# (Contradiction) All symplectic manifolds are holomorphic

I’m studying symplectic manifolds and almost complex structures. This lead to two propositions:

Proposition 1 (from da Silva’s Lectures on Symplectic Geometry): If $$J_0$$ and $$J_1$$ are almost complex structures compatible with a symplectic manifold $$(M,\omega)$$, then there is a family of almost complex structures $$J_t, t \in [0,1]$$ such that $$J_t$$ is compatible with $$(M,\omega)$$.

Proposition 2 (from Hatcher’s K-theory book): Let $$E \twoheadrightarrow B \times [0,1]$$ be a (topological) vector bundle. Then the restrictions of $$E$$ to $$B \times \{0\}$$ and $$B \times \{1\}$$ are isomorphic.

I think proposition 2 can be extended to smooth bundles, though maybe this is the source of my confusion below.

The confusion: I think I can prove the following absurdity using the above two propositions.

Absurdity: Let $$(M^{2n},\omega)$$ be a symplectic manifold. Then $$M$$ admits a holomorphic atlas.

Proof: Let $$\phi: U \to \mathbb{R}^{2n}$$ be a Darboux chart (the pullback of the standard symplectic form on $$\phi(U)$$ is $$\omega$$) of a neighborhood $$U \subset M$$ around an arbitrary point $$p \in U$$. Let $$J$$ be an almost complex structure on $$TM$$. We have two almost complex structures on $$\phi(U)$$. The first one, call it $$j_0$$, comes from $$J$$:

$$j_0 \equiv \phi_* \circ j \circ \phi^{-1}_*.$$

The second one, call it $$j_1$$, comes from identifying $$\mathbb{R}^{2n}$$ with $$\mathbb{C}^n$$. Since $$\phi$$ is a Darboux chart, $$j_0$$ is compatible with the standard symplectic form. $$j_1$$ is also compatible with the standard form. Thus, Proposition 1 tells us that we have a family $$j_t$$ of almost complex structures on $$\phi(U)$$ compatible with the standard symplectic form.

We now construct a complex vector bundle $$\pi: E \twoheadrightarrow \phi(U)$$ as follows. For a point $$(x,t) \in \phi(U) \times [0,1]$$, the fiber $$\pi^{-1}(x,t)$$ is the vector space $$T_x \phi(U)$$ equipped with the almost complex structure $$j_t$$. From Proposition 2, we obtain an isomorphism $$\psi: \left(\phi(U),j_0\right) \to \left(\phi(U),j_1\right)$$. Notice $$(\phi(U),j_1)$$ is a neighborhood in $$\mathbb{C}^n$$. Therefore, $$\psi \circ \phi$$ is a holomorphic chart around $$p$$.

$$p$$ is arbitrary, so we obtain our atlas. $$\blacksquare$$

I’ve been perusing this proof for a while, but I cannot clearly see where it fails. What am I missing?

• I don't think this proof guarantees that the transition functions are holomorphic, no?
– dhy
Jan 24 '21 at 20:11
• The Darboux charts are pseudoholomorphic with respect to $j_0$. Shouldn’t this tell us that the transition maps are holomorphic? Jan 24 '21 at 20:16
• From Proposition 2, you conclude the tangent bundle of $\phi(U)$ is isomorphic with tangent bundle of $\phi(U)$, i.e. homeomorphism on the base and fibrewise linearly isomorphic. But the isomorphism needs not transport $j_0$ to $j_1$, i.e. inequality generally holds $\psi_* \circ j_0 \neq j_1 \circ \psi_*$ along the fibres.
– JHM
Jan 24 '21 at 20:43
• I was mistaken. It is incorrect to write $\psi_*$, since we are speaking of the restriction of $\psi$ to fibres. What is needed is a self map $f$ of the base $\phi(U)$ for which $f_*\circ j_0 = j_1 \circ f_*$. The bundle morphism from Prop. 2 is identity map $f=id$ on the base. So Prop. 2 does not yield a holomorphic map $f$.
– JHM
Jan 24 '21 at 22:23
• Let me elaborate a bit on @JHM's comment. Suppose you have almost complex manifolds $(M_1, J_1)$ and $(M_2, J_2)$, a diffeomorphism $\varphi \colon M_1 \to M_2$, and a complex isomorphism $F\colon TM_1 \to TM_2$ covering $\varphi$. Although this latter map identifies the a. c. structures, it doesn't identify their Nijenhuis tensors, since the definition of the latter involves the commutator of vector fields. In particular, you may have $(M_1, J_1)$ a complex manifold ($N_{J_1} = 0$) but $(M_2, J_2)$ not. The problem here is that in general $F \ne d\varphi$, so $F$ may not preserve commutators. Jan 24 '21 at 22:35

The comments of @JHM and @Ivan Solonenko contain an answer to OP's question.

(1) The OP is correct that $$j_0:=\phi_* \circ J \circ \phi^{-1}_*$$ is an almost-complex structure on the image $$V:=\phi(U) \subset \mathbb{R}^{2n}$$ which is compatible with the standard symplectic form $$\omega_{std}$$ satisfying $$\omega=\phi^* \omega_{std}$$, where $$(U, \omega, J)$$ is the initial pseudo-holomorphic structure.

(2) Since $$\phi$$ is a Darboux chart, then $$\omega_{std}$$ is also compatible with the standard almost complex structure $$j_1$$ on $$\mathbb{R}^{2n}$$ restricted to $$V$$.

So we have two $$\omega_{std}$$-compatible a.c. structures $$j_0, j_1$$ on $$V$$, and can even find a smooth $$1$$-parameter family of compatible a.c structures $$j_t$$ on $$V$$, for $$0\leq t \leq 1$$.

(3) The OP constructs a complex vector bundle $$E \to V \times [0,1]$$, where the fibre over $$(p,t)$$ is the $$j_t$$-complex vector space over $$T_{p} V$$. Now the OP correctly applies a form of Hatcher's Prop.2, obtaining a complex vector bundle isomorphism $$F: E|_{V \times \{0\}} \to E|_{V \times \{1\}}$$.

(4) As clearly stated by @Ivan Solonenko, here the OP errs in interpreting $$F$$ as the differential of an isomorphism between the base space $$V=\phi(U)$$. Indeed $$F$$ is the identity map on the base $$V$$. The existence of fibrewise complex isomorphisms does not imply that the base spaces $$(V, j_0)$$ and $$(V, j_1)$$ are holomorphic. If the isomorphism $$F$$ produced by Prop.2. was induced by a diffeomorphism between the bases $$f: V \to V$$, then we would have $$F=df$$ on the fibres, and would find $$df\circ j_0=j_1\circ df$$. But this is not the case.

In conclusion, fibrewise deformations like $$F$$ are not induced by maps on the base. $$F$$ is not a differential $$df$$ except if integrability conditions are satisfied. For a.c. structures, the tensorial form of the integrability condition is given by Nijenhuis tensor. E.g., the $$\pm i$$-eigenspaces of a.c. structures $$j|_p$$ define distributions on $$V$$, and which are almost never integrable for generic a.c. structures.

This can be recurring point of frustration in studying symplectic geometry, and is basically the subject of Gromov-Eliashberg $$h$$-principles. You might find Eliashberg--Mishachev's AMS textbook "Introduction to $$h$$-principle" interesting -- it contains several such examples as above.

• Thank you for the literature recommendation as well as this answer! Jan 26 '21 at 22:39