Difference between revisions of "Talk:Aufgaben:Problem 13"
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It seems to be shorter, and thus is probably better for the exam... [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 23:16, 13 June 2015 (CEST) | It seems to be shorter, and thus is probably better for the exam... [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 23:16, 13 June 2015 (CEST) | ||
| − | + | ---- | |
| − | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) | + | Did anyone succeed in getting a shorter solution for a) by using Schur's lemma (as suggested by Carl)? |
| + | |||
| + | - [[User:Snick|Snick]] ([[User talk: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. | ||
| + | |||
| + | - [[User:Rayan|Rayan]] ([[User talk: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 | ||
| + | |||
| + | --[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 17:01, 4 August 2015 (CEST) | ||
| + | |||
| + | Very nice, that does same some time. | ||
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| + | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 17:21, 4 August 2015 (CEST) | ||
| + | |||
| + | ---- | ||
| + | |||
| + | I get slightly confused, that you act \(L^{W}\) on \(\chi_{sy}\) although \(\chi_{sy}\) may not be in \(W\) (Claim 9). Does it come from the fact, that linearity of \(L^{W}\) is only assured in \(W\) and not in \(V_s\) so you would have to use \(L^{V_s}\) here and make the restriction afterwards? | ||
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| + | --[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 07:56, 5 August 2015 (CEST) | ||
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| + | No \(V_s\) is not always one-dimentional. This would mean that every irreducible rep. is one dimensional. \(V_s = W\) is only true for \(\text{dim}V_s = 1\) | ||
| + | |||
| + | Your right that is not correct. I just changed the \(L^{V_s}\) to \(L^W\) in claim 9, without much thought, after they changed the exercise. Not sure if applying the restriction afterwards works. I guess you can bypass the step by just writing it out a bit more: | ||
| + | |||
| + | $$\sum_{i= 1}^n (e_i, L^W(g) e_i)_G =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) (L^W(g) e_i)(h)^* =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) e_i(g^{-1}h)^* = \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sy}(g^{-1}h)^*$$ | ||
| + | |||
| + | $$= \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sgy}(h)^* = \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \dots$$ | ||
| + | |||
| + | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 15:44, 5 August 2015 (CEST) | ||
| + | |||
| + | ---- | ||
| + | |||
| + | I messed that up with dimensions, I was thinking in terms of conjugacy classes, but that wouldn't imply W=Vs. | ||
| + | |||
| + | Nevertheless your bypass seems quite accurate. | ||
| + | |||
| + | --[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 22:59, 5 August 2015 (CEST) | ||
Latest revision as of 20:59, 5 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)
I get slightly confused, that you act \(L^{W}\) on \(\chi_{sy}\) although \(\chi_{sy}\) may not be in \(W\) (Claim 9). Does it come from the fact, that linearity of \(L^{W}\) is only assured in \(W\) and not in \(V_s\) so you would have to use \(L^{V_s}\) here and make the restriction afterwards?
--Brynerm (talk) 07:56, 5 August 2015 (CEST)
No \(V_s\) is not always one-dimentional. This would mean that every irreducible rep. is one dimensional. \(V_s = W\) is only true for \(\text{dim}V_s = 1\)
Your right that is not correct. I just changed the \(L^{V_s}\) to \(L^W\) in claim 9, without much thought, after they changed the exercise. Not sure if applying the restriction afterwards works. I guess you can bypass the step by just writing it out a bit more:
$$\sum_{i= 1}^n (e_i, L^W(g) e_i)_G =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) (L^W(g) e_i)(h)^* =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) e_i(g^{-1}h)^* = \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sy}(g^{-1}h)^*$$
$$= \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sgy}(h)^* = \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \dots$$
Carl (talk) 15:44, 5 August 2015 (CEST)
I messed that up with dimensions, I was thinking in terms of conjugacy classes, but that wouldn't imply W=Vs.
Nevertheless your bypass seems quite accurate.
