Fredholm module over $\mathbb R^{N \times N}$ and over matrix-valued measures The following is taken from Compact metric spaces, Fredholm modules,
and hyperfiniteness by Alain Connes, cf. also his paper Non-commutative differential geometry.
Definition.
Let $A$ be a unital $C^*$ algebra. An unbounded Fredholm module $(H, D)$ over $A$ is a Hilbert space $H$, which is a left $A$-module (i.e. a $*$ representation $\pi$ of $A$ in $H$ is given) together with an unbounded self-adjoint operator $D$ in $H$ such that

*

*$\{ a \in A: [D, a] \text{ is bounded} \}$ is norm dense in $A$.

*$(1 + D^2)^{-1}$ is a compact operator.

Proposition 4.
Let $A$ be a $C^*$ algebra and $(H, D)$ an unbounded Fredholm module over $A$ such that: $(^*\hspace{-.75mm})\{ a \in A; \| [ D, a ] \| \le 1 \} / \mathbb{C} 1$ is bounded.
Then
$$
d(\phi, \psi)
:= \sup\{| \phi(a) - \psi(a) |; a \in A, \| [D, a] \| \le 1 \}
$$
is a metric on the state space of $A$.

My question.
I have no background in Algebra, but I want to understand this definition (which involves a sensible choice for $(H, D)$) for the simple case of $A = \mathbb R^{N \times N}$, such that, if I am not mistaken the state space of $A$ can be identified with the positive semidefinite matrices with unit trace ("density matrices").
Are the following ideas of mine correct?
After reading about representation of C$^*$-algebras, I think we should choose $H = \mathbb R^N$, because the canonical isomorphism (a faithful representation)
$$
\pi \colon \mathbb{R}^{N \times N} \to B(\mathbb{R}^N), \qquad A \mapsto (x \mapsto A x),
$$
where $B(H)$ are the bounded linear operators $H \to H$, is a homomorphism of C$^*$ algebras.
Hence we can consider $D \in B(H)$ and $a \in A$ to live in the same space.
I suppose that the commutator $[D, a] := D \pi(a) - \pi(a) D$ is an (unbounded) linear operator $H \to H$.
As $H$ is finite-dimensional, $(1 + D^2)^{-1}$ will be compact.
Using the C$^*$ identity and the bijectivity of $\pi$, I think that $\| [D, a] \|_A \le 2 \| D \|_H \| a \|_A$, so that $\{ a \in A: [D, a ] \text{ is bounded } \} = A$.
Hence also the additional hypothesis (... is bounded) should be fulfilled.
(Can we write down $(^{\star})\{ a \in A: \| [D, a ] \| \le 1 \} / \mathbb{C}1$? I am a bit confused whether the algebras are taken to be over $\mathbb C$, because for an algebra over $\mathbb R$, this quotient doesn't make much sense to me.)
Furthermore, I can choose $D$ to be the identity, such that $[D, a] = 0$ for all $a \in A$, right?
I suspect that the metric depends on the choice of $D$, but as far as I can tell, this is not explicitly stated in the paper cited above.
Lastly, does the star in the additional condition posed in proposition 4 mean that the image of that set under the involution $*$ of $A$ is bounded?
Question 2.
How does the metric look if I choose $A := \mathcal C(X; \mathbb R^{N \times N})$ to be the unital non-commutative real C$^*$-algebra of $M_N(\mathbb R) := \mathbb R^{N \times N}$-valued continuous function on a compact connected Riemannian $d$-manifold without boundary $X$? By this answer the corresponding state space $S(A)$ be be identified with the set of symmetric positive semidefinite matrix-valued measures $u$ with $\text{tr}(u (X)) = 1$.
Update
Since we can write $A = \mathcal C(X; \mathbb R) \otimes M_N(\mathbb R)$ as product of two unital C$^*$ algebras, with known representations, the we can tensor them to get a representation of $A$.
If I am not mistaken, $(\pi_N, \mathbb R^N)$ is a faithful representation of $M_N(\mathbb R)$, where
$$
\pi_N \colon M_N(\mathbb R) \to B(\mathbb R^N), \qquad
A \mapsto (\mathbb R^N \ni x \mapsto A x)
$$
and a representation of $\mathcal C(X; \mathbb R)$ is $(\pi_u, L^2(X; u))$, where $u$ is a probability measure on $X$ and $\pi_u$ is induced via the GNS construction by the state $\tau_u(f) := \int_{X} f(x) \, \text{d}u(x)$:
$$
\pi_u \colon \mathcal C(X; \mathbb R) \to B\big(L^2(X, \mu)\big), \qquad
f \mapsto \big( g \mapsto f \cdot g \big).
$$
Hence the unique *-homomorphism $$\pi \colon \mathcal C(X; \mathbb R) \otimes M_N(\mathbb R) \to B\big(\mathbb R^N \otimes L^2(X, \mu)\big)$$ with $\pi(f \otimes A) = \pi_u(f) \otimes \pi_N(A)$ for all $f \in \mathcal C(X; \mathbb R)$ and all $A \in M_N(\mathbb R)$
should be something like
$$
\pi(f \otimes A)[g, x] = (f \cdot g) \otimes (A x)
$$
I think we can identify $\mathbb R^N \otimes L^2(X, \mu)$ with the vector valued square integrable functions $L^2(X, \mu; \mathbb R^N)$.

Applying those identifications, how does $\pi \colon \mathcal C\big(X; M_N(\mathbb R)\big) \to B\big(L^2(X, \mu; \mathbb R^N)\big)$ look like (not just on elements of the form $f \otimes A$)?

 A: Several remarks are in order:

*

*When talking about $C^\ast$-algebras, all vector spaces are taken to be complex unless explicitly stated otherwise. There are real $C^\ast$-algebras, but they are very much outside the mainstream of research. So the algebra $A$ in this definition is complex, the Hilbert space $H$ is complex etc. so you should take complex matrices instead of real ones to get an example.

*The metric depends on the choice of $(H,D)$. In your case you can take $H=\mathbb C^N$, but there are also other reasonable choices (like $H=\mathbb C^N\otimes\mathbb C^N)$.

*In the definition, $D$ is required to be unbounded, which cannot happen for finite-dimensional $H$. But I don't think this assumption is crucial. But if you take $D=I$, then $(\ast)$ will not be satisfied.

*The $(\ast)$ before the condition in Proposition 4 is just a label, no mathematical symbol.

*The answer to your second question is yes, this is one possible way to represent the states on $C(X;M_n(\mathbb C))$.

Given your questions, it seems good to explain the (commutative) example which motivates these constructions of Connes. Let $X=[0,1]$ for simplicity, $A=C(X)$, $H=L^2(X)$ with the canonical left action and $D=i\frac{d}{dx}$. Then $[D,f]=if'$ for $f\in C^1([0,1])$, and $(1+D^2)^{-1}$ is a compact operator by the Sobolev embedding theorem.
The set $\{f\in C^1([0,1])\mid \lVert f'\rVert_\infty\leq 1\}/\mathbb C1$ is bounded because
$$
\lvert f(x)-f(0)\rvert\leq\int_0^1 \lvert f'(x)\rvert\,dx\leq 1,
$$
which implies $\lVert f-f(0)1\rVert_\infty\leq 1$.
The states on $A$ are (canonically identified with) the Borel probability measures on $[0,1]$ and the metric $d$ defined by Connes is given by
$$
d(\mu,\nu)=\sup_{\lVert f'\rVert_\infty\leq 1}\left\lvert\int f\,d\mu-\int f\,d\nu\right\rvert,
$$
which is exactly the dual formulation of the $1$-Wasserstein (or Kantorovich-Rubinstein) metric.
Already in this case, the metric depends on the choice of $D$ (for example, we could rescale $D$, which would result in a scaling of the metric $d$). Moreover, $[D,a]$ is thought to be the noncommutative analog of the derivative of $a$, so taking $D=I$ (which results in zero commutators) is certainly not productive.
A: I now found that there is an explicit construction in the complex case to answer my first question in section 3.2 of A dual formula for the spectral distance in noncommutative geometry by Francesco D'Andrea and Pierre Martinetti, where the larger representing Hilbert space $H = \mathbb C^N \otimes \mathbb C^m$ is used, see also my answer here.
