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.



Dense subspace of $\operatorname{Ind}_{H_1 \times H_2}^{G_1 \times G_2} \chi$
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 $\chi$.


