Talk:Aufgaben:Problem 14
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)
I did some reading about this stuff: The hat function is the discrete Fourier transform on an abelian group (no surprise). And (a) is the Pontryagin dual theorem. All the proofs that I found in the literature use or first prove the fact that a finite abelian group is isomorphic to a direct sum of cyclic groups (long an not so easy). Maybe there still is a shorter proof, especially since we can easily prove the fact that the characters are on the unit circle, but I couldn't quite get there...
Regarding the dimension theorem and #irr. rep. = #conjugacy classes: we obviously need them in Exercise 7, so I think we can use them after all.
Carl (talk) 12:20, 16 June 2015 (CEST)
Lemma 4 isn't that long though. It's really only showing, that the characters are on the unit circle, as they are a \(n^{th}\) root of 1. Maybe we can use the fact about groups, which proof I've put in brackets. That hasn't much to do with MMP, that's general geometry stuff.
It will be much easier, if we are allowed to use those facts. We should probably ask.
--Brynerm (talk) 17:21, 16 June 2015 (CEST)
What I wonder is, if there's a simple proof for Lemma 3. A version of another group says, that the regular representations are faithful (they only send e to the identitiy matrix). And since they are a direct orthogonal sum of irreducible representations, the statement is proven. But I can't remember, that we learned such a thing about the regular representations, and I can't think of a way to prove. And what I did in the alternative solution with the basis, is so ugly and it feels like solving the task backwards.
--Brynerm (talk) 17:43, 16 June 2015 (CEST)
No I don't think we did at least not while I was there. However I think the proof we have right now is fine. It's a bit long but manageable not like my proof of Exercise 13 :)
Carl (talk) 18:46, 16 June 2015 (CEST)
I believe that we will still need a trivial proof of \(e \mapsto (\chi \mapsto 1)\) to show that our map is a group homomorphism. My proof would be
$$O(e) = (\chi \mapsto \chi(e)) = (\chi \mapsto dim(\mathbb{C}*)) = (\chi \mapsto 1)$$
I'd like someone to check this before I'll include it to the solution. This also shows that \(e\) is in the kernel of the group homomorphism. Perhaps we also have to show that \(\chi \mapsto 1\) is the identitiy of \(\hat{\hat{G}}\), but this should be sort of clear.
--Lilit (talk) 10:08, 15 July 2015 (CEST)
Agreed, it is probably better include this then not. I would write: \(\chi\) is a group homomorphism therefore \(\chi(e) = e = 1\) and because \(\hat{\hat{G}}\) are the irreducible characters of the abelian group \(\hat{G}\) the trivial representation is again the identity and \(e \mapsto (\chi \mapsto 1) = e\)
Carl (talk) 13:31, 15 July 2015 (CEST)
I also think, it's enough to write: "from a) fallows that the characters are homomorphism, therefore \(e\) has to map to \(1\)." Then as Carl said, because the given map is homomorphous \(\chi \mapsto 1\) must be the neutral element of \(\hat{\hat{G}}\)
--Brynerm (talk) 14:02, 15 July 2015 (CEST)
That is not exactly what I said, but I guess there are a number of ways to argue and in the end they are all more or less the same. Everyone happy with the edit?
Carl (talk) 18:32, 19 July 2015 (CEST)
I think this is fine now.
--Lilit (talk) 12:51, 23 July 2015 (CEST)
Okay, so I have another issure here. We've proven that \(\mathcal{F}^{-1} \mathcal{F} f(x) = f(x)\), but we should also prove that \(\mathcal{F} \mathcal{F}^{-1} \hat{f}(\chi) = \hat{f}(\chi)\). This is just another calculation, but we have to do it to show that the inverse is as it is.
--Lilit (talk) 12:51, 23 July 2015 (CEST)
By the giving calculation, iductivity certainly is showed. You could argue like in the alternative solution by dimension equality and the the fact that it's homomorphous or
$$\mathcal{F}\mathcal{F}^{-1}(\hat{f})(\chi) =\sum\limits_{g}{\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)\rho(g)\chi^*(g)}}=\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)\sum\limits_{g}{(\rho\cdot\chi^{-1})(g)}}$$
$$=\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)|G|\delta_{\rho\chi^{-1},\tau}}=\hat{f}(\chi)$$
Maybe then the alternative solution becomes shorter, but I still find it ugly --
Brynerm (talk) 12:23, 2 August 2015 (CEST)
About the invers in c)
We show that (*):
$$\mathcal{F}^{-1}(\hat{f}) = f $$
When we now start with the definition:
$$ \mathcal{F}(f) =\hat{f} $$
Then it follows both, that (by applying the invers on both sides):
$$ \mathcal{F}(f) = \hat{f} \Rightarrow \mathcal{F}^{-1}(\mathcal{F}(f)) =\mathcal{F}^{-1}(\hat{f}) \stackrel{(*)}{=} f$$
As well as that (by substituting (*) for f):
$$ \mathcal{F}(f) = \hat{f} \Rightarrow \mathcal{F}(\mathcal{F}^{-1}(\hat{f})) = \hat{f}$$
So I think it should be fine, since we show (*). Do you agree? Or am I missing something here...?
Jo (talk) 12:41, 4 August 2015 (CEST)
That only works if surjectivity is showed, so you know that for any \(\hat{f}\) you can find a \(f\) ... the given equation shows injectivity while the backwards equation would show surjectivity and therfore with both, it's bijective. I will insert that into the solution
--Brynerm (talk) 14:11, 4 August 2015 (CEST)
About your previous argument, it would actually be \(|\hat{G}| \leq |G|\) for both cases: If \(|\hat{G}| |Z_g| = |G|\) and if \(\hat{G}\) are linear independent over \(L(G)\), so there seems to be no possibility to show that \(|\hat{G}| = |G|\) other then with the argument #conjugacy classes = #irreducible rep.
Carl (talk) 08:59, 3 August 2015 (CEST)
:-( I noticed... I hate inequalities!
