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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.

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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

1 Answer 1

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.

This was developed in D Handelman [me], Positive Polynomials, Convex Integral Polytopes, and a random walk problem, 1987, SLN, in order to study random walks on lattices, among other things. Some of the quotients by order ideals (which are also ideals) may be closer to what you are looking for.

More work is done in D Handelman, Effectiveness of an affine invariant for indecomposable integral polytopes, JPAA 66 (1990) 165--184, and there is further work later on.

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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
Oh, I see. Thanks, this looks very helpful. –  Peter Kravchuk Jun 22 '14 at 4:45

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