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Suppose $M$ is a compact, orientable surface with boundary, $N$ is a manifold with symplectic form $\omega$, and $F:I\times M\to N$ is a homotopy fixing the boundary $\partial M$. I'd like to show that $\int_M F_0^* \omega=\int_M F_1^* \omega.$ Here's what I've tried. We know $F_1^* \omega-F_0^* \omega$ is exact. Then $$ \int_M F_1^* \omega-F_0^* \omega=\int_M d\eta=\int_{\partial M}\iota_{\partial M}^* \eta. $$ We also know that $\iota_{\partial M}^* \eta$ is closed since $$ d(\iota_{\partial M}^* \eta) =\iota_{\partial M}^* d \eta =\iota_{\partial M}^* (F_1^* \omega-F_0^* \omega) =(F_1\circ \iota_{\partial M})^* \omega-(F_0\circ \iota_{\partial M})^* \omega =0. $$ Unfortunately I think I need $\iota_{\partial M}^* \eta$ to be exact. I'm wondering if maybe I'm not using the full strength of the condition that the boundary is fixed throughout the homotopy?

I think Interpreting the closed condition of a symplectic form and Homotopy invariance of integrals are related.

subrosar
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1 Answers1

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Let $\iota_t:M\to I \times M$ be given by $\iota_t(x)=(t,x)$, so that $F_1=f\circ \iota_1$ and $F_2=f\circ \iota_2$. Let $h:\Omega^2(I\times M)\to \Omega^1(M)$ be the homotopy operator given by contracting with $\partial_t,$ pulling back, and then integrating from $0$ to $1$. It is a standard result that $\iota_1^*-\iota_0^*=dh+hd$. Because the boundary is fixed, $hF^*(\omega)=0$ along $\partial M$. Then $$ \int_M F_1^* \omega-F_2^* \omega =\int_M (\iota_1^*-\iota_2^*)F^*(\omega) =\int_M (dh+hd)F^*(\omega) =\int_{\partial M} h F^*(\omega) =0. $$

subrosar
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