3
$\begingroup$

I am am very confused about a fundamental result in representation theory of finite groups. Please let me first introduce the setting.

Let $G$ be a finite group. The group algebra $\mathbb{C}[G]$ is a semisimple ring by Maschke's theorem. Hence, there is an isomorphism $$ \mathbb{C}[G]\cong U_1\oplus\ldots\oplus U_n \hspace{20pt}\text{(1)} $$ for simple $\mathbb{C}[G]$ (always ignored in this question: left-)modules $U_i$ and those correspond to (wlog non-trivial) irreducible complex representations of $G$. In particular, we get a formula $|G|=|U_1|+\ldots+|U_n|$ by taking the dimension as complex vector spaces.

On the other hand the Artin-Wedderburn implies that the semisimple ring $\mathbb{C}[G]$ decomposes as $$ \mathbb{C}[G]\cong M_1\oplus\ldots\oplus M_m\hspace{20pt}\text{(2)} $$ where $M_i$ are (wlog non-trivial) matrix rings over division rings $D_i$. It can be seen that $m$ equals the number of conjugacy classes of $G$ and the number of non-equivalent (and non-trivial) irreducible representations of $G$ and one has the formula $|G|=d_1^2+\ldots+d_m^2$ where the $d_i$ are the dimensions of the irreducible representations of $G$.

My question is basically:

How do the two decompositions (1) and (2) relate?

My guess is the following: In the decomposition (1), some of the $U_i$ may be isomorphic as $\mathbb{C}[G]$-modules. Group them together to obtain $$ \mathbb{C}[G]\cong (U_{1}\oplus\ldots\oplus U_{j_1})\oplus\ldots\oplus(U_{j_{k-1}+1}\oplus\ldots\oplus U_{j_k}) $$ (to keep the indexing simple, we assumed that in (1) the modules already ordered properly.). Then one can perhaps (?) show that $$ U_{i_{k-1}+1}\oplus\ldots\oplus U_{i_k}\cong M_i $$ as $\mathbb{C}[G]$ modules. Is this the right way to relate (1) and (2)? Thank you.

$\endgroup$
3
$\begingroup$

If $M_n(D)$ is a matrix over a division ring $D$, then it is semisimple. If you consider the column space of the matrix then it is $D^{n\times 1}$ and hence is semisimple. Then you can write $M_n(D)=M_1\bigoplus M_2\bigoplus...\bigoplus M_n $, where $M_i$ is the matrix which has non-zero entries only in it's $i$-th column. Hence $M_n(D)$ is semi-simple. So by Maschke's theorem

$\mathbb{C}[G]=M_1^{n_1}\bigoplus ...\bigoplus M_n^{n_1}\bigoplus M_{1}^{n_2}\bigoplus ...\bigoplus M_n^{n_2}\bigoplus...\bigoplus M_1^{n_s}\bigoplus ...\bigoplus M_n^{n_s}$, where $M_i^{n_j}$ is the matrix corresponding to the matrix having only non-zero entries in the $i$-th column corresponding to the $n_j$-th matrix in the statement of the Artin-Wedderburn theorem and $s$ denote the number of conjugacy classs of $G$.

If you just compare the number of matrices then you will see that the expresion for $|G|$ is same from both the theorems.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Just that I understand you correctly: The connection between your answer and my "suggestion" (at the end of the question) is that your answer gives a reason why my "suggestion" is correct, does it? $\endgroup$ – user8463524 Dec 13 '15 at 12:51
  • $\begingroup$ @jeffrey yes, that's correct. and in fact, now you have the concrete description of your $U_i$'s. $\endgroup$ – seeker Dec 13 '15 at 16:40
  • $\begingroup$ please let me again ask about the logic of your proof: We start with the Artin-Wedderburn decomposition (this is (2) in my question). Then we split the summands $M_n(D)$ further into summands $D^{n\times 1}$. Do you mean $D^{n\times 1}$ is simple instead of semisimple? If you mean this, then we have a decomposition of $\mathbb{C}[G]$ into simple summands in this way. Now you can show that each simple module over $\mathbb{C}[G]$ is isomorphic to one of the summands above. Does this reproduce your line of reasoning? Thanks. $\endgroup$ – user8463524 Dec 16 '15 at 15:01
  • 1
    $\begingroup$ @jeffrey Yes. I wanted to say that $D^{n\times 1}$ is simple and hence $M_n(D)$ is semi-simple. Rest of the reasoning is same as you spelled out in your comment. $\endgroup$ – seeker Dec 16 '15 at 17:26
3
$\begingroup$

The Artin-Wedderburn decomposition can thought of as a decomposition

$$\mathbb{C}[G] \cong \bigoplus_V V \otimes V^{\ast}$$

with respect to two $G$-actions, namely both the action on the left and the action on the right (although I may be off by either an inverse or a dual). Here $V$ runs over all complex irreps of $G$. When you only pay attention to the left action, each of the right action bits $V^{\ast}$ just become vector spaces, and you get

$$\mathbb{C}[G] \cong \bigoplus_V (\dim V) V.$$

$\endgroup$

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.