8
$\begingroup$

I came up with a "proof" for the Maschke's Theorem in the representation theory of finite groups that seems to make sense. But it doesn't use the fact that the group being represented is finite. So I suspect there must be something wrong with it. Can someone please help me find where it goes wrong?

In the following, the underlying field is assumed to have characteristic $0$, and all the vector spaces involved are assumed to be finite dimensional.

The following form of the theorem is taken from Fulton&Harris' Representation Theory - A First Course (Proposition 1.5).

Maschke's Theorem. If $W$ is a subrepresentation of a representation $V$ of a finite group $G$, then there is a complementary invariant subspace $W'$ of $V$, so that $V=W\oplus W'$.

Proof. Firstly, since the action of every $g\in G$ is invertible, if a subspace $V'\subset V$ is $g$-invariant, then $g.V'=V'$ without loss of dimensions. Hence $V'$ is also $g^{-1}$-invariant and $g^{-1}.V'=V'$.

Suppose for now $W\subset V$ is an arbitrary subspace, not necessarily invariant under the actions of any $g\in G$. Let $\pi: V\to W$ be the corresponding linear projection, which induces a direct sum decomposition $V=W\oplus \text{Ker}(\pi)$.

Define $p_{g,\pi}:=g^{-1}\pi g$. Then it is easy to verify $p_{g,\pi}^2=g^{-1}\pi gg^{-1}\pi g=p_{g,\pi}$ and so it is also a linear projection. This would induce an alternative direct sum decomposition $V= \text{ran}(p_{g,\pi})\oplus \text{Ker}(p_{g,\pi})$. We further have $g. \text{ran}(p_{g,\pi})\subset W$. In fact, for every $x\in \text{ran}(p_{g,\pi})$, we have $x=p_{g,\pi}(x)=g^{-1}\pi gx$ and so $gx=\pi gx\in W$.

In case $W$ is $g$-invariant, by first paragraph, $W$ is also $g^{-1}$-invariant. Therefore, since $\pi|_W$ is the identity map, for every $w\in W$, we have $\text{ran}(p_{g,\pi})\ni p_{g,\pi}(w)=g^{-1}\pi gw=g^{-1}gw=w$. Hence $W\subset \text{ran}(p_{g,\pi})$.

Then by both previous paragraphs, $g. \text{ran}(p_{g,\pi})\subset W\subset \text{ran}(p_{g,\pi})$ and so $ W= \text{ran}(p_{g,\pi})$, since $g$ is invertible. This gives $\text{Ker}(p_{g,\pi})=\text{Ker}(\pi)$ and $p_{g,\pi}=\pi$. That is, the two direct sum decompositions of $V$ become identical.

Then, we can check that $g$ and $g^{-1}$ both preserve the two identical kernels. In fact, for every $y\in \text{Ker}(p_{g,\pi})=\text{Ker}(\pi)$, we have $p_{g,\pi}(g^{-1}y)=g^{-1}\pi gg^{-1}y=g^{-1}\pi y=0$, and so $g^{-1}y\in \text{Ker}(p_{g,\pi})=\text{Ker}(\pi)$. By the invertibility of $g^{-1}$, the two identical kernels are both $g$-invariant and $g^{-1}$-invariant.

Lastly, if $W$ is $G$-invariant, then every $g\in G$ induces the same projection $p_{g,\pi}=\pi$ as reasoned above, and the subspace $W':=\text{Ker}(\pi)$ is then also $G$-invariant and complementary to $W$ such that $V=W\oplus W'$. ${\rm \square}$

As you can see, nowhere in the "proof" above is the fact used that $G$ is a finite group, and that's horrifying to me...

UPDATE:

According to Eric's answer, the issue is that, once we have $W= \text{ran}(p_{g,\pi})$, we cannot conclude the two projections are identical because their kernels might still be different. However, we would still have direct sum decomposition $V=W\oplus \text{Ker}(p_{g,\pi})$. This would be true for every $g\in G$. But the corresponding kernel would change as $g$ runs through $G$. This is where the "averaging technique" comes into play, and how the finiteness of $G$ plays a key role.

$\endgroup$
1
  • 5
    $\begingroup$ Why not take a counterexample to (infinite) Maschke's Theorem and go through your argument line by line to find the error? A cheap counterexample: Consider the integers embedded into $SL_2( \mathbb C)$ as upper triangular matrices, with invariant subspace $\mathbb C\times 0$. $\endgroup$
    – lulu
    Feb 19 at 21:08

1 Answer 1

Reset to default
13
$\begingroup$

You seem to be assuming that there is a unique projection onto a subspace of a vector space. That is not true. For instance, if $W$ is two-dimensional space with basis $\{e_1,e_2\}$, then for any scalar $c$, $\{e_1,e_2+ce_1\}$ is also a basis, and there is a projection onto the span of $e_1$ corresponding to this basis. These projections are all different, since their kernels are the spans of $e_2+ce_1$ which are all different subspaces of $W$.

So, in your argument, while the projections $p_{g,\pi}$ and $\pi$ have the same range, this does not imply that they are equal, or that they have the same kernel.

$\endgroup$
5
  • $\begingroup$ That's indeed the crux of the matter! Thanks a lot. $\endgroup$
    – user760
    Feb 19 at 21:21
  • 7
    $\begingroup$ What is true is that, in the presence of an inner product and working over $\mathbb{R}$ or $\mathbb{C}$, there's a unique orthogonal projection (whose kernel is the orthogonal complement). If the inner product is $G$-invariant then so is this projection (and in particular so is its kernel). This is how one proves Maschke's theorem for compact groups, by averaging over an inner product to obtain an invariant inner product. $\endgroup$
    – Qiaochu Yuan
    Feb 19 at 21:23
  • 1
    $\begingroup$ @QiaochuYuan This makes Fulton&Harris' proof intuitive for me. Thanks! $\endgroup$
    – user760
    Feb 19 at 21:43
  • 1
    $\begingroup$ @QiaochuYuan Is that why, in von Neumann algebras, the topic of lattice of projections only concerns orthogonal projections on a Hilbert space? This is of course off topic a bit $\endgroup$
    – user760
    Feb 20 at 11:23
  • 1
    $\begingroup$ Orthogonal projections correspond exactly to closed subspaces for this reason, yes. $\endgroup$
    – Qiaochu Yuan
    Feb 20 at 19:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .