2024-01-24--notes-RT.md

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title: (Representation Theory) Invariant subspaces & complete reducibility
date: 2024-01-22

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Invariant subspaces

Let $(\rho,V)$ be a representation of the group $G$ on the $F$-vector space $V$.

If $W$ is a subspace of $V$ [^1], one says that $W$ is a sub-representation of $V$, or that $W$ is an invariant subspace, provided that $$\rho(g)W \subseteq W \quad \forall g \in G.$$

If $W$ is a sub-representation, then $W$ "is" itself a representation of $G$ in a natural way, since $\rho$ determines a group homomorphism $$g \mapsto \rho(g)_{\mid W}:G \to \operatorname{GL}(W).$$

[^1]: The term "subspace" means "vector subspace". One might also say "$F$-subspace" to emphasize the scalars.

Proposition: : If $(\rho,V)$ and $(\psi,W)$ are $G$-representations and if $\Phi:V \to W$ is a homomorphism of $G$-representations then $\ker \Phi$ is a subrepresentation of $V$ and $\Phi(V)$ is a subrepresentation of $W$.

Recollections on vector subspaces and direct sums

Let $W_1$ and $W_2$ be $F$-vector subspaces of the vector space $V$. We can form the direct sum $W_1 \oplus W_2$.

And the defining property of the direct sum tells us that we get a linear mapping $$\phi:W_1 \oplus W_2 \to V$$ by the rule $\phi(w_1,w_2) = w_1 + w_2$.

Suppose the following hold:

• $W_1 + W_2 = V$ -- i.e. $\phi$ is surjective, and

• $\ker \phi = 0$ -- i.e. $W_1 \cap W_2 = \{0\}$.

Under these conditions, $\phi$ determines an isomorphism $W_1 \oplus W_2 \simeq V$, and one says that $V$ is the internal direct sum of the subspaces $W_1$ and $W_2$.

Remark: : More generally, if $W_1,W_2,\cdots,W_n$ are subspaces of $V$, suppose that

- $V = \sum_{i=1}^n W_i$, and
- for each $i$ we have $W_i \cap \left(\sum_{j\ne i} W_j\right) = 0.$

Then $V$ is the *internal direct sum* $V = W_1 \oplus W_2 \oplus
\cdots \oplus W_n$.

Example: : Let $\phi:V \to V$ be a linear mapping with $\dim V < \infty$, and suppose that $\phi$ is diagonalizable i.e. that $V$ has a basis consisting of eigenvectors for $\phi$.

Let $\lambda_1, \cdots, \lambda_k \in F$ be the *distinct eigenvalues* of $\phi$,
and let 
$$V_i = \{x \in V \mid \phi(x) = \lambda_i x\}$$
be the $\lambda_i$-eigenspace.

Then $$V = V_1 \oplus V_2 \oplus \cdots \oplus V_n$$
i.e. $V$ is the internal direct sum of the eigenspaces for $\phi$.

Proposition: : Let $W$ be a subspace of $V$ where $\dim V < \infty$. Then there is a subspace $W'$ of $V$ for which $V$ is the internal direct sum of $W$ and $W'$.

Remark : The analogue of the property described by the Proposition fails for abelian groups in general. Consider $A = \mathbb{Z}/4\mathbb{Z}$. For the subgroup $B = 2\mathbb{Z}/4\mathbb{Z}$ (of order 2), there is no subgroup $B'$ for which $A = B \oplus B'$.

Sketch of proof of Proposition: : Choose a basis $\beta_1,\dots,\beta_\ell$ for the $F$-vector space $V/W$. Now choose vectors $b_1,\cdots,b_\ell \in V$ so that $$\beta_i = b_i + W \in V/W.$$

Let $W'$ be the *span* of $b_1,\cdots,b_\ell$; i.e
$$W' = \sum_{i=1}^\ell Fb_i.$$

We are going to show that $V$ is the internal direct sum of $W$ and $W'$.

For $v \in V$, viewing $v+W$ as an element of $V/W$ we may write
$$v+W = a_1 \beta_1 + \cdots + a_\ell \beta_\ell$$
for scalars $a_i \in F$.

Let $w' = \sum_{i=1}^\ell a_i b_i \in W'$. It is clear that
$w=v - w' \in W$. Since $v = w + w'$ we have showed that $W + W' = V$.

Finally the linear independence of the $\beta_i$ shows the only element
of $W'$ contained in $W$ is $0$; thus $W \cap W' = \{0\}$ so that
$V = W \oplus W'$.

With notation as in the statement of the Proposition, one says that the subspace $W'$ is a complement to the subspace $W$.

Complements and projections

Given a subspace $W \subset V$ and a complement $W'$, so that $V = W \oplus W'$, we get a projection operator $$\pi:V = W \oplus W' \to V = W \oplus W' \quad \text{via} \quad \pi(x,y) = (x,0)$$

The mapping $\pi$ satisfies the following properties:

P1. $\pi^2 = \pi$, and

P2. $\pi(V) = W$.

We say that a linear mapping $\pi:V \to V$ is a projection onto $W$ provided that conditions P1 and P2 hold.

Lemma: : Suppose that $\pi:V \to V$ is a linear mapping. Then $\pi$ is projection onto $W$ if and only if $\pi(W) = W$ and the restriction of $\pi$ to $W$ is the identity mapping $\operatorname{id}_W$.

Proof of Lemma: : For a linear map $\pi:V \to V$ for which $\pi(V) = W$, we must show that $\pi^2 = \pi$ if and only if $\pi_{\mid W} = \operatorname{id}_W$.

$(\Rightarrow)$: Suppose that $w \in W$. Since $\pi(V) = W$, we
may write $w = \pi(v)$ for some $v \in V$. Then $\pi^2 = \pi$
shows that $\pi^2(v) = \pi(v) \implies \pi(\pi(v)) = \pi(v)$ so
that $\pi(w) = w$.

$(\Leftarrow)$: Suppose that $v \in V$. We have $\pi(v) \in W$,
and since $\pi_{\mid W}$ is the identity, we find $\pi^2(\pi(v)) =
\pi(v)$. Since this holds for every $v$, we have $\pi^2 = \pi$ as
required.

Proposition: : There is a bijection between the following:

- *complements* $W'$ to $W$ in $V$

- projections $\pi:V \to V$ onto $W$

Proof: : We've already described how to build a projection $\pi$ from a complement $W'$.

Given a projection $\pi$, take $W' = \ker \pi$. We must argue that
$W'$ is a complement to $W$ in $V$.

Suppose $x \in W \cap W'$. Since $x \in W$, the Lemma shows that
$x = \pi(x)$.  But on the other hand since $x \in W' = \ker \pi$
we find that $x = \pi(x) = 0$. This proves that $W \cap W' =
\{0\}$.

Finally we must show that $V = W + W'$. Let $v \in V$. Then $w =
\pi(v) \in W$ by P2. Now, $$v = \pi(v) + (v - \pi(v)) = w + (v - \pi(v))$$ and it
just remains to see that $v - \pi(v) \in W'$. But by P1,
$$\pi(v-\pi(v)) = \pi(v) - \pi^2(v) = \pi(v) - \pi(v) = 0.$$

Complete reducibility of $G$ representations.

Let $G$ be a finite group and $(\rho,V)$ a representation of $G$.

Definition: : We say that $(\rho,V)$ is completely reducible if for every subrepresentation $W \subseteq V$, there is a subrepresentation $W' \subseteq V$ such that $V$ is the internal direct sum of $W$ and $W'$ as vector spaces.

Theorem: : Let $F$ be a field of char. 0 and let $G$ be a finite group. Then every representation of $G$ on a finite dimensional $F$-vector space is completely reducible.

Proof: : Let $(\rho,V)$ be a (finite dimensional) representation of $G$ and let $W \subset V$ be a subrepresentation.

We *choose* a vector space complement, which by the Proposition
above amounts to the choice of a projection operator $\pi:V \to V$
onto the subspace $W$.

We form a new linear mapping $$\widetilde \pi:V \to V$$ by the
rule $$\widetilde \pi = \dfrac{1}{|G|}\sum_{g \in G} \rho(g) \circ
\pi \circ \rho(g)^{-1}.$$

We are going to prove: 

(i) $\widetilde \pi$ is a homomorphism of $G$-representations, and

(ii) $\widetilde \pi$ is a *projection*.

Together (i) and (ii) imply the Theorem. Indeed, if (ii) holds,
one knows that $W' = \ker \widetilde \pi$ is a *complement to $W$*. Since
$\widetilde \pi$ is a homomorphism of $G$-representations, one knows
that its kernel $W'$ is a subrepresentation.

To prove (i), let $h \in G$ and $v \in V$ and observe

\begin{align*}
\widetilde \pi(\rho(h) v) &= \dfrac{1}{|G|} \sum_{g \in G} 
\rho(g) \circ
\pi \circ \rho(g)^{-1} (\rho(h) v) \\	
&= \dfrac{1}{|G|} \sum_{g \in G} 
\rho(g) \circ
\pi \circ \rho(g^{-1}h) (v) \\
&= \dfrac{1}{|G|} \sum_{x \in G} 
\rho(hx) \circ \pi \circ \pi(x^{-1})(v) \\
&= \dfrac{1}{|G|} \rho(h) \sum_{x \in G} \rho(x) \circ \pi \circ \rho(x)^{-1} (v) \\
&= \rho(h) \widetilde \pi(v)
\end{align*}

Thus $\widetilde \pi$ is indeed a homomorphism of $G$-representations.

To prove (ii), we observe that for each $g \in G$, the mapping
$\rho(g) \circ \pi \circ \rho(g)^{-1}$ is also a projection onto
$W$.  Indeed, since $W$ is a subrepresentation, $\rho(g)W = W$, so
that $\rho(g) \circ \pi \circ \rho(g)^{-1}(V) \subseteq W$. On the
other hand, since $\pi_{\mid W}$ is the identity mapping, $\rho(g)
\circ \pi \circ \rho(g)^{-1}(w) = w$ for any $w\in W$ so the Lemma
above shows that $\rho(g) \circ \pi \circ \rho(g){-1}$ is a
projection onto $W$.

Now it is clear that $\sum_{g \in G} \rho(g) \circ \pi \circ
\rho(g^{-1})$ maps $V$ to $W$.  Since each mapping $\rho(g) \circ
\pi \circ \rho(g^{-1})$ is the identity on $W$, it follows that
$$\widetilde \pi = \dfrac{1}{|G|} \sum_{g \in G} \rho(g) \circ \pi
\circ \rho(g^{-1})$$ is the identity mapping on $W$, so
$\widetilde \pi$ is a projection by the Lemma above.

This completes the proof of the Theorem.