Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi! I'm getting stuck with a proposition, please can someone help me? Let $E$ be a holomorphic vector bundle with hermitian metric $h$ over a connected compact kahler manifold $X$ with kahler form $\omega$ and associated kahler metric $g$. Let $s\in H^{0}(X,E)$ a non zero section. Consider the current $T_{s}=\frac{i\partial\bar{\partial}log(h(s,s))}{2\pi}$. Let's write $F(h)$ for the curvature of the chern connection respect to the metric $h$, and another current $R_s=\frac{ih(F(h)s,s)}{2\pi h(s,s)}$, how can i prove that $0 \leq T_{s}+R_{s}$ in the following sense: (regarding those currents as forms) the hermitian form associated to $T_{s}+R_{s}$ is semipositive definite. The Hermitian form associated to a (1,1)-real form $\delta$ is the one given in local coordinates by the matrix of coefficients of $-i\delta$ and is (semi-)positive if this matrix is.

Thank you in advance.

share|improve this question
    
First: about $F_s$, I think there's no need to speak of currents: as you say, it's just a (1,1)-form. Second: about $R_s$, as it is written it is just a smooth function on $M$, rather than a form (or a current). So, I don't understand how $T_s+R_s$ should be interpreted: as it's written even as a current it doesn't have homogeneus (bi)degree (or (bi)dimension). –  Qfwfq May 14 '10 at 16:06
    
*sorry, I meant $T_s$ above. –  Qfwfq May 14 '10 at 16:08
    
Also, it's not completely obvious to me that $R_s$ -as it's written- doesn't have singularities along the zero set of the section $s$ (but it doesn't sound absurd, since $F(h)$ is $C^{\infty}$-linear). –  Qfwfq May 14 '10 at 16:11
    
Hi! Both $T_{s}$ and $R_{s}$ can have singularities because there's no hypothesis on $s$, this is why i said currents,anyway, you can assume that $s$ doesn't vanish at any point. $T_{s}$ and $R_{s}$ belong to $H^{1,1}(X)\cap H^2(X,\mathbb{R})$, $T_{s}$ because the operator $\partial\bar{\partial}$ sends smooth complex functions to (1,1) forms, $R_{s}$ because $F(h)$ is a $End(E)$-valued (1,1) form so $F(h)s$ is a $E$-valued (1,1) form and finally $h(F(h)s,s)$ is a (1,1) form. –  Italo May 14 '10 at 17:43
    
Ah, I see: they're currents cause you allow $s$ to have zeroes (I suppose the resulting "form" should be checked to be in $L^1_{loc}$). –  Qfwfq May 14 '10 at 20:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.