# question about kahler cone of a compact kahler manifold

Hi to all!

I'm studying complex geometry from Huybrechts book "Complex Geometry" and i have problems with an exercise, please can anyone help me?

I define the kahler cone of a compact kahler manifold X as the set

$K_X \subseteq H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ of kahler classes. I have to prove that $K_X$ doesn't contain any line of the form $\alpha + t \beta$ with $\alpha , \beta\in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ and $\beta\neq 0$ (i identify classes with representatives).

This is what i thought: i know that a form $\omega \in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ that is positive definite (locally of the form $\frac{i}{2}\sum_{i,j} h_{ij}(x)dz^i\wedge d \overline{z}^{j}$ and $(h_{ij}(x))$ is a positive definite hermitian matrix $\forall x\in X$) is the kahler form associated to a kahler structure. Supposing $\alpha$ a kahler class i want to show that there is a $t\in\mathbb{R}$ such that $\alpha + t \beta$ is not a kahler class. Since $\beta\neq0$ i can find a $t\in\mathbb{R}$ such that $\alpha + t \beta$ is not positive definite any more, now i want to prove that there is no form $\omega \in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ such that $\omega=d\lambda$ with $\lambda$ a real 1-form and $\omega=\overline{\partial}\mu$ with $\mu$ a complex (1,0)-form (what i'd like to prove is: correcting representatives of cohomology classes with an exact form i don't get a kahler class). From $\partial\overline{\partial}$-lemma and a little work i know that $\omega=i\partial\overline{\partial}f$ with f a real function. And now (and here i can't go on) i want to prove that i can't have a function f such that $\alpha + t \beta+i\partial\overline{\partial}f$ is positive definite.

Please, if i made mistakes, or you know how to go on, or another way to solve this, tell me.

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As a feeble partial answer, I can explain why this is true for Kaehler surfaces. By the Hodge index theorem, if $P$ is a 2-dimensional subspace of $H^{1,1}(X)\cap H^2(X;\mathbb{R})$ containing a Kaehler class, the signature of the wedge-product quadratic form on $P$ is zero. Therefore there is no affine line in $P$ on which the wedge-square takes only positive values, and hence no line of Kaehler classes. – Tim Perutz Apr 4 '10 at 22:35

WLOG, we assume $\alpha$ is kahler, fix it as a metric on $M$. Assume $\alpha+t\beta$ is kahler for every $t$. So $\int(\alpha+t\beta)\wedge \alpha^{n-1}=\int \alpha^n+t\int\beta\wedge\alpha^{n-1}>0$ for every $t$. It then follows $\int\beta\wedge\alpha^{n-1}=0$. In a same manner, by considering $\int(\alpha+t_1\beta)\wedge(\alpha+t_2\beta)\wedge\alpha^{n-2}$, we have $\int \beta^2\wedge\alpha^{n-2}=0$. By Lefschetz decomposition, we can write $\beta=\beta_1+c\alpha$, where $\beta_1$ is a primitive cohomology class. Then $\beta_1\wedge\alpha^{n-1}=0$. By the fact $\int\beta\wedge\alpha^{n-1}=0$, we conclude $c=0$ and $\beta$ itself is primitive class. Then it is a contradiction that $\int \beta^2\wedge\alpha^{n-2}=0$ unless $\beta=0$ by Hodge-Riemann bilinear relation for primitive classes.
How do you see that the wedge product of $\alpha^{n-1}$ and $\alpha+t\beta$ has positive integral? On a complex surface, if the Kaehler classes span a subspace of dim > 1, won't some pairs of Kaehler classes necessarily be orthogonal to one another? – Tim Perutz Apr 3 '10 at 20:04
we assume $\alpha+t\beta$ is kahler. – lemega Apr 3 '10 at 22:55
Hi Lemega! It's a very clean proof and i don't see mistakes.Thank you! Only one question:if $\alpha,\omega$ are kahler classes then $\int{\beta\wedge{\alpha}^{n-1}}>0$?I tried to prove this way:for each point $x\in X$ i can find a chart such that in x the kahler metric of $\alpha$ is the usual euclidean metric,so i check that in x $\beta\wedge {alpha}^{n-1}$ is in coordinates $\tr(-2i\beta){\alpha}^{n}$ with $tr(-2i\beta)=\sum_{j}-2i\beta_{jj}$ but since $-2i\beta$ is hermitian positive definite $tr(-2i\beta)>0$ since i can do it for all $x\in X$ $\int{\beta\wedge{\alpha}^{n-1}}>0$.Am i wrong? – Italo Apr 5 '10 at 14:17