Difference between revisions of "Talk:Aufgaben:Problem 13"

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Maybe a little shortening. Instead taking \(S^{(k,l)}\) and \(T^{(k,l)}\), take a more generall \(U^{(k,l),[\rho,\sigma]}_{ij}=(\rho_{ki},\sigma_{lj})_G\) for any to unitary representations \(\rho, \sigma \) and with \(U^{(k,l),[\rho,\sigma]}\sigma(g)=\rho(g) U^{(k,l),[\rho,\sigma]}\) also \([T^{(k,l)},\rho]\) is showed
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Maybe a little shortening. Instead taking \(S^{(k,l)}\) and \(T^{(k,l)}\), take a more generall \(U^{(k,l),[\rho,\sigma]}_{ij}=(\rho_{ki},\sigma_{lj})_G\) for any two unitary representations \(\rho, \sigma \) and with \(U^{(k,l),[\rho,\sigma]}\sigma(g)=\rho(g) U^{(k,l),[\rho,\sigma]}\) also \([T^{(k,l)},\rho(g)]=0\) is showed.
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\(\sum\limits_{i,j}{U^{(i,j),[\rho,\sigma]}_{i,j}}=(ch(\rho),ch(\sigma))_G\) may also save some time
  
 
--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 17:01, 4 August 2015 (CEST)
 
--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 17:01, 4 August 2015 (CEST)

Revision as of 17:42, 4 August 2015

There is an alternative way of proving (a) in the Felder Script:

https://people.math.ethz.ch/~felder/mmp/mmp2/

see chapter 3 -> Satz 3.1. and chapter 2 -> Satz 2.6 for a prove of Schur's lemma.

It seems to be shorter, and thus is probably better for the exam... Carl (talk) 23:16, 13 June 2015 (CEST)


Did anyone succeed in getting a shorter solution for a) by using Schur's lemma (as suggested by Carl)?

- Snick (talk) 13:06, 30 July 2015 (CEST)

I don't think you can make it any shorter. Shur's Lemma is used implicitly in the proof (claim 2), and to it you must first show that T is a homomorphism of representations (i.e. claim 1). The Felder Skript provesClaim 2 and Claim 5, but then you must still show the rest.

- Rayan (talk) 11:36, 31 July 2015 (CEST)


Maybe a little shortening. Instead taking \(S^{(k,l)}\) and \(T^{(k,l)}\), take a more generall \(U^{(k,l),[\rho,\sigma]}_{ij}=(\rho_{ki},\sigma_{lj})_G\) for any two unitary representations \(\rho, \sigma \) and with \(U^{(k,l),[\rho,\sigma]}\sigma(g)=\rho(g) U^{(k,l),[\rho,\sigma]}\) also \([T^{(k,l)},\rho(g)]=0\) is showed.

\(\sum\limits_{i,j}{U^{(i,j),[\rho,\sigma]}_{i,j}}=(ch(\rho),ch(\sigma))_G\) may also save some time

--Brynerm (talk) 17:01, 4 August 2015 (CEST)

Very nice, that does same some time.

Carl (talk) 17:21, 4 August 2015 (CEST)