Theorem. Finite spectral triples and the Connes distance [fdip-0095]
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  • November 2, 2026

The naive lattice Dirac operator does not reproduce lattice spacing.

[bimonte1994lattice, equation (45)] computes the distances along one axis. For one, two, three and 4 steps they are \(2a\), \(2\sqrt {2}\,a\), \(4a\) and \(2\sqrt {6}\,a\). Those are not the multiples \(a\), \(2a\), \(3a\) and \(4a\).

A Wilson term multiplies every one of those distances by \((1+r^{2})^{-1/2}\).

[bimonte1994lattice, the paragraph containing (47)] records that rescaling.

Let \(D\) be a real symmetric \(3\times 3\) matrix, and let a real function on its three indices act by the diagonal matrix \(f\). The distance from index \(1\) to index \(2\) is the supremum of \(|f_2-f_1|\) over all \(f\) with \(|[D,f]|\le 1\). Write \(D_{ij}\) for the entries of \(D\) and \[\Delta =D_{12}^{2}D_{13}^{2}+D_{12}^{2}D_{23}^{2}+D_{13}^{2}D_{23}^{2}.\] If \(\Delta >0\), the distance is \[\sqrt {\frac {D_{13}^{2}+D_{23}^{2}}{\Delta }}.\] If \(D_{13}=D_{23}=0\) and \(D_{12}\ne 0\), the distance is \(1/|D_{12}|\). In every remaining case there is no finite upper bound. The distances among the other pairs are the same statements with the indices cycled.

Proof. 

  • November 2, 2026

The commutator has entries \([D,f]_{ij}=D_{ij}(f_j-f_i)\). Adding a constant to \(f\) does not change it, so set \(f_1=0\), \(f_2=x\) and \(f_3=y\). The entries above the diagonal are \(a=D_{12}x\), \(b=D_{13}y\) and \(c=D_{23}(y-x)\). Let \(\omega =(-c,b,-a)\) and let \(\sigma ^{2}=a^{2}+b^{2}+c^{2}\). Write \(M\) for the matrix of \([D,f]\). It is skew-symmetric, so its square is the negative of \(M^{\mathrm T}M\). Expanding that product gives \(M^{\mathrm T}M=\sigma ^{2}I-\omega \omega ^{\mathrm T}\). Thus \(M^{\mathrm T}M\) fixes the line through \(\omega \) by \(0\) and acts as \(\sigma ^{2}\) on the orthogonal plane. The operator norm of \([D,f]\) is therefore \(\sigma \). The constraint \(\sigma \le 1\) reads \[(D_{12}^{2}+D_{23}^{2})x^{2}-2D_{23}^{2}xy+(D_{13}^{2}+D_{23}^{2})y^{2}\le 1.\]

Write \(A=D_{13}^{2}+D_{23}^{2}\). If \(A>0\), the left side is a quadratic polynomial in \(y\) opening upwards. The discriminant of this polynomial set equal to \(1\) is \(4A-4\Delta x^{2}\). A real \(y\) exists exactly when this discriminant is nonnegative. If \(\Delta >0\), that condition is \(x^{2}\le A/\Delta \), so the largest value of \(|x|\) is the displayed square root. If \(\Delta =0\), the discriminant is positive for every \(x\), and there is no finite upper bound.

If \(A=0\), then \(D_{13}=D_{23}=0\) and the constraint reduces to \(D_{12}^{2}x^{2}\le 1\). A nonzero \(D_{12}\) gives the reciprocal bound. If \(D_{12}\) vanishes as well, every \(x\) is allowed. Since \(A=0\) forces \(\Delta =0\), this case and the case \(A>0\) with \(\Delta =0\) cover every vanishing \(\Delta \).

On the path with vertices \(0,1,2,3\) and the adjacency matrix as \(D\), the same commutator norm gives distance \(2\) from \(0\) to \(3\). The graph distance along the three edges is \(3\). Let \(a=f_1-f_0\), \(b=f_2-f_1\) and \(c=f_3-f_2\), so \(f_3-f_0=a+b+c\). The commutator is zero except for these three adjacent pairs. Replacing each difference by its absolute value does not change the operator norm: the signs are absorbed by changing the signs of the last three basis vectors. Thus it is enough to take \(a,b,c\ge 0\).

Suppose \(a+b+c>2\). If \(a\ge 2\), the first basis vector is sent to a vector of length \(a\), so the norm exceeds \(1\). If \(a<2\), then \(b+c>2-a\). The vector \((0,1,0,-1)\) has length \(\sqrt {2}\) and is sent to \((a,0,-(b+c),0)\), whose square length is \(a^{2}+(b+c)^{2}\). That square exceeds \(a^{2}+(2-a)^{2}=2(a-1)^{2}+2\ge 2\), so the operator norm exceeds \(1\). Therefore every admissible triple has \(a+b+c\le 2\).

The values \(a=c=1\) and \(b=0\) give \(f=(0,1,1,2)\). The commutator is the orthogonal sum of two copies of the block with entries \(0,1,-1,0\), and each block has norm \(1\). The difference of the endpoint values is \(2\), so the distance is exactly \(2\).

[iochum1999distances, Theorem 9] proves a complementary fact for a general real symmetric matrix on 4 indices: the distance is a root of a polynomial of degree at most \(12\), and that polynomial is generically not solvable by radicals. The path above is one matrix for which the distance can still be read off directly. This note does not reproduce the degree count.