# Monotone vs. Fano

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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### 5 responses to “Monotone vs. Fano”

1. Going up a dimension, $\mathbb{CP}^2 \# k \overline{\mathbb{CP}^2}$ for $k \geq$ 8, 9 with an appropriate symplectic form should be symplectic Fano and monotone but not Fano (i.e. Del Pezzo).

2. natebottman

I guess you mean $k \geq 9$. I agree this isn’t Fano, but are you sure there’s a compatible $\omega$ so that this is monotone? It’s projective, so that would contradict what I wrote.

3. I intended to phrase the above comment as a question, not a statement. Monotonicity is a very rigid condition and I haven’t thought carefully about whether you can construct a monotone $\omega$. I was just extrapolating and assuming that each (symplectic) blowup can be done in a way that preserves monotonicity. Looking back, the induction might fail because there’s no reason why you should be able to fit in all those -1 spheres all of the same size.

I noticed that this might contradict your statement about projective Kahler manifolds, but if such a monotone $\omega$ existed, would a compatible almost-complex structure necessarily be (homotopic to) the obvious one?

4. natebottman

Well, if $i$ is the usual complex structure on $X := \text{Bl}_9(\mathbb{P}^2)$, and there is a monotone $i$-compatible $\omega$, then $(X, i)$ would be Fano on the nose. On the other hand, for all I know there could be another complex structure $J$ on $X$ that is compatible with a monotone $\omega$, but I don’t know why $(X, J)$ would not be Fano.

So it’s indeed the case that given a monotone $(X, \omega)$ and $p \in X$, it isn’t always possible to choose $\epsilon > 0$ so that the $\epsilon$-blowup of $X$ at $p$ is monotone. The polytope criterion for monotonicity of toric manifolds (I think due originally to Entov and Polterovich; there’s a proof in Dusa’s first “probes” paper) gives us reason to not expect monotonicity to be preservable under blowups of points.

5. @natebottman: Nice post!

@PLC: In answer to your question, $\mathbf{CP}^2$ blown up nine times doesn’t admit a monotone symplectic form. Indeed the first Chern class has square zero, so if the symplectic class were positively proportional to the first Chern class it would no longer be symplectic.

Actually something much stronger is true: it follows from combining the work of Seiberg-Witten/Taubes/Li-Liu that all monotone symplectic 4-manifolds are symplectic deformation equivalent to (in particular diffeomorphic to) Del Pezzo surfaces (I think this is explained in McDuff-Salamon).

Also, many (or at least some!) people use “symplectic Fano” synonymously with “monotone”. I’d be suspicious of any definition which calls a complex torus “Fano”…