2012arXiv (Cornell University)Open access

An index formula for simple graphs

Oliver Knill

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Abstract

Gauss-Bonnet for simple graphs G assures that the sum of curvatures K(x) over the vertex set V of G is the Euler characteristic X(G). Poincare-Hopf tells that for any injective function f on V the sum of i(f,x) is X(G). We also know that averaging the indices E[i(f,x)] over all functions gives curvature K(x). We explore here the situation when G is geometric of dimension d: that is if each unit sphere S(x) is geometric of dimension d-1 and that X(S(x))=0 for even d and X(S(x))=2 for odd d. The dimension of G is inductively defined as the average of 1+dim(S(x)) over all S(x) assuming the empty graph has dimension -1. We prove that any odd dimensional geometric graph G has zero curvature. This is done with the help of an index formula j(f,x) = 1-X(S(x))/2-X(B(f,x))/2, where j(x)=[i(f,x)+i(-f,x)]/2. The graph B(f,x) is the discrete level surface {y | f(y) = f(x)} intersected with S(x). It is a subgraph of the line graph of G and geometric if G is geometric. The index formula simplifies for geometric graphs: for even d it is j(f,x) = 1-X(B(f,x))/2, where B(f,x) is a (d-2)-dimensional graph. For odd d it becomes j(f,x) =-X(B(f,x))/2, where B(f,x) is an odd dimensional graph. Because by induction with respect to d, the X(B(f,x))=0 we know now that that j(f,x) is zero for all x and so, by taking expectation over f that curvature K(x) is zero for all x. We also point out that all these results hold almost verbatim for compact Riemannian manifolds and actually are much simpler there. The same integral geometric index formula is valid if f is a Morse function, i(f,x) is the index of the gradient vector field and if S(x) is a sufficiently small geodesic sphere around x and B(f,x) which is S(x) intersected with the level surface {y | f(y)=f(x)}. Also in the continuum, the symmetric index j(f,x) is constant zero everywhere if d is odd.

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Gauss-Bonnet for simple graphs G assures that the sum of curvatures K(x) over the vertex set V of G is the Euler characteristic X(G). Poincare-Hopf tells that for any injective function f on V the sum of i(f,x) is X(G). We also know that averaging the indices E[i(f,x)] over all functions gives curvature K(x). We explore here the situation when G is geometric of dimension d: that is if each unit sphere S(x) is geometric of dimension d-1 and that X(S(x))=0 for even d and X(S(x))=2 for odd d. The dimension of G is inductively defined as the average of 1+dim(S(x)) over all S(x) assuming the empty graph has dimension -1. We prove that any odd dimensional geometric graph G has zero curvature. This is done with the help of an index formula j(f,x) = 1-X(S(x))/2-X(B(f,x))/2, where j(x)=[i(f,x)+i(-f,x)]/2. The graph B(f,x) is the discrete level surface {y | f(y) = f(x)} intersected with S(x). It is a subgraph of the line graph of G and geometric if G is geometric. The index formula simplifies for geometric graphs: for even d it is j(f,x) = 1-X(B(f,x))/2, where B(f,x) is a (d-2)-dimensional graph. For odd d it becomes j(f,x) =-X(B(f,x))/2, where B(f,x) is an odd dimensional graph. Because by induction with respect to d, the X(B(f,x))=0 we know now that that j(f,x) is zero for all x and so, by taking expectation over f that curvature K(x) is zero for all x. We also point out that all these results hold almost verbatim for compact Riemannian manifolds and actually are much simpler there. The same integral geometric index formula is valid if f is a Morse function, i(f,x) is the index of the gradient vector field and if S(x) is a sufficiently small geodesic sphere around x and B(f,x) which is S(x) intersected with the level surface {y | f(y)=f(x)}. Also in the continuum, the symmetric index j(f,x) is constant zero everywhere if d is odd.

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Available abstract

Gauss-Bonnet for simple graphs G assures that the sum of curvatures K(x) over the vertex set V of G is the Euler characteristic X(G). Poincare-Hopf tells that for any injective function f on V the sum of i(f,x) is X(G). We also know that averaging the indices E[i(f,x)] over all functions gives curvature K(x). We explore here the situation when G is geometric of dimension d: that is if each unit sphere S(x) is geometric of dimension d-1 and that X(S(x))=0 for even d and X(S(x))=2 for odd d. The dimension of G is inductively defined as the average of 1+dim(S(x)) over all S(x) assuming the empty graph has dimension -1. We prove that any odd dimensional geometric graph G has zero curvature. This is done with the help of an index formula j(f,x) = 1-X(S(x))/2-X(B(f,x))/2, where j(x)=[i(f,x)+i(-f,x)]/2. The graph B(f,x) is the discrete level surface {y | f(y) = f(x)} intersected with S(x). It is a subgraph of the line graph of G and geometric if G is geometric. The index formula simplifies for geometric graphs: for even d it is j(f,x) = 1-X(B(f,x))/2, where B(f,x) is a (d-2)-dimensional graph. For odd d it becomes j(f,x) =-X(B(f,x))/2, where B(f,x) is an odd dimensional graph. Because by induction with respect to d, the X(B(f,x))=0 we know now that that j(f,x) is zero for all x and so, by taking expectation over f that curvature K(x) is zero for all x. We also point out that all these results hold almost verbatim for compact Riemannian manifolds and actually are much simpler there. The same integral geometric index formula is valid if f is a Morse function, i(f,x) is the index of the gradient vector field and if S(x) is a sufficiently small geodesic sphere around x and B(f,x) which is S(x) intersected with the level surface {y | f(y)=f(x)}. Also in the continuum, the symmetric index j(f,x) is constant zero everywhere if d is odd.

Key concepts: Combinatorics, Mathematics, Injective function, Vertex (graph theory), Graph, Dimension (graph theory)

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