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Let $H = H_1 \times H_2$ be a closed subgroup of a second-countable locally compact Hausdorff group $G = G_1 \times G_2$, with $H_i \leq G_i$. Let $\chi = \chi_1 \otimes \chi_2$ be a unitary character of $H$, for $\chi_i$ a character of $H_i$. The induced representation

$$I(\chi) = \operatorname{Ind}_H^G\chi$$

is the Hilbert space consisting of all classes of measurable functions $f: G \rightarrow \mathbb C$ satisfying $f(hg) = \chi(h)f(g)$ and the norm condition

$$\int\limits_{H \backslash G} |f(g)|^2 \, dg < \infty$$

where $dg$ is any quasi-invariant measure on $H \backslash G$.

We can also consider the induced representations $I(\chi_i) = \operatorname{Ind}_{H_i}^{G_i} \chi_i$. The space of continuous functions $f_i \in I(\chi_i)$ is norm-dense in $I(\chi_i)$.

Let $V \subset I(\chi)$ be the linear space spanned by functions of the form $(g_1,g_2) \mapsto f_1(g_1)f_2(g_2)$, for continuous $f_i \in I(\chi_I)$. Is $V$ always norm-dense in $I(\chi)$? I am certain this should be true, but I am wondering if there is a simple reason why.

Most likely, one can make some analogue of the details of the case where $\chi$ is trivial.

Proof of the case where $\chi$ is trivial:

In this case, $I(\chi) = L^2( H \backslash G) = L^2( [H_1 \backslash G_1] \times [H_1 \backslash G_2])$. It's a standard result in functional analysis that $L^2(H_1 \backslash G_1) \otimes_{\mathbb C} L^2(H_2 \backslash G_2)$ is norm dense in this last space. And continuous functions $f_i: H_i \backslash G_i \rightarrow \mathbb C$ are of course norm dense in $L^2(H_i \backslash G_i)$. It's easy to put these two facts together to get the result.

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  • $\begingroup$ Just to check terminology: since we are dealing with infinite groups here, and possibly infinite-dimensional representations, what does "group character" mean in this context? $\endgroup$
    – Yemon Choi
    Commented Apr 15, 2021 at 19:41
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    $\begingroup$ Here I just mean a continuous homomorphism into $S^1$ $\endgroup$
    – D_S
    Commented Apr 15, 2021 at 19:42

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Yes, $V$ is always norm dense.

You have $V=V_1\otimes V_2$ for obvious choices of dense subspaces $V_i<I(\chi_i)$ and you ask whether $V$ is dense in $I(\chi)$. Equivalently, you may ask directly whether $I(\chi_i) \otimes I(\chi_i)$ is dense in $I(\chi)$.

Why reformulating in this way? Because now one can use another model for the induction in which the positive answer is transparent. The model I am having in mind is the one realizing the induction space as $L^2(G/H)$ itself given e.g in Folland's "A course in abstract harmonic analysis", remark 2 after proposition 6.1. Then the reasoning becomes literally the one you used for trivial $\chi$.

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