Talk:Aufgaben:Problem 14
--Brynerm (talk) 10:09, 15 June 2015 (CEST) is this accurate?
Define \(Z_{\hat{G}}:=\{g\in G:\chi(g)=1\; \forall \chi \in \hat{G}\}\)
- Lemma 5' : \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x \in Z_{\hat{G}}} \)
- Proof of Lemma 5': let \(\rho \in \hat{G}\) be arbitrary
$$\sum\limits_{\chi}{\chi(x)}=\sum\limits_{\sigma=\rho^{-1}\chi}{\sigma(x)\rho(x)}=\rho(x)\sum\limits_{\sigma \in \hat{G}}{\sigma(x)}$$ $$\Rightarrow \rho(x)=1 \;\forall \rho \in \hat{G}\; or \; \sum\limits_{\chi}{\chi(x)}=0$$ $$\Rightarrow \sum\limits_{\chi}{\chi(x)}=\delta_{x \in Z_{\hat{G}}}\sum\limits_{\chi}{1} =|G|\delta_{x \in Z_{\hat{G}}}$$
- Lemma 6: \(Z_{\hat{G}}=\{e\}\)
- Proof of Lemma 6: Use Lemma 1
$$ \sum\limits_{\chi}{\sum\limits_{g}{\chi(g)}}=\sum\limits_{\chi}{|G|\delta_{\chi, \tau}}=|G|$$ $$ \sum\limits_{g}{\sum\limits_{\chi}{\chi(g)}}=\sum\limits_{g}{|G|\delta_{g \in Z_{\hat{G}}}}=|G| \cdot |Z_{\hat{G}}| $$
- \( \Rightarrow |Z_{\hat{G}}|=1 \), and as \(e \in Z_{\hat{G}} \Rightarrow Z_{\hat{G}}=\{e\}\)
Therefore It's proven that \(\forall \chi \in \hat{G}, \; \chi(g)=1 \Leftrightarrow g=e\) and Lemma 5' becomes Lemma 5: \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x,e} \)
So you can do the whole exercise without proving first that \(\{\chi\}\) is a basis of \(L(G)\).
Not sure of we are allowed some of the arguments in you proof of (a) in the exam, so here is my attempt for an alternative proof:
Let \(\rho\) be an irred. rep. of G on \(V\Rightarrow\) for any \(g\in G\) there \(\exists \lambda \in \mathbb{C}, v \in V\) such that \( \rho(g) v = \lambda v\). \( \forall h \in G\):
$$ (\rho(g) - \lambda \mathbb{I}) \rho(h)v = \rho(g)\rho(h)v - \lambda \mathbb{I}\rho(h)v$$
because G is abelian, and the identity commutes:
$$ = \rho(h)\rho(g)v - \lambda\rho(h)\mathbb{I}v = \rho(h)\lambda v - \lambda\rho(h)v = 0$$
\( \Rightarrow Ker(\rho(g) - \lambda \mathbb{I})\) is invariant and therfore equal to V \( \Rightarrow \rho(g) = \lambda \mathbb{I}\).
This implies that every one dimensional subspace of V is invariant, and thus V has to be one dimensional itself.
\(\Rightarrow ch(\rho)(g) = \lambda\) and the forward dircetion is proven.
Back: The homomorphisms \(\chi:G \rightarrow \mathbb{C}^* = GL(\mathbb{R}) \) are one-dimensional and therefore irreducible representations of G on \(\mathbb{R}\).
Carl (talk) 09:37, 15 June 2015 (CEST)
--Brynerm (talk) 13:33, 15 June 2015 (CEST)
You're right. My proof is inaccurate. Instead of \(End(\mathbb{C}^G)\) I actually wantet to use the space of all unitary representations \(u: G \rightarrow U(\mathbb{C}^G) \). But the dimension formula is completly wrong. One could use the dimension theorem \(|G|=\sum{dim(\chi)^2}\) which we didn't prove though.
Yes, but we where allowed to use it in one of the exercises, so I don't know. Same goes for \(|\{\chi\}|=|\{C_k\}| = |G| \) which you need for (b) if not for (a)
About (b): how do you get \(|Im(f)|=dim(\hat{\hat{G}})\)?
Carl (talk) 15:38, 15 June 2015 (CEST)
Yes, that exactly was my thought. I was quite sure, that we've had in the lecture, that the character table is quadratic. But maybe we did this only for the first definition of the characters.
--Brynerm (talk) 16:14, 15 June 2015 (CEST)
I can't find any mistakes in the alternative proof or lemma 5' and 6. Seems to work out very nicely. Where do you see any problems?
Carl (talk) 18:13, 15 June 2015 (CEST)
It's just, that it is so short, and I'd been struggling with finding a proof for this for so long ;-) So I will now seperate the alternative proof completly. Otherwise it's getting messy.
Maybe Lemma 6 can be used to show \(|\{\chi\}|=|G|\): If you differ \(|G|\) and \(|\hat{G}|\) strictly, Lemma 6 Looks like \(|G|=|Z_{\hat{G}}|\cdot |\hat{G}|\) which leads to \(|G|\leq|\hat{G}|\). But as the deltafunctions are a basis for \(L(G)\) with \(|\{\delta_g\}|=|G|\) there can't be more than \(|G|\) different linear independent functions in \(L(G)\). So \(|G|=|\hat{G}|\)
