeline score: 106 | |
| Mar 9, 2022 at 3:43 | answer | added | Kapil | timeline score: 4 | |
| Mar 9, 2022 at 3:36 | comment | added | Timothy Chow | Let us continue this discussion in chat. | |
| Mar 9, 2022 at 3:33 | answer | added | paul garrett | timeline score: 7 | |
| Mar 9, 2022 at 3:33 | answer | added | Timothy Chow | timeline score: 27 | |
| Mar 9, 2022 at 3:32 | comment | added | Noah Snyder | @TimothyChow If you don’t know it’s all rotations how do you get closure? | |
| Mar 9, 2022 at 3:17 | comment | added | Timothy Chow | @NoahSnyder No, I don't think so. I'm building up a group by multiplying by $n$ at each step. I don't need to prove that I'm getting "all" rotations, just that I'm getting a group of order $n!/2$. | |
| Mar 9, 2022 at 2:44 | comment | added | Noah Snyder | @TimothyChow: Don't you need both implications or else it could be that there are n! rotations. | |
| Mar 9, 2022 at 2:43 | answer | added | Dan Ramras | timeline score: 27 | |
| Mar 9, 2022 at 2:35 | comment | added | Timothy Chow | @NoahSnyder Almost. The second step is, rotations one dimension lower fix your favorite vertex. But I gather that Tim Campion is not asking for a proof that is "simpler" or that somehow avoids a step that other proofs take. He's asking for a way of looking at it that makes things seem "inevitable" or "not miraculous." I'd argue that this point of view makes $A_n$ seem like an obvious entity in its own right, rather than something that has to extricate itself from inside $S_n$. That it is a subgroup of index 2 then follows by noting that $n!/2$ is half of $n!$. | |
| Mar 9, 2022 at 2:24 | comment | added | Noah Snyder | @TimothyChow: Is the induction step you had in mind to first show that the group of rotations acts transitively on vertices and then say that rotations fixing your favorite vertex are rotations one dimension lower? | |
| Mar 9, 2022 at 2:10 | comment | added | Timothy Chow | @TimCampion I don't understand what troubles you about Will Brian's argument. What if we phrase it this way: Rotations give you $n!/2$ of the automorphisms of a simplex. This is clear by induction; rotate vertex 1 to any of $n$ vertices and then apply a rotation of one dimension lower. So this is a subgroup of index 2, which is your first equivalent formulation. Right? | |
| Mar 9, 2022 at 0:59 | history | became hot network question | |||
| Mar 8, 2022 at 23:49 | answer | added | R. van Dobben de Bruyn | timeline score: 10 | |
| Mar 8, 2022 at 23:43 | comment | added | Noah Snyder | Theo’s first suggestion is essentially the proof that the first stable stem is Z/2Z. Which is just to say maybe we should accept that this is a deep fact and so all proofs are saying something quite interesting. | |
| Mar 8, 2022 at 19:59 | comment | added | LSpice | @BenjaminSteinberg, re, if $E$ equalled $-E$, then $L \mathrel{:=} 2{\bigwedge}^n\mathbb Z^n$ would be trivial; but $L$ admits a sign-free, non-trivial determinant map to $2^n\mathbb Z/2^{n + 1}\mathbb Z$. | |
| Mar 8, 2022 at 19:50 | comment | added | Benjamin Steinberg | @LSpice, How do you rigorously prove that $E\neq -E$ on the top exterior power of Z without using determinants or the sign? | |
| Mar 8, 2022 at 19:39 | comment | added | Will Brian | @TimCampion: I suppose I was taking it as intuitively obvious. For evidence in my favor, I just asked my 4-year-old daughter, and she said that she knows what it means. I would count my previous comment as a conceptually clear explanation (to me, anyway), but I would not necessarily find it easy to turn the explanation into a rigorous proof. | |
| Mar 8, 2022 at 19:34 | comment | added | LSpice | @BenjaminSteinberg, re, since the top exterior power is clearly spanned by the image $E$ of $e_1 \otimes \dotsb \otimes e_n$ (where $\{e_1, \dotsc, e_n\}$ is a basis of $\mathbb Z^n$), and since the top exterior power is non-$0$, we must have $E \ne 0$; but (now working over $\mathbb Z$) it is clear that any transposition takes $E$ to $-E$. That is, we may show the top exterior power is non-$0$ by working over $\mathbb F_2$, but that the action of $\Sigma_n$ is non-trivial by working over $\mathbb Z$; no a priori signum character required. | |
| Mar 8, 2022 at 19:30 | comment | added | Benjamin Steinberg | @LSpice but one still has to show the action of $S_n$ on the top exterior power is non-trivial. That would seem to me to require some form of the sign map. You can't see that over F_2. | |
| Mar 8, 2022 at 19:12 | comment | added | LSpice | @BenjaminSteinberg, re, for working over $\mathbb Z$ it suffices to show that the top exterior power over $\mathbb F_2$ is non-$0$, and that is exhibited by the fact that the mod-2 determinant, which requires no sign in its definition, factors through the top exterior power. (Thinking along these lines is why I made my silly comment about $\mathbb F_2$.) | |
| Mar 8, 2022 at 18:32 | comment | added | Benjamin Steinberg | Usually the sign is used to show that the top exterior power is nonzero | |
| Mar 8, 2022 at 18:28 | comment | added | YCor | @TimCampion the symmetric group $\mathfrak{S}_k$ acts on $(\mathbf{Z}^k)^{\otimes n}$ in the obvious way. This action obviously preserves the subspace generated by all the tensors $v_1\otimes \dots v_n$ such that $v_i=v_j$ for some $i\neq j$. Hence it factors through an action on the quotient, which is $\Lambda^n(\mathbf{Z}^k)$. Now specify to $n=k$. | |
| Mar 8, 2022 at 18:24 | comment | added | Tim Campion | @YCor Thanks, tags changed accordingly. That's a good point about the alternating power -- probably the "correct" definition of the determinant. I wonder if one can see that $\Sigma_k$ acts on $\Lambda^k V$ without knowing that the sign of a permutation is well-defined... | |
| Mar 8, 2022 at 18:23 | history | edited | Tim Campion |
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| Mar 8, 2022 at 18:21 | comment | added | YCor | The definition I learnt of determinant when studying was not based on row operations (invariance under row operations was proved afterwards), but essentially on the basis that the space of alternating $n$-forms on an $n$-dimensional space is 1-dimensional. So here the sign would be the action of a matrix in $\mathrm{Aut}(V)$, $V$ a free $\mathbf{Z}$-module of rank $n$, on $\Lambda_\mathbf{Z}^n(V)$. | |
| Mar 8, 2022 at 18:19 | comment | added | YCor | It seems to me that soft-question is not suitable here (the tag is usually not meant in the case of a heuristic question, but rather of a question peripheral to mathematics). I'd rather see gm.general-mathematics or intuition instead. | |
| Mar 8, 2022 at 18:17 | answer | added | Abdelmalek Abdesselam | timeline score: 17 | |
| Mar 8, 2022 at 18:02 | answer | added | LSpice | timeline score: 9 | |
| Mar 8, 2022 at 17:57 | review | Close votes | |||
| Mar 9, 2022 at 11:03 | |||||
| Mar 8, 2022 at 17:56 | comment | added | Tim Campion | The proofs mentioned by Theo and by Phil are both of the form "exhibit an explicit nontrivial action of $\Sigma_n$ on a set with two elements". Ultimately, I suppose any proof will have to do this. Maybe one can't ultimately do better than this. | |
| Mar 8, 2022 at 17:54 | comment | added | Tim Campion | @WillBrian True. Is there a way to see that "inside out" is well-defined without knowing already that the sign of the determinant is? | |
| Mar 8, 2022 at 17:52 | comment | added | Tim Campion | @NathanielJohnston Indeed, and that's actually how I'm teaching it this semester. What I'm actually hung up on is that the proof, using this definition, that you have a homomorphism, though not bad, is still delicate enough that it doesn't leave me feeling enlightened. Perhaps this is indeed too subjective. | |
| Mar 8, 2022 at 17:47 | comment | added | LSpice | While I'm cluttering up the comments, I can't help noticing that (6) is false when $k = \mathbb F_2$ (and of course (1), (3), (5) are false for $n = 1$). | |
| Mar 8, 2022 at 17:44 | comment | added | LSpice | Re, of course different people will find a different 'real' answer as to why, but you seem to have explicitly ruled out many people's 'real' candidates for why. If the goal is to satisfy you and not a student, then, for example, it's hard for me to imagine a much more convincing proof than @TheoJohnson-Freyd's above. | |
| Mar 8, 2022 at 17:44 | answer | added | David E Speyer | timeline score: 78 | |
| Mar 8, 2022 at 17:43 | history | edited | LSpice | CC BY-SA 4.0 |
Oops, missed one `sgn` -> `\operatorname{sgn}`
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| Mar 8, 2022 at 17:42 | comment | added | Tim Campion | @LSpice Although the motivation is partly pedagogical, at this point I'm not primarily interested in finding a proof which will optimally satisfy my students, but rather one which will satisfy me. Right now, I know that the sign of a permutation is a well-defined homomorphism, but I don't know "why, really". | |
| Mar 8, 2022 at 17:40 | history | edited | LSpice | CC BY-SA 4.0 |
TeX; name of article
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| Mar 8, 2022 at 17:40 | comment | added | Nathaniel Johnston | If you think about the sign of a permutation as counting (the parity of) the number of inversions, rather than (the parity of) the number of transpositions, then I think #3 becomes much more clear. You still have to prove that this function is a homomorphism (see, e.g., this question), but at least you don't have to deal with any of the well-definedness issues. | |
| Mar 8, 2022 at 17:39 | comment | added | LSpice | This seems like a great MESE question, but, surely, given its explicitly elementary nature, it is not an MO question? | |
| Mar 8, 2022 at 17:37 | comment | added | Will Brian | Take the vertices of an $n$-simplex, and apply the permutation. If this turns the simplex inside out, then the permutation is odd; otherwise it is even. (This is basically equivalent to the comment by markvs, if one thinks of the determinant geometrically.) | |
| Mar 8, 2022 at 17:36 | comment | added | Phil Tosteson | For a topological construction, you can take the action of $S_n$ on $H_n(\mathbb R^n, \mathbb R^n-0) = \mathbb Z$. For this, you need to know that the top homology of a sphere is one dimensional. | |
| Mar 8, 2022 at 17:35 | review | Low quality posts | |||
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| Mar 8, 2022 at 17:29 | comment | added | Theo Johnson-Freyd | So when you ruled out things about polynomials, I guess you had in mind something like the following? Pick $n$ real numbers, all distinct, $x_1,\dots,x_n$, and write down $D := \prod_{i<j}(x_j-x_i)$. Now discover that a permutations act by $D \mapsto \pm D$. | |
| Mar 8, 2022 at 17:27 | comment | added | Tim Campion | @TheoJohnson-Freyd I guess my goal is not precisely to be elementary, but really to be conceptually illuminating, and it's hard for me to imagine the "invariance mod 2" part of the squiggly lines argument looking like more than a contingency. I wonder if there's a proof using unoriented bordism or something... | |
| Mar 8, 2022 at 17:22 | comment | added | Theo Johnson-Freyd | @TimCampion I agree --- I don't know a completely elementary calculation of $\pi_0 O(n)$. Certainly not if the benchmark for "elementary" is at or below "existence of a sign representation of $S_n$". Then again, I still think "draw some wiggly lines and count crossings" is rather elementary... | |
| Mar 8, 2022 at 17:22 | comment | added | Tim Campion | @markvs Ok sure, but now we're back in the territory where it feels like magic that the determinant should be well-defined, because the proof boils down to checking that it's invariant under row operations. This feels particularly like magic to me because you have to put in "by hand" the prescription that row-swapping changes the sign of the determinant. | |
| Mar 8, 2022 at 17:18 | comment | added | Tim Campion | @TheoJohnson-Freyd If there's a proof that $O(n)$ or $GL_n(\mathbb R)$ has two components which doesn't implicitly require one to know that the sign is well-defined (which as far as I can see, includes the case of relying on the existence of the determinant), that would be interesting. | |
| Mar 8, 2022 at 17:17 | comment | added | Tim Campion | @markvs I don't actually know how to define the determinant (or else characterize it and prove its existence) without knowing that the sign of a permutation is well-defined. | |
| Mar 8, 2022 at 17:13 | comment | added | Theo Johnson-Freyd | Are you willing to take for granted that the orthogonal group has two components? Or the real general linear group? Because that's what you need in order to know something about determinants of matrices... but maybe that's more sophisticated than the sign representation of $S_n$, not less. | |
| Mar 8, 2022 at 17:12 | comment | added | Theo Johnson-Freyd | So I guess your first and third bullet points have ruled out a proof of the form: "Draw the permutation by connecting dots on either sides of a strip. Use wiggly lines if you like, but only use normal crossings. Now count the number of crossings, and argue that this count mod 2 is a topological invariant." | |
| Mar 8, 2022 at 16:55 | history | asked | Tim Campion | CC BY-SA 4.0 |