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I am learning about the so-called "Mackey Machine" for unitary irreps of semidirect products of locally compact groups. Let $G = N \rtimes H$ where $N$ is a closed normal abelian subgroup and $H$ is a closed subgroup which acts on $N$ by $\phi: H \to Aut(N)$. Given an unitary irrep $\sigma$ of $H$ and a irreducible representation $p$ of $N$, I have found two (I believe equivalent) ways in which an irrep of $G$ is constructed by different authors.

If $H_p$ is the semigroup of $H$ which stabilizes $p$, and $G_p = N \rtimes H_p$, then we have the following commutative diagram of inclusions. $\require{AMScd}$ \begin{CD} H_p @>>> H\\ @VVV @VVV\\ G_p @>>> G \end{CD}

If I am understanding it correctly, Folland, in A Course in Abstract Harmonic Analysis (see the discussion leading to Theorem 6.43, on pp. 199-201 of the second edition) goes "first down then right": defines a representation of $G_p$ as $p\otimes \sigma$, then takes the representation of $G$ induced by it (i.e. the same way it is presented in this question). On the other hand, Etingof et al. - Introduction to representation theory (section 4.26 on pp. 76) takes the route "right then down": first they consider the representation of $H$ induced by $\sigma$, then they extend it to $G$ with a skew-product.

Question 1: How to express these two constructions in terms of modules?

Question 2: Since the two constructions are equivalent, the modules resulting from the first and the second path should be isomorphic. Can we see that isomorphism explicitly?

Let me pretend that the groups are finite, since I believe it should work equally well without any topological complicacies.

What I tried

Let $\mathcal{H}_\sigma$ the $\mathbb{C}(H_p)$-module given by $\sigma$. Since $p$ is a 1-dimensional representation of $N$, we can give to $\mathbb{C}$ the structure of a $\mathbb{C}(N)$ module, which I will denote with $\mathbb{C}_p$. Then Folland's construction is (I am keeping the structure of the diagram above, to match each module to the corresponding group, but the arrow have no further meaning):

\begin{CD} \mathcal{H}_\sigma \\ @VVV \\ \mathbb{C}_p \otimes_{\mathbb{C}(H_p)} \mathcal{H}_\sigma @>>> \mathbb{C}(G) \otimes_{\mathbb{C}(G_p)} \mathbb{C}_p \otimes_{\mathbb{C}(H_p)} \mathcal{H}_\sigma \end{CD}

Let me explain what I think should be going on: $\mathbb{C}_p$ can be made into a $(\mathbb{C}(G_p), \mathbb{C}(H_p))$-bimodule, thanks to the action $\phi$ (which stabilizes $p$).

But I am not sure what module is given by the other route \begin{CD} \mathcal{H}_\sigma @>>> \mathbb{C}(H) \otimes_{\mathbb{C}(H_p)} \mathcal{H}_\sigma \\ @. @VVV\\ @. ? \end{CD}

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  • $\begingroup$ Although the conceptual meaning is pretty clear from the context, it may be worth clarifying that your arrows between Hilbert spaces are "construct from" arrows, not Hilbert-space morphisms. $\endgroup$
    – LSpice
    Commented Jun 2, 2020 at 4:44
  • $\begingroup$ Also, (1) where do these two constructions occur? (The closest I can find is Folland, Theorem 6.38; Etingof et al., Section 4.26.) (2) Where are the parentheses in $\mathbb C_p \rtimes \mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$? (3) In the notation $a\phi_h(b)$, I think that $b$ is a complex number, but $\phi_h$ is an automorphism of $N$; so what does $\phi_h(b)$ mean? $\endgroup$
    – LSpice
    Commented Jun 2, 2020 at 5:01
  • $\begingroup$ If I'm right to look at Folland, Theorem 6.38, and Etingof et al., Section 4.26, then the two constructions they describe, in 'module language', are $\mathbb C(G) \otimes_{\mathbb C(G_p)} (\mathbb C_p \otimes_{\mathbb C} \mathcal H_\sigma)$ (where the $N$ piece of $G_p = N \rtimes H_p$ acts only on $\mathbb C_p$, and the $H_p$ piece acts only on $\mathcal H_\sigma$); and $\mathbb C_p \otimes_{\mathbb C} (\mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma)$ (similar action convention). It is clear that these are the same construction. $\endgroup$
    – LSpice
    Commented Jun 2, 2020 at 5:06
  • $\begingroup$ ($(n \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v) \mapsto p(n)z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v)$ and $(1 \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v) \leftarrow z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v)$ (stuck with $\leftarrow$ because we don't seem to have \mapsfrom).) $\endgroup$
    – LSpice
    Commented Jun 2, 2020 at 5:11
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    $\begingroup$ @LSpice You are right, I made a mistake: the skew-product I wrote did not make any sense. I have added some more specific references, and clarified the meaning of the arrows as you suggested. Since my construction was flawed, I have modified my answer so that you can post your both comments as an answer. $\endgroup$
    – Angelo Lucia
    Commented Jun 2, 2020 at 23:22

1 Answer 1

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By request, I post my comments (1 2). I think that they answer the revised question, but let me know if not.

Question 1. Folland's construction corresponds to extending a $\mathbb C(H_p)$-module $\mathcal H_\sigma$ of $H_p$ across $N$ by $p$ to $G_p = N \rtimes H_p$, giving $\mathbb C_p \otimes_{\mathbb C} \mathcal H_\sigma$; and then inducing up to $G$, giving $\mathbb C(G) \otimes_{\mathbb C(G_p)} (\mathbb C_p \otimes_{\mathbb C} \mathcal H_\sigma)$. (I'm not sure how to make sense of your proposed $\mathbb C_p \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$, since $\mathbb C_p$ is not a $\mathbb C(H_p)$-module in any obvious-to-me way; notice that, since $\mathbb C_p$ is $1$-dimensional, specifying the module structure would be equivalent to specifying a homomorphism $H_p \to \mathbb C^\times$. The tensor product over $\mathbb C(H_p)$ will collapse to a $1$-dimensional representation, on which $H_p$ acts trivially (if $\mathcal H_\sigma$ is not isomorphic to $\mathbb C_p$ as $\mathbb C(H_p)$-modules) or by the relevant character (if they are isomorphic).)

Etingof et al.'s construction corresponds to inducing $\mathcal H_\sigma$ up to $H$, giving $\mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$; and then extending across $N$ by $p$ to $G = N \rtimes H$, giving $\mathbb C_p \otimes_{\mathbb C} (\mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma)$.

Question 2. The isomorphism from Folland's to Etingof's construction is given by $(n \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v) \mapsto p(n)z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v)$. The inverse isomorphism is given by $z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v) \mapsto (1 \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v)$.

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  • $\begingroup$ Regarding your question about the $\mathbb{C}(H_p)$ structure of $\mathbb{C}_p$: I think that every character $p:N \to \mathbb{C}^{\times}$ (I am abusing notation and calling it $p$ again) could extended to a character of $G_p$ (and thus by restriction of $H_p$), as $\tilde{p}(nh) = p(n)$ for $n\in N$ and $h\in H_p$. This is an homomorphism due to the fact that $G_p$ leaves $p$ invariant, so if $n,m\in N$ and $h,k \in H_p$ then $\tilde{p}(nh)\tilde{p}(mk) = \tilde{p}(n) \tilde{p}(m)$ is the same as $\tilde{p}(nhmk) = \tilde{p}(n\phi_h(m)hk)=\tilde{p}(n)\tilde{\phi_h(m)}$. $\endgroup$
    – Angelo Lucia
    Commented Jun 3, 2020 at 19:46
  • $\begingroup$ @AngeloLucia, you are right that it is a homomorphism; it corresponds to choosing the trivial action of $H_p$ on $\mathbb C_p$. Then we always have $\mathbb C_p \cong \mathbb C_p \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$ as $\mathbb C(G_p)$-modules via any map $z \mapsto z \otimes_{\mathbb C(H_p)} v$ with $v \ne 0$, which doesn't seem to be what you want. $\endgroup$
    – LSpice
    Commented Jun 3, 2020 at 21:06

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