Let $ \mathcal A$ be a finite von Neumann algebra with finite trace $ \tau$ and let $ l^2(\mathcal A)$ be the Hilbert space completion of $ \mathcal A$ with respect to the inner product induced by $ \tau$ , as usual. Further, let $ H$ a sepearbale Hilbert space and let $ \mathcal H$ be the free Hilbert $ \mathcal A$ -module $ \mathcal H := l^2(\mathcal A) \otimes H$ , where $ \otimes$ denotes the tensor product of Hilbert spaces.

Denote by $ \mathcal B_{\mathcal A}(\mathcal H)$ the space of endomorphisms on $ \mathcal H$ that commute with the diagonal *left-regular representation* $ \mathcal A \hookrightarrow \mathcal B(\mathcal H)$ . Then $ \mathcal B_{\mathcal A}(\mathcal H)$ is a Type II von-Neumann-algebra, such that $ \tau$ induces a $ [0,\infty]$ -valued trace $ Tr^+$ on the submonoid $ \mathcal B_{\mathcal A}^+(\mathcal H)$ of positive elements.

An operator $ A \in \mathcal B_{\mathcal A}(\mathcal H)$ is said to be *$ \mathcal A$ -Hilbert Schmidt* if $ $ Tr^+(A^{*}A) < \infty$ $ Denote by $ \mathcal B_{\mathcal A}^2(\mathcal H)$ the subset of $ \mathcal A$ -Hilbert Schmidt operators. As in the classic case $ \mathcal A = \mathbb C$ , one shows that $ \mathcal B_{\mathcal A}^2(\mathcal H)$ is a $ *$ -closed subalgebra and a $ 2$ -sided ideal in $ \mathcal B_{\mathcal A}(\mathcal H)$ , equipped with the inner product $ $ \langle A,B \rangle_2 := \sum_{i \in I} \langle A(e \otimes x_i), B(e \otimes x_i) \rangle, $ $ where $ e \in \mathcal A$ is the unit element and $ (x_i)_{i \in I}$ is an orthonormal base of $ H$ . Making use of the polarization identity, one shows that this inner product does not depend on the explicit choice of $ (x_i)_{i \in I}$ , endowing $ \mathcal B_{\mathcal A}^2(\mathcal H)$ with the canonical strucutre of a pre-Hilbert space. *Unlike* the classic case $ \mathcal A = \mathbb C$ , however, $ \mathcal B_{\mathcal A}^2(\mathcal H)$ is usually *not* complete and there is *no* inequality of the form $ $ ||A|| < C||A||_2,$ $ for all $ A \in \mathcal B_{\mathcal A}^2(\mathcal H)$ and some $ C > 0$ , where $ ||\:.\:||$ denotes the operator norm . This is already witnessed in the case $ \dim_{\mathbb C}(H) = n < \infty$ , since we then have an obvious isometric isomorphism of pre-Hilbert spaces $ $ \mathcal B_{\mathcal A}^2(\mathcal H) \cong \mathcal A^{n^2}.$ $ Thus, we see that in this instance, $ \mathcal B_{\mathcal A}^2(\mathcal H)$ is complete as a pre-Hilbert space if and only if the same is true for $ \mathcal A$ .

**Question 1: What is the structure of $ \mathcal B_{\mathcal A}(\mathcal H)$ when $ \dim_{\mathbb C}(H) = \infty$ ? Can one give an explicit orthonormal base for $ \mathcal B_{\mathcal A}(\mathcal H)$ (in the pre-Hilbert sense) ?**

A *possible* candidate for an orthonormal base *could* be constructed as follows: For a fixed ONB $ (x_i)_{i \in \mathbb N}$ of $ H$ and ONB $ (b_k)_{k \in \mathbb N} \subset \mathcal A$ of $ l^2(\mathcal A)$ containing $ e$ , denote for $ i,j,k \in \mathbb N$ and by $ P_{ij}^k$ the partial isometry from $ \mathcal l^2(\mathcal A) \otimes \mathbb Cx_i$ onto $ \mathcal l^2(\mathcal A) \otimes \mathbb Cx_j$ , followed by right multiplying $ b_k$ . Then one easily verifies that $ (P_{ij}^k)_{i,j,k \in \mathbb N}$ is an orthonormal subset of $ \mathcal B_{\mathcal A}^2(\mathcal H)$ . For an operator $ A \in \mathcal B_{\mathcal A}^2(\mathcal H)$ , one *could* then try to establish an equality (in some sense) of the form $ $ A = \sum_{i,j,k} \lambda_{i,j}^k P_{i,j}^k $ $ with $ \lambda_{i,j}^k := \langle A(e \otimes x_i), b_k \otimes e_j \rangle$ , so that $ \sum_{i,j,k} |\lambda_{i,j}^k|^2 = ||A||_2^2 < \infty$ . However, I am unsure whether this idea is fruitful.

Knowing a little bit more about the structure of $ \mathcal B_{\mathcal A}(\mathcal H)$ could also help me solve my second Question: For $ i =1,2$ , let $ M_i$ be a closed Riemannian manifold and let $ \mathcal H_i$ be a free, finite Hilbert $ \mathcal A$ -module, let $ L^2(M_i,\mathcal H_i)$ be the Hilbert space of Bochner-integrable functions from $ M_i$ to $ \mathcal H_i$ . The identity $ $ L^2(M_i, \mathcal H_i) = L^2(M_i) \otimes \mathcal H_i$ $ estabishes that $ L^2(M_i, \mathcal H_i)$ is in fact a free (infinite) Hilbert $ \mathcal A$ -module.

**Question 2: Does there exist an isometric isomorphism of pre-Hilbert spaces $ $ \mathcal B^2_{\mathcal A}(L^2(M_1, \mathcal H_1),L^2(M_2,\mathcal H_2)) \cong L^2(M_1 \times M_2, \mathcal B^2_{\mathcal A}(\mathcal H_1,\mathcal H_2))?$ $ **

Again, this is well-known in the classic case $ \mathcal A = \mathbb C$ , but i am completely uncertain whether this works in this generality.