Let $G$ be a locally compact Hausdorff group and $H$ a closed subgroup of $G$. If $\sigma: H \to B(\mathcal{K}_\sigma)$ is a unitary representation of $G$, we can associate an "induced representation", denoted by $\operatorname{Ind}_H^G(\sigma)$.

I'll briefly recall a possible construction for this representation. All details can be found in appendix E of the book "Kazhdan's property (T)" by B. Bekka, P. de la Harpe and A. Valette or in Folland's book "A course on abstract harmonic analysis" in chapter 6. Let $p: G \to G/H$ the canonical quotient map. Start by defining the space $$\mathcal{A}:= \{f \in C(G, K_\sigma): p(\operatorname{supp}(f)) \text{ is compact}, f(gh) = \sigma(h^{-1})f(g) \text{ for } g\in G, h \in H\}.$$ Fix a strongly quasi-invariant measure $\mu$ on $G/H$ and consider the associated continuous function $$\lambda: G\times (G/H)\to (0, \infty)$$ such that $\lambda(x,yH)= \frac{d\mu_x}{d\mu}(yH).$ The space $\mathcal{A}$ becomes an inner product space for $$\langle f,g\rangle := \int_{G/H} \langle f(x), g(x)\rangle_\sigma d\mu(xH) $$ and we denote the Hilbert space completion of $\mathcal{A}$ by $\mathcal{H}_\mu$. Then the map $$\pi_\mu: \mathcal{A}\to \mathcal{A}$$ defined by $$(\pi_\mu(x)f)(y) = \sqrt{\lambda(x,yH)}f(x^{-1}y)$$ extends to a representation and this is $\operatorname{Ind}_H^G(\sigma): G \to B(\mathcal{H}_\mu)$.

Consider now the following theorem in the book "Kazhdan’s Property (T)" by B. Bekka, P. de la Harpe and A. Valette:

enter image description here

There are a lot of details to be filled in, but the main strategy is clear: we define $$U: \mathcal{H}_\pi\odot \mathcal{A}\to \mathcal{L}$$ (where $\mathcal{A}$ is defined as above) and we want to show that this map extends to a unitary $$U: \mathcal{H}_\pi\otimes \mathcal{H}\to \mathcal{L}$$ that implements the equivalence of the two representations. However, one thing is not clear to me in this proof: Why is $U$ surjective? I see that $U$ is isometric on the algebraic tensor product, so it extends uniquely to an isometry $$U: \mathcal{H}_\pi\otimes \mathcal{H}\to \mathcal{L}$$ and it therefore has closed range. Hence, it suffices to show that range of $U$ is dense in the codomain, but I did not manage to prove this.

Lemma E.1.3 in the aforementioned book might be relevant, but I could not make it work (I also believe that the proof of this lemma contains a mistake, but it can be fixed using a partition of unity argument).

Any help/insights is highly appreciated!

EDIT: In reply to comment:

enter image description here

  • 1
    $\begingroup$ This is Lemma 4.2 of Fell, J. M. G., Weak containment and induced representations of groups, Can. J. Math. 14, 237-268 (1962). ZBL0138.07301, there attributed to Mackey, George W., Induced representation of locally compact groups. I, Ann. Math. (2) 55, 101-139 (1952). ZBL0046.11601, Theorem 12.2. $\endgroup$ Jun 20, 2022 at 22:10
  • $\begingroup$ BTW, I had a look at Lemma E.1.3 and it seems okay to me. There are lots of fiddly topological things used which, for a really complete proof, would be a pain to add it. But overall it seems okay? $\endgroup$ Jun 22, 2022 at 7:31
  • $\begingroup$ @MatthewDaws The problematic part in the proof (I think) is the boxed inequality (in the image I just included). It looks like the author uses that $\|\eta(x)-\eta(x_i)\| = \|\eta(x)-v_i\|\le \epsilon$ whenever $f_i(x)\ne 0$. But I don't see how to justify that. Maybe I miss something here, but it is here that a partition of unity argument comes to the rescue because you then have control in which open sets the $f_i$'s are supported. What do you think? $\endgroup$
    – Andromeda
    Jun 22, 2022 at 7:47
  • $\begingroup$ @MatthewDaws The partition argument I talk about can be found in Folland's book "A course in abstract harmonic analysis" p170, proposition 6.8 (in the proof below), though the context is slightly different (but exactly the same proof applies). $\endgroup$
    – Andromeda
    Jun 22, 2022 at 7:56

1 Answer 1


I don't know if the following is a "standard" argument, but it's what I came up with. Introduce some notation: $\newcommand{\mc}{\mathcal}\mc A_\sigma$ comes from forming $\operatorname{Ind}_H^G\sigma$ while $\mc A_1$ comes from forming $\operatorname{Ind}_H^G(\pi|_H\otimes\sigma)$. As you note, $$ U:\mc H_\pi\odot\mc A_\sigma\rightarrow\mc A_1; \quad U(\theta\otimes\xi)(x) = \pi(x^{-1})\theta\otimes\xi(x) $$ forms an isometry, and so this extends to the completions.

Now pick $\xi' \in \mc A_1$ so $\xi'$ is continuous, has "compact support" in $G/H$ and $\xi'(xh) = (\pi(h^{-1})\otimes\sigma(h^{-1}))\xi'(x)$ for $x\in G, h\in H$. Define $$ \xi''(x) = (\pi(x)\otimes\operatorname{id})\xi'(x) \qquad (x\in G). $$ As $\pi$ is strongly continuous and unitary, a triangle-inequality argument shows that $\xi''$ is still continuous. It has the same "support". Also $$ \xi''(xh) = (\pi(xh)\otimes\operatorname{id})\xi'(xh) = (\pi(x)\otimes\sigma(h^{-1}))\xi'(x) = (\operatorname{id}\otimes\sigma(h^{-1}))\xi''(x). $$ So $\xi''$ is in the $\mc A$-space for $1\otimes\sigma$. (We could explore this more, but I'll just quickly finish). This construction can be reversed, so any $\xi''$ satisfying these properties comes from some $\xi'\in\mc A_1$. Obviously this bijection $\xi' \leftrightarrow \xi''$ is linear. For similarly $\eta',\eta''$, we see that $$ (\xi''(x)|\eta''(x)) = (\xi'(x)|\eta'(x)) \qquad (x\in G), $$ and so $(\xi''|\eta'') = (\xi'|\eta')$ and hence our bijection is a unitary.

We now do use Lemma E.1.3. Let $\theta\in\mc H, v\in\mc K_\sigma$, let $f\in C_c(G)$, and form $\xi''_{f,\theta\otimes v}$ using $1\otimes\sigma$, that is, $$ \xi''_{f,\theta\otimes v}(x) = \int_H f(xh) (\theta\otimes\sigma(h)v) \ dh = \theta \otimes \xi_{f,v} \in \mc H_\pi \odot \mc A_\sigma \qquad (x\in G). $$ The lemma says that such $\xi''$ are total. However, notice that $$ \xi' = U(\theta\otimes\xi) \ \implies \ \xi''(x) = \theta\otimes\xi(x), $$ so putting these together gives $\xi''_{f,\theta\otimes v} = U(\theta\otimes\xi_{f,v})$. Hence $U$ has total range, as required.

Edit: There was also a question asked about the proof of Lemma E.1.3. I think this is fine. I freely use the notation in the proof from the book. By the choice of $U$ and uniform continuity we know that for $x\in Ux_i$, say $x=ux_i$ for some $u\in U$, $$ \|\eta(x)-v_i\| = \|\eta(x) - \eta(x_i)\| = \|\eta(ux_i)-\eta(x_i)\| \leq\epsilon. $$ As $f_i$ is chosen supported on $Ux_i$, then $$ f_i(x)>0 \ \implies\ x\in Ux_i \ \implies\ \|\eta(x)-v_i\|\leq\epsilon. $$ Thus in the sum $\sum_i f_i(x) \|\eta(x)-v_i\|$, fix $x$ and chosen each term. If $f_i(x)=0$ then there is no contribution; if $f_i(x)>0$ then the term is $\leq f_i(x)\epsilon$. This gives exactly $$ \sum_i f_i(x) \|\eta(x)-v_i\| \leq \epsilon \sum_i f_i(x) $$ for each $x$, as claimed.

  • $\begingroup$ Thanks, works perfectly! $\endgroup$
    – Andromeda
    Jun 22, 2022 at 20:20
  • $\begingroup$ Concerning the edit: in my version of the book, $f_i = \frac{1}{n} f$, so the support of all $f_i$'s is the same, namely the support of $f$. Thus "As $f_i$ is supported on $Ux_i$" is false. This is what I meant with my comment above, that a partition of unity argument saves this argument, but like it is written in the book I think it doesn't quite work. $\endgroup$
    – Andromeda
    Jun 24, 2022 at 20:15
  • $\begingroup$ Ah... in the published version the argument is different and the $(f_i)$ are a partition of unity. $\endgroup$ Jun 24, 2022 at 21:16
  • $\begingroup$ I checked, and this seems to be a different difference between the published version of the book, and a "draft" version which I also have in PDF format. I hadn't realised there were any major differences, but apparently not... $\endgroup$ Jun 25, 2022 at 4:31
  • $\begingroup$ Alright! Thanks! Then that clears everything up. $\endgroup$
    – Andromeda
    Jun 25, 2022 at 10:11

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