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One way to define toric varieties is as quotients of affine $n-$space by the action of some torus. However, this is not strictly true as we need to throw away "bad points" which ruin this construction.

For example consider the construction of projective space as a toric variety. Let $\mathbb{G}_m$ act on $\mathbb{A}^n$ in the obvious way. Then the quotient of $\mathbb{A}^n$ by this action is a single point, as "everything is rescaled to the origin". More rigously the only functions invariant under this action are the constants, thus the quotient is the spectrum of the ground field. To fix this we of course we remove the origin and then take the quotient and we get projective space as required.

So given an action of some torus on affine space, how do we know which points to remove before we take the quotient to make sure we get a toric variety?

My first naive guess is to remove the points which are fixed under the action, but Im wary it may be more subtle than that, as I know GIT can get quite technical.

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Thanks for this question! I have been confused by this in the past as well. –  Kevin H. Lin Jul 7 '10 at 18:33
Thanks for all the answers! Most useful. –  Daniel Loughran Jul 8 '10 at 17:55
It's important to note that this Cox quotient need not be a GIT quotient. Indeed, all toric varieties may be thus obtained, including proper non-projective ones. But a GIT quotient is always projective (or at least quasi-projective). –  Michael Thaddeus Jul 12 '10 at 19:32

3 Answers 3

up vote 5 down vote accepted

You want to read Section 2 of The homogenous ring of a toric variety, by David Cox. Nick Proudfoot has written an expository note on the projective case, which you might find helpful as well.

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Another survey by Antonio Laface and Mauricio Velasco which may be helpful: math.berkeley.edu/~velasco/Survey.pdf –  Steven Sam Jul 7 '10 at 21:31
Or look it up in Miller and Sturmfels, "Combinatorial commutative algebra". –  Victor Protsak Jul 7 '10 at 21:54

Let me advertise the polytopal point of view.

Start with $T^k$ acting on ${\mathbb A}^n$ linearly, i.e. we have a map $T^k \to T^n$. The $T^n$-moment polytope of ${\mathbb A}^n$ is just the orthant ${\mathbb R}_+^n$. The $T^k$-moment polytope is the projection of that orthant to ${\mathbb R}^k = Lie(T^k)$* under the transpose $Lie(T^n){}^* \to Lie(T^k){}^*$ of the Lie algebra map $Lie(T^k) \to Lie(T^n)$.

To do GIT, you need to pick a point $\mu$ (technically, an integral point, but it doesn't much matter) in $Lie(T^k)^*$. The fiber over $\mu$ will be some (not necessarily compact) polyhedron in the orthant. Not every face of the original orthant will meet that fiber: the ones that don't are the images of the unstable set for the GIT quotient. (Well, this is only true if the point chosen was generic, i.e., if $\mu$ is in the interior of each face whose image contains $\mu$.) The polyhedron can be thought of as living in $Lie(T^n/T^k)^*$, and is the moment polytope for the GIT quotient, itself a toric variety.

Here's an example. Let $T^1$ act on ${\mathbb A}^2$ by $t\cdot (x,y) = (tx, t^{-1} y)$. Then the corresponding projection can be pictured by turning the first quadrant $45^\circ$ degrees counterclockwise, and dropping it onto the $x$-axis. The genericity condition says that we shouldn't take $\mu=0$. Then depending on the sign of $\mu$, GIT says we must leave out either all $\{ (x,0) \}$ or all $\{ (0,y) \}$ in ${\mathbb A}^2$. Quotienting either open set obviously gives ${\mathbb A}^1$, whose polytope is the half-line fiber over $\mu$.

Note that your naive guess would have left out just the origin, and the quotient would have had two origins. Some entire axis must be left out, or some further identification (not just quotienting by the group) must be made.

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There is a notion of a GIT quotient with respect to a linearization of the action. I am sure you can find the definition in the Mumford's book. Basically, a linerization of the action is a choice of a line bundle and an extension of the action to this line bundle. Given these you can consider the graded algebra of invariant sections of the powers of this line bundle, $R = \oplus_{k=0}^\infty \Gamma(X,L^k)^G$, and take its projective spectrum. In the case of the torus action on the affine space, the only line bundle is the trivial one, but you can extend the action by any character of the torus. So, basically, there are three possibilities --- either you choose a negative character, or a positive, or zero, In the first case $R_k = K[z_1,\dots,z_n]_{\chi^k} = 0$ for $k > 0$ ($K$ is the base field), so $R = R_0 = K$ and $Proj R = \emptyset$. In the second case $R$ is just the polynomial algebra (considered as a graded ring), or its Veronese subalgebra (depending whether the character is primitive or not), so $Proj R$ is the projective space. In the last case, $R_k = K$ for all $k$, so $Proj R = Spec K$ is a point which coincides with the usual quotient (without linearization).

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