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Let $\Sigma$$\Sigma = S^2$ be thought of as a Riemann surface, and let $L$ be a Hermitian line bundle on $\Sigma$ with curvature $2$-form $-2 \pi i \Omega \in \Omega^2(\Sigma, \mathbb{R})$. Then $L$ is a positive line bundle iff $\Omega$ is cohomologous to a positive $2$-form. If $\Omega$ is a positive $2$-form then the Hermitian bilinear form on $H^0(L)$ $$ (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?

Let $\Sigma$ be 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?

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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skr
skr
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Source Link
skr
skr
  • 512
  • 2
  • 9

Inner product on global sections of positive line bundle

Let $\Sigma$ be 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?