Calculus without limits

I looked at a quadratic cohomology example. For theoretical backgroun, see the ArXiv paper “Fusion inequality for quadratic cohomology”. It is the case when U is a union of two disjoint smallest open sets in a 2-sphere for which I take the Icosahedron, one of the Platonic solids and a 2-sphere, meaning a 2-manifold which becomes contractible if one of the points are taken away. A 2-manifold is a complex for which every unit sphere is a 1-sphere. The Icosahedron is a 2-dimensional simplicial complex G a set of sets with 62 elements. 12 vertices, 30 edges and 20 triangles. With the code posted last week’s blog or on the ArXiv (download the LaTeX source), we can get G=Whitney[PolyhedronData["Icosahedron","Skeleton"]] . Now add the procedure for getting the star of an element in G. It is obtained with the procedure OpenStar[G_,x_]:=Select[G,SubsetQ[#,x] &]. In the example of the code, I took now the union of two open sets U=Union[OpenStar[G,{1}],OpenStar[G,{2}]] and the subsimplicial complex K=Complement[G,U]. This gives for the linear cohomology (simplicial cohomology) the data in the first table below. The Betti data in the first column are topological (even homotopy invariants), while the f-vector in the second column of course depends and just tells how many elements there are from each dimension. The f-vector of the icosahedron G for example is f(G)=(12,30,20), numbers which Rene Descartes already noted (in his secret notebook) to super sum to 2.

The Betti vectors are the same for all 2-spheres and situations U, where the complement K of U in G is a closed annulus, a manifold with boundary.

Case      Betti       F-vector       Euler
----------------------------------------------------
U         {0, 0, 2}   {2, 10, 10}    2
K         {1, 1, 0}   {10, 20, 10}   0
G         {1, 0, 1}   {12, 30, 20}   2
-----------------------------------------------------
Compare   {0, 1, 1}   {0, 0, 0}      0


I should maybe have taken the set U={{1,3,9},{2,4,12}} (because this is the level set of a function $f:G \rightarrow \{1,2,3\}$ and which is a union of two facets. But for linear and quadratic cohomology, the Betti vectors would have been the same. We essentially deal with a zero dimensional sphere modeled as a delta set in a 2-manifold. Open sets U are not simplicial complexes unless U={} or U=G, but still delta sets. So, here is the table which was written down on the blackboard:

Case      Betti             F-vector                   Wu
---------------------------------------------------------
U         {0, 0, 0, 0, 2}   {2, 20, 70, 100, 50}       2
K         {0, 0, 1, 1, 0}   {10, 80, 200, 180, 50}     0
UK        {0, 0, 0, 0, 0}   {0, 10, 60, 100, 50}       0
KU        {0, 0, 0, 0, 0}   {0, 10, 60, 100, 50}       0
UU        {0, 0, 0, 0, 0}   {0, 0, 0, 0, 0}            0
G         {0, 0, 1, 0, 1}   {12, 120, 390, 480, 200}   2
---------------------------------------------------------
Compare   {0, 0, 0, 1, 1}   {0, 0, 0, 0, 0}            0


If we would have gone with U={{1,3,9},{2,4,12}}, then the table would have been

Case      Betti             F-vector                   Wu
---------------------------------------------------------
U         {0, 0, 0, 0, 2}   {0, 0, 0, 0, 2}            2
K         {0, 0, 1, 1, 0}   {12, 120, 378, 432, 162}   0
UK        {0, 0, 0, 0, 0}   {0, 0, 6, 24, 18}          0
KU        {0, 0, 0, 0, 0}   {0, 0, 6, 24, 18}          0
UU        {0, 0, 0, 0, 0}   {0, 0, 0, 0, 0}            0
G         {0, 0, 1, 0, 1}   {12, 120, 390, 480, 200}   2
---------------------------------------------------------
Compare   {0, 0, 0, 1, 1}   {0, 0, 0, 0, 0}            0


In the code, I compared the situation with U=Union[OpenStar[G,{1}],OpenStar[G,{4}]], which is a situation, where U is still the union of two open sets but where K=Complement[G,U] is no more a manifold with boundary. It is still homotopic to a circle but it is an annulus in which part of the annulus has degenerated to a one-dimensional “neck”. This is topologically no more equivalent. It is no more a manfold with boundary. The boundary of the simplicial complex K is a figure 8 object and not the union of two circles. I drew both examples onto the board. The linear simplicial cohomology is the same but the quadratic cohomology now gives different Betti vectors. Here is the second table you see on the board

Case      Betti             F-vector                   Wu
---------------------------------------------------------
U         {0, 0, 0, 0, 2}   {2, 20, 70, 100, 50}       2
K         {0, 0, 2, 0, 0}   {10, 80, 204, 196, 64}     2
UK        {0, 0, 0, 2, 0}   {0, 10, 56, 84, 36}        -2
KU        {0, 0, 0, 2, 0}   {0, 10, 56, 84, 36}        -2
UU        {0, 0, 0, 0, 2}   {0, 0, 4, 16, 14}          2
G         {0, 0, 1, 0, 1}   {12, 120, 390, 480, 200}   2
---------------------------------------------------------
Compare   {0, 0, 1, 4, 3}   {0, 0, 0, 0, 0}            0


One can see come to live three additional delta sets UK,U,UU. For UU for example, we look at all pairs (x,y) of elements x,y in U x U, such that $c \cap y$ is in K. There are 34 elements of this type. In the code, one can list the elements with WuComplex[U,U,"Open"]. It contains for example the pair of edges x={4,6} and y={1,6} or the pair of triangles x={1,5,6}, y={4,5,6}. With Betti[U,U,"Open"] the Betti vector of this situation is computed. By accident, it is the same than Betti[U,U,"Closed"] which is the intrinsic quadratic cohomology of U and which in the paper I denote with b(U). In the