2
$\begingroup$

Let $H$ be an open subgroup in a locally compact group $G$, $\iota:H\to G$ the embedding of $H$ into $G$, $\pi:H\to B(X)$ a unitary representation of $H$ in a Hilbert space $X$, and $\rho:G\to B(Y)$ the corresponding induced representation of $G$. When does there exist a (continuous in a proper sense) involutive homomorphism of algebras $\varphi:B(X)\to B(Y)$ such that $$ \varphi\circ\pi=\rho\circ\iota \quad ? $$ This seems to be true at least when $H$ is finite and $\pi$ is irreducible. Is it possible that this is always true?

$\endgroup$
1
  • 2
    $\begingroup$ I tried to input a diagram, but they seem to do not work here. Is it possible to use diagrams in MO? $\endgroup$ Sep 15, 2014 at 18:00

2 Answers 2

5
$\begingroup$

This is false in many cases where $G$ is finite. Let $\rho\circ \iota$ and $\pi$ denote the corresponding maps of group algebras $\mathbb{C}[G]$. The equation above can only hold if any element killed by $\pi$ must also be killed by $\rho\circ \iota$, that is $\mathrm{ker}(\pi)\subset \mathrm{ker}(\rho\circ \iota)$. If $\pi$ is irreducible, then $\mathrm{ker}(\pi)$ is the sum of all the other matrix algebra summands in the group algebra, so $\mathrm{ker}(\rho\circ \iota)$ can only contain $\mathrm{ker}(\pi)$ if $\rho\circ \iota$ is just a bunch of copies of the same irrep. Of course, this could happen, but it doesn't have to; for example, let $\pi$ be trivial, and $H$ any non-normal subgroup. The restriction of the permutation rep of $G/H$ to $H$ is non-trivial, so it can't factor through the trivial rep.

$\endgroup$
3
  • $\begingroup$ Ben, can this be true when $H$ is normal? $\endgroup$ Sep 18, 2014 at 16:44
  • $\begingroup$ When $H$ is normal, then $\rho\circ \iota$ is a bunch of copies of $\pi$, so there the desired factoring is true. $\endgroup$
    – Ben Webster
    Sep 19, 2014 at 2:34
  • $\begingroup$ My $H$ is normal. Ben, I need a reference. I must refer to a textbook or to a paper in my text. $\endgroup$ Sep 19, 2014 at 13:21
2
$\begingroup$

In addition to @BenWebster's points, my own reaction would be to see that the hoped-for property is fighting against any/all of several natural characterizations of "induction", e.g., as right (or left) adjoint to restriction... That is, some "type" violation is hoped-for, which, especially in light of counter-examples, is hard to see the reason to believe-in.

Thus, I'd ask if there's a context in which something of this sort arises, so that perhaps what might really suffice/work/solve-the-problem is different enough so as to avoid the "un-naturality"/type-violation in the first place, and avoid the tangible counter-example schema mentioned by Ben W... ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.