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Let $\Sigma = S^2$ be thought of as a Riemann surface, and let $L$ be a Hermitian line bundle on $\Sigma$ with curvature $2$-form $-2 \pi i \Omega \in \Omega^2(\Sigma, \mathbb{R})$. Then $L$ is a positive line bundle iff $\Omega$ is cohomologous to a positive $2$-form. If $\Omega$ is a positive $2$-form then the Hermitian bilinear form on $H^0(L)$ $$ (f, g) \mapsto \int_\Sigma \langle f, g\rangle\Omega $$ is positive definite on $H^0(L)$. Is this bilinear form still positive definite without the assumption that $\Omega$ is positive, i.e. while assuming only that $L$ is positive?

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This is in general false, if I am not missing anything. Suppose you start from the form $\Omega_0$ given by the Chern connection on $O(1)$ with respect to the standard metric coming from $\mathbb{C}^2.$ Up to a constant multiple this is the standard area form on $S^2.$

If you multiply the metric $h$ by $f=e^{\phi}$ then the curvature changes by $\frac{1}{4} \Delta \phi.$ This means that for a section $s$ its norm squared becomes: $$ \int_{S^2} |s|^2 e^{\phi} (1+\frac{1}{4} \Delta \phi) \Omega_0.$$

The idea is to make $e^{\phi} (1+\frac{1}{4} \Delta \phi)$ negative where $|s|^2$ is large.

Take the section corresponding to the homogeneous polynomial $z_2^d$ on $\mathbb{C}^2.$ Then $$|s|^2 = \left(\frac{|z_2|^2}{|z_1|^2+|z_2|^2}\right)^d.$$ Note that in suitable real coordinates $(x,y,z)$ coming from an orthonormal basis on $\mathbb{R}^3$ this is a constant multiple of $(z+1)^{d}.$

Take for simplicity $d=1.$ Let $\phi = 4 z.$ Passing to polar coordinates $$(\sin(\theta)\cos(\varphi),\sin(\theta)\sin(\varphi),\cos(\theta))$$ for $(\theta,\varphi) \in (0,\pi) \times (0,2\pi),$ and using the formula $$\Delta = \cot(\theta) \partial_{\theta} + \partial^2_{\theta} + \sin^{-2}(\theta) \partial_{\varphi}^2$$ for the Laplacian in spherical coordinates the norm squared of $s$ becomes: $$2\pi \int_0^{\pi} (\cos(\theta)+1) e^{4 \cos(\theta)} (1-2\cos(\theta)) \sin(\theta) d\theta =$$ $$ = -\pi \sinh(4) < 0,$$ the computation having been done by Wolfram Alpha.

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