March | 2013 | The Electric Handle Slide

Hi everyone! Peter, Allison, and Laura have graciously allowed me to join them as an author.

I thought I would start by hashing out the relation between monotone and Fano. It is not a subtle relationship, but somehow I end up working it out for myself every six months. The idea is that they are basically the same for projective Kaehler manifolds:

Theorem. If $(X, \omega, J)$ is a projective monotone Kaehler manifold, then it is Fano. If $(X, J)$ is a Fano complex manifold, then there exists a symplectic form $\omega$ on $X$ so that $(X, \omega, J)$ is a monotone Kaehler manifold.

Recall that a symplectic manifold $(X, \omega)$ is monotone if $[\omega] = \lambda c_1(TX)$ for $\lambda > 0$ , where the first Chern class is defined using an almost-complex structure compatible with $\omega$ . A complex manifold $(X^n, J)$ is Fano if there is an immersion $\varphi: X \hookrightarrow \mathbb{CP}^N$ and a positive integer $k$ so that $(K_X^{-1})^{\otimes k} = \varphi^*\mathcal{O}(1)$ , where $K_X^{-1} = \Lambda^{n,0}TX$ is the dual of the canonical bundle.

Let’s start with the Fano-implies-monotone direction. Fix $(X^n, J)$ and $\varphi: X \hookrightarrow \mathbb{CP}^N$ as above. If $\omega_{\text{FS}}$ is the Fubini-Study form on $\mathbb{CP}^N$ , then $c_1(T\mathbb{CP}^N) = [\omega_{\text{FS}}]$ . Define $\omega_X := \varphi^*\omega_{\text{FS}}$ ; I claim that $(X, \omega_X)$ is monotone. The compatibility of $\omega_X$ with $J$ is obvious. The first Chern class of a line bundle equals the first Chern class of its top exterior power more-or-less by definition, so: $k c_1(TX) = k c_1(K_X^{-1}) = c_1((K_X^{-1})^{\otimes k}) = c_1(\varphi^*O(1)) = [\omega_X]$ . But is $\omega_X$ a symplectic form? It is closed since $\omega_{\text{FS}}$ is closed. The nondegeneracy of $\omega_X$ follows from the fact that if $Y$ is a Kaehler manifold and $Z \subset Y$ is a complex submanifold, then $\omega|_Z$ is nondegenerate. This is easy: given $v \in T_pZ$ , the Kaehler condition gives $\omega(v, Jv) > 0$ .

The monotone-implies-Fano direction is less elementary. It relies on the Nakai-Moishezon-Kleiman criterion (see Lazarsfeld’s first positivity book):

Theorem. Let $L$ be a line bundle on a projective scheme $X$ . Then $L$ is ample if and only if $\int_V c_1(L)^{\text{dim}(V)} > 0$ for every positive-dimensional irreducible subvariety $V \subset X$ .

So, let $(X, \omega, J)$ be a projective monotone Kaehler manifold. To show that $(X, J)$ is Fano it is enough to show that for every positive-dimensional complex submanifold $Y^l$ , $\int_Y c_1( K_X^{-1} ) > 0$ . But $c_1(K_X^{-1}) = c_1(TX) = \lambda[\omega_X]$ , so this integral is equal to $\lambda^l \int_Y \omega_X^l$ , which is positive.

Things are this straightforward only in the projective Kaehler setting. E.g. there are various generalizations of “Fano” to the symplectic setting. McDuff–Salamon say that an almost-Kaehler manifold $(M, \omega, J)$ is symplectic Fano if $\langle c_1(TM), A\rangle > 0$ for every nonzero $A \in H_2(M; \mathbb{Z})$ that can be represented by a $J$ -curve. Clearly monotone implies symplectic Fano, but the converse is false: e.g. there are 2-dimensional complex tori $\mathbb{C}^2/\Lambda$ that have no holomorphic curves (hence are symplectic Fano) but are not projective (hence are not Fano) and do not admit an $i$ -compatible monotone symplectic form (thanks John!).

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Monotone vs. Fano

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