Suppose $G$ is a locally compact group and $H$ is an open subgroup for simplicity. Further suppose $\pi$ is a representation of $H$ on some Hilbert space $\mathcal{H}_{\pi}$, i.e. $\pi(h)$ is unitary for all $h\in H$, $\pi$ is weakly continuous and $\pi$ is also a homomorphism.

To induce a representation on $G$ from $H$, it is natural to consider the usual action of $G$: as translation of functions on $G$. $H$ however acts naturally on $\mathcal{H}_{\pi}$ by $\pi$. The way to marry these two notions in a self-consistent manner in order to induce a representation on $G$ is to consider those $f:G\to\mathcal{H}_{\pi}$ such that $f(gh) = \pi(h^{-1})f(g)$. Such $f$ are "constant" on cosets of $H_{\pi}$ in $G$ - more precisely, $\| f(\cdot)\|$ is constant on cosets.

From this, we can define a Hilbert space $\mathcal{H}(G,\pi)$:

$$\mathcal{H}(G,\pi) = \{f:G\to\mathcal{H}_{\pi}:f(gh) = \pi(h^{-1})f(g),\; \sum_{g'H\in G/H}\|f(g')\|^2<\infty\}$$

with the natural inner product given by

$$\langle f_1,f_2\rangle = \sum_{gH\in G/H}\langle f_1(g),f_2(g)\rangle.$$

From here we can finally define the induced representation $\text{ind}_H^G\pi:G\rightarrow U(\mathcal{H}(G,\pi))$:

$$\text{ind}_H^G\pi(g)f(\cdot) = f(g^{-1}\cdot).$$

It is not hard to see that this is indeed a homomorphism and that it is unitary. However it is not necessary that $\pi$ be a homomorphism in order for $\text{ind}_H^G\pi$ to be a homomorphism. This is not surprising since we are just acting by left translation. The weak continuity of $\pi$ seems necessary to keep in order for $\text{ind}_H^G\pi$ to be weakly continuous since weak continuity is not guaranteed for a unitary homomorphism (pathological things may yet happen).

The only place in which $\pi$ being a homomorphism plays a role is in showing that $\text{ind}_H^G\pi$ is in fact weakly continuous. The role is also quite minor in the proof. Is it possible to omit that the assumption that $\pi$ be a homomorphism and still have $\text{ind}_H^G\pi$ be a representation? Just because it is used to prove that $\text{ind}_H^G\pi$ is weakly continuous does not mean it is a necessary condition.

  • $\begingroup$ Are you sure? What happens if you take G to be finite - are you claiming that inducing an arbitrary function on a subgroup will give you a representation of G? $\endgroup$ – Yemon Choi Jul 13 '14 at 0:56
  • $\begingroup$ Without looking at the details, I suspect one needs \pi to be a homomorphism in order for Ind^G_H \pi(g) to map ${\mathcal H}(G,\pi)$ to itself $\endgroup$ – Yemon Choi Jul 13 '14 at 0:59
  • $\begingroup$ @YemonChoi That is surprisingly not the case. The norm condition is not hard to check (it only relies on unitarity of $\pi$). As for the algebraic condition: $$\text{ind}_H^G\pi(g')f(gh) = f(g'^{-1}gh) = \pi(h^{-1})f(g'^{-1}g) = \pi(h^{-1}) \text{ind}_H^G(g')f(g).$$ $\endgroup$ – Cameron Williams Jul 13 '14 at 1:07
  • $\begingroup$ @YemonChoi Also I'm a little bit confused by your first comment. Could you elaborate? Did I say something false in my post? $\endgroup$ – Cameron Williams Jul 13 '14 at 1:08
  • 1
    $\begingroup$ OK, having done some calculations, it seems that if $\pi$ is not a homomorphism then ${\mathcal H}(G,\pi)$ could be unreasonably small, probably even zero. Another related problem: one usually wishes to identify ${\mathcal H}(G,\pi)$ with $L^2(G/H, {\mathcal H}_\pi)$ whenever $G/H$ is a well-behaved quotient, and the usual way of making this identification seems to rely on $\pi$ being a homomorphism. However, I did this in a rush so perhaps I have overlooked something. $\endgroup$ – Yemon Choi Jul 13 '14 at 2:00

I don't believe that weak continuity is "the only place in which $\pi$ being a homomorphism plays a role".

In fact, before you can even talk about weak continuity of the resulting representation, you want $\mathcal H(G, \pi)$ to be a vector subspace of $(\mathcal H_\pi)^G$ and be $G$-invariant under left translation of functions, right?

But if so, then barring any stupid mistake on my part, one checks without trouble that 1º) $$ \mathcal V_\pi:=\{v\in \mathcal H_\pi:\text{$v$ is the value of some $f\in\mathcal H(G, \pi)$ at some $g\in G$}\} $$ is a vector subspace of $\mathcal H_\pi$, and is invariant under $\pi(h)$ for all $h\in H$. And 2º) if $v\in\mathcal V_\pi$, then choosing $f$ and $g$ such that $v=f(g)$ one can write: \begin{align} \pi(hh')v &= \pi(hh')f(g) \\ &= f(gh'^{-1}h^{-1}) \\ &= \pi(h)f(gh'^{-1}) \\ &= \pi(h)\pi(h')f(g) \\ &= \pi(h)\pi(h')v. \end{align} So $\pi$ is a homomorphism after restriction to $\mathcal V_\pi$. Which is obviously all that matters, as vectors in $\mathcal H_\pi\setminus\mathcal V_\pi$ play no role anywhere.

  • $\begingroup$ So your point is that, in fact, we get that $\pi$ is a homomorphism for free? $\endgroup$ – Cameron Williams Jul 13 '14 at 3:11
  • $\begingroup$ Rather, that: the space on which $\pi$ is a homomorphism is all that matters. That space could be zero, but then so is your $\mathcal H(G,\pi)$. $\endgroup$ – Francois Ziegler Jul 13 '14 at 3:14
  • $\begingroup$ Hmm. That's interesting if rather unsettling. Haha. I like this way of thinking about the problem. So correct me if I'm wrong but the answer seems to me that: $\pi$ acts as a homomorphism on the space of functions for which we are interested. Since the homomorphism property is irrespective of what functions we are acting on, $\pi$ is actually a homomorphism? $\endgroup$ – Cameron Williams Jul 13 '14 at 3:20
  • $\begingroup$ @CameronWilliams Francois's argument is what I was trying to get at in my comment on your original question. The point is that the induced rep only sees the part on which $\pi$ is a homomorphism, and if you choose $\pi$ NOT to be a homomorphism then the induced rep may not be large enough to be of any use or interest $\endgroup$ – Yemon Choi Jul 13 '14 at 3:25
  • $\begingroup$ @YemonChoi Oh I understand your comment now. Makes sense to me. $\endgroup$ – Cameron Williams Jul 13 '14 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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