Timeline for Why do these finite group Dedekind matrices seem to have integer spectrum when specialized to the order of group elements?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Aug 17 at 18:51 | vote | accept | mathoverflowUser | ||
| Aug 17 at 16:21 | history | became hot network question | |||
| Aug 17 at 16:05 | answer | added | Benjamin Steinberg | timeline score: 12 | |
| Aug 17 at 15:27 | comment | added | mathoverflowUser | @LSpice: It is inteded to mean what mme wrote. | |
| Aug 17 at 15:25 | comment | added | mme | @LSpice Take $p < q$ and extend $f_p$ by the identity. | |
| Aug 17 at 15:08 | comment | added | LSpice | Everyone else seems to understand, so I guess I'm just being dense, but what does $f_p \circ f_q$ mean when $f_p$ is a permutation of $1, \dotsc, p - 1$ and $f_q$ is a permutation of $1, \dotsc, q - 1$? | |
| Aug 17 at 14:50 | answer | added | Sean Eberhard | timeline score: 8 | |
| Aug 17 at 13:37 | comment | added | Benjamin Steinberg | @StevenStadnicki he is looking at the matrix in the regular representation of the element of the center if the group algebra that adds all the elements of the group weighted by their order. This particular combination maybe always has rational coefficients even when the splitting field is bigger but the formula is correct | |
| Aug 17 at 10:11 | comment | added | mathoverflowUser | @KasperAndersen: thanks, that is interesting. | |
| Aug 17 at 9:34 | comment | added | Kasper Andersen | I just checked that all groups of order less than 128 have integral spectrum, so this seems to hold not only for dihedral groups! | |
| Aug 17 at 6:41 | history | edited | mathoverflowUser | CC BY-SA 4.0 |
corrected bug in code and added new data
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| Aug 17 at 6:00 | comment | added | Steven Stadnicki | @BenjaminSteinberg Note that OP's matrix isn't a character table of $G$, so the formula you're using may not apply. | |
| Aug 17 at 5:56 | comment | added | Steven Stadnicki | @BenjaminSteinberg The dihedral group in the last example ($p=7, q=5$) is $D_{2\cdot10}$ which doesn't have $\mathbb{Q}$ as a splitting field but the eigenvalues are shown as integral, so either there's something off in the calculation or something else is going on? | |
| Aug 17 at 1:53 | comment | added | mathoverflowUser | @BenjaminSteinberg: Thanks for your insight in this question. How do I see your formula for the "corresponding eigenvalue"? | |
| Aug 17 at 1:46 | comment | added | Benjamin Steinberg | I suspect that your integer eigenvalues are due to perhaps not checking big enough dihedral groups. The general formula for the eigenvalues for any finite group $G$ is there is one eigenvalue of each irreducible character $\chi$ of $G$. The corresponding eigenvalue is $\sum_{g\in G}\frac{ord(g)\chi(g)}{\chi(1)}$. If your group has Q as a splitting field, then these will take on integer values. Very small order dihedral groups have Q as a splitting field. | |
| Aug 16 at 21:28 | comment | added | mathoverflowUser | @StevenStadnicki: I have only a similar question where the representation theory of finite groups as presented by Keith Conrad is applied but about abelian finite groups: mathoverflow.net/questions/369941/… | |
| Aug 16 at 21:24 | comment | added | Steven Stadnicki | Do you have any pointers to more info on these order matrices? I'd like to know more about them but all I'm finding for Dedekind group matrices seem to be ones that treat the elements of the group as variables and take the determinant of the multiplication table as a polynomial in the group members. | |
| Aug 16 at 20:43 | comment | added | mathoverflowUser | @StevenStadnicki: I have not tested it for all dihedral groups. | |
| Aug 16 at 19:59 | comment | added | Steven Stadnicki | Is the statement about the matrix of orders having integer spectra specific to the dihedral groups being generated by inversion permutations in this way, or does it seem to be the case for all dihedral groups? | |
| Aug 16 at 17:04 | history | edited | mathoverflowUser | CC BY-SA 4.0 |
corrected title by suggestion of SamHopkins.
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| Aug 16 at 16:33 | comment | added | Sam Hopkins | Well, it's a specialization of the group matrix, I'm not sure it has a particular name. | |
| Aug 16 at 16:31 | comment | added | mathoverflowUser | @SamHopkins: Thanks for your comment. How would you call the matrix in this case? | |
| Aug 16 at 16:15 | comment | added | Sam Hopkins | Your question seems interesting, but just in terms of terminology: usually the "group matrix" of a group has an abstract symbol for each group element (so it is basically the same as the multiplication table of the group), rather than the orders like you have here. | |
| Aug 16 at 16:04 | history | edited | mathoverflowUser | CC BY-SA 4.0 |
corrected typo
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| Aug 16 at 13:41 | history | asked | mathoverflowUser | CC BY-SA 4.0 |