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 \[g \mapsto \rho(g)_{\mid W}:G \to \operatorname{GL}(W).\]

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:

1. \(\widetilde \pi\) is a homomorphism of \(G\)-representations, and

2. \(\widetilde \pi\) 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.