User jonathan fine - MathOverflow

Answer by Jonathan Fine for Polytopes with few vertices.

2013-03-27T18:23:02Z

I've got some information about a lower bound.

We know how many binary codes there are of length 5 on n words - http://oeis.org/A034190. Each corresponds to a subset of the vertices of a cube and thus a convex polytope. For length 5 with 16 words we get 169112, and a total of 1,226,525 up to symmetry.

I've computed the flag vectors for all such polytopes in dimension 5 - http://arxiv.org/abs/1011.4269. Including polars I get 688,298 flag vectors. (I did this for a good reason.)

From this calculation I can conclude that in dimension 5 with up to 32 vertices there are at least 344,149 such polytopes that are combinatorially distinct.

What is known about the locus of zero-divisors in the group ring of a (non-abelian) finite group?

2011-11-16T21:12:48Z

Let $G$ be a finite group and $\mathbb{C}G$ its group ring. Left multiplication by $\alpha\in \mathbb{C}G$ is a linear map $\alpha:\mathbb{C}G \to \mathbb{C}G$, and so $\alpha$ has a left determinant $\det(\alpha)$. If $\alpha\beta = 0$ for some non-zero $\beta$ (i.e. $\alpha$ is a left zero-divisor in $\mathbb{C}G$) then $\det(\alpha)=0$. The converse is also true.

For $G$ abelian $\det(\alpha)$ is zero just in case $\alpha$ lies on a coordinate hyperplane associated with the character idempotents. (This is an easy consequence of this PlanetMath page.)

What is known about the locus $\det(\alpha)= 0$ in $\mathbb{C}G$ when $G$ is finite but not abelian? Are there any similar statements?

Answer by Jonathan Fine for Algorithm for finding the volume of a convex polytope

2010-02-20T21:25:07Z

Matthias Beck and Dennis Pixton used almost 17 GigaHertz years to compute the value of the 10-dimensional Birkhoff polytope. Read about it, and find the answer, here.

If you can get the same result quicker, I'm sure they'd be delighted to know how you did it.

Answer by Jonathan Fine for Computing fundamental groups and singular cohomology of projective varieties

2010-02-20T21:13:26Z

Here's an interesting special case. If X is a simple convex polytope then the Betti numbers h(X) of the associated toric variety can be computed from the face vector f(X). In fact, h(X) = Cf(X) where C is a matrix of binomial coefficients. This is closely related to Peter McMullen's shelling argument for proving the Dehn-Sommerville equations.

Here's another special case. For certain general convex polytopes the Betti numbers are not a linear function of the flag vector. This was done using explicit calculations (using Macaulay as I recall) by Mark McConnell. However, the middle perversity intersection homology (mpih) Betti numbers of the associated toric variety are a linear function of the flag vector.

The ring structure on the homology of the toric variety associated with a simple polytope is closely associated with the volume of the polytope, as it varies when the facets are moved in and out.

Finally, if the variety is defined over say the rationals then one can reduce mod p and start counting points, and then apply the Weil conjectures to determine the Betti numbers. In fact, this is a quick way to determine the Betti numbers of a smooth toric variety.

Comment by Jonathan Fine

2011-11-17T21:11:13Z

Formanek and Sibley (Proc AMS 112 (3)) and Mansfield (Proc AMS 116 (4)) are articles to the effect that the group determinant of a finite group determines the group. The first paper gives the complete factorization of the group determinant, along the lines in Angelo's comment. All useful information. My interest is more in the geometry of the zero locus of the group determinant than in the algebra of the equation defining it.