Hi, Let $G_1, G_2$ be topological groups with $G_1 \subset G_2$ is closed. Let $\rho:G_1 \to Aut(V)$ be a smooth irreducible representation. Can anyone tell me if there is a criterion/example/idea about when $\rho$ cannot be extended to $G_2$, i.e., there does not exist smooth representation $\rho':G_2 \to Aut(V)$ such that $\rho'|_{G_1}=\rho$.
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$\begingroup$ I haven't thought this all the way through, but many examples of non-normal abelian $G_1$ inside non-abelian $G_2$ give obstructions. For example, let $G_1$ be strictly upper-triangular matrices in $G_2=GL_2(\mathbb R)$. All irreducible representations of $G_1$ are $1$-dimensional, but the only $1$-dimensional representations of $G_2$ factor through the determinant, so are trivial on $G_1$. This depends on what you mean by "extending/restricting" and the result being "equal to" the original. For example, Frobenius reciprocity says that induction and restriction are adjoint functors. $\endgroup$– B RCommented Mar 30, 2012 at 18:43
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1$\begingroup$ Take $\mathbb Z$ (discrete topology) as closed subgroup of $\mathbb R$, suppose $V$ is of dimension one over $F$, when $F$ is not "big enough", there are only trivial character of $Z$ to $F^{*}$ which could be extended to $\mathbb R$, e.g $F=\mathbb Q_p$. $\endgroup$– user20421Commented Mar 30, 2012 at 19:07
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The above comments show that in general linear representations do not extend, so a better question is to ask when linear representations do extend! This leads rather quickly to Margulis' super-rigidity and the subsequent programme generated by this result.