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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 $X$?

Is the trivial line bundle $\mathcal{O}_X$ Nakano semi-positive as a vector bundle?

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Of course it is Nakano semi-positive : the curvature of $\mathcal O_X$ (equipped with its standard metric) is zero! –  Henri Jun 14 '10 at 12:43
    
Nakano's positivity is just the ordinary positivity for line bundles. –  diverietti Dec 15 '10 at 22:13
    
Gianni, doesn't my answer satisfy you? –  diverietti Jan 25 '11 at 23:08
    
Yes, it does. I just forgot to accept it. Grazie Simone! –  Gianni Bello Jan 27 '11 at 18:08
    
Of course, it wasn't a sake of reputation... Just I wanted to know if my answer was satisfactory to you... Thanks anyway! Prego caro! –  diverietti Jan 30 '11 at 10:04
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2 Answers 2

up vote 2 down vote accepted

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.

I would like also to add some precisions to Sándor's answer.

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

Theorem (Nakano, 1955). Let $X$ be a compact connected Kähler manifold of dimension $n$ and $E\to X$ a hermitian vector bundle. Then

  • if $E\ge_{\text{Nak}}0$, strictly in one point, $H^{n,q}(X,E)=0$, for $q\ge 1$;

  • if $E\le_{\text{Nak}}0$, strictly in one point, $H^{p,0}(X,E)=0$, for $p<n$.

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 $H^*$ is Nakano (semi)negative but $H$ is neither Nakano (semi)positive nor Nakano (semi)negative.

If we look to short exact sequences of vector bundles $$ 0\to S\to E\to Q\to 0, $$
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).

Of course Nakano's positivity implies the Griffiths' one; the following "partial converse" is due to Demailly and Skoda:

Theorem (Demailly-Skoda, 1979). For any hermitian vector bundle $E$, $E>_{\text{Grif}}0$ implies $E\otimes\det E>_{\text{Nak}}0$.

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Nakano (semi-)positivity is not an algebraic notion. It implies Griffiths positivity, which implies ampleness. It is known that Griffiths positivity does not imply Nakano positivity (an example is the tangent bundle of the complex projective space), but it is not known whether ampleness implies Griffiths positivity.

For more on this see 6.1.D of Lazarsfeld's book. The fact that the tangent bundle of the complex projective space is not Nakano positive follows from a vanishing theorem p.97 of ibid. that fails for it.

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