# Universal property of induced representation

Let $$H$$ be a closed subgroup of the compact Lie group $$G$$. Let $$E$$ be a continuous representation of $$H$$. In the book "Representations of compact Lie groups" by Bröcker and Dieck the induced representation of $$E$$ is defined as the vector space $$iE$$ of all continuous functions $$f:G\to E$$ satisfying $$f(g\cdot h)=h^{-1}f(g)$$ for all $$g\in G$$ and $$h\in H$$. They show that as in the finite case this construction satisfies the Frobenius reciprocity theorem.

Now I wonder whether this construction also satisfies the universal property that we know from the case of finite groups (or, more generally, finite index), i.e., my question is whether the following is true:

There exists an $$H$$-linear map $$j:E\to iE$$ such that for all $$H$$-linear maps $$g:E\to E'$$ to a $$G$$-module $$E'$$ there is a unique map $$G$$-linear map $$g':iE\to E'$$ such that $$g'\circ j=g$$.

Moreover, is $$g'$$ continuous if $$g$$ is? If the answer is "No", is there a better notion of induced representation that makes this true? Or does it help when we restrict to unitary representations?

• By unwinding the Frobenius reciprocity theorem, which says the functors of restriction and induction are adjoint, you’ll get exactly this sort of universal property. However, some care is needed - since this definition of induced representation imposes a continuity condition, it will only have the universal property among continuous representations. – dorebell Jun 15 '20 at 17:11
• In the case of finite groups, induction is both left and right adjoint to restriction. Is that true in general? In the book I am looking at, they only prove one of these two. In particular, I only get a natural map $iE\to E$ from that theorem. What is the natural map $E\to iE$? – Hans Jun 15 '20 at 17:19
• Oh sorry, I was being careless! The definition you give is naturally the right adjoint to restriction (in the category of continuous representations). However, since the category of continuous representations of a compact Lie group is semisimple (i.e. there’s always a unitary structure), you can apply Frobenius reciprocity to the duals and then dualize back to get the other universal property. This isn’t as natural as the other statement, because it’s relying crucially on semisimplicity - for more general groups, the right and left adjoints are indeed sometimes different. – dorebell Jun 15 '20 at 17:42
• Maybe I am overlooking something but it seems to me that one needs that $iE$ is reflexive or something in that direction. Is that true? – Hans Jun 15 '20 at 18:24

You are writing a right adjoint to restriction so you have a natural $$H$$-module map $$iE\rightarrow E, \ (f(x):G\rightarrow E)) \mapsto f(1) .$$ To cook up a map in the opposite direction, you need to use the fact that the category of $$H$$-modules is semisimple and choose a splitting map.

Now you use the fact the category of $$G$$-modules is semisimple. Because of this $$E\rightarrow iE$$ gives your left adjoint "locally", for this particular $$E$$ only. This is so called SSC (solution set condition) in Freyd's Theorem.

At this point you will need to work slightly harder. Essentially you will need to use Freyd's Theorem. You can choose $$E\rightarrow iE$$ for each simple module but your task is to extend it functorially to all modules. Each module is canonically a direct sum of simples $$V = \oplus_{S} Hom(S,E) \otimes S$$ but it does not help because it is a coporoduct and you will need products. So it boils down to understanding limits in the category of continuous modules and whether restriction preserves them. My guess is that the left adjoint (that you are looking for) exists if and only if $$H$$ is of finite index in $$G$$ (this means $$H$$ is open, not closed).

Here is a recent paper that I can find where a similar question has been treated. It has no answer to your question but has all the necessary techniques to attack it.

Your question can be rephrased as "When is the induction the same as coinduction?" This has appeared on MathOverflow before and fancy answer to your question can be found here: When are induction and coinduction of representations of Lie groups isomorphic? When they are compact? Semisimple?

A direct elementary proof could be perhaps gleaned from https://math.stackexchange.com/questions/225730/left-adjoint-and-right-adjoint-nakayama-isomorphism/226493#226493 as it mentions averaging over group which works equally well for compact Lie groups as it does for finite groups.

• Sorry but you are in a wrong category. What you say will work in finite dimensional representations but not in continuous representations. To disprove you just need a family of continuous representations $V_i$ of a compact group $G$ such that $\prod_i V_i$ is no longer continuous. The the restriction to the trivial group does not preserve limits. – Bugs Bunny Jun 16 '20 at 14:29
• @BugsBunny You mean the averaging trick? – Vít Tuček Jun 16 '20 at 14:40
• No, if only I knew what I meant:-)) Suppose there is a left adjoint to RES. Then RES must preserve limits. Consider restriction from G to 1. A product of vector spaces is a limit in the category of representations of the trivial group. Hence there ought to be a continuous on any product of representations. This boils down to a definition of a continuous representation. A "standard" definition is a representation such that every representable function $G\rightarrow{\mathbb C}$ is continuous. With this definition there will be no continuous action. – Bugs Bunny Jun 16 '20 at 17:35