# Inequality on Kähler classes

Let $X$ be a compact Kähler manifold of complex dimension $n$, and let $\omega_1, \omega_2$ be Kähler classes on $X$. Denote the Lefschetz operator of a Kähler class $\omega$ by $\Lambda_{\omega}$. Then $$\Lambda_{\omega_1} \omega_2 + \Lambda_{\omega_2} \omega_1 \geq 2n$$ with equality if and only if $\omega_1 = \omega_2$.

Has anyone seen this inequality? I checked in the usual places like Lazarfeld's Positivity, but didn't find it. It seems simple enough that someone must have come across it before.

(Aside one: We can rewrite this as an inequality on intersection numbers if we want, since $\Lambda_{\omega_1} \omega_2 = n \int_X \omega_2 \cup \omega_1^{n-1} / \int_X \omega_1^n$.)

(Aside two: On a complex torus, everything reduces to linear algebra. There one can whittle the statement of the inequality down to that $\operatorname{tr}(H) + \operatorname{tr}(H^{-1}) \geq 2n$ for any Hermitian positive-definite matrix $H$, which follows from the arithmetric-geometric inequality.)

The proof I found is a little amusing. The Kähler cone of $X$ has a natural Riemannian metric $g$, which at a point $\omega$ is given by the inner product that $\omega$ induces on real $(1,1)$-classes. If $u,v$ are such classes, then $$g(u,v)_{\omega} = \Lambda(u) \Lambda(v) - \tfrac12 \Lambda^2(u \cup v).$$ Given two points $\omega_1, \omega_2$ in the Kähler cone, we can estimate the Riemannian distance between them by calculating the length of any path between them. Doing so for the segment of the Euclidean line that runs through the two gives the estimate $$d(\omega_1,\omega_2)^2 \leq \Lambda_{\omega_1} \omega_2 + \Lambda_{\omega_2} \omega_1 - 2n,$$ from which the inequality follows.

• very nice proof! – diverietti Feb 18 '16 at 15:16
• Thanks! The proof obviously came first and the inequality second. ;) – Gunnar Þór Magnússon Feb 18 '16 at 17:30

Here is a simple proof using Theorem 1.6.1 in Lazarsfeld book, which is the following:

Theorem (Demailly)

If $H_1,\ldots,H_n$ are Kähler classes in a compact Kähler manifold of dimension $n$, then the following inequality holds:

$$(H_1 \cdots H_n)^n \ge (H_1^n)\cdots(H_n^n).$$

Let $H_1$ and $H_2$ be Kähler classes. By the AM–GM inequality,

$$\left[\frac{1}{2}\left( \frac{{H_1^{n-1}H_2}}{H_1^{n}} + \frac{H_2^{n-1}H_1}{H_2^{n}} \right)\right]^2 \ge\frac{{H_1^{n-1}H_2}}{H_1^{n}} \cdot \frac{H_2^{n-1}H_1}{H_2^{n}}.$$

Using the above theorem, it is easy to see that

$$({H_1^{n-1}H_2}) \cdot ({H_2^{n-1}H_1}) \ge {H_1^{n}}{H_2^{n}},$$

which finishes the proof.

• ... except that in Lazarsfeld's book this is stated for nef classes on a complete algebraic variety, and the proof given works only in that set-up. – abx Feb 19 '16 at 5:50
• Yes, I should've mentioned that originally Demailly proved this inequality for pseudo-effective classes in his paper "A numerical criterion for very ample line bundles" (see Proposition 5.2). – HYL Feb 19 '16 at 7:17

It is a consequence of the Khovanskii-Teissier inequality for Kähler classes (which was proved by Gromov and Demailly on Kähler manifolds, the algebraic case is also in Lazarsfeld's book):

$$\int_X \omega_1^{n-1}\wedge\omega_2 \geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{1}{n}},$$

which implies

$$\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)\geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n+1}{n}}.$$

Take this inequality and the same inequality with the roles of $\omega_1$ and $\omega_2$ exchanged, and sum them up:

$$\begin{split}&\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)+\left(\int_X \omega_2^{n-1}\wedge\omega_1 \right)\left(\int_X\omega_1^n\right)\\ &\geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n+1}{n}}+\left(\int_X \omega_1^n\right)^{\frac{n+1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n-1}{n}}.\end{split}$$

Now use the Young inequality $$2xy \leq x^{\frac{n-1}{n}}y^{\frac{n+1}{n}}+x^{\frac{n+1}{n}}y^{\frac{n-1}{n}},$$

and get

$$\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)+\left(\int_X \omega_2^{n-1}\wedge\omega_1 \right)\left(\int_X\omega_1^n\right)\geq 2\left(\int_X\omega_1^n\right)\left(\int_X\omega_2^n\right),$$

which is equivalent to your inequality. If equality holds, then equality must hold in Khovanskii-Teissier, hence $\omega_1$ and $\omega_2$ are proportional by Boucksom-Favre-Jonsson, but equality also holds in Young, hence $\omega_1=\omega_2$.