# Inner product on global sections of positive line bundle

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?

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.