# The Unapologetic Mathematician

## Subspaces from Irreducible Representations

Because of Maschke’s theorem we know that every representation of a finite group $G$ can be decomposed into chunks that correspond to irreducible representations: $\displaystyle V\cong\bigoplus\limits_{i=1}^km_iV^{(i)}$

where the $V^{(i)}$ are pairwise-inequivalent irreps. But our consequences of orthonormality prove that there can be only finitely many such inequivalent irreps. So we may as well say that $k$ is the number of them and let a multiplicity $m_i$ be zero if $V^{(i)}$ doesn’t show up in $V$ at all.

Now there’s one part of this setup that’s a little less than satisfying. For now, let’s say that $V$ is an irrep itself, and let $m$ be a natural number for its multiplicity. We’ve been considering the representation $\displaystyle mV=\bigoplus\limits_{i=1}^mV$

made up of the direct sum of $m$ copies of $V$. But this leaves some impression that these copies of $V$ actually exist in some sense inside the sum. In fact, though inequivalent irreps stay distinct, equivalent ones lose their separate identities in the sum. Indeed, we’ve seen that $\displaystyle\hom_G(V,mV)\cong\mathrm{Mat}_{1,m}(\mathbb{C})$

That is, we can find a copy of $V$ lying “across” all $m$ copies in the sum in all sorts of different ways. The identified copies are like the basis vectors in an $m$-dimensional vector space — they hardly account for all the vectors in the space.

We need a more satisfactory way of describing this space. And it turns out that we have one: $\displaystyle mV=\bigoplus\limits_{i=1}^mV\cong V\otimes\mathbb{C}^m$

Here, the tensor product is over the base field $\mathbb{C}$, so the “extra action” by $G$ on $V$ makes this into a $G$-module as well.

This actually makes sense, because as we pass from representations to their characters, we also pass from “plain” vector spaces to their dimensions, and from tensor products to regular products. Thus at the level of characters this says that adding $m$ copies of an irreducible character together gives the same result as multiplying it by $m$, which is obviously true. So since the two sides have the same characters, they contain the same number of copies of the same irreps, and so they are isomorphic as asserted.

Actually, any vector space of dimension $m$ will do in the place of $\mathbb{C}^m$ here. And we have one immediately at hand: $\hom_G(V,mV)$ itself. That is, if $V$ is an irreducible representation then we have an isomorphism: $\displaystyle mV\cong V\otimes\hom_G(V,mV)$

As an example, if $V$ is any representation and $V^{(i)}$ is any irrep, then we find $\displaystyle m_iV^{(i)}\cong V^{(i)}\otimes\hom_G(V^{(i)},V)$

We can reassemble these subspaces to find $\displaystyle V\cong\bigoplus\limits_{i=1}^km_iV^{(i)}\cong\bigoplus\limits_{i=1}^kV^{(i)}\otimes\hom_G(V^{(i)},V)$

Notice that this extends our analogy between $\hom$ spaces and inner products. Indeed, if we have an orthonormal basis $\{e_i\}$ of a vector space of dimension $k$, we can decompose any vector as $\displaystyle v=\sum\limits_{i=1}^ke_i\langle e_i,v\rangle$

November 10, 2010