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I am reading Hitchin's beautiful paper "The Geometry of Three-forms in Six Dimensions". Everything goes smooth up to now except for a tiny problem in Section 6.2, which can be formulated as follows. Let $X$ be a complex compact 3-fold with trivial canonical bundle such that the $\partial\bar{\partial}$-lemma holds, (for simplicity we may just think of $X$ as a Calabi-Yau 3-fold). Fix an Hermitian metric on $X$ and define $E=\{\alpha\in\Lambda^3(M,\mathbb{R}):\mathrm{d}\alpha=0,\mathrm{d}^*\alpha\textrm{ has no (2,0) and (0,2) parts}\}$. By Hodge decomposition, $E=E_1\oplus E_2$, where $E_1$ is the space of $\mathrm{d}$-harmonic 3-forms, and $E_2$ consists of $\mathrm{d}$-exact forms. In order to apply implicit function theorem, Hitchin needs to show that the linear map $Q:E_2\to E_2$ is surjective, where $Q(\alpha)=P_2(*J\alpha)$. Here $J$ is the complex structure, $*$ is the Hodge star, and $P_2$ is the orthogonal projection onto $E_2$.

My interpretation of Hitchin's argument goes as the following: Let $L=\{\mathrm{d}\beta:\beta \textrm{ is a real (1,1)-form}\}$, and it is easy to see that $E_2\subset L$. Therefore $E_2=P_2(E_2)=P_2(L)$. Hitchin showed that for any $\mathrm{d}\beta\in L$, one can find a real (1,1)-form $\mu$ such that $Q(\mathrm{d}\mu)=P_2(\mathrm{d}\beta)$. But it seems that this is not enough to prove the surjectivity since one does not know $\mathrm{d}\mu\in L$ falls into $E_2$.

Hitchin's explanation is the following: "Note that if $\beta$ is of type (2,0)+(0,2), $P_2(*J\mathrm{d}\beta)=0$, so surjectivity for $\mathrm{d}\beta\in L^2_k(\Lambda^3)$ implies surjectivity on the transversal $E_2$."

Here $L^2_k$ just means Sobolev space. My questions are: 1) why $P_2(*J\mathrm{d}\beta)=0$ given that $\beta$ is of type (2,0)+(0,2); 2) why is this enough to prove that $Q$ is surjective?

Thank you!

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  • $\begingroup$ I think I understand the argument now. $\endgroup$ – Piojo Oct 8 '14 at 4:38

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