Given a right Hilbert $A$-module $E$, and a right Hilbert $B$-module $F$, together with non-degenerate $*$-homomorphism $\phi:A \to \mathcal{L}_B(F)$, we can form the tensor product $$ E \otimes_{\phi} F, $$ by completing the $A$-balanced tensor product of $E$ and $F$ and completing with respect to the obvious norm.

For two adjointable maps $L \in \mathcal{L}_A(E)$, and $K \in \mathcal{L}_B(F)$, such that $K$ is in addition a bimodule map, will their tensor product again be adjointable? (I am assuming here, that just as in the Hilbert space case, there is no difficulty with tensoring bounded maps to produce bounded maps.)

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    $\begingroup$ $K$ needs to be a bimodule map, i.e. you will need some compatibility between $K$ and $\phi$, otherwise then tensor product will not exist. $\endgroup$ – Ulrich Pennig Jul 29 '19 at 14:05
  • $\begingroup$ Sure, of course, I have edited $\endgroup$ – Dave Shulman Jul 29 '19 at 14:34

That $K$ is a bimodule map means that $\phi(a)K(x) = K(\phi(a)(x))$ for all $a\in A, x\in F$. That is, $K \in \phi(A)' \subseteq \mathcal{L}_B(F)$.

I am here following the ideas of Lance's little Hilbert $C^*$-modules book, Chapter 4. Let $z = \sum_i x_i\otimes y_i \in E\odot F$, so $(z|z) = \sum_{i,j} (y_i|\phi((x_i|x_j))y_j) = (y|\phi_n(X)y)$. Here $X\in M_n(A)$ is the matrix with entries $(x_i|x_j)$, $\phi_n$ is the $n$th amplification (as $\phi$ is a $*$-homomorphism, it is completely positive, and so $\phi_n(X) \in M_n(\mathcal{L}_B(F))$ is positive), and $y$ is column with entries $(y_n)$. Here we have turned $F^n$ (as column vectors) into a Hilbert $C^*$-module over $B$ in the obvious way.

Now consider $z'=(\iota\otimes K)z$ which has $(z'|z') = (y|K^*\phi_n(X)Ky)$ where $K$ acts on $F^n$ pointwise. As $K$ commutes with $\phi_n(X)$, $K^*\phi_n(X)K = K^*K \phi_n(X)$. As $\phi_n(X)$ is positive and $\phi_n(M_n(A)) = M_n(\phi(A))$ is a $C^*$-subalgebra of $M_n(\mathcal{L}_B(F)) = \mathcal{L}_B(F^n)$, there is $T\in M_n(A)^+$ with $\phi_n(T^2) = \phi_n(X)$. Then $K^*K\phi_n(T^2) = \phi_n(T) K^*K \phi_n(T)$. It follows that \begin{align*} (z'|z') &= (\phi_n(T)y|K^*K \phi_n(T)y) \leq \|K\|^2 (\phi_n(T)y|\phi_n(T)y) \\ &= \|K\|^2 (y|\phi_n(X)y) = \|K\|^2 (z|z). \end{align*} Hence $(\iota\otimes K)$ is bounded (with norm $\|K\|$).

As always $L\otimes\iota$ is bounded (see Lance's book) we can compose to see that $L\otimes K$ is bounded, as required. (Contrary to the original question, I think it is far from obvious that $L\otimes K$ is bounded).

Once we have established that $L\otimes K$ exists, it follows from just a calculation that it is adjointable, with adjoint $L^*\otimes K^*$. Indeed, $(x\otimes y| (L\otimes K)(x'\otimes y')) = (y|\phi((x|Lx')) Ky') = (y|K \phi((x|Lx')) y') = (K^*y|\phi((L^*x|x')) y') = ((L^*\otimes K^*)(x\otimes y)|x'\otimes y')$.

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