Today I Learned

Archive

(back up to January)

January 23rd

Today I learned a proof of Maschke’s Theorem: every representation of a finite group $\rho:G\to\op{GL}(V)$ can be decomposed into a direct sum of irreducible representations. (An irreducible representation is one that has no $G$-invariant subspaces.) The main claim is that $\rho$ preserves exists a positive-definite Hermitian form on $V$ preserved by $\rho(G)$; that is $\rho$ is always unitary for some form.

This is somewhat clever. We remark that certainly some positive-definite Hermitian form $\{\bullet,\bullet\}$ on $V$ exists—for example, choose an arbitrary basis to write $V\cong\CC^n$ for dimension $n,$ and then use the induced Hermitian form from $\CC^n.$ For this, we "average'' $\{\bullet,\bullet\}$ over $G,$ by writing\[\langle v,w\rangle=\frac1{|G|}\sum_{g\in G}\{\rho_gv,\rho_gw\}.\]We claim that this $\langle\bullet,\bullet\rangle$ is a positive-definite Hermitian form that is also $G$-invariant.

  • This is positive-definite because \[\langle v,v\rangle=\frac1{|G|}\sum_{g\in G}\{\rho_gv,\rho_gv\}\ge0,\] inherited from the positive-definite nature of $\{\bullet,\bullet\}.$ Additionally, equality holds if and only if $v=0$ because $\rho_gv=0$ is equivalent to $v=0.$ (In particular, $\rho_g\in\op{GL}(V)$ is invertible.)

  • This is bilinear because \[\langle au+bv,w\rangle=\frac1{|G|}\sum_{g\in G}\{a\rho_gv+b\rho_gv,\rho_gw\}=a\langle u,w\rangle+b\langle v,w\rangle\] after expanding out and distributing $\{\bullet,\bullet\}.$

  • This is Hermitian because \[\langle w,v\rangle=\frac1{|G|}\sum_{g\in G}\{\rho_gw,\rho_gv\}=\overline{\langle v,w\rangle},\] again by inheriting $\{\bullet,\bullet\}$ being Hermitian.

  • Finally, this is $G$-invariant because for any $h\in G,$ we have that $g\mapsto gh$ is a bijection over $G$ because of left-cancellation. Thus, we may write \[\langle\rho_hv,\rho_hw\rangle=\frac1{|G|}\sum_{g\in G}\{\rho_{gh}v,\rho_{gh}w\}=\frac1{|G|}\sum_{g\in G}\{\rho_gv,\rho_gw\}=\langle v,w\rangle\] by re-indexing the sum. We have used the fact that $\rho_\bullet$ is a homomorphism.

We remark that this means any finite subgroup of $\op{GL}(V)$—those that are the images of some representation—are made of unitary transformations, up to a change of basis.

Now that we know our representation gives unitary transformations, it's not terribly difficult to give our decomposition; equip $V$ with a fixed form $\langle\bullet,\bullet\rangle$ to make $\rho$ unitary. Decomposition is done inductively: the claim is that if $\rho$ has a $G$-invariant subspace named $W,$ then $\rho$ is the direct sum of its restriction to $W$ and its restriction to the orthogonal complement $W^\perp$ of $W.$ In this way, we can continually subdivide $\rho$ into direct sums until no parts have a $G$-invariant subspace.

Quickly, we recall that\[W^\perp=\{v\in V:\langle v,w\rangle=0\text{ for all }w\in W\}.\]It is true that $V=W\oplus W^\perp.$ For example, if we build an orthonormal basis $\{w_1,\ldots,w_m\}$ for $W$ (say, using Gram–Schmidt), then we can look at\[v-\sum_{k=1}^m\langle v,w_k\rangle w_k\in W^\perp.\]Indeed, for any basis vector $w_\ell,$ most of the dot products $\langle w_k,w_\ell\rangle$ are $0,$ unless $w_\ell=w_k,$ in which case $\langle v,w_\ell\rangle$ will properly cancel with it. A basis is probably unnecessary for this result, but whatever.

We note also that the restriction of $\rho$ to $W^\perp$ makes sense. Indeed, for $v\in W^\perp$ and any $w\in W,$ we have that\[\langle\rho_\bullet v,w\rangle=\langle v,\rho_\bullet^{-1}w\rangle=0\]because $\rho_\bullet^{-1}w\in W.$ Thus, $W^\perp$ is also $G$-invariant, so we may restrict $\rho$ to $W^\perp.$

Thus, because $V=W\oplus W^\perp,$ and $\rho$ restricts properly to each subspace, we may say that $\rho$ is the direct sum of its restrictions, say $\rho^\parallel$ to $W$ and $\rho^\perp$ to $W^\perp.$ To show how this works, we note that for any $v\in V,$ we can write $v=w^\parallel+w^\perp$ for $w^\parallel\in W$ and $w^\perp\in W^\perp,$ so\[\rho_\bullet v=\rho_\bullet w^\parallel+\rho_\bullet w^\perp=\rho^\parallel_\bullet w^\parallel+\rho^\perp_\bullet w^\perp,\]which is what we wanted. This completes the proof of Maschke’s Theorem.