2
$\begingroup$

Let $k$ be an algebraically closed field of characteristic $0$ and $G$ a finite group. The representation theory of $G$ is essentially the same the as the module theory over the (non commutative) algebra $k[G]$. Now, it is a well known fact that there is an isomorphism as algebras: $$k[G] \to \bigoplus_i M_{n_i}(k)$$ where $i$ runs over the irreducible representations of $G$ and $n_i$ is the dimension of the $i$-th representation, denoted by $V_i$. The problem is, I know two distinct ways of defining the above map. I suspect they are the same thing but can't show it.

The First Way: Consider $R = k[G]$ as a simple module over $k[G]$. There is an isomorphism, as modules: $$k[G] \cong \bigoplus_i V_i^{n_i}.$$ Applying the function $End_{k[G]}( -)$ on both sides, we obtain and isomorphism: $$\mu: k[G]^{opp} \to \bigoplus_i M_{n_i}(k).$$ On the right, the element $g$ corresponds to the $G-$ endomorphism of $R$ that sends $a \to ag$ for $a \in R$. On the right, the isomorphism with the matrix algebras is because of Schur's lemma and an element in $M_{n_i}(k)$ acts on $V_i^{n_i}$ by treating each $V_i$ as a one dimensional vector space essentially.

The second way: Now, for each $V_i$, the representation gives us a map $\rho_i: G \to GL_{n_i}(k)$ where $GL_n(k)$ is the invertible linear maps on $V_i$ (just one copy this time!). We can extend linearly to $\rho_i: k[G] \to M_{n_i}(k)$ and put everything together to get a map: $$\rho: k[G] \to \bigoplus_i M_{n_i}(k).$$

Now, it isn't clear to me at all that these two maps $\mu, \rho$ are the same thing. In particular, we need to show:

Fix a module isomorphism $\theta: kG \to \bigoplus_i V_i^{n_i}$. Then, define the map $R_h$ on $kG$ that sends $g \to gh$. This will translate into an endomorphism on the $V_i^{n_i}$ and since it is equivariant, we can represent it by an element in $\bigoplus_i M_{n_i}(V_i)$ where in the $i-th factor$, the entries of the matrix correspond to the trace of $R_h$ restricted to various factors. The question is whether on the $i-th$ factor, the matrix thus defined is equal to $\rho_i(h)$, the matrix defining the representation

For one, there is the small detail that $\mu$ has domain $k[G]^{opp}$ while $\rho$ has domain $k[G]$ but we can consider $\mu^{opp}$ instead since matrix algebras are self dual.

Second, there is the bigger issue that the two matrix algebras on the right act on different things in different ways. The first matrix algebra acts on $V_i^{n_i}$ by $G$-linear endomorphisms. The second matrix algebra acts on $V_i$ by simply linear endomorphisms.

Third: If we remove condition that $k$ be algebraically closed, then in the case of $\mu$, the fields $k$ are replaced by appropriate division algebras (and the number of direct summands = number of representations goes down).

However, in the case of $\rho$, the number of direct summands goes down as before but I don't think the field $k$ gets replaced by a division algebra.

Given this three points, I am doubtful wether $\mu$ and $\rho$ are really the same but it somehow seems even more improbable that we get distinct maps like this. What is going on here?

Of course, there is quite a bit of choice involved in defining these maps so the best we can hope for is that $\rho,\mu$ are equivalent, not equal on the nose.

$\endgroup$
  • 1
    $\begingroup$ You mean $\mu: k[G]^{opp} \to \bigoplus_{i} M_{n_i}(k)$, right? $\endgroup$ – AnalysisStudent0414 Jul 9 '18 at 10:02
  • 1
    $\begingroup$ Yea, sorry for the oversight. $\endgroup$ – Asvin Jul 9 '18 at 10:02
1
$\begingroup$

Let $R$ be the regular representation, $h$ an element of $G$, $R_h$ the $G$-equivariant endomorphism or $k[G]$ that sends $g \to gh$. Let $\theta: kG \to \bigoplus_i V_i^{n_i}$ be an isomorphism and fix $i$.

Also, let $\pi: k[G] \to V_i^{n_i}$ be the projection map induced by $\theta$. Through $\theta$, $R_h$ acts on the $V_i^{n_i}$ since it is $G$-equivariant. Moreover, the action is by a scalar on each component by Schur's lemma.

That is, we can represent $R_h$ by a scalar $A = [a_{ij}]$ so that for a vector $v = (v_1,v_2,\dots,v_{n_i}) \in V_i^{n_i}$, $R_h$ sends it to $Av$. Explicitly, $v_l \to \sum_k a_{lk}v_k$.

On the other hand, consider the image of the identity element $\pi(e)$ in $V_i^{n_i}$. Let $\pi(e) = (v_1,v_2,\dots,v_n)$. Since $\pi$ is a surjection and $G\pi(e)$ generates all of $V_i^{n_i}$, the $v_1,\dots, v_n$ are linearly independent in $V_i$ and hence form a basis for it.

Moreover, $\pi(R_h(e)) = \pi(h) = h\pi(e) = (hv_1,hv_2,\dots,hv_n)$ but this is also equal to the image of $(v_1,\dots,v_n)$ under $R_h$ (or $A$). Let $h = [h_ij]$ as a matrix on $V_i$ in the basis $v_1,\dots,v_{n_i}$.

Therefore, $\sum_kh_{lk}v_k = hv_{l} = \sum_k a_{lk}v_k$ and since the $v_k$ are linearly independent, this implies that $h_{lk} = a_{lk}$.

That is, $R_h$ acts on $V_{n_i}$ by the matrix given by $h$.

$\endgroup$
0
$\begingroup$

I hope I understand your doubts correctly: here's what I think.
Forget for a moment about the matrix algebras: if you go one step behind, and look at where this comes from, you have the map $$kG \longrightarrow \bigoplus_i (V_i)^{n_i}$$ as $G$-modules, that is, $G$ is acting on $kG$ via the regular representation, and this decomposes into a sum of irreducibles, each with its multiplicity (this is Maschke's theorem). Of course, you can extend the $G$-action linearly and get a $kG$ action on $kG$.

In the first case, you are looking exactly at this action: you pick all the $kG$-linear endomorphisms of $kG$, and "translate" this action on the right. If $k$ is algebraically closed, Schur's lemma tells you exactly what you get (scalars basically), so you do get a matrix algebra on the right.

In the second case, you project the $G$-action on each component $(V_i)^{n_i}$, and you get the map $G \to \operatorname{GL}_{n_i}(V_i)$. This is still a $G$-module, so when you extend linearly to get an action of $kG$ what you get is still a bunch of $G$-linear endomorphisms (of course, you can choose to "only" view this as a linear map, but it still is $G$-linear). Again, if $k$ is algebraically closed Schur's lemma now tells you that the thing on the right is a matrix algebra $M_{n_i}(V_i)$. Now you put them back together again, getting an action $$kG \longrightarrow \bigoplus_i M_{n_i}(V_i)$$

Essentially, what it looks like to me is that in the first step you project on a factor, identify the matrix algebra there and then lift it back up to the direct sum, while in the first factor you do it "all at once" (only considering the direct sum structure when you apply Schur's lemma).
Another small difference is that in the first case you use the $\operatorname{End}$ functor, which gives you the $^{opp}$ (which, as you correctly point out, doesn't matter if the endgame is identifying things as matrix algebras), while in the second case you look at the action of every $g \in G$ identifying it as an invertible matrix in $\operatorname{GL}$. If you try to use the functor also in the second case, on each summand, you will get the same thing as in the first, without the $^{opp}$.

$\endgroup$
  • $\begingroup$ In the second case, I believe we are projecting to the component $V_i$ instead of $V_i^{n_i}$ and the representation is then defined by a map $G \to Aut_k(V_i) = GL_n(k)$ that defines the representation on $V_i$. $\endgroup$ – Asvin Jul 9 '18 at 10:20
  • $\begingroup$ Well, let me ignore the non algebraically closed case for now. In the second case, I make no mention of the group algebra even. All I do is pick for each irreducible representation the map $G \to Aut(V_i)$ defining the representation. I don't see why this has anything to do with the map defined in the first case where we only talk about endomorphisms that preserve the group action. $\endgroup$ – Asvin Jul 9 '18 at 10:27
  • $\begingroup$ Let me try to clarify: Fix a module isomorphism $\theta: kG \to \bigoplus_i V_i^{n_i}$. Then, define the map $R_h$ on $kG$ that sends $g \to gh$. This will translate into an endomorphism on the $V_i^{n_i}$ and since it is equivariant, we can represent it by an element in $\bigoplus_i M_{n_i}(V_i)$ where in the $i-th factor$, the entries of the matrix correspond to the trace of $R_h$ restricted to various factors. The question is whether on the $i-th$ factor, the matrix thus defined is equal to $\rho_i(h)$, the matrix defining the representation $\endgroup$ – Asvin Jul 9 '18 at 10:37
  • $\begingroup$ Yes, it does, and it also preserves the group action (it is a direct sum, you are going down and back up). In the first case you choose to highlight this first, in the second case you focus on something else, but it is the same map. $\endgroup$ – AnalysisStudent0414 Jul 9 '18 at 10:47
  • 1
    $\begingroup$ Okay, I figured it out and wrote up an answer. I am not sure if you were suggesting the same thing as I wrote up but I did need to do some computations to figure out what was happening. $\endgroup$ – Asvin Jul 9 '18 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.