Let $(M, \omega)$ be a holomorphic symplectic manifold of (complex) dimension $2n$. Let $x$ be a point in $M$. My understanding from the discussion and answers to this MO question is that there exists a neighborhood $U \subseteq M$ of $x \in M$, and a neighborhood $V \subseteq T^* \mathbb{C}^n$, such that $V$ is symplecto-bi-holomorphic to $U$, where $T^*\mathbb{C}^n$ is given its standard homomorphic symplectic form. In other words there is a holomorphic symplectic version of Darboux's theorem. If I have miss-understood that discussion please correct me!

Now given a holomorphic function $f: \mathbb{C}^n \to \mathbb{C}$, the graph of $df$ in $T^*\mathbb{C}^n$ is a holomorphic Lagrangian submanifold.

Question: Given any holomorphic Lagrangian submanifold $L \subseteq M$, is it locally isomorphic to a holomorphic Lagrangian of this form?

More specifically, given any $x \in L$, do there exist a neighborhood $U \subseteq M$ of $x$ and a neighborhood $V \subseteq T^*\mathbb{C}^n$ and a symplecto-bi-holomorphism $U \cong V$, as above, such that under this isomorphism $L \cap U$ coincides with the graph of $df$ for some holomorphic function $f$ (where $f$ is defined on, say, $V \cap \mathbb{C}^n$)?

Remark: We know (see here) that there is no holomorphic version of the Weinstein Lagrangian Neighborhood theorem. This question is morally asking if there is a much weaker local version that still holds holomorphically.


1 Answer 1


The answer is 'yes'. Specifically, the holomorphic version of the Darboux-Weinstein theorem holds, just as it does in the smooth category. In particular, if $L\subset M$ is a holomorphic Lagrangian submanifold where $M$ is a holomorphic symplectic complex manifold and $x\in L$ is specified, then there is an open $x$-neighborhood $U\subset M$ on which there exist holomorphic coordinates $z^1,\ldots,z^{2n}$ such that $L\cap U$ is defined by $z^{n+1} = z^{n+2} = \cdots = z^{2n}=0$ and the symplectic form $\omega$ on $U$ has the form $$ \omega = \mathrm{d}z^1\wedge\mathrm{d}z^{n+1} + \cdots +\mathrm{d}z^n\wedge\mathrm{d}z^{2n}. $$ The proof is the same as in the classical case, because one only uses ODE existence and uniqueness, which is true in the holomorphic category just as it is in the smooth category.

  • $\begingroup$ Excellent! Now what if I have two Lagrangians $L_1$ and $L_2$ and a point $x \in L_1 \cap L_2$ in the intersection. Is it possible to find coordinates around $x$ so that $L_1$ is the zero section (as in your answer) and $L_2$ is the graph of $df$ for some $f$? It would be great if that is the case. $\endgroup$ Oct 6 at 19:49
  • 1
    $\begingroup$ @ChrisSchommer-Pries: Yes, that's certainly true. For any point $x\in L_1\cap L_2$, there is a Lagrangian subspace $V\subset T_xM$ that is transverse to both $T_xL_1$ and $T_xL_2$. Hence it is possible to choose Darboux coordinates $z^i$ on an $x$-neighborhood $U$ as above so that the $n$-form $\mathrm{d}z^1\cdots\wedge\cdots\mathrm{d}z^n$ is nonvanishing on $L_i\cap U$. Then $L_i$ will be defined by equations of the form $z^{n+k} = \partial f_i/\partial z^k$ for some functions $f_i(z^1,\ldots, z^n)$. Now replace $z^{n+k}$ by $z^{n+k}-\partial f_1/\partial z^k$ to arrange $f_1=0$. $\endgroup$ Oct 6 at 20:22

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