question about kahler cone of a compact kahler manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:06:02Z http://mathoverflow.net/feeds/question/19632 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19632/question-about-kahler-cone-of-a-compact-kahler-manifold question about kahler cone of a compact kahler manifold Italo 2010-03-28T15:39:44Z 2013-01-02T12:23:11Z <p>Hi to all!</p> <p>I'm studying complex geometry from Huybrechts book "Complex Geometry" and i have problems with an exercise, please can anyone help me?</p> <p>I define the kahler cone of a compact kahler manifold X as the set </p> <p>$K_X \subseteq H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ 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).</p> <p>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. </p> <p>Please, if i made mistakes, or you know how to go on, or another way to solve this, tell me.</p> <p>Thank you in advance. </p> http://mathoverflow.net/questions/19632/question-about-kahler-cone-of-a-compact-kahler-manifold/20256#20256 Answer by lemega for question about kahler cone of a compact kahler manifold lemega 2010-04-03T19:08:44Z 2010-04-03T19:08:44Z <p>Here is another try:</p> <p>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.</p>