MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is just as in the box. Is every smooth projective toric variety diffeomorphic to a quotient of $\prod_i S^{n_i} \times T^k$ (I know torus is a one-sphere but I just wanted to make clear I allow this) by a free torus action? If not which ones can be realized like this? Maybe it can be proven using the Geometric Invariant theory construction. Namely, when is the fiber of the corresponding torus action on $\mathbb{C}^l$ diffeomorphic to the one of the above spaces?

share|cite|improve this question
up vote 5 down vote accepted

Let us consider the case of toric varieties of real dimension $4$ and prove they can not be represented as such a quotient unless they have second Betti number $1$ or $2$.


Let us introduce some notations. Let $B$ be the toric manifold of real dimension $4$, $n=b_2(B)$. Denote by $E$ the product $\Pi_i S^{n_i}\times T^k$. Let $k_2=b_2(E)$, $k_3=b_3(E)$. Finally, denote by $m$ the dimension of the torus that is acting on $E$.

Suppose that $T^m$ is acting freely on $E$ and $B=E/T^m$. Then we have the following obvious relation on dimensions:

$$4=dim(E)-m\ge k+2k_2+3k_3-m$$

We will explain now that we must have $n\le 2$ (recall $n=b_2(B)$). For this purpose we will consider the long exact sequence of homotopy groups for the fibration $E\to B$.

$$0\to \pi_3(E)\to \pi_3(B)\to 0 \to \pi_2(E)\to \pi_2(B)\to \mathbb Z^m\to \mathbb Z^k\to 0$$

Since $\pi_2(B)=\mathbb Z^n$, from the second half of the sequence we get


substituting this in the inequality on dimensions we get

$$4\ge 3k_2+3k_3-n$$

To prove finally that $n\le 2$ we use the classical statement that $\pi_3(B)$ contains sub-group $\mathbb Z^{(n^2+n)/2-1}$. It follows that

$$rk (\pi_3(E))=k_3+k_2\ge (n^2+n)/2-1$$


$$4\ge 3((n^2+n)/2-1)-n$$

share|cite|improve this answer
Dmitri, thank you very much for this answer. To make it clear, you are saying the converse is true for surfaces because P^1 \times P^1 and P^2 obviously have such a representation but also P^2 blown up at a point? – Eleanor Von Hohlandsbourg Jan 7 '13 at 5:45
Dear Eleanor, I did not check yet if $\mathbb CP^2$ blown up in one point can be represented as such a quotient. I would first need to calculate its homotopy groups to see if this could be true. – Dmitri Jan 7 '13 at 6:08
@Eleanor and Dmitri: For every Hirzebruch surface $\Sigma_d = \mathbb{P}_{\mathbb{CP}^1}(\mathcal{O}\oplus \mathcal{O}(-d))$, there is a differentiable fiber bundle over $\Sigma_d$ with fibers $\mathbb{S}^1\times \mathbb{S}^1$ whose total space is $\mathbb{S}^3 \times \mathbb{S}^3$. To see this, consider the projection $\pi:\Sigma_d \to \mathbb{CP}^1$ and the Hopf fibration $p:\mathbb{S}^3\to \mathbb{CP}^1$. Since $H^2(\mathbb{S}^3,\mathbb{Z})$ is zero, the pullback of $\mathcal{O}\oplus \mathcal{O}(-d)$ by $p$ is the trivial (differentiable) complex vector bundle. contd. – Jason Starr Jan 7 '13 at 15:50
... Thus the pullback of $\pi$ by $p$ is a trivial $\mathbb{CP}^1$-bundle, $\mathbb{S}^3 \times \mathbb{CP}^1$. Thus the Hopf bundle over $\mathbb{CP}^1$ gives $\mathbb{S}^3 \times \mathbb{S}^3$. This total space is the same as the $\mathbb{S}^1\times \mathbb{S}^1$-bundle over $\Sigma_d$ coming from the "universal torsor" / Cox construction. Thus $\mathbb{S}^3\times \mathbb{S}^3$ is the total space of a $\mathbb{S}^1\times\mathbb{S}^1$-principal bundle over $\Sigma_d$. – Jason Starr Jan 7 '13 at 15:58
... and the blowing up of $\mathbb{CP}^2$ in one point equals $\Sigma_1$, so that handles the last case. – Jason Starr Jan 7 '13 at 20:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.