Kahler metric on projectivised bundle - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:03:51Z http://mathoverflow.net/feeds/question/68843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68843/kahler-metric-on-projectivised-bundle Kahler metric on projectivised bundle math_donk 2011-06-26T09:11:43Z 2011-06-26T11:04:01Z <p>Let $E\rightarrow M$ be a holomorphic bundle over a Kahler manifold. Does its projectivisation $\mathbb{P}(E)$ always admit a Kahler metric? If yes, how to see that?</p> http://mathoverflow.net/questions/68843/kahler-metric-on-projectivised-bundle/68844#68844 Answer by Francesco Polizzi for Kahler metric on projectivised bundle Francesco Polizzi 2011-06-26T09:39:25Z 2011-06-26T10:06:50Z <p>As pointed out by Georges Elencwajg, the answer is <strong>yes</strong>. </p> <p>However, if one substitutes the assumption "holomorphic vector bundle" with the weaker "complex vector bundle", the answer is <strong>no</strong>.</p> <p>In fact, there is the following result proven by C. Voisin.</p> <p>Start with a complex Kähler manifold $X$ having a given class $\alpha \in H^4(X, \mathbb{Q})$ such that, for any given compatible Hodge decomposition on $H^*(X)$, $\alpha$ is not of type $(2,2)$. </p> <p>Then if $E$ is any complex vector bundle on $X$ satisfying $$c_1(E)=0, \quad c_2(E)=\alpha,$$ the projective bundle $\mathbb{P}(E)$ admits no Kähler metric (even better, it is not homeomorhic to any Kähler manifold).</p> <p>The simplest example of such a pair $(X, \alpha)$ is obtained by choosing for $X$ a complex torus of dimension $4$ and for $\alpha$ a class satisfying the property that the cup product map $$\alpha \cup \colon H^1(X, \mathbb{Q}) \longrightarrow H^5(X, \mathbb{Q})$$ has odd rank.</p> <p>See <a href="http://www.math.jussieu.fr/~voisin/Articlesweb/verbanianotes.pdf" rel="nofollow">these notes</a> by C. Voisin for more details.</p> http://mathoverflow.net/questions/68843/kahler-metric-on-projectivised-bundle/68845#68845 Answer by Georges Elencwajg for Kahler metric on projectivised bundle Georges Elencwajg 2011-06-26T09:43:07Z 2011-06-26T09:43:07Z <p>If $M$ is compact ( the usual assumption in Kähler manifold theory) the answer is "yes". You can look it up in Claire Voisin's <a href="http://books.google.com/books/about/Hodge_Theory_and_Complex_Algebraic_Geome.html?id=AhSqPwAACAAJ" rel="nofollow">book </a>Proposition 3.18, page 78.</p> http://mathoverflow.net/questions/68843/kahler-metric-on-projectivised-bundle/68848#68848 Answer by diverietti for Kahler metric on projectivised bundle diverietti 2011-06-26T11:04:01Z 2011-06-26T11:04:01Z <p>Let me recall you briefly how to obtain such a Kähler metric on the total space of the projectivized bundle.</p> <p>Start with any given hermitian metric $h$ on $E$ and consider on the projectivized bundle $\mathbb P(E)$ of hyperplanes of $E$ the tautological line bundle $\mathcal O_E(1)\to\mathbb P(E)$ of rank one quotients of $E$. Then, on $\mathcal O_E(1)$ you have a natural induced quotient hermitian metric, which I shall call again $h$.</p> <p>If you compute the Chern curvature $\Theta(\mathcal O_E(1))$ of $\mathcal O_E(1)$ with respect to $h$, you will find a closed $(1,1)$-form on $\mathbb P(E)$ which is positive along the relative tangent bundle (after all, the restricion of $\mathcal O_E(1)$ to fibers $\simeq\mathbb P^{\operatorname{rk}E-1}$is just the usual $\mathcal O(1)$, so that its curvature restricted to fibers is just the usual Fubini-Study metric). No more can be said along "horizontal" directions.</p> <p>Now, suppose that $M$ is compact Kähler, with Kähler form $\omega$ and call $\pi\colon\mathbb P(E)\to M$ the natural projection. Since $E$ is holomorphic (see Francesco's answer), $\pi$ is holomorphic as well. Thus, $\pi^*\omega$ is again a closed $(1,1)$-form on $\mathbb P(E)$ which is zero on vertical directions (as a pull-back) and strictly positive on horizontal ones.</p> <p>Finally, by compactness of $M$, and so of $\mathbb P(E)$, you can find a large constant $C$ such that the closed $(1,1)$-form <code>$$C\,\pi^*\omega+i\,\Theta(\mathcal O_E(1))$$</code> is positive definite everywhere (such a large multiple is chosen in such a way that $\omega$ compensates the possible lack of positivity of $\Theta(\mathcal O_E(1))$ along horizontal directions, and observe that $\omega$ does not interfere with the vertical ones).</p> <p>This gives you the desired Kähler metric on $\mathbb P(E)$.</p>