Sphere Spectrum

Sphere Spectrum

We have had a great time on Block Island. The new off-shore wind-farm near the island are well visible from various places at the island. For example from the Mohegan Bluffs. Here is part of a panorama done during this winter vacation:
Photo near Mohegan Bluffs, Wind farm before Block Island, Photo: Oliver Knill, December, 2016 For me, this wind farm is a symbol of hope that we can solve our energy problems with clean energy, sun, wind and better appliances needing less power. The out-time at Block island was also good for tanking new ideas. The three bladed windmill in the Block island farm for example will lead to a graph which is a counter example to something I first thought to be true. More to this in the end. The following story is essentially a working draft for an update to that paper.
[January 9, 2017. I decided the topic is interesting and rich enough for a new paper as there is some interesting geometry involved in the proofs. ]

I think to be able to prove now (or be very close to be able to prove [January 9, 2017. The proof is complete]) that

the diagonal Green function entry values (1+A’)-1xx are a combinatorial invariant of a simplicial complex G with connection graph G’ having the adjacency matrix A’.

The key is a dual proposition to the key proposition proven in the unimodularity theorem. It tells what happens if we remove a cell, rather than add a cell. It is better to formulate this now in two separate proposition and keep it as a twin proposition as it can be a bit confusing first to see both together.

As in the paper, we call i(x) the Poincaré-Hopf index of a vertex x. It is 1-\chi(S(x)), where S(x) is the unit sphere of x in the Barycentric refinement of G. The following proposition is already more general than the proposition proven in the paper. It does not assume that the cell which is added is the largest cell. It can also be intermediate. We can for example add a new edge to a graph. We simply write G+x for the complex where x has been added. Now G’+x = (G+x)’ as we did not add x as a new vertex to a graph but as a new cell to a generalized CW complex. The proposition is now more general:

Proposition : If G is a simplicial complex and G’ its connection graph. If we chose a subgraph H’ of G’ and connect a new cell x to it leading to a new complex G’+x, then the Fredholm characteristic of G+x gets multiplied by a factor i(x)=1-\chi(H').

Note that in the new CW complex, S(x) is the unit sphere in the Barycentric refinement graph of G+x.
This first proposition has now got an other sibling (born a bit later although). Let us write G-x for the complex, where x has been removed as a cell. Now what matters is the sphere S(x) in G1 and not the sphere in G’. The reason is that if we remove the cell x, we remove all links to sub-simplices and super simplices. That this is important became apparent first when doing experiments, but the proof will make it even more clear:

Proposition: If G is a simplicial complex and G’ its connection graph. If we chose a vertex x in G’ and remove x, leading to the new complex G’-x = (G-x)’, then the Fredholm characteristic of G is multiplied by i(x)=(1-\chi(S(x))).

Seeing the two propositions together can raise suspiciaon at first. It appears not to be right that both adding as well as removing a cell produces a multiplication by an integer i(x) or -i(x). The solution to the conundrum is that these two operations are not inverse operations to each other except in the case when we stay in the class of connection graphs, where the multiplication factor is 1 or -1 and where division or multiplication is the same. But it has to make sense because the Fredholm characteristic is an integer, and the Poincaré-Hopf index i(x) is an integer which in general is not -1 or 1. So, we can not divide by it. Indeed, i(x) can be zero or larger than 1 or smaller than -1. The proof however will make the story crystal clear. What happens in the first proposition is that we start with a connection graph G’ getting to a complex G’+x which is no more a connection graph in general. In the second case, we start with a connection graph G’ and add up with a complex which is no more a connection graph in general. But in both cases, when either increasing the complex or decreasing the complex, we end up with a situation where the proposition does no more allow to go backwards.

Why is the second proposition important? Because of Gabriel Cramer who found an explicit formula for the inverse of a matrix in terms of determinants. This inversion formula is nowadays a bit neglected in courses as it is not a very efficient way to compute the inverse but it is beautiful and explicit and shows that determinants are not only natural but also useful. The formula tells that the (i,j) entry of the inverse of a matrix A is A-1ij= det(A(j,i)) (-1)i+j/det(A), where A(j,i) is the matrix where row j and column i were deleted. For the diagonal entries this simplifies to A-1iidet(B(i))/det(A), where B(i) is the matrix A in which row i and column i were deleted.

[January 10: the Swiss Gabriel Cramer was an interesting mathematician. Yes, he contributes to the perception that mathematical talent develops early: (Born in Geneva in 1704, he got his doctorate with 18). But maybe his biggest achievement, the Cramer rule, he got only shortly before his death. This was 1750. (Extrapolating to today, that would correspond to a mathematician today getting his or her best results in the 70ies or 80ies). An other interesting historical conundrum is the fact that long before matrices were introduced, mathematicians worked with determinants. The notion of matrix (the mother of determinants) came only 100 years later. Some sources about that. This question actually had motivated me to look closer at the Cauchy-Binet formula. The result obtained here which is related to the work of Cramer.]

Before we come to the punch line, just a side remark about the use of determinants:

[By the way, there is an infamous article “Down with Determinants” by Sheldon Axler, which in my opinion is best advertisement for using determinants because it shows how difficult linear algebra becomes when determinants are avoided. There is no better way to see this when looking at the printed book, where the reader quickly gets convinced that it does not work well. But it is a way to get a mathematician trained to look into alternative approaches and to learn about originality (the book itself is highly original and therefore valuable). It is like climbing the Matterhorn north face rather than taking the standard route which can be climbed by pretty everybody. Axler asks the student to climb the north face. They should get a lot out of it (strength, overcoming fear, knowledge), if they don’t fall …]

Now, in the case of the adjacency matrix A of a graph, if we look at cell x and delete it, then this produces a complex for which the adjacency matrix is B(x). The quotient B(x)/A is now just

Corollary (Green function Value): If G is a finite abstract simplicial complex and G’ its connection graph. The Green function value (1+A’)xx is equal to i(x)=1-\chi(S(x)), where S(x) is the unit sphere in the Barycentric refinement graph G1 of G.

That is the reason why we are now interested in the Sphere spectrum which is the set of in a graph. Is the sphere spectrum a combinatorial invariant? No! There are various things which can happen. First of all, even if a graph has no Euclidean structure at first (with Euclidean structure meaning that there are parts where unit spheres are discrete Evako spheres) they will develop after refinement. The next picture shows what happens with the triangle. There are initially no spheres with index -1 or 1 as in a simplex, there is no interior point, every point is a boundary point for which the unit sphere is contractible. But this changes after a refinement:

This is not a big issue. We can either ignore the sphere spectrum part which belongs to values 0,1,-1 as this comes from refinements anyway.

But it is still not ok: an other thing which can happen and this is where the 3-fold blades of the Block island wind farm come in. It can happen that we have initially only indices 0,1,-1 (which is a Morse situation as for a Morse function, the Poincare-Hopf indices are always 0 or 1 or -1), but then after a refinement the singularity starts to show. In other words, only after refinement, we see indices different from -1,0,1 popping up. Here is the picture:

We believe however currently that this is no more possible after having done one refinement. [January 9. We are now sure.]
The only thing which remains to be shown in order to prove the conjecture that the Green function values are a combinatorial invariant is that the Sphere spectrum of a graph is a combinatorial invariant, provided that the graph is already a Barycentric refinement. Indeed, we think something a bit stronger is true (we also computed the Betti number spectrum for example by computing the cohomology groups) and it is very reasonable that:

Conjecture (January 5, 2017, still work in progress): Given a graph G which is the Barycentric refinement of a simplicial complex or cell complex. The unit sphere topologies are the same than the sphere topologies of the Barycentric refinement. Especially, the set of Poincaré-Hopf indices of G (=the sphere spectrum of G) is a combinatorial invariant and stays the same under Barycentric refinement.

This means that it can not happen that some topology which is present as a unit sphere starts to disappear in the next refinement step, nor is it possible that a new type of unit sphere starts to develop.

The statement is quite easy to see for a one-dimensional graph (a graph equipped with the 1-dimensional skeleton complex, this is the situation which most graph theorists in the 20’th century have considered when talking about graphs). The reason is that in that case the degree spectrum the set of vertex degrees encodes completely the sphere spectrum. The reason is that for one-dimensional graphs, the unit spheres are by definition zero dimensional in which case the Euler characteristic of the unit sphere is equal to the degree of the vertex. Now, under Barycentric refinements, we get topologically homeomorphic (again in the classical sense of 20th century topological graph theory) graphs as we just make edge refinements. But in this one-dimensional situation, where graphs are “curves”, also unimodularity is quite simple. In that case also, it is clear what the Green function values are as they describe what happens if we take away a vertex or edge, then we either take a way loops or then add connected components.

So, here is the quest I work on now:

(*) Analyze the sphere spectrum of graphs in general and show that for already Barycentric refined graphs, the sphere spectrum stays invariant when doing further refinements.

For the vertex corresponding to a maximal simplex the story is clear as then the unit sphere is a discrete sphere which therefore has index 1 or -1. For geometric graphs also, graphs for which all unit spheres are discrete spheres, the story is simple. In that case, all new unit spheres are again discrete spheres of the same dimension. For geometric graphs (the discrete analogue of manifolds), the sphere spectrum is quite boring. It is either constant 1 or constant -1. Also for geometric graphs with boundary (the discrete analogue of manifolds with boundary), the sphere spectrum is quite boring. It is 1 or -1 in the interior (again depending on the dimension of the discrete manifold) or then 0, which happens at the boundary where the unit spheres are contractible so that the index is 0. As in the windmill graph seen above, we see that only the case of singularities is really interesting and the quest is to show that the unit spheres at such singularities do not change topology after a refinement. Its not difficult to show for new vertices which originally had been vertices. But the Windfarm graph showed that it had been the edge which produced a new sphere type.
I currently have a hunch that only the original vertex sphere remains interesting and that all other spheres close by will be trivial Euclidean type spheres or half spheres. A first step is to show (SG(x))1 = SG1(x) which translates that the Barycentric refinement of the unit sphere is the unit sphere of the vertex in the refined graph. Such a formula of course would prove (*) as Euler characteristic is trivially invariant under Barycentric refinement (we have shown it last year in a fancy way as a corollary of Kuenneth as the Barycentric refinement is nothing else than G x K1 so that we have isomorphic cohomology groups. It is one of the simplest cases of the de Rham theorem, which deals with the cohomology of a product graph. (See the corresponding article).

[January 9 update: the proof is now clear: the spheres S(x) in a Barycentric refinement are always the join of two graphs S(x) and S+(x), where the former is the set of simplices which are contained in x and the later is the set of simplices which contain x. Now we can show that the join operation commutes with taking Barycentric refinements (up to topological equivalence). In the simplest case, when x is a simplex which had been a vertex in the original graph,l then the Barycentric refinement of that sphere is the same than the sphere in that Barycentric refinement (in that case S(x) is the (-1)-dimensional sphere (the empty graph, which is the zero element in the monoid on sphere graphs defined by the join operation). There are however lots of more spheres. The sphere around an original edge for example is the join of a zero dimensional sphere S and an other graph S+(x). It is a general fact that the graph S(x) is always a discrete sphere because the ball S(x) +x is a simplex. [We have used this exactly one year ago (in this article) to give a Poincaré-Hopf proof of the fact that Euler characteristic of the Barycentric refinement is the Euler characteristic of the original graph: the definition \chi(G_1) = \sum_x \omega(x) of the Euler characteristic of G is just the Poincaré-Hopf formula in G1 because \omega(x) = (-1)^{{\rm dim}(x)} is the index with respect to the function dim on the vertex set of G1, (which is locally injective and so a coloring on the Barycentric refinement graph G1). ] In any case, to show that the unit spheres of the Barycentric refinement G2 have always the same topologies than the unit spheres of G1 we can use induction in the dimension. The proof also clearly shows why things can fail if we have a general graph G, which is not a Barycentric refinement. The unit spheres S(x) in G1, where x has not been a vertex in G do factor in the Barycentric refinement G1 but this factorization can not be related to a factorization in G.