Difference between revisions of "Talk:Aufgaben:Problem 14"

Revision as of 19:39, 15 June 2015

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 3 (Solution 1) 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 \(|G|=|\hat{G}|\)

That would be a circle proof: when proving that \(\{\chi\}\) is a basis of \( L(G)\) you already used the fact that \(|G|=|\{\chi\}|\).

Carl (talk) 19:48, 15 June 2015 (CEST)

I only used that \(\{\delta_g\}\) is a basis and that \(\{\chi\}\) are linearly independent. Therefore \(|\{\chi\}|\leq|\{\delta_g\}|=|G|\)

--Brynerm (talk) 20:05, 15 June 2015 (CEST)

You're right, it should work. But then you would have to use the longer proof with linear independence...

Carl (talk) 20:27, 15 June 2015 (CEST)

Indeed, the independence would have to be showed too. Maybe there's a much shorter proof then the one with the convolution. As those lemmas are not in the alternative solution, both solution will get about equally longer.

--Brynerm (talk) 20:36, 15 June 2015 (CEST)

alternative for Lemma 4:

\(\chi\) is a character:

$$\chi(g)^* = \chi(g^{-1})$$

\(\chi\) is homomorphic

$$ = \chi(g)^{-1} \Rightarrow 1 = \chi(g)\chi(g)^* = |\chi(g)|$$

Carl (talk) 21:39, 15 June 2015 (CEST)