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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 $\varphi:\mathbb{R}^n\rightarrow\mathbb{R}^m$ with $m<n$. 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), $y_j$ -- the exponents of basis vectors in $\mathbb{R}^m$. 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?

The above is what I'm chiefly after. However, another interesting trait of my polytopes is that the summands which don't vanish, turn into neat products superficially resembling integer point transfroms of simplicial cones, namely $F$ transforms them into $$\exp(\varphi(v))\prod\limits_{i=1}^n \frac 1{1-\exp(\varphi(e_i))},$$ where $\{e_i\}$ is a subset of edges of the (non-simplicial) vertex cone at vertex $v$.

Any known reasons for such behaviour to take place? Unfortunately, I can't grasp any geometrical sense of the set $\{e_i\}$.

Update. Yes, here is a small nontrivial example. Pyramid with vertices $(0,0,0)$,$(2,0,0)$,$(0,2,0)$,$(2,2,0)$ and $(1,1,1)$, map $(a,b,c)\rightarrow(a,b)$ (projection onto the base). Then the integer point transform of the cone at the apex is $$f=\frac{(1+xy)(1-z^{-2})}{(1-xyz^{-1})(1-x^{-1}y^{-1}z^{-1})(1-xy^{-1}z^{-1})(1-x^{-1}yz^{-1})}$$ and $F$ is given by $x\rightarrow t,y\rightarrow w,z\rightarrow 1$. Thus, $f$ is seen to vanish under F.

Update 2. Actually, just like in the above example, each of the vertex cones $C$ the integer point transfrom of which vanishes, has two edges $e_1$ and $e_2$ such that $\varphi(e_1)=-\varphi(e_2)$. This forces $\varphi(C)$ to have an apex of positive dimension and sometimes gives me the feeling that the rest should be obvious via some formal argument...

Update 3. Maybe, I didn't make clear enough what $F$ is... The definition I give is a bit formal as compared to natural. The natural way might be to first introduce $F$, substituting each variable with a Laurent monomial in some other (smaller) set of variables. Then the matrix $φ$ can be defined via $F$.

And, just in case, in general, $\mathrm{IPT}(φ(C))\not=F(\mathrm{IPT}(C))$! (Integer Point Transform.)

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Could you add a small, nontrivial, example? Thanks! –  David Speyer Jul 10 '13 at 19:59
    
Hm... I wouldn't call any of the polytopes I'm working with "small". But maybe I can come up with one, let me see... –  Igor Makhlin Jul 10 '13 at 20:08
    
Eureka. But, honestly, I am really hoping for someone to see some general result they are familiar with in the situation described above. –  Igor Makhlin Jul 10 '13 at 20:51
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I actually have been interested in very similar (perhaps the same) question some years ago. I think I can give an algebro-geometric (not combinatorial) argument for why the vertices that are not mapped to the vertices of $\phi(P)$ do not appear in the formula, at least for the case of $P =$ a Gelfand-Zetlin polytope. I would be very much interested to know if you have a combinatorial argument in this case. –  Kiu Feb 17 at 18:30
    
@Kiu Wow, I knew there must be someone out there! Interestingly enough, I did come up with a combinatorial proof for the GT a few months back. –  Igor Makhlin Feb 18 at 1:04

1 Answer 1

It seems that when your map $\phi$ maps a vertex of $P$ to a non-vertex of $\phi(P)$, you're looking at a cone in the image of $\phi$ that now contains a whole line. Thus, by general Brion-type tangent cone analysis, any cone generating function containing a whole line must vanish.

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In case I have understood you correctly, this is the argument I first comforted myself with, but now believe it to be inaccurate. The reason is that, in general, $f(\varphi(C))$ is not $F(f(C))$, where $C$ is a cone, $f$ denotes integer point transform and $\varphi$ and $F$ are from the question. –  Igor Makhlin Jul 11 '13 at 23:25
    
Example, just in case. $C=\{(a\ge 0,b\ge 0)\}$ and $\varphi:(a,b)\rightarrow (a-b)$. Then $\varphi(C)$ is the line, $f(\varphi(C))=0.$ However, $f(C)=\dfrac{1}{(1-x)(1-y)}$ and $F(f(C))=\dfrac{1}{(1-t)(1-t^{-1})}\not\equiv 0$. (Since $\varphi=(1\text{ }-1)$, so $F$ substitutes $x$ by $t$ and $y$ by $t^{-1}$) –  Igor Makhlin Jul 11 '13 at 23:42
    
Maybe, I didn't make clear enough what $F$ is? The definition I give is a bit formal as compared to natural. The natural way might be to first introduce $F$, substituting each variable with a Laurent monomial in some other (smaller) set of variables. Then the matrix $\varphi$ can be defined via $F$. –  Igor Makhlin Jul 11 '13 at 23:52
    
Aha, thanks for the clarification, this makes the definition of your $F$ clearer and indeed it is different than the geometry that came to mind from simply projecting down the polytope and comparing the relevant tangent cones. –  Sinai Robins Jul 12 '13 at 0:13
    
Alright then! Also, your post is exactly what's behind the words "gives me the feeling that the rest should be obvious via some formal argument" in update 2. –  Igor Makhlin Jul 12 '13 at 1:10

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