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 Kähler manifold $X$ with Kähler form $\omega$ and associated Kähler 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.
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$\begingroup$ 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). $\endgroup$– QfwfqCommented May 14, 2010 at 16:06
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$\begingroup$ *sorry, I meant $T_s$ above. $\endgroup$– QfwfqCommented May 14, 2010 at 16:08
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$\begingroup$ 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). $\endgroup$– QfwfqCommented May 14, 2010 at 16:11
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$\begingroup$ 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. $\endgroup$– ItaloCommented May 14, 2010 at 17:43
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1$\begingroup$ Hi Henri! Take the hermitian product in the E-section part and the form part beheaves like scalar function. More explicitly in coordinates: choose a local trivialization so you have $UxV$ with $V$ complex vector and $U$ open set of ${\mathbb{C}}^{r}$, choose a frame ${v_i}_i$ on $V$, you can express the metric respect to this frame $(h_{lm})_{lm}$ so you have $h(dz^{\alpha}\otimes v_{j}, v_{k})=dz^{\alpha}h_{jk}$. Is it more clear? $\endgroup$– ItaloCommented May 15, 2010 at 8:42
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This is a consequence of the generalized Lelong-Poincaré formula for vector bundles: denoting $D'$ the $(1,0)$ part of the Chern connection of $(E,h)$ and $\Theta_h(E)$ its Chern curvature, one has:
$$dd^c \log |s|^2=\frac{1}{|s|^2}\cdotp \left( |D's|^2-\frac{|\langle D's,s\rangle|^2}{|s|^2}- \langle \Theta_h(E)s, s\rangle\right)$$
and essentially by Cauchy-Schwartz inequality, one has $$|s|^2\cdotp |D's|^2-|\langle D's,s\rangle|^2 \ge 0.$$
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6$\begingroup$ See Lemma 2.10 of the lectur note of Takuro Mochizuki, KOBAYASHI-HITCHIN CORRESPONDENCE FOR TAME HARMONIC BUNDLES AND AN APPLICATION math.kyoto-u.ac.jp/preprint/2005/15mochizuki.pdf and take $θ=0$ $\endgroup$– user21574Commented Jun 7, 2017 at 20:38