Induction and restriction of unitary representations $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Res{Res}$Given a locally compact group $G$ and a closed subgroup $H\subset G$,
let $\Rep(G)$ and $\Rep(H)$ denote their categories of (strongly continuous) unitary representations on Hilbert spaces.
Let $NG$ be the group von Neumann algebra of $G$ (the weak closure of the left action of $G$ on $L^2G$),
and let $NH$ be the group von Neumann algebra of $H$.
We have full subcategories $\Rep(NG)\subset \Rep(G)$, and $\Rep(NH)\subset \Rep(H)$. (The category $\Rep(NG)$ can also be described as the full subcategory of $\Rep(G)$ consisting of reps which are direct sums of direct summands of the regular representation.)

It is well-known that the restriction and induction functors
$\Res: \Rep(G) \to \Rep(H)$and
$\Ind: \Rep(H) \to \Rep(G)$               defined by   $\Ind(V) = L^2(G/H , G\times_HV)$
are NOT each other's adjoints in any sense (see for example this previous MO question).

But maybe I'll be more lucky if I restrict these functors to the subcategories $\Rep(NG)$ and $\Rep(NH)$?

$\Res: \Rep(NG) \to \Rep(NH)$and
$\Ind: \Rep(NH) \to \Rep(NG)$

Is there any relationship between the two latter functors which can be formulated in category-theoretic terms?
More precisely, is there an $NH$-$NG$-bimodule $X$ such that $\Res: \Rep(NG) \to \Rep(NH)$ is given by tensoring by $X$, and $\Ind: \Rep(NH) \to \Rep(NG)$ is given by tensoring by the complex conjugate of $X$?
One result in that direction is this old theorem of Mackey (from this paper):
 A: Yes, you can take $\overline X = {}_{N(G)}L^2(G)_{N(H)}$ using the canonical inclusion of $N(H)$ into $N(G)$ to restrict the "trivial" correspondence $L^2(G)$ to a $N(G)-N(H)$ correspondence.
(I am assuming here "bimodule" is a "correspondence" in the sense of Connes, i.e. a Hilbert space with commuting, normal actions of $N(G)$ and $N(H)^o$.  The (tensor) product is the Connes fusion, aka the Sauvageot relative tensor product (Sur le produit tensoriel rélatif d’espaces de Hilbert).  See also Takesaki, Volume 2, Chapter IX, Section 3.)
Restriction is easy to check.  As $N(G)$ acts in standard form on $L^2(G)$, we know that ${}_{N(G)}L^2(G)_{N(G)}$ is self-dual, so $X \cong {}_{N(H)}L^2(G)_{N(G)}$.  From Takesaki, Prop 3.19, we know that if ${}_{N(G)}H$ is any left $N(G)$-module then ${}_{N(G)}L^2(G)_{N(G)} \otimes {}_{N(G)}H \cong {}_{N(G)}H$.  So ${}_{N(H)}X_{N(G)} \otimes {}_{N(G)}H \cong {}_{N(H)}H$, which is restriction.

Checking that we get induction seems harder.  I think if $G$ is discrete, or more generally $H$ open, this can be done with a long but elementary computation.  I want to give an indication of the general case, which involves analytic difficulties.
First a sanity check: if we induce the left-regular representation $\lambda_H$ we get $\lambda_G$.  So we wish to induce $L^2(H)$ to $L^2(G)$ as $N(H)$ and $N(G)$ modules.  But $\overline X_{N(H)} \otimes {}_{N(H)}L^2(H)$ can indeed be shown to be isomorphic to ${}_{N(G)} L^2(H)$ (see below).  As any left module of $N(H)$ is obtained by direct sum and subrepresentation of $L^2(H)$, and as induction respects direct sums, this would seem to indicate that we get what we want.  However, I see no "soft" way of checking that all the isomorphisms involved really do commute.  That is, if $\pi_1 \oplus \pi_2 \cong \lambda_H$ then both "induction" methods give modules of $N(G)$ which sum to $\lambda_G$, but I do not see why the individual components are automatically isomorphic.
Perhaps there is some way to use imprimivity, but I also do not see this.

Instead, I want to build "models" of induction, and tensoring by $\overline X$, which are manifestly the same.  I will use the model of induction which works by using a (suitably measurable) cross section $G/H \rightarrow G$.  This is "Realization III" in Kaniuth and Taylor's book (page 79-80).  We can deal with any locally compact $G$ by using results of Kehlet - Cross sections for quotient maps of locally compact groups.  Following Kehlet and page 3494 of Daws et al. - Closed quantum subgroups of locally compact quantum groups (I thank in particular my coauthor Pawel here) we pick a quasi-invariant measure $\lambda$ on $G/H$ and then we find a unitary
$$ T: L^2(G/H,\lambda) \otimes L^2(H) \rightarrow L^2(G), $$
with $T(1\otimes \rho^H(h))T^* = \rho^G(h)$ for $h\in H$, where $\rho^H$ and $\rho^G$ are the right-regular reps of $H$ and $G$, respectively.
If you perform the calculation, you'll also find that
$$ T U^{\lambda_H}(g) T^* = \lambda^G(g) \qquad (g\in G), $$
where $\lambda^G$ is the left-regular representation of $G$, and $U^{\lambda_H}$ is the induced representation of the left-regular representation of $H$, up to $G$.  Thus we have our "model" of induction.  Clearly, if $\pi$ is a sub-representation of $\lambda^H$ then we get the "same" model by restricting $T$ to $L^2(G/H,\lambda) \otimes K$ where $K\subseteq L^2(H)$ is the $N(H)$-invariant subspace associated to $\pi$.  (Really "clearly" here means follow the isomorphisms through.)
What about tensoring with $\overline X$?  I don't want to compute with the Sauvageot tensor product, as that would involve consideration of vectors in $L^2(G)$ bounded with respect to the Plancheral measure on $N(H)$ etc.  Instead, I'll use a realisation of correspondences as self-dual Hilbert $C^*$-bimodules (see work of Reiffel and Paschke).  For von Neumann algebras $M$, $N$ this identifies ${}_M H_N$ with
$$ \hom_N(L^2(N), {}_MH_N) $$
the bounded linear maps $t:L^2(N)\rightarrow H$ with $t(\xi n) = t(\xi)n$ for $n\in N$.  Composition of operators defines a left $M$ and right $N$ module structure, and we get an $N$-valued inner-product by setting $(t|s) = t^*s$.  Note that $t^*s$ is an operator on $L^2(N)$ which commutes with the right $N$-action on $L^2(N)$, that is, can be identified with an element of $N$.
Conversely, given a bimodule ${}_MY_N$ we take the balanced tensor product $Y_N \otimes_N L^2(N)$ which gives a Hilbert space, and this space inherits the left $M$ action from ${}_M X$ and the right $N$ action from $L^2(N)$, so becoming a correspondence.
Our $\overline X = {}_{N(G)} L^2(G)_{N(H)}$ is identified with
$$ \hom_{N(H)}(L^2(H), L^2(G)) $$
and if $K\subseteq L^2(H)$ is a sub-rep of $\lambda_H$, then $\overline X \otimes K$ is identified with the left $N(G)$-module
$$ \hom_{N(H)}(L^2(H), L^2(G)) \otimes_{N(H)} K. $$
This, by definition, is the separation-completion of the algebraic tensor product for the scalar-valued inner-product
$$ (t_1\otimes\xi_1 | t_2\otimes\xi_2) = (\xi_1 | t_1^*t_2 \xi_2)_{K}, $$
where $t_1^*t_2 \in N(H)$ and so acts on $K$.  However, this is simply
$$ (t_1(\xi_1)| t_2(\xi_2))_{L^2(G)}, $$
which makes sense as $K\subseteq L^2(H)$.  Thus $\overline X \otimes K$ is identified with a subspace of $L^2(G)$; the left $N(G)$-action works as expected.  The space we get is the closed linear span of $\{t(K) : t\in \hom_{N(H)}(L^2(H), L^2(G)) \}$.
We want to understand $\hom_{N(H)}(L^2(H), L^2(G))$.  Apply the unitary $T$ to identify $L^2(G)$ with $L^2(G/H)\otimes L^2(H)$.  Due to the above commutation of the right-regular representations, the right $N(H)$ action is intertwined by $T$.  So we get
$$ \hom_{N(H)}(L^2(H), L^2(G/H)\otimes L^2(H)) = \{ t:L^2(H) \to L^2(G/H)\otimes L^2(H) : tn' = (1\otimes n')t \ (n'\in N(H)')\}. $$
As $K$ is the range of a projection $e\in N(H)'$ we see that $t(K) = te(L^2(H)) = (1\otimes e)t(L^2(H))$ and as $t$ varies we obtain all of the image of $1\otimes e$, that is, $L^2(G/H) \otimes K$.
Conclude: both operations of "induction" respect sub-representations of $\lambda_H$ in the same way.  By taking direct sums, this shows that both versions of induction agree on all $N(H)$-modules, as required.
