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Suppose two toric varieties are isomorphic as abstract varieties. Does it follow that there exists a toric isomorphism between them?


Edit: the comments below lead me to believe that I'm using the terms "toric variety" and "toric morphism" in a non-standard way, so let me clarify. Here are the definitions I have in mind.

Definition: A toric variety is a normal variety $X$ together with (1) a dense open subvariety $T\subseteq X$, and (2) a group structure on $T$ making it a torus, such that the action of $T$ on itself extends to an action on $X$. (Note: I do not include the data of an isomorphism between $T$ and $\mathbb G_m^{\dim X}$, but only require that such an isomorphism exists.)

Definition: If $(X,T_X)$ and $(Y,T_Y)$ are toric varieties (group structures on $T_X$ and $T_Y$ implicit), a toric morphism between them is a morphism $f:X\to Y$ which restricts to a group homomorphism $f|_{T_X}:T_X\to T_Y$.

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I think there is a trivial class of counterexamples, take a given action and compose it with an automorphism of the torus. This gives a new toric variety in general different, as what is usually meant by "toric variety", from the original. Taking that into account Dave's remark below gives indeed an equivalent formulation. –  Torsten Ekedahl Jun 23 '11 at 5:23
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Thanks for point this out, Torsten -- for a very easy example, take a fan and its negative! (Corresponding to $t \mapsto t^{-1}$.) So until you factor out automorphisms of $T$, even the affine line has multiple toric structures. –  Dave Anderson Jun 23 '11 at 5:37
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I don't follow. The question does not ask if every isomorphism is toric (of course not), just if a toric isomorphism exists. For example, you could translate the usual toric structure on $\mathbb A^1$ so that the origin is in the big torus orbit and 1 is torus-invariant. This would be a different toric structure on $\mathbb A^1$, and $id_{\mathbb A^1}$ is a non-toric isomorphism. But there also exists a toric isomorphism: the translation we used to define the toric structure. –  Anton Geraschenko Jun 23 '11 at 13:55
    
Anton, here is another way of thinking of it for $\mathbb{A}^1$: the two structures given by the standard action and by composing with $t\mapsto t^{-1}$ are two different equivariant line bundles on a point. (Their Chern classes are different, so there can be no equivariant isomorphism between them.) –  Dave Anderson Jun 23 '11 at 15:51
    
@Dave: I still don't follow. It sounds to me like you're saying that there is no way to extend the automorphism of the torus $t\mapsto t^{-1}$ to an automorphism of $\mathbb A^1$. I agree with that, but I don't see how it is relevant. For any two toric structures you put on $\mathbb A^1$, there is an automorphism of $\mathbb A^1$ sending one to the other. A toric structure does not include a choice of isomorphism between the torus and $\mathbb G_m^n$. –  Anton Geraschenko Jun 23 '11 at 16:21
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3 Answers

This is a partial answer. Let $X$ be the given abstract variety. I think the question is equivalent to asking whether all maximal tori in the group $\mathrm{Aut}(X)$ are conjugate. When $X$ is complete, this is a linear algebraic group, so all maximal tori are conjugate, and the answer is affirmative. (See Cox's famous paper on the homogeneous coordinate ring.)

If $X$ is not complete, the automorphism group may of course be infinite-dimensional, but perhaps you can argue by compactifying.

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Can you explain why the question is equivalent to showing that maximal tori of $Aut(X)$ are conjugate? If $T\subseteq X$ is a torus whose action on itself extends to an action on $X$, then we get $T\subseteq Aut(X)$, but is it necessarily a maximal torus? –  Anton Geraschenko Jun 23 '11 at 2:49
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Let $T$ be the torus coming from the toric variety, and let $T \subseteq T'$ be tori in the automorphism group. Then $T'$ is abelian and contains $T$. So the $T'$ action commutes with the $T$ action. But it is standard (meaning that I can't remember a reference) that the group of automorphisms of a toric variety commuting with the torus action is precisely the torus action. –  David Speyer Jun 23 '11 at 3:34
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Thanks David. Such an automorphism must respect the open subset $T\subseteq X$, so after adjusting it by the action of $T$, we may assume it sends the identity element to itself. Now since it commutes with the action of $T$, it must fix all of $T$, so it must fix all of $X$ (since $X$ is separated). –  Anton Geraschenko Jun 23 '11 at 4:11
    
David and Anton, that seems like the best proof of this fact. (I don't know anything about conjugacy of tori in wild beasts like $Aut(X)$ in general though -- do you?) –  Dave Anderson Jun 23 '11 at 6:01
    
Is there any hope in general with this line of though? $Aut(\mathbb A^n)$ is pretty wild... (literally :) ) –  Mariano Suárez-Alvarez Jun 23 '11 at 19:44
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I'm pretty sure the answer is yes, and that the proof will use Cox rings. However, I haven't been able to find references for all the facts about Cox rings I want, so I'll leave this as CW and hope someone else fills in the gaps.

Let $\Sigma$ and $\Sigma'$ be two fans, and $X$ and $X'$ the corresponding toric varieties, with $X \cong X'$. For simplicity, I'll assume that the unit groups of $X$ and $X'$ are both trivial (i.e. equal to $k^*$). This is equivalent to assuming that $\Sigma$ (respectively $\Sigma'$) is not contained in a hyperplane.

Let $H$ be the class group of $X$ (Weil divisors modulo principal divisors). Let $G$ be the character group $\mathrm{Hom}(H, \mathbb{G}_m)$. Cox's construction gives: A variety $Y$; an action of $G$ on $Y$; a $G$-invariant open subspace $U$; and a map $U \to X$ which makes $X$ into the categorical quotient $U/G$. This is sometimes called the universal torsor of $X$, although it is only a torsor when $X$ is smooth (earlier false statement corrected).

To help get you oriented, if $X = \mathbb{P}^2$, then $H = \mathbb{Z}$, $G = \mathbb{G}_m$, $Y= \mathbb{A}^3$ and $U$ is the complement of the origin in $Y$.


There are two ways to think about this construction. The one I understand well is the one from Cox's original paper. Let $\rho_1$, ..., $\rho_N$ be the rays of $\Sigma$. Let $Y = \mathbb{A}^N$. Each ray represents a class in $H$, and hence a character of $G$. Let $G$ acts on the $i$-th coordinate of $Y$ by the character of $G$ corresponding to $\rho_i$. Let $(y_1, \ldots, y_N)$ be a point of $Y$. Let $i_1$, \ldots, $i_k$ be the coordiantes for which $y_{i_j} =0$. Then $(y_1, \ldots, y_N)$ is in $U$ if and only if there is a cone of $\Sigma$ containing the rays $\rho_{i_1}$, ..., $\rho_{i_k}$. I'll cite you to Cox's paper for the map $U \to X$.


That contsruction is very explicit, but it uses the fan. There is a more abstract construction, which I don't understand as well, which works for any normal variety and does not use the torus action. Specifically, for every $h \in H$, choose a specific Weil divisor $D(h)$ representing $h$. For any $h_1$ and $h_2$, choose a rational function $u(h_1, h_2)$ with divisor $D(h_1+h_2) - D(h_1) - D(h_2)$, subject to conditions to be named later.

Define the ring $$R := \bigoplus_{h \in H} H^0(X, \mathcal{O}(D(h)))$$ with multiplication $H^0(X, \mathcal{O}(D(h_1))) \times H^0(X, \mathcal{O}(D(h_2))) \to H^0(X, \mathcal{O}(D(h_1 + h_2)))$ given by $f_1 \times f_2 = f_1 f_2 u(h_1, h_2)$. My understanding is that (1) it is always possible to choose the $u$'s such that this is a commutative and associative ring and (2) $R$ is independent of these choices, up to nonunique isomorphism. However, you should check these claims before using them.

We set $Y = \mathrm{Spec} \ R$. This comes with an obvious action of $G$, where $G$ acts on $H^0(X, \mathcal{O}(D(h)))$ by the character $h$. I am not sure how to define $U$ or the map $U \to X$, but I think there is a way. Note that none of this uses the torus action.


Now, let $(\Sigma, X)$ and $(\Sigma', X')$ be as above, and let $(Y, G, U, U \mapsto X)$ and $(Y', G', U', U' \mapsto X')$ be the results of the above constructions. If I am right that the Cox construction is sufficiently natural, then there should be isomorphisms $Y \cong Y'$ and $U \cong U'$, making the obvious diagrams commute. (Note that the construction of $G$ is intrinsic, so we definitely have $G \cong G'$.)

From the first description of the Cox ring, we see that $Y \cong Y' \cong \mathbb{A}^N$, so that $\Sigma$ and $\Sigma'$ both have $N$ rays.

A priori, we have $Y \cong Y'$ as $G$-varieties, and we know that each of $Y$ and $Y'$ can be identified with $\mathbb{A}^N$ so that the $G$-action becomes linear. My next claim is that the two representations of $G$ are isomorphic; i.e., we can arrange that the composition $\mathbb{A}^n \cong Y \cong Y' \cong \mathbb{A}^N$ is linear. Proof: Find a point $y$ in $Y$, fixed by the $G$-action. There is at least one such point, because the origin is fixed. Let $\phi: Y \to Y'$ be any $G$-equivariant isomorphism. Then the $G$-action on $Y$ is isomorphic to the $G$-action on the tangent space $T_y Y$, and the $G$-action on $Y'$ is isomorphic to the $G$-action on $T_{\phi(y)} Y'$. But the differential $D \phi : T_y Y \to T_{\phi(y)} Y'$ gives a linear isomorphism between the tangent spaces. Transporting structure, we have the claim.

So, $G$ acts on $Y$ and on $Y'$ by the same characters. This means that $\Sigma$ and $\Sigma'$ have the same rays. Now, looking at the combinatorics of $Y \setminus U$ and $Y' \setminus U'$, you should be able to read off the corresponding fans.

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I don't know if you are still interested in a reference for this, but I recently found that this is proved in section 4 of "Lifting of morphisms to quotient presentations" by F. Berchtold. The proof uses the Cox construction and it seems to me that the argument is the one that David suggested above, but I am not done reading the paper. I hope this helps.

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