Spectral triple for a (real) full matrix algebra

Question

Let $\mathcal A = \mathbb R^{N \times N}$ be the real full matrix algebra, $N \in \mathbb N_{> 1}$, which is represented by the Hilbert space $H := \mathbb R^N$ (that is, $\mathcal A \to B(H)$, $A \mapsto (x \mapsto A x)$ is a $*$-isomorphism, where $B(H)$ are the linear operators $H \to H$). I am searching for a good / canonical choice for an Dirac-type operator $D$, such that $(\mathcal A, H, D)$ becomes a spectral triple, that is, a (linear) self-adjoint operator $D \colon H \to H$ such that ($(H, D)$ is a Fredholm module, that is) $$\{ a \in \mathcal A: [D, a] \le 1 \} / \mathbb R 1 $$ is bounded (in particular, it means that no element of $\mathcal A$ except scalars commutes with $D$), hopefully chosen such that Connes spectral metric $$ d \colon S(\mathcal A) \times S(\mathcal A), \qquad (\psi, \phi) \mapsto \sup_{\substack{a \in \mathcal A \\ [D, a] \le 1}} | \psi(a) - \phi(a) | $$ becomes something meaningful, where $S(\mathcal A)$ is the state space of $\mathcal A$, which can be identified with the positive semidefine matrices with unit trace, because any linear functional on $\mathcal A$ is of the form $A \mapsto \text{tr}(A H)$, where $\text{tr}$ denotes the usual (unnormalised) trace from linear algebra.

I don't have any background in Algebra, so I haven't been able to find any papers online describing sensible $D$ or $d$ and I am grateful for any pointers.

Remark. If this question only makes sense with $\mathbb C$ instead of $\mathbb R$, this is fine.

Remark 2. If I am not mistaken, $[D, \cdot]$ is an inner derivation on $\mathcal A$ (with respect to the left bimodule $B(H)$) and all derivations on $\mathcal A$ are inner, so we can equivalently ask for a sensible choice of derivation on $\mathcal A$ instead of searching for $D$. Furthermore, the vector space of inner derivations on $\mathcal A$ can be identified with the traceless skew-symmetric matrices.

Answer 1

In the definition of spectral triple, $D$ is also assumed to be self-adjoint. In this case, such an operator cannot exist (if $N>1$):

If the only operators commuting with $D$ were scalar multiples of the identity, then the unital $\ast$-algebra generated by $D$ would be the full matrix algebra by the bicommutant theorem. However, as $D$ is self-adjoint, it generates a commutative $\ast$-algebra, and $B(H)$ is not commutative.

Since I am not sure if the bicommutant theorem holds over the reals, here is a direct argument: It is easy to see that spectral projections commute with $D$. If $D$ is not a constant multiple of the identity, then it has a non-trivial (not equal to $0$ or $1$) spectral projection. If $D$ is a constant multiple of the identity, it commutes with everything. In any case, $\{D\}^\prime\neq \mathbb CI$.

Answer 2

As @MaoWao pointed out, there is no spectral triple for this choice of $H$, but if we make $H$ larger, we can get a spectral triple.

This is an explicit construction in the complex case from section 3.2 of A dual formula for the spectral distance in noncommutative geometry by Francesco D'Andrea and Pierre Martinetti for $H := \mathbb C^N \otimes \mathbb C^m$ for $m \ge 1$: for Hermitian matrices $\left(L^{(k)}\right)_{k = 1}^{m} \subset \mathbb C^{N \times N}$ let $$ D_L := \sum_{k = 1}^{m} L^{(k)} \otimes E_{kk} \in \mathbb C^{N \times N} \otimes \mathbb C^{m \times m}, $$ where $E_{i, j} \in \mathbb C^{m \times m}$ for $i, j \in \{1, \ldots, m\}$ is the $(i, j)$-th matrix unit. We identify the operator algebra $(B(H), \circ, *)$, where the multiplication $\circ$ is composition and the involution $*$ means taking the adjoint operator, with $\mathbb C^{N \times N} \otimes \mathbb C^{m \times m}$ by the same $*$-isomorphism used in the question applied to both factors of $H$ individually. The C$^*$-algebra $\mathcal A := (\mathbb C^{N \times N}, \cdot, \ {}^{\mathsf{H}})$ with the usual matrix multiplication and the Hermitian adjoint as involution, acts on the first factor of $H$, which means the representation of $\mathcal A$ in $H$ is $$ \pi \colon \mathcal A \to B(H), \qquad A \mapsto \big( x \otimes y \mapsto (A x) \otimes y \big). $$

Verification of the properties. Indeed $\pi$ is a *-homomorphism: for $A, B \in \mathcal A$ we have $$ \pi(A \cdot B) = \big( x \otimes y \mapsto (A B x) \otimes y \big) = \big( x \otimes y \mapsto (A x) \otimes y \big) \circ \big( x \times y) \mapsto (B x) \otimes y \big) = \pi(a) \circ \pi(b) $$ and $$ \pi(A^*) = \big( x \otimes y \mapsto (A^{\mathsf{H}} x) \otimes y \big) $$ and $\langle \pi(A) h_1 \otimes h_2, k_1 \otimes k_2 \rangle_H = \langle h_1 \otimes h_2, \pi(A)^* (k_1 \otimes k_2) \rangle_H$ for all $h_1 \otimes h_2, k_1 \otimes k_2 \in H$, that is \begin{align*} \langle (A h_1) \otimes h_2, k_1 \otimes k_2 \rangle_H & = \langle A h_1, k_1 \rangle_{\mathbb C^N} \cdot \langle h_2, k_2 \rangle_{\mathbb C^m} \\ & = \langle h_1, [\pi(A)^*(k_1, k_2)]_1 \rangle_{\mathbb C^N} \cdot \langle h_2, [\pi(A)^*(k_1, k_2)]_2 \rangle_{\mathbb C^m} \\ & = \langle (h_1, h_2), \pi(A)^* (k_1, k_2) \rangle_H. \end{align*} Hence $\pi(A)^*(h \otimes k) = (A^{\mathsf H} h) \otimes k$ and thus $\pi(A^*) = \pi(A)^*$.

The operator $D_L$ is self-adjoint: for $h := h_1 \otimes h_2, j := j_1 \otimes k_2 \in \mathbb C^{N} \otimes \mathbb C^{m}$ we have \begin{align*} \langle D_L h, k \rangle_{H} & = \left\langle \sum_{k = 1}^{m} (L^{(k)} h_1) \otimes (E_{kk} h_2), j \right\rangle_H \\ & = \left\langle \sum_{k = 1}^{m} (L^{(k)} h_1) \otimes (E_{kk} h_2), j \right\rangle_H \\ & = \sum_{k = 1}^{m} \left\langle (L^{(k)} h_1) \otimes (E_{kk} h_2), j \right\rangle_H \\ & = \sum_{k = 1}^{m} \langle L^{(k)} h_1, j_1 \rangle_{\mathbb C^N} \langle (E_{kk} h_2), j_2 \rangle_{\mathbb C^m} \\ & = \sum_{k = 1}^{m} \langle h_1, L^{(k)} j_1 \rangle_{\mathbb C^N} \langle h_2, E_{kk} j_2 \rangle_{\mathbb C^m} \\ & = \ldots = \langle h, D_L k \rangle_{H}. \end{align*}

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The spectral distance between the density matrices $p_1$ and $p_2$ induced by $D_L$ is equal to (proposition 5) $$ d(p_1, p_2) := \inf\left\{ \text{tr}(\sqrt{u^* u}): u = (u_k)_{k = 1}^{m} \in \mathbb (C^{N \times N})^m, \sum_{k = 1}^{m} \big[ L^{(k)}, u_k \big] = \rho_1 - \rho_2 \right\}, $$ and the initial definition by Connes reads $$ d(p_1, p_2) = \sup\left\{ \big| \text{tr}\big(A (p_1 - p_2)\big) \big|: A \in \mathcal A, \ \| [D_L, \pi(A) ] \|_H \le 1 \right\}, $$ where \begin{align} \| [D_L, \pi(A) ] \|_H & = \| D_L \pi(A) - \pi(A) D_L \|_H \\ & = \left\| \sum_{k = 1}^{m} (L^{(k)} A) \otimes E_{kk} - (A L^{(k)}) \otimes E_{kk} \right\|_H \\ & = \left\| \sum_{k = 1}^{m} [L^{(k)}, A] \otimes E_{kk} \right\|_H \\ & = \sqrt{\left\langle \sum_{k = 1}^{m} [L^{(k)}, A] \otimes E_{kk}, \sum_{j = 1}^{m} [L^{(j)}, A] \otimes E_{jj} \right\rangle_H} \\ & = \sqrt{\sum_{j = 1}^{m} \sum_{k = 1}^{m} \left\langle [L^{(k)}, A], [L^{(j)}, A] \right\rangle_{\mathbb C^{N \times N}} \cdot \left\langle E_{kk}, E_{jj} \right\rangle_{\mathbb C^{m \times m}}} \\ & = \sqrt{\sum_{k = 1}^{m} \left\langle [L^{(k)}, A], [L^{(k)}, A] \right\rangle_{\mathbb C^{N \times N}}} \\ & = \sqrt{\sum_{k = 1}^{m} \left\| [L^{(k)}, A] \right\|_{\mathbb C^{N \times N}}^2}, \end{align} as $$ D_L \pi(A) = \sum_{k = 1}^{m} (L^{(k)} \otimes E_{kk}) \cdot (A \otimes \text{id}_m) = \sum_{k = 1}^{m} (L^{(k)} A) \otimes E_{kk}. $$