## 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 2010 at 22:35

Here is another try:

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.

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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 2010 at 20:04
we assume $\alpha+t\beta$ is kahler. – lemega Apr 3 2010 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 2010 at 14:17
Yes. You can interpret this way. – lemega Apr 6 2010 at 17:16
@Tim: No, two Kahler classes are never orthogonal to each other on any compact manifold. You can see this by representing both by smooth forms and calculating their scalar product in coordinates that diagonalize both forms at a point. What comes out will necessarily be positive, so its integral over the manifold (ie the scalar product of the Kahler classes) will be positive. – Gunnar Magnusson Jan 2 at 13:48
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