Nakano semipositivity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:02:10Z http://mathoverflow.net/feeds/question/28109 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28109/nakano-semipositivity Nakano semipositivity Gianni Bello 2010-06-14T10:27:56Z 2010-12-15T22:13:36Z <p>Let $X$ be a compact Kaehler manifold. What is a good, possibly algebraic-geometric, way to think to Nakano semipositivity of holomorphic vector bundles on $X$?</p> <p>Is the trivial line bundle $\mathcal{O}_X$ Nakano semi-positive as a vector bundle?</p> http://mathoverflow.net/questions/28109/nakano-semipositivity/49563#49563 Answer by Sándor Kovács for Nakano semipositivity Sándor Kovács 2010-12-15T20:42:35Z 2010-12-15T20:42:35Z <p><em>Nakano (semi-)positivity</em> is not an algebraic notion. It implies <em>Griffiths positivity</em>, which implies <em>ampleness</em>. It is known that Griffiths positivity <em>does not imply</em> Nakano positivity (an example is the tangent bundle of the complex projective space), but it is not known whether ampleness implies Griffiths positivity.</p> <p>For more on this see 6.1.D of <a href="http://www.springer.com/mathematics/algebra/book/978-3-540-22534-8" rel="nofollow">Lazarsfeld's book</a>. The fact that the tangent bundle of the complex projective space is not Nakano positive follows from a vanishing theorem p.97 of <a href="http://www.springer.com/mathematics/algebra/book/978-3-540-22534-8" rel="nofollow">ibid</a>. that fails for it.</p> http://mathoverflow.net/questions/28109/nakano-semipositivity/49574#49574 Answer by diverietti for Nakano semipositivity diverietti 2010-12-15T22:13:36Z 2010-12-15T22:13:36Z <p>I would say more properly that nowadays it is not known any satisfactory algebraic description or characterization of the concept of Nakano's positivity for a hermitian vector bundle.</p> <p>I would like also to add some precisions to Sándor's answer.</p> <p>First, the positivity in the sense of Nakano is a good notion to obtain vanishing theorems for vector bundles. For instance, we have the following result due to Nakano</p> <p><strong>Theorem (Nakano, 1955).</strong> Let $X$ be a compact connected Kähler manifold of dimension $n$ and $E\to X$ a hermitian vector bundle. Then</p> <ul> <li><p>if $E\ge_{\text{Nak}}0$, strictly in one point, $H^{n,q}(X,E)=0$, for $q\ge 1$;</p></li> <li><p>if $E\le_{\text{Nak}}0$, strictly in one point, $H^{p,0}(X,E)=0$, for <code>$p&lt;n$</code>.</p></li> </ul> <p>On the other hand, Nakano's positivity is not well-behaved with respect to taking duals: we have that $E$ is Griffiths positive if and only if $E^*$ is Griffiths negative, but if we take $H\to\mathbb P^n$ to be the vector bundle of rank $n$ defined by $$0\to\mathcal O(-1)\to\underline{\mathbb C}^{n+1}\to H\to 0,$$ where $\underline{\mathbb C}^{n+1}$ si the trivial vector bundle over $\mathbb P^n$ with fiber $\mathbb C^{n+1}$, then $H$ is Griffiths (semi)positive and <code>$H^*$</code> is Nakano (semi)negative but $H$ is neither Nakano (semi)positive nor Nakano (semi)negative.</p> <p>If we look to short exact sequences of vector bundles $$0\to S\to E\to Q\to 0,$$<br> then the Nakano's negativity of $E$ implies the Nakano's negativity of $S$, but nothing can be said about the Nakano's positivity of $Q$ when $E$ is Nakano positive (the desired property holds if we look instead to Griffiths' positivity).</p> <p>Of course Nakano's positivity implies the Griffiths' one; the following "partial converse" is due to Demailly and Skoda:</p> <p><strong>Theorem (Demailly-Skoda, 1979).</strong> For any hermitian vector bundle $E$, $E>_{\text{Grif}}0$ implies <code>$E\otimes\det E&gt;_{\text{Nak}}0$</code>.</p>