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Say I have two normal lattice polytopes $P$ and $Q$ in $\mathbb{R}^n$ (lattice $\mathbb{Z}^n$) with the same number of lattice points $N+1$. Then they define two toric varieties $X_P$ and $X_Q$ which are embedded via $P$, resp. $Q$, into a projective space $\mathbb{P}^N$.

I'm pretty sure that it is true that $X_P$ and $X_Q$ are projectively equivalent if and only if $P$ and $Q$ are affinely isomorhpic as lattice polytopes (so you get one from the other by a composition of a translation and a lattice automorphism in ${\rm GL}(n,\mathbb{Z})$).

I couldn't find a reference for that statement. Can anyone point me in the right direction? Or does anyone know a short and concise proof?

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First of all, the assertion is not quite true if $X_P$ and $X_Q$ are not embeddings of the torus $T=\mathbf G_m^n$. Let, e.g., $n=3$, $P=conv(0,e_1,e_2,e_3)$, $Q=conv(0,e_1,e_2,e_1+e_2+2e_3)$. Then $N+1=4$ and $X_P=X_Q=\mathbf P^3$.

So assume that $X_P$ and $X_Q$ are embeddings. This means that the lattice $\mathbb Z^n$ is generated by the differences of lattice points. . Then the corresponding homomorphisms $p,q:T\to PGL(N+1)$ are injective. Let $G=\{g\in PGL(N+1)\mid g(X_P)=X_P\}$. Then $G\to Aut(X_P)$ is injective since $X_P$ spans $\mathbb P^N$. Thus $p(T)$ is a maximal torus of $G$. Indeed $T$ is already a maximal torus in the entire Cremona group (see Demazure's paper).

Now assume that $X_P$ and $X_Q$ are projectively equivalent. Then there is $a\in PGL(N+1)$ with $a(X_Q)=X_P$. Then, we get two injective homomorphisms $T\to PGL(N+1)$ namely $p$ and $\bar q:=aqa^{-1}$. Since both $p(T)$ and $\bar q(T)$ are maximal tori of $G$, there is $b\in G$ with $p(T)=b\bar q(T)b^{-1}=cq(T)c^{-1}$ with $c=ba$. It follows that there is an automorphism $\alpha$ of $T$ with $p(\alpha(t))=cq(t)c^{-1}$. Now replace $P$ by $\alpha(P)$. Then we may assume that $\alpha=id_T$. Then $p$ and $q$ are conjugated by an element $c\in PGL(N+1)$.

Finally observe that the projective representation $p:T\to PGL(N+1)$ uniquely determines the weights of $T$ on $\mathbb C^{N+1}$ up to a translation. This set of weights is nothing else than the lattice points in $P$. So the lattice points of $P$ and $Q$, and therefore $P$ and $Q$ themselves, differ by a translation.

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  • $\begingroup$ Thank you for your answer. I will go through the representation theory carefully. I swept the fact that the torus is actually embedded in the toric variety under the rug. If the polytope is normal, that is the case. The polytope Q in your example is not a normal lattice polytope. $\endgroup$ – Rainer Sinn Jul 28 '16 at 7:55

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