Quest for a Green Function Formula

A simplicial complex G, a finite set of non-empty sets closed under the operation of taking finite non-empty subsets, has the Laplacian $L(x,y) = {\rm sign}(|x \cap y|)$. It is natural as it is always unimodular so that its inverse $g(x,y)$ is always integer valued.

In a potential theoretical setup, the Green function values $g(x,y)$ measure a potential energy between the simplices x and y. We know now that the sum $\sum_x \sum_y g(x,y)$ is the Euler characteristic $\chi(G) = \sum_x \omega(x)$ of $G$, where $\omega(x)=(-1)^{dim(x)}$ is the Poincaré-Hopf index of the energy functional $dim(x)$. In the proof of that energy theorem, it was essential to see that $g(x,x)$, the self-energy of a simplex, is equal to $1-\chi(S(x))$, where $S(x)$ is the unit sphere of a simplex $x$. We would like to have a formula for all entries $g(x,y)$. This is important. To see why, one has to look at the history of classical calculus:

The prototype of all potential theory is the Gauss Laplacian $-\Delta/(4\pi)$ in Euclidean 3 space $R^3$, where $g(x,y)=1/|x-y|$ is the Newton potential. The mathematics is due to Gauss: we have the gravitational potential $V(x) = g \rho(x)=\int_{R^3} \rho(y)/|x-y| dy$ from a mass density $\rho$. Because $\Delta = div (grad)$, the formula $\Delta V(x) = -4\pi \rho(x)$ is equivalent to ${\rm div}(F) = -4 \pi \rho$ if $F=\nabla V$ is the gravitational force field. This is the Gauss picture to gravity.

The transition from the Newton picture with ordinary differential equations to the Gauss picture with linear partial differential equations can not be underestimated because the Gauss description is not only more powerful in Euclidean space, it also allows to formula a {\bf classical gravitational theory} in any geometric space with a Laplacian. For any geometric space $G$ with an exterior derivative $d$ one has a Laplacian $(d+d^*)^2 = d d^* + d^* d$ which on the lowest scalar level is just the scalar Laplacian $d_0^* d_0 = {\rm div} ( {\rm grad})$. The new picture also shows why the inverse square field is so natural in our three dimensional world. It appears in gravity. It appears as the electric field strength of an electron.

So, if we have any geometry with Laplacian, we also have a natural potential energy $g(x,y)$. One can do this especially for the Hodge Laplacian H of a simplicial complex which has, when restricted to 0-forms has as the Laplacian the Kirchhoff Laplacian of the graph $G_1$ which as a simplicial complex is the Barycentric refinement of the simplicial complex G. As in the classical Newton case, there is a problem with the self interaction. In the Newton case, $g(x,x)$ is infinite at every point in the finite case. Also in the Hodge case, the Laplacian has to be inverted on the orthogonal complement of its kernel, the constant functions, which are the harmonic scalar functions on a finite geometry.

The spectral theorem in linear algebra tells how to invert $L$: by diagonalization one has $g(x,y) = \sum_k \psi_k(x) \psi_k(y)/\lambda_k$, where $\lambda_k$ runs over all non-zero eigenvalues of $H$ and $\psi_k$ form an orthonormal basis of the ortho complement of the kernel. The formula is self-evident as the orthonormality condition immediately gives $g(x,y) \psi_k(y) = (1/\lambda_k) \psi_k(x)$. This formula holds for any Laplacian. As this already gives a formula for $g(x,y)$ why look further? The reason is that we want a geometric formula, a classical geometry formula. The above formula using eigenvectors is a quantum formula which tells how the quantum incarnations of the simplices interact.

In the case of the connection Laplacian $L$ we have a couple of results already. Let $W^+(x,x)$ denote the unstable manifold and $W^-(x,x)$ the stable manifold. Let $W^+(x,y)$ be the intersection of the stable manifolds of $x$ and $y$. Let $W^0(x,y) = x \cap y$ denote the central manifold of $x$ and $y$. Here is what we know so far

  1. $g(x,x) = 1-\chi(S(x)) = (1-\chi(W^+(x,x))) (1-\chi(W^-(x,x))$.
  2. If $x$ is maximal and $y \subset x$ then $g(x,y) = (1-\chi(W^+(x,y))) (1-\chi(W^-(x,y)))$.
  3. If the stable, center and unstable manifolds of $x$ and $y$ are disjoint, then $g(x,y)=0$.
  4. In particular, if the maximal simplices of $x$ and $y$ are disjoint then $g(x,y)=0$.

Over thanksgiving, I was running pretty heavy computer assisted searches for a general formula involving intersections of stable and unstable manifolds of x and y. I would really like to see g(x,y) as some kind of index. So far there is no luck. But there are many possibilities still to be explored. This is the Baconian side of the story. A Descartian approach is to use Cramer and try to make sense of the minors of the Laplacian $L$. We obviously want to understand the matrix $L$ in which column x and row y are deleted. This is some kind of partition function over all one dimensional submanifolds, where one connected component goes from x to y and the others are closed loops. This is the Fredholm determinant picture. So, the Green function g(x,y) is some kind of path integral. But this does not help yet to get a geometric formula. I still hope for a formula which involves stable manifolds $W^+(x)=\{ y | x \subset y, y \neq x \}$ and unstable manifolds $W^-(x) = \{ y | y \subset x, y \neq x \}$. It might not be only the intersections which matter but how they intersect. I tried with the Wu intersection number $\omega(A,B) = \sum_{x \in A, y \in B} \omega(x) \omega(y)$ but so far no luck. Since often the formulas start to hold in almost all cases but fall in a few make it likely that something subtle is missing. Especially with formulas involving the Euler characteristic of locally defined sets, we get for small networks often contractible sets or empty sets there, which make render some formulas true for large sets. The data fitting techniques then lead to things which are often true but not always. Aesthetic formula attempts like $g(x,y) = (1-\chi(W^{++}(x,y))) (1-\chi(W^{–}(x,y))) – (1-\chi(W^{+-}(x,y) )(1-\chi(W^{-+}(x,y))$ involving all kind of stable and unstable manifold intersection did not work so far.