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.


1 Answer 1


If I understand correctly, the action of $\mathcal{A}$ on $H$ is trivial. So the algebra $\mathcal{B}_{\mathcal{A}}(\mathcal{H})$ is exactly equal to the commutant of the algebra $L(\mathcal{A})\otimes \mathrm{Id}$, hence equal to $R(\mathcal{A})\otimes \mathcal{B}(H)$, where $L(\mathcal{A})$ ($R(\mathcal{A})$) denotes the image of $\mathcal{A}$ under the left (right) action of $\mathcal{A}$ in $\ell^2(\mathcal{A})$. It follows that the Hilbert space associated to $Tr$ may be identified with $\ell^2(\mathcal{A})\otimes \mathrm{HS}(H)$ and the procedure you suggested for obtaining an orthonormal basis works. Namely, take any ONB $(b_n)$ of $\ell^2(\mathcal{A})$ that is contained in $R(\mathcal{A})$ and any ONB $(x_i)$ of $H$ and the operators $P_{ij}^{k}:= b_k \otimes |x_i\rangle\langle x_j|$ form an orthonormal basis.

  • $\begingroup$ Thanks for your reply. I realised that I should've been more precise in the stating of my problem: In terms of the ONB $(P_{ij}^k)$, is there a way to distinguish the Fourier-series presentation of elements in $\mathcal B_{\mathcal A}(\mathcal H)$ from the elements that appear first in the completion ? $\endgroup$
    – H1ghfiv3
    Feb 18, 2018 at 14:06
  • 1
    $\begingroup$ Definitely no. In the very classical case where $A = L^{\infty}(\mathbb(T))$ and $H$ is just the complex numbers, you are asking whether one can tell that an $L^2$-function on the unit circle is bounded just by looking at Fourier coefficients; there is no way to achieve that. In particular, the Kahane-Katznelson-de Leeuw theorem states that for any sequence $(a_n)$ in $\ell_2$ there is a continuous function with Fourier coefficients greater than this given sequence (in terms of absolute value). $\endgroup$ Feb 18, 2018 at 19:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .