I have a trouble with understanding the connection between polynomials and Newton polytopes. I will try to make a short introduction to my problem. Hope you will catch. In the end I will ask questions.

I have a game with n players: $ 1,2,…,n$ .

i-th player has $ d_i$ strategies and he allocates probabilities to them which we denote by $ (p_1^{(i)},…,p_{d_i}^{(i)})$ . So, it holds $ \forall i,j : p_j^{(i)} \geq 0 \quad$ and $ \quad \forall i : p_1^{(i)} + p_2^{(i)} + \cdots + p_{d_i}^{(i)} = 1 $ .

We use $ p_{d_i}^{(i)} = 1 – \sum_{j=1}^{d_i-1} p_j^{(i)}$ .

i-th player has a payoff matrix $ X^{(i)}$ , which is n-dim of format $ d_1 \times … \times d_n$ and enteries are rational numbers.

I have following polynomials:

$ \sum_{j_1=1}^{d_1} \cdots \sum_{j_{i-1}=1}^{d_{i-1}} \sum_{j_{i+1}=1}^{d_{i+1}} \cdots \sum_{j_n=1}^{d_n} \Big( X_{j_1 … j_{i-1} k j_{i+1} j_n}^{(i)} – X_{j_1…j_{i-1}1j_{i+1}…j_n}^{(i)} \Big) \cdot p_{j_1}^{(1)} \cdots p_{j_{i-1}}^{(i-1)} p_{j_{i+1}}^{(i+1)} \cdots p_{j_n}^{(n)} $

where $ i=1,2,…n$ i $ k=2,3,…,d_i$ .

Consider the $ d_i−1$ polynomials for a fixed upper index i. They share the same Newton polytope, namely, the product of simplices $ $ \Delta^{(i)} = \Delta_{d_1 – 1} \times \cdots \Delta_{d_i-1} \times \{0\} \times \Delta_{d_i+1} \times \cdots \times \Delta_{d_n-1}. $ $ Here $ \Delta ^{(i)}$ is the convex hull of the unit vectors and the origin in $ \R^{d_i-1}$ . Hence the Newton polytope $ \Delta^{(i)}$ is a polytope of dimension $ \delta – d_i + 1$ , where $ \delta=d_1+ \cdots + d_n – n$ .

My questions are:

- What does it mean that this polynomials share Newton polytope, what does it even mean that a polynomial is supported ( or whatever is the word) by Newton polytope?
- And in this case why is $ \Delta_{d_i-1} \subset \R^{d_i-1}$ . Should it not be subset of $ \R^{d_i}$ because $ (p_1^{(i)},…,p_{d_i}^{(i)})$ is a point in that polytope.
- Why is the dimension if $ \Delta^{(i)}$ $ \delta-d_i+1$ . I thought it is: $ (d_1-1)+ \cdots + (d_{i-1}-1) + 1 + (d_{i+1}-1) + \cdots + (d_n-1) = (d_1 + \cdots + d_n) – d_i – (n-1)+1=(d_1+ \cdots + d_n – n) – d_i + 2= \delta – d_i + 2$ .

Thanks for your answers.