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I'm reading a paper that uses representation theory, and I'm stuck on something simple. Say $\Delta$ is an abelian group, and $\hat{\Delta}$ its group of irreducible characters. Say $M$ is a $\mathbb{C}[\Delta]$-module. The paper uses the decomposition $M=\oplus_{\chi \in \hat{\Delta}} M^{\chi}$, where $M^{\chi}:=e_{\chi}M$ and $e_{\chi}:=\frac{1}{|\Delta|} \sum_{\delta \in \Delta} \chi(\delta)\delta^{-1}$.

The question, roughly speaking, is: how should I think of the $M^{\chi}$'s? What is their meaning, and function? How else are they characterized?

I've taken a course in representation theory a few years back, so it's all a little vague, but if I remember correctly, $\mathbb{C}[\Delta]$ is the direct sum of $|\Delta|$ many simple submodules, in bijection with the $\chi$'s. Then the $e_{\chi}$'s can be interpreted as being the decomposition of $1$ ($1=\sum_{\chi} e_{\chi}$, where $e_{\chi}$ is in the simple submodule of $\mathbb{C}[\Delta]$ corresponding to $\chi$).

But I don't see how this translates to a decomposition of some random module $M$. $M$ is a direct sum of simple submodules. But there's reason to think that there are $|\Delta|$ many (for example: what if $M$ is simple?). So what is the interpretation of the $M^{\chi}$'s?

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I guess you are referring to finite abelian group since there's a sum involved in $e_{\chi}$. Then every irreducible representation of $G$ is 1-dimensional, i.e. they are the irreducible characters. Decompose your $M$ as a direct sum of irreducible representations, each corresponding to a character, and put those corresponding to the same character together. That's your $M^{\chi}$. – Soarer Apr 5 '11 at 21:35
So it's possible that a lot of these $M^{\chi}$'s are going to be trivial? – user9170 Apr 5 '11 at 21:50
Dear Nicole, Yes, many of the $M^{\chi}$s could be trivial. The extreme case (other then when $M$ is trivial itself!) would be if $M$ were just one-dimensional, with $\Delta$ acting by a single character $\chi$. Then $M^{\chi} = M$, while $M^{\chi'} = 0$ if $\chi' \neq \chi$. Regards, – Matt E Apr 6 '11 at 4:16

3 Answers 3

The $M^{\chi}$ are called isotypic components in the literature. You should think of $M^{\chi}$ as the subspace of $M$ consisting of elements which "transform like $\chi$" under the action of $\Delta$.

This will make much more sense to you once you master a simple example. Consider the group $\mathbb{Z}/2\mathbb{Z}$ acting on, say, the space of continuous functions $\mathbb{R} \to \mathbb{R}$ via $f(x) \mapsto f(-x)$. This space decomposes into two isotypic components: the even functions and the odd functions (corresponding to the trivial and nontrivial character of $\mathbb{Z}/2\mathbb{Z}$). Any function is uniquely a sum of an even and an odd function since

$$f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}.$$

The statement you are asking about is just a natural generalization of this idea.

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First fix an isomorphism between the characters on $A$ and $A$. Are you saying that $M^{\chi}$ is {$m \in M| am=m$} where $a$ corresponds to $\chi$? – user9170 Apr 6 '11 at 2:52
You don't need to fix such an isomorphism. $M^{\chi}$ is precisely the submodule of elements $m$ such that $gm = \chi(g) m$. – Qiaochu Yuan Apr 6 '11 at 3:13
Excellent! That sort of description is what I was looking for. – user9170 Apr 6 '11 at 3:22

The category of representations of a finite group over $\mathbb{C}$ (or more generally a field whose characteristic is prime to the order of the group) is a semisimple category: every object splits as a direct sum of finitely many simple objects. In fact, one can directly show that, if the group $G$ is abelian, then $\mathbb{C}[G]$ is isomorphic to $G$ copies of $\mathbb{C}$ as a ring; to do this, note that $\mathbb{C}[G]$ can be identified with complex-valued functions on $G$ with convolution the product. Now the Fourier transform induces an isomorphism of complex-valued functions on $G$ with complex-valued functions on $G$, that takes convolution to multiplication.

So if we think of $\mathbb{C}[G]$ as a commutative ring, it is isomorphic (via the Fourier transform) to the ring of functions on $G$ with the multiplication being pointwise multiplication. In other words, this is just the direct product ring $\prod_G \mathbb{C}$.

What are the modules over $\prod_G \mathbb{C}$? Well, any such splits as a finite direct sum of modules over which the $i$th factor acts by multiplication by complex numbers, and where the other factors act by zero. In general, a module over a product ring $R_1 \times R_2$ decomposes uniquely as a direct sum of an $R_1$-module (on which $R_2$ acts trivially) and an $R_2$-module (on which $R_1$ acts trivially) simply because we have idempotents $e_1, e_2$ in the product ring, and can take their images.

So what does this correspond to if we have a $\mathbb{C}[G]$-module and want to decompose it? Well, the above construction shows that we need to find the primitive idempotents. Since the Fourier transform gave us an isomorphism with $\mathbb{C}^G$, and the latter has an obvious set of idempotents, we see that there is a correspondence of primitive idempotents. This gives that the idempotents in $\mathbb{C}[G]$ correspond precisely to the characters (since the characters define the Fourier transform) and in the way you describe.

So the decomposition is just a special instance of the general fact alluded to above about modules over a direct product of rings. (This works even when $G$ is non-abelian, though then the group algebra decomposes as a product of matrix algebras, by general structure theory for semisimple algebras.)

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Although I'm sure Akhil is aware of this, for other readers it might be best to clafiry:

Finite-dimensional $\mathbb{C}[G]$-modules are always semi-simple, as is the case when $\mathbb{C}$ is replaced by any other field of characteristic $0$. The category of representations over a field of prime characteristic, however, is not semi-simple. Any group whose order is divisible by the characteristic of the field will have a a representation which does not decompose into irreducible representations.

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Yes, you are right; thanks for pointing that out. – Akhil Mathew Apr 6 '11 at 6:04

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