# Is it necessary for $\pi:H\to U(\mathcal{H}_{\pi})$ to be a homomorphism in order for $\text{ind}_H^G\pi$ to be weakly continuous?

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.

• 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? – Yemon Choi Jul 13 '14 at 0:56
• 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 – Yemon Choi Jul 13 '14 at 0:59
• @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: $$\text{ind}_H^G\pi(g')f(gh) = f(g'^{-1}gh) = \pi(h^{-1})f(g'^{-1}g) = \pi(h^{-1}) \text{ind}_H^G(g')f(g).$$ – Cameron Williams Jul 13 '14 at 1:07
• @YemonChoi Also I'm a little bit confused by your first comment. Could you elaborate? Did I say something false in my post? – Cameron Williams Jul 13 '14 at 1:08
• 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. – 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.
• So your point is that, in fact, we get that $\pi$ is a homomorphism for free? – Cameron Williams Jul 13 '14 at 3:11
• Rather, that: the space on which $\pi$ is a homomorphism is all that matters. That space could be zero, but then so is your $\mathcal H(G,\pi)$. – Francois Ziegler Jul 13 '14 at 3:14
• 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? – Cameron Williams Jul 13 '14 at 3:20
• @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 – Yemon Choi Jul 13 '14 at 3:25