This is, probably, a question for those knowledgeable on the subject of Brion's theorem and its applications.

Lately, I've been dealing with situations of the following sort. Suppose we are given a polytope $P\subset\mathbb{R}^n$ as well as a linear map $\varphi:\mathbb{R}^n\rightarrow\mathbb{R}^m$, $m<n$. Bases are fixed in both spaces and $\varphi$ maps integral points to integral points (in other words, has an integral matrix with respect to these bases). Also $P$ is rational and thus subject to Brion's theorem.

Now, $\varphi$ can be viewed as a change of variables $F$: $$F:x_i\rightarrow\prod\limits_{j=1}^m y_j^{\varphi_{ji}}, 1\le i\le n.$$ Here $x_i$ are the exponents of basis vectors in $\mathbb{R}^n$ (the variables in Brion's formula), $y_i$ -- the exponents of basis vectors in $\mathbb{R}^n$. Thus, here $\exp(\bar v)$ is simply substituted by $\exp(\varphi(\bar v))$.

In my case $F$ is applicable to the identity provided by Brion's theorem (the denominators do not vanish identically) and I noticed the following to hold. The summands (vertex cones' integer point transforms) corresponding to vertices of $P$ not mapped to vertices of $\varphi(P)$ vanish under $F$!

I see that the naive generalization is far from being correct... But isn't there, just maybe, a known theorem stating something of the sort?

Another interesting trait of my polytopes is that the summands which don't vanish, turn into neat products identical to the integer point transfroms of certain simplicial cones, namely $F$ transforms them into $$\prod\limits_{i=1}^n \frac 1{1-\exp(\varphi(e_i))},$$ where $\{e_i\}$ is a linearly independent subset of edges of the (non-simplicial) vertex cone.

Any known reasons for such behaviour to take place?