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Let $X$ be a locally-compact, Hausdorff space, and let $\mathcal{A}:=C_{0}(X)$ the $C^{*}$-algebra of complex-valued, continuous functions vanishing at infinity. Let $\mathcal{B}:=M_{n}(\mathbb{C})$ the $C^{*}$-algebra of $n\times n$ complex matrices. The tensor product $\mathcal{A}\otimes\mathcal{B}$ is well-defined as a $C^{*}$-algebra and it is isomorphic to $C_{0}(X,\mathcal{B})$ (see for instance thm: II.9.4.4 in Blackadar's book).

Since the dual space $\mathcal{A}'$ of $\mathcal{A}$ may be identified with the space of complex-valued Radon measures on $X$, and the dual space $\mathcal{B}'$ may be identified with $\mathcal{B}$ itself, I was wondering if it is possible to identify the dual space $(\mathcal{A}\otimes\mathcal{B})'\cong(C_{0}(X,\mathcal{B}))'$ with the space of $\mathcal{B}$-valued Radon measures on $X$, and thus the space of states of $\mathcal{A}\otimes \mathcal{B}$ with the space of $S$-valued Radon measures on $X$, where $S$ is the space of positive semidefinite matrices with unit trace.

Since it is just a mere curiosity triggered by a casual conversation I had yesterday, I did not yet really try to prove anything, but I have the impression that this could be a well-known result for experts in $C^{*}$-algebra theory.

Hanno
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fmc2
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  • Does the answer provided also prove the conjecture you posed about the state space of $\mathcal A \otimes \mathcal B$? – ViktorStein Jan 16 '23 at 10:38
  • @Ramanujan, I think it does because if you consider elements of the type $\mathbf{x}\otimes\chi_{A}$ with $\mathbf{x}$ positive in $\mathcal{B}=\mathcal{M}{n}(\mathbb{C})$ and $\chi{A}$ a characteristic function on $X$, you can check that the dual pairing forces every positive linear functional to be a positive measure with values in positive matrices, and then the converse should follow from direct computation. However, I do not have a rigorous proof. At the time I was thinking about this problem it felt true and that was enough for me :-) – fmc2 Jan 17 '23 at 11:51
  • Of course you can ask! I tend to be interested in the differential geometry of (suitable subsets of) the space of states/positive linear functionals of C/W/Jordan algebras, essentially because of my interest in the so-called field of Information Geometry. Therefore, I have to deal with functional-analytic things even if "I am not academically trained" for these things. I bounce back the question to you :-) – fmc2 Jan 18 '23 at 12:12
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    Note: in this answer and this answer it is stated that the state space can be identified with the symmetric positive semidefinite matrix-valued measures $u$ with $\text{tr}(u(X)) = 1$, which is different from your conjecture. – ViktorStein Jan 21 '23 at 21:35
  • My interest stems from understanding how the Connes metric on state space looks like for matrix-valued measures. – ViktorStein Jan 21 '23 at 21:36
  • I think what explained in this answer is exactly what I conjectured. Perhaps, the way I wrote it ($S$-valued Radon measure on $X$ where $S$ is the space of positive semidefinite matrices with unit trace) was ambiguous but I meant exactly what written there. – fmc2 Jan 21 '23 at 23:34

1 Answers1

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It is true that if $A$ is any $C^*$-algebra, then $M_n(A)^*$ is isomorphic to $M_n(A^*)$. In your case, if $R_X$ denotes the space of Radon measures on $X$ \begin{align} C_0\big(X,M_n(\mathbb C)\big)^* & \cong (A\otimes B)^* = \big(C_0(X)\otimes M_n(\mathbb C)\big)^* \\ & = M_n(C_0(X))^* \cong M_n(C_0(X)^*) \cong M_n(R_X). \end{align} Now I am no expert in measure theory but I don't feel confident about matrices with measure-entries being the same as measures with matrix values. That's only my guess though.

As requested: let $\zeta_{i,j}:A\to M_n(A)$ denote the map $\zeta_{i,j}(a)=a\otimes e_{i,j}$, where $e_{k,l}$ are matrix units (i.e. $a\otimes e_{i,j}$ is the matrix with $0$'s everywhere and $a$ in the i,j slot). This is easily seen to be a linear isometry. Define $T:M_n(A)^*\to M_n(A^*)$ by $T(\varphi)=[\zeta_{i,j}^*(\varphi)]_{i,j}\equiv[\varphi\circ\zeta_{i,j}]_{i,j}$. This is easily seen to be a vector space isomorphism. This is for example demonstrated in C. Lance's paper "On nuclear $C^*$-algebras" (1973), but the details are easy to fill in.

ViktorStein
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Just dropped in
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  • thank you for the details. I thought a little bit more about the issue "matrix-valued measures VS matrix of measures", and I somehow changed idea. I think they are the same because a matrix-valued measure gives you as much measures as there are entries in the matrix, and from these measures you can build a matrix of measures, and "viceversa". However, I do not have a rigorous proof. – fmc2 Sep 22 '20 at 13:25
  • @Ittiolo well, if that's true I am pretty sure that the details will be easy to work out – Just dropped in Sep 22 '20 at 13:27