# Continuation of homomorphisms of representations…

Hi all.

If $G$ is a finite group and $\varrho : G \to \text{GL}(V), \eta : G \to \text{GL}(W)$ are finite dimensional representations, $V_0$ is a $G$-invariant subspace of $V$ and $f : V_0 \to W$ is a homomorphism of representations, i.e. a homomorphism of vector spaces satisfying $f(\varrho(g)x) = \eta(g) f(x)$,

Question: is there any theory on the question whether or not this homomorphism can be continued to whole $V$?

Phrasing differently: How 'many' values of a homomorphism of representations can one prescribe?

Clearly, there is the theory of induced representations but unfortunately, in my case, it is not true that $V = \oplus_{g \in G} ~ \varrho(g)V_0$, neither is the sum direct nor does equality hold.

Providing more details: In my case, $G$ is $\text{SL}_2(\mathbb{Z}/N\mathbb{Z})$, say $\text{SL}_2(\mathbb{F}_p)$ for the beginning. $V$ is a certain group ring $\mathbb{C}[D]$ where $D$ is a discriminant form, i.e. $D=L'/L$ for some lattice $L$ that satisfies some properties (non degenerate, even, even signature, determinant is a $p$-power for the beginning, ... [i.e. all simplifications that seem senseful/necessary]), $\varrho$ is the Weil representation and $W$ is a certain subspace of $X = \mathbb{C}[D] \oplus \mathbb{C}[D'] \oplus ...$ i.e. a finite direct sum of the original group ring with some other group rings of other discriminant forms. The representation on $X$ is the direct sum of the Weil representations of the single discriminant forms.

Fabian Werner

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You don't explicitly say your representation is complex but I think your example shows that this is the case you're interested in. If so, then $V_0$ has a $G$-invariant complement $V_1$ by Maschke's Theorem http://en.wikipedia.org/wiki/Maschke%27s_theorem, and the unique linear extension of $f$ whose kernel contains $V_1$ will be a homomorphism of representations $V\to W$ extending $f$.

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No need to restrict to complex representations. One just has to assume that the characteristic of the field does not divide the order of $G$. – Vít Tuček May 15 '13 at 15:21
That's right. – Simon Wadsley May 16 '13 at 9:40
@Simon: Thank you very much. That was more simple than i thought. – Fabian Werner May 16 '13 at 11:08