Let $G$ be a locally profinite (i.e., locally compact Hausdorff and totally disconnected) topological group, $H \le G$ a closed subgroup, and $(W, \sigma)$ a representation of $H$ over $\mathbb{C}$ such that $W$ is smooth, i.e., every $w \in W$ is fixed by a compact open subgroup $K \le H$. 


One may form compact induction $\mathrm{ind}\_H^G W$ as the $\mathbb{C}$-vector space of all functions $f\colon G \rightarrow W$ such that $f(hg) = \sigma(h) f(g)$ for all $h \in H$ and $g \in G$, that there is a compact open subgroup $K \le H$ such that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$, and that the support of $f$ has compact image in $H\setminus G$. This is a smooth representation of $G$ by setting $(g\cdot f)(x) = f(xg)$.


On the other hand, one may form the smooth $G$-representation $(\mathbb{C}G \otimes\_{\mathbb{C}H} (\delta\_G/\delta\_H) W)^{\infty}$, where the superscript indicates the subspace of smooth vectors and $\delta\_G$ (resp., $\delta\_H$) is the modulus character of $G$ (resp., $H$) defined by $\delta\_G(g) = [gKg^{-1} : K]$ (fractional index) for (no matter which) compact open subgroup $K \le G$ (resp., analogous thing for $H$).


My question is: are these representations isomorphic, i.e., is
$$
\mathrm{ind}\_H^G W \cong (\mathbb{C}G \otimes\_{\mathbb{C}H} (\delta\_G/\delta\_H) W)^{\infty}
$$
as smooth $G$-representations?



Compact induction as a tensor product
No. Take $H$ the trivial subgroup, $W$ the trivial rep. The left hand side is 
all smooth functions on $G$ with compact support. The right hand side is zero. The reason for this is that elements of $\mathbb C G$ are linear combinations of finitely 
many elements of $G$, but it is easy to see that such element cannot be fixed by an open subgroup.


