Timeline for Is there a non-constant function on the sphere that diagonalizes all rotations simultaneously?
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2023 at 16:53 | vote | accept | Giuseppe Negro | ||
| Mar 11, 2023 at 10:44 | comment | added | Dabed | @მამუკაჯიბლაძე thanks!, ok so if I got it right then when I'm counting colorings I'm counting element of X that are left unchanged by an action g whereas when I'm counting the permutations of a polynomial what I'm really counting is rather the actions of G that that make no change for an element x, and I was confusing both of them. | |
| Mar 10, 2023 at 18:58 | comment | added | მამუკა ჯიბლაძე | @Dabed While there are lots of representations, you have only so many stabilizers of points since they are all subgroups of your group. So in general only knowing the collection of stabilizers that occur does not provide enough information. This can be also seen from the formula in Burnside's lemma: it needs information about the sets of the form $X^g$ (collection of elements of $X$ stabilized by any given $g\in G$) rather than about $G^x$ (collection of elements of the group stabilizing any given $x\in X$). Unlike $G^x$ I don't see how to relate those $X^g$ for polynomials and for colorings | |
| Mar 10, 2023 at 9:24 | comment | added | Dabed | @მამუკაჯიბლაძე ah I see I just thought maybe the permutations that left a particular polynomial invariant corresponded with a particular orbit of the colorings or something like that, I (clearly) don't understand much so will have to learn and think more about it, thanks a lot. | |
| Mar 10, 2023 at 8:40 | comment | added | მამუკა ჯიბლაძე | @Dabed I am not sure but I guess connection is like this. You may view colorings as functions constant inside each face separately and along all edges collectively. Clearly your group acts on such functions, and the task is to enumerate orbits of this action. Now any function can be obtained, in a certain sense uniquely, from eigenfunctions. But the eigenfunctions I listed are polynomial, while for colorings you need piecewise constant ones. How to get the latter from the former, I don't know. | |
| Mar 9, 2023 at 22:44 | comment | added | Dabed | @მამუკაჯიბლაძე (How) do the eigenfunctions of the rotation symmetry group of the cube $xyz, (x^2-y^2)(x^2-z^2)(y^2-z^2), x^4+y^4+z^4$ relate with the number of ways we can color the cube using the Burnside's lemma? | |
| Mar 9, 2023 at 17:45 | comment | added | paul garrett | @GiuseppeNegro, thanks for the positive feedback! :) :) | |
| Mar 9, 2023 at 16:51 | comment | added | Giuseppe Negro | @paulgarrett: (This is not totally related, but let me take the opportunity to tell you that I loved your lecture notes on "functions on circles", when I first studied Harmonic Analysis. Thanks a lot for making those public). More on point, I like the idea of looking at finite non-abelian groups. That sure avoids unnecessary complications and makes available algorithmical exercises. It gives a really algebraic flavor to the course, though. | |
| Mar 8, 2023 at 21:06 | comment | added | paul garrett | Offering what I think is an accurate, and not toooo onerous, reframing: I tell my beginning students that the generalization to non-abelian (e.g., finite, if we want to avoid some complications, both analytical and algebraic) groups, of the notion of "simultaneous eigenvector" is "irreducible repn". I use this as a motivation for (a bit of) repn theory, since, after all, it's the answer to a question. The sphere case is probably the simplest with a physical sense, although there is a bit of analysis... I tell my students that people have known this case for 200+ years. :) | |
| Mar 8, 2023 at 18:41 | comment | added | მამუკა ჯიბლაძე | @WillSawin Yes, thanks! I should clarify that I have in mind the same action when restricted to $SO(3)$ | |
| Mar 8, 2023 at 18:35 | comment | added | Will Sawin | @მამუკაჯიბლაძე It depends on what space $SO(3) \times \mathbb R$ is acting on. If it's acting on $\mathbb R^3$, with $\mathbb R$ acting by scaling, then the functions $x \mapsto |x|^{\alpha}$ would be examples - not incredibly nontrivial, but still useful for many purposes in mathematics. | |
| Mar 8, 2023 at 18:34 | answer | added | Will Sawin | timeline score: 12 | |
| Mar 8, 2023 at 18:09 | answer | added | მამუკა ჯიბლაძე | timeline score: 8 | |
| Mar 8, 2023 at 10:34 | comment | added | Giuseppe Negro | @მამუკაჯიბლაძე: thank you very much for these beautiful examples. If you could make those into an answer I would gladly upvote it. That would not be precisely an answer to the present question, but I think it would be extremely interesting anyway. | |
| Mar 6, 2023 at 11:24 | comment | added | მამუკა ჯიბლაძე | Another example of such subgroup in $SO(3)$ is the (finite) rotation symmetry group of the (standardly situated) cube: it has common eigenfunctions $xyz$, $(x^2-y^2)(x^2-z^2)(y^2-z^2)$, $x^4+y^4+z^4$, ... | |
| Mar 6, 2023 at 10:26 | comment | added | მამუკა ჯიბლაძე | No, if the group contains $SO(3)$ the common eigenfunction would be also common for $SO(3)$. However $SO(3)$ has nonabelian subgroups with common eigenfunctions. Generate e. g. the subgroup by arbitrary rotations around the $z$ axis together with $(x,y,z)\mapsto(y,x,-z)$. For it, $f(x,y,z)=x^2+y^2$ and $f(x,y,z)=z^2$ are common eigenfunctions, for example. | |
| Mar 6, 2023 at 0:57 | history | became hot network question | |||
| Mar 5, 2023 at 22:08 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing (formula hyperlinking)
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| Mar 5, 2023 at 18:51 | answer | added | B K | timeline score: 27 | |
| Mar 5, 2023 at 18:08 | comment | added | Giuseppe Negro | @მამუკაჯიბლაძე: I see. So a non-abelian group with non-trivial abelianization could support such simultaneous eigenfunctions, I gather. Could you make a concrete example, as simple as possible? Would something like $SO(3)\times \mathbb R$ work? | |
| Mar 5, 2023 at 17:27 | comment | added | მამუკა ჯიბლაძე | In other words, just noncommutativity does not suffice to rule out existence of such functions. | |
| Mar 5, 2023 at 17:25 | comment | added | მამუკა ჯიბლაძე | Sorry I confused things. What I should actually say is this: if there would be such a function, it would provide a nontrivial 1-dimensional representation of $SO(3)$, hence a nontrivial homomorphism from $SO(3)$ to the multiplicative group, which does not exist since abelianization of $SO(3)$ is trivial. | |
| Mar 5, 2023 at 16:56 | history | asked | Giuseppe Negro | CC BY-SA 4.0 |