Timeline for Schur's Theorem about immanants
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 9, 2018 at 15:39 | vote | accept | Denis Serre | ||
| Nov 9, 2018 at 11:19 | comment | added | Suvrit | @DenisSerre -- I'll expand with the details as soon as I get a chance. The heavy notation is repeated for instance in §2 of this note of ours here: arxiv.org/pdf/1410.1958.pdf ; alternatively, this paper develops notation and details: ams.org/journals/tran/2002-354-02/S0002-9947-01-02785-4/… | |
| Nov 9, 2018 at 8:22 | comment | added | Denis Serre | @Suvrit. You should develop point 1, because otherwise, one does not see the role of the assumption that $\chi$ is a character. I see the definition of $z$, whose coordinate $z_{i_1\cdots i_n}$ vanishes unless $k\mapsto i_k$ is a permutation, in which case it equals $\chi(i)$. But still the formula depends upon a non trivial identity, valid only for characters. | |
| Nov 9, 2018 at 4:26 | comment | added | Suvrit | I have fixed the proof now (to use CS correctly!) and undeleted. | |
| Nov 9, 2018 at 4:25 | history | undeleted | Suvrit | ||
| Nov 9, 2018 at 4:24 | history | deleted | Suvrit | via Vote | |
| Nov 9, 2018 at 4:21 | history | undeleted | Suvrit | ||
| Nov 9, 2018 at 4:19 | history | edited | Suvrit | CC BY-SA 4.0 |
applied cauchy-schwarz correctly this time!
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| Nov 9, 2018 at 1:27 | history | deleted | Suvrit | via Vote | |
| Nov 9, 2018 at 1:27 | comment | added | Suvrit | @DenisSerre indeed, you are right I flipped the inequalities, and the "proof" is incorrect as written. Time to dig into multilinear algebra to get a clean proof. Apologies for the rushed incorrect answer. Deleting it now. | |
| Nov 8, 2018 at 19:51 | comment | added | Denis Serre | I am dubious about the use of Cauchy-Schwarz. First, $B$ is not symmetric (or Hermitian). Second, you write an inequality whose sense is opposite to that of "Schur's Theorem". | |
| Nov 8, 2018 at 18:56 | comment | added | Mark Wildon | To get the immanent inequality from Schur's Theorem, take $x_\sigma = \chi(\sigma)$. The coefficient of $a_{1\rho(1)}\ldots a_{n\rho(n)}$ in the left-hand side is then $\sum_{\sigma, \tau : \tau\sigma^{-1} = \rho} \chi(\sigma) \chi(\tau) = \sum_{\sigma} \chi(\sigma)\chi(\rho\sigma) = \sum_{\sigma} \chi(\sigma^{-1})\chi(\rho\sigma) = |G| \chi(\rho) / \chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/\chi(1)$ times the immanent sum, and the right-hand side is $|G| \mathrm{det}(A)$ again by character orthogonality. | |
| Nov 8, 2018 at 16:11 | history | answered | Suvrit | CC BY-SA 4.0 |