I have no background in Algebra, but want to understand matrix-valued measures and matrix-valued continuous functions from the C$^*$-algebra perspective to identify what definition of matrix-valued probability measure makes sense and why.
Here is what I understood so far (some things are verified through examples 1.1.3 and 1.1.8 in Murphy's C$^*$-algebras and operator theory).
Let $X$ be a compact Hausdorff space with Borel sigma algebra $\Sigma$.
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The set $C(X; \mathbb R)$ of real-valued continuous functions on $X$ becomes a commutative unital C$^*$-algebra (but not a von Neumann algebra) with pointwise addition and (scalar) multiplication. The involution is the identity, the unit is the constant one function on $X$, $1_X$ and the norm is the sup-norm. By the Riesz-Markov-Kakutani theorem, for every positive linear functional $\tau$ on $C(X; \mathbb R)$ there exists a regular measure $\mu$ with $\| \mu \| = \| \tau \|$ (identification via $\tau_{\mu}(f) := \int_X f(x) \; \text{d}(\mu)(x)$, which is lower semicontinuous by Fatou's lemma). Hence the state space of $C(X; \mathbb R)$ (i.e. the set of positive linear functionals with unit norm) can be identified with the probability measures on $X$ equipped with the total variation norm $\| \mu \| := | \mu |(X)$. (cf. II.6.2.3(ii) in Operator algebras by Blackadar). The GNS-representation space for any such state $\tau_{\mu}$ is $L^2(X; \mu)$.
All the elements of $C(X; \mathbb R)$ are symmetric. The positive elements of $C(X; \mathbb R)$ are the nonnegative continuous functions $X \to \mathbb R$.
What would be a trace on $C(X; \mathbb R)$? For $X = \mathbb R$, there is $f \mapsto \int_{\mathbb R} f(x) \, \text{d}x$. -
The ring of real $N \times N$ matrices $M_N(\mathbb R)$ is a non-commutative unital C$^*$-algebra (even a von Neumann algebra) with normal matrix addition and multiplication and unit $\text{id}_N$, the unit matrix.
The involution is transposition $^{\mathsf{T}}$.
We choose the spectral norm $\| A \|_2 := \sqrt{\lambda_{\max}(A^{\mathsf{T}} A)}$ (we have to choose this norm, because we identify $M_N(\mathbb R)$ with the bounded linear operators from $\mathbb R^N$ to $\mathbb R^N$, $B(\mathbb R^N)$, and on $\mathbb R^n$ we choose the Euclidean norm $\| \cdot \|_2$ (the only one which makes it into a Hilbert space) and the corresponding operator norm on $B(\mathbb R^N)$ is the spectral norm).
This choice of norm makes $M_N(\mathbb R)$ into a Banach space, but not into a Hilbert space (which it would be with the Frobenius inner product and induced norm).
The symmetric elements are the symmetric matrices and the positive elements are the symmetric positive semidefinite matrices.
We equip $M_N(\mathbb R)$ with the usual trace from Linear Algebra, which is a faithful normal tracial state.
The pure states of $M_N(\mathbb R)$ are $A \mapsto x^{\mathsf{T}} A x$ for $\| x \|_2 = 1$ and thus can be identified with the sphere $\{ x \in \mathbb R^N: \| x \|_2 = 1 \}$. The state space $S(M_N(\mathbb R))$ is the closed convex hull of these pure states (is there a nice closed form?). -
The set $\mathbb M(X; \mathbb R)$ of real-valued measures $u$ on $X$ (provided $X$ is a locally compact abelian group) with $\| u \| < \infty$ together with usual linear addition / scalar multiplication and convolution as multiplication is a unital commutative Banach algebra, but not a C$^*$ algebra unless the group is trivial (cf. Algebra of Measures – Encyclopedia of Math or II.10.1.5 in Blackadar's book). For a Banach space $B$, let $\mathbb M(X; B)$ denote the finite measures on $(X, \Sigma)$ with values in $B$. We (hopefully) have $\mathbb M(X; \mathbb R) \otimes M_N(\mathbb R) \cong \mathbb M\big(X; M_N(\mathbb R)\big)$ via $u \otimes A \longleftrightarrow (a_{i, j} u)_{i, j = 1}^{N}$.
Question 1. Is everything up to now correct? I am especially interested in the choice of norm on $M_N(\mathbb R)$ and trace on $C(X; \mathbb R)$.
We have $C(X; \mathbb R) \otimes M_N(\mathbb R) \cong C\big(X; M_N(\mathbb R)\big)$ via $(f \otimes A)(x) \longleftrightarrow f(x) A$ for $f \in C(X; \mathbb R)$ and $A \in M_N(\mathbb R)$ (cf. II.9.4.4 in Operator Algebras by Blackadar). Is the multiplication $(f \otimes A) \cdot (g \otimes B) = (f \cdot g) \cdot (A \cdot B)$?
This non-commutative unital C$^*$ algebra has the involution $(f \otimes A)^* = f \otimes A^{\mathsf{T}}$ and the unit $1_X \otimes \text{id}_N$.
I would expect the norm on this tensor $C^*$ algebra to be the following combination of both norms: $\| f \otimes A \|_{C(X; \mathbb R) \otimes M_N(\mathbb R)} := \| A \|_2 \cdot \max_{x \in X} | f(x) |$.
(Indeed, in Murphy's book cited above, on p. 190 it is stated that for the spatial C$^*$ norm on $A \otimes B$ we have $\| a \otimes b \|_* = \| a \| \| b \|$ for all $a \in A$, $b \in B$.)
In this Math.SE answer it is stated that the dual space of $C\big(X; M_N(\mathbb R)\big)$ can be identified with $M_N\big(\mathbb M(X; \mathbb R)\big)$, the space of $N \times N$ matrices whose entries are from $\mathbb M(X; \mathbb R)$, which should be the same as $\mathbb M\big(X; M_N(\mathbb R)\big)$.
By this Math.SE post the only norm on $M_N\big(\mathbb M(X; \mathbb R)\big)$ is $\| X \|_{M_N\big(\mathbb M(X; \mathbb R)\big)} := \sup_{ \| a \|_* \le 1 } \| X a \|_*$, where $\| (a_k)_{k = 1}^{n} \|_*^2 = \sum_{k = 1}^{n} \| a_k \|_{\mathbb M(X; \mathbb R)}^2$.
I am interested in the state space of $C\big(X; M_N(\mathbb R)\big)$.
Formally, the state space of $C\big(X; M_N(\mathbb R)\big)$ is
$$
\bigg\{ \tau \in L\big(C\big(X; M_N(\mathbb R)\big); \mathbb R\big): \tau(\text{id}_N \cdot 1_X) = 1 = \| \tau \| \bigg\},
$$
where $L(Y; Z)$ denote the linear operators from $Y$ into $Z$.
Question 2. How can we, along the lines of the answer mentioned above, identify $\tau \in L\big(C\big(X; M_N(\mathbb R)\big); \mathbb R\big)$ with $\mu_{\tau} \in \mathbb M(X; M_N(\mathbb R))$ and what do the conditions $\tau(\text{id}_N \cdot 1_X) = 1 = \| \tau \|$ become, then?
I hope that we can identify the state space of $C\big(X; M_N(\mathbb R)\big)$ with the symmetric positive (semi?)definite matrix-valued measures $\mu$ on $X$ with $\text{tr}(\mu(X)) = 1$.
Question 3. Can we write $\mathbb M(X; M_N(\mathbb R))$ as a tensor product and is there a canonical way to define a trace on it? Would $u \mapsto \text{tr}(u(X))$ be a trace on it?
Best Answer
First a comment that one usually does the theory of Banach algebras, C$^*$-algebras, and von Neumann algebras over $\mathbb C$. Restricting to $\mathbb R$ generates some problems (induced by the possibility of having an empty spectrum) and while there is theory about it, it is not mainstream.
What you wrote is mostly correct. Here are some comments.
It is possible for $C(X,\mathbb R)$ to be a von Neumann algebra, for special enough $X$ (the term is that the space $X$ is hyperstonean).
On $C(X,\mathbb R)$, any state is a trace. If you want a faithful trace, then for the general case you might be out of luck. If $X$ is sufficiently big, a faithful trace will not exist.
To get a state in $M_n(\mathbb R)$ you need the normalized trace, so it is not the one from Linear Algebra.
The state space is easy to identify, after you notice that any linear functional on $M_n(\mathbb R)$ is of the form $$\tag1 A\longmapsto \operatorname{Tr}(AH) $$ (usually written with the non-normalized trace) for some $H\in M_n(\mathbb R)$. To get a state, you need $H\geq0$ and $\operatorname{Tr}(H)=1$.
When you consider a tensor product $\mathcal A\otimes\mathcal B$, the multiplication is indeed defined as $(A_1\otimes B_1)(A_2\otimes B_2)=A_1A_2\otimes B_1B_2$, and indeed the norm on elementary tensors is $\|A\otimes B\|=\|A\|\,\|B\|$. But a crucial point is that a tensor product is not the set of elementary tensors; is the closure of the linear span of the elementary tensors.
Another thing to consider when doing tensor products of C$^*$-algebras is that in general there is not a single possible norm on the tensor product. In your case, though, your algebras are nuclear, which means that when tensored with any other C$^*$-algebra they admit a single norm.
Note that states are often characterized in the more useful way "unital and positive".
It is canonical that $C(X,M_n(\mathbb R))$ can be identified, as you mentioned, with $C(X)\otimes M_n(\mathbb R)$. And it is also canonical that the latter can be identified with $M_n(C(X))$. With the same kind of duality as in $(1)$, $$M_n(C(X))^*=M_n(C(X)^*).$$ So a linear functional $\Phi=[\varphi_{kj}]$ maps $$ [f_{kj}]\longmapsto \sum_{k,j}\varphi_{kj}(f_{kj}). $$ Each $\varphi_{kj}$ has a corresponding measure $\mu_{kj}$. So we can identify $\Phi$ with the matrix-valued measure $\mu=[\mu_{kj}]$. When $\Phi$ is positive, the measure $\mu$ is positive (in the sense that its values are positive semidefinite). Indeed, \begin{align} \sum_{k,j}\mu_{kj}(E)\,\omega_k\omega_j &=\sum_{k,j}\int_X\omega_k\omega_j\,1_E\,d\mu_{kj} =\sum_{k,j}\varphi_{k,j}(\omega_k\omega_j\,1_E)=\Phi([\omega_k\omega_j 1_E]). \end{align} The matrix $[\omega_k\omega_j 1_E]$ can be seen as $$ \omega\omega^T\,(1_E\,I_n), $$ so it is positive (positive semidefinite). So the above equation shows that the matrix $\mu(E)=[\mu_{kj}(E)]$ is positive semidefinite. For $\Phi$ to be unital means that $$ 1=\Phi(I_n)=\sum_k\varphi_{kk}(1)=\operatorname{Tr}(\mu(X)). $$ That is, the states are exactly as your characterization.