# Vertices of a polytope as algebra generators

I am wondering if the following kind of objects has some name, or are there any studied examples. I apologize for perhaps too specific definition, this is an adoptation of a situation that arises in physics.

Consider a $n$-polytope $P$ with a distinguished vertex $v_0$. Suppose there exist a map $f:\,V(P)\to A$, where $A$ is a unital commutative algebra over $\mathbb{C}$ and $V(P)$ is the set of vertices of $P$. Moreover, assume that $f(v_0)=1$ and that the image of $f$ forms a basis for $A$ (and in particular is a linearly-independent set).

Edit (in the older version I made a stupid mistake and disregarded a very important grading):

Now we employ the structure of the polytope. We put $v_0$ at the origin of $\mathbb{R}^n$ and introduce the 'grading' on the basis elements that forces the product of two vertices as elements of $A$ to be proportional to the vertex given by their sum as of elements of $\mathbb{R}^n$. And the product is zero if there is no such vertex.

As an example, consider the square with vertices labeled clockwise as 1,a,ab,b and a map to $\mathbb{C}[a,b]/(a^2,b^2)$ given by the names of the vertices.

• The algebra may not be associative, is that OK with you? The trouble is that you can have $a$,$b$,$a+b$,$c$,$a+b+c$ vertices, but $b+c$ not be a vertex. – Lev Borisov Jun 21 '14 at 15:41
• @LevBorisov, thanks for your comment. I do not know yet if the product is associative. In any case, it is not required that the product should be non-zero, there is just a selection rule that forces it to be zero in some cases (by proportional I meant that the coefficient can still be zero). – Peter Kravchuk Jun 21 '14 at 22:56

In your formulation, do you assume that if there is an integral affine relation among the vertices, say $\sum n(i)v_i = \sum m(j)w_j$, where $n(i), m(j)$ are nonnegative integers, and $v_i, w_j$ are vertices, and $\sum n(i) = \sum m(j)$, then $\prod f(v_i)^{n(i)} = \prod f(w_j)^{m(j)}$?
If so, there is a construction, although it may not be what you are looking for. Let $F$ be a polynomial in $k$ variables with nonnegative coefficients, say $F = \sum r_{\alpha} x^{\alpha}$ (using monomial notation; $\alpha \in Z^k$ and $r_{\alpha} > 0$). Form the subring of the function field in $\left\{x_i\right\}$ generated by $\left\{x^{\alpha}/F\right\}_{r_{\alpha} \neq 0}$ together with whatever subring of the reals you want that contains all the $r_{\alpha}$. This does not satisfy your condition of zero multiplication, etc.
• Thank you for the answer! In your construction, how are $r_\alpha$ related to the polytope? – Peter Kravchuk Jun 21 '14 at 23:03
• [By the way, I interpreted vertices as lattice points; you may possibly have meant vertex to mean extreme point; it doesn't change much here.] The polytope corresponding to $F$ is its Newton polytope. Varying the nonzero coefficients, the $r_{\alpha}$, changes the (ordered) ring so constructed, but this can be avoided by taking the direct limit over all choices for $F$ with the same Newton polytope, or localizing at the strictly positive elements (as is done in the cited references). This creates a single (ordered) ring attached to the polytope. It contains a lot of information about it. – David Handelman Jun 21 '14 at 23:22