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

Is there an example of a smooth projective hypersurface in $\mathbb{P}^n_k$ ($k=\overline{k}$) that does not contain any projective toric varieties (edit: of positive dimension)? Or is it the case that every such hypersurface will contain a projective toric variety (edit: of positive dimension)?

share|cite|improve this question
A point is an example of a toric variety. Maybe you would like to refine your question? – David Speyer Dec 29 '11 at 21:51
Yes, thanks, I edited. – Mahdi Majidi-Zolbanin Dec 29 '11 at 21:53
Let $n=2$, and let the hypersurface be an elliptic curve. Its only subvariety of positive dimension is itself, and it is not toric since all toric varieties are rational. – Alexander Woo Dec 29 '11 at 21:57
up vote 14 down vote accepted

As Alexander Woo said in a comment, toric varieties are rational. Now, it turns out that projective hypersurfaces have strong hyperbolicity-type properties. This properties have been established by several authors in the last decades.

First, in 1986 Clemens showed that if $X$ is a generic hypersurface of degree $d \ge 2$ in $\mathbb P^{n+1}$, then $X$ does not admit an irreducible family $f\colon\mathcal C\to X$ of immersed curves of genus $g$ and fixed immersion degree $\deg f$ which cover a variety of codimension less than $D = ((2 -2g)/ \deg f) + d - (n + 2)$. As an immediate consequence, one gets, for example, that there are no rational curves on generic hypersurfaces $X$ of degree $d \ge2n + 1$ in $\mathbb P^{n+1}$.

Two years later, Ein studied the Hilbert scheme of $X \subset G$, a generic complete intersection of type $(m_1,\dots,m_k)$ in the Grassmann variety $G = G(r,n+2)$. As a remarkable corollary one gets that any smooth projective subvariety of $X$ is of general type if $m_1 + m_2 +\cdots+ m_k \ge\dim X + n + 2$. It is also proved that the Hilbert scheme of $X$ is smooth at points corresponding to smooth rational curves of "low" degree.

In 1996, Voisin had the idea of regarding the hypersurfaces in family and to use the positivity property of the tangent bundle of the family itself. Her main result is the following theorem which improves Ein's result in the case of hypersurfaces:

Let $X\subset\mathbb P^{n+1}$ be a hypersurface of degree $d$. If $d\ge 2n-\ell+ 1$, $1 \le\ell\le n - 2$, then any $\ell$-dimensional subvariety $Y$ of $X$ has a desingularization $\widetilde Y$ with an effective canonical bundle. Moreover, if the inequality is strict, then the sections of $K_{\widetilde Y}$ separate generic points of $\widetilde Y$.

The bound is now sharp and, in particular, the theorem implies that generic hypersurfaces in $\mathbb P^{n+1}$ of degree $d\ge 2n$, $n\ge 3$, contain no rational curves. The method also gives an improvement of a result of Xu as well as a simplied proof of Ein's original result.

Lastly, let me cite a result by Pacienza in 2004: this paper gives the sharp bound $d\ge 2n$ for a general projective hypersurface $X$ of degree $d$ in $\mathbb P^{n+1}$ containing only subvarieties of general type, for $n\ge 6$. This result improves the aforesaid results of Voisin and Ein.

share|cite|improve this answer
This is very helpful, thank you. – Mahdi Majidi-Zolbanin Dec 30 '11 at 4:42
Just an addendum, as long as I know, what I wrote is in the case $k=\mathbb C$. I don't know if it is true in an arbitrary algebraically close field... – diverietti Dec 30 '11 at 9:45

It was shown by H. Clemens (1986) that a general hypersurface of degree $d$ in $\mathbb P^n$ does not contain any rational curves if $d$ is sufficiently large, specifically for $d \geq 2n-1$ for $n \geq 3$. Such a hypersurface can never contain a toric subvariety since any such subvariety must contain a rational curve.

share|cite|improve this answer
Thank you Parsa. – Mahdi Majidi-Zolbanin Dec 30 '11 at 4:41

If $X$ is an abelian variety then it contains no rational curves. Indeed, $\Omega_X$ is generated by global sections, hence if $f:P^1 \to X$ is a map then the image of the morphism $f^*\Omega_X \to \Omega_{P^1}$ is generated by global sections as well. But $\Omega_{P^1}$ has no global sections at all, so the image is $0$, which means that the map $f$ is constant.

share|cite|improve this answer
Even easier (and a little bit more general), if $X=\mathbb C^n/\Lambda$ is a complex torus, then $X$ does not admit any non-constant map from $\mathbb P^1$. In fact, any such map would lift to a proper non-constant map $\mathbb P1\to\mathbb C^n$. In particular, its image would be a positive dimensional compact analytic subset of $\mathbb C^n$, impossible. – diverietti Dec 31 '11 at 2:35
This may be a silly question, but what is an example of an abelian variety of dimension $n$ that can be embedded as a hypersurface in some $\mathbb{P}^{n+1}$ for $n≥4$? – Mahdi Majidi-Zolbanin Jan 2 '12 at 0:54
Sorry, I didn't notice the hypersurface condition. Of course you cannot get an abelian variety as a hypersutface in $P^{n+1}$ unless $n = 1$, this follows immediately from Lefschetz Theorem. – Sasha Jan 2 '12 at 11:24

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.