# two almost complex structure define isomorphic complex manifolds

First,let me state the question:

Let $$M$$ be the real manifold described as a hypersurface $$x_0^4+x_1^4+x_2^4+x_3^4=0$$ in $$\mathbb P^3$$.We denote the naturally induced complex structure by $$I$$.Show that $$(M,I)$$ and $$(M,-I)$$ define isomorphic manifolds.

As far as I know,the oriented dimension two differentiable manifolds (Riemannian surface) have naturally almost complex structures(and also integrable,then determine complex structures).So,

1.Should I show the orientability of $$M$$?If yes,I notice this post: Hypersurface orientable if it admits a smooth normal vector field .I think it can work.

2.How can I show two two almost complex structure define isomorphic complex manifolds?

Any advice and references will be appriecated.Thanks a lot.

• Dear @MoisheKohan,thank you.So in this situation,the defining equation $x_0^4+x_1^4+x_2^4+x_3^4=0$ just to guarantee the orientability of $M$? Commented Sep 24, 2019 at 3:46
• Not orientability but the existence of an almost complex structure. Commented Sep 24, 2019 at 10:50
• Dear @MoisheKohan,what the meaning of "holomorphic"?Should I show that $-J\circ id_*=id_*\circ J$?I tried but fail to prove your claim.Here is what I think:If $z_i\in T^{1,0}(M)$,then $J(z_i)=iz_i$,$-J(z_i)=-iz_i$.However,$-J\circ id_*(z_i)=-id_*\circ J(z_i)$.Where did I make mistakes?Can you explain it more explicitly?Thanks a lot. Commented Sep 27, 2019 at 7:28
• So a nontrivial homogeneous polynomial of even degree defines an orientable surface in $\Bbb RP^3$, which of course has a complex structure when it's smooth. But go back to your original polynomial. The locus in $\Bbb RP^3$ is empty. What did you mean to type? (If we're in $\Bbb CP^3$, then we get a perfectly lovely $2$-dimensional complex submanifold.) Commented Sep 30, 2019 at 22:35
• Dear@TedShifrin,you are absolutely right!I forgot the locus in $\mathbb RP^3$ is zero.So,now the problem reduces to :$M$ is a 2-dimension complex manifold which admits natural almost complex structure $I$.Similarly,$-I$ is also the a.c.s.It seems like $(M,I)$ and $(M,-I)$ define $M$ so naturally,but how can we prove this strictly?Should we build a map $f:(M,I)\rightarrow (M,-I)$ satisfying the pseduo-holomorphic condition or use other ways to prove it?Again,thanks for your time and patience,You are so kind. Commented Oct 1, 2019 at 3:28

## 1 Answer

Here's the best answer I can come up with. The problem as you posed it is vacuous. If $$M=\Bbb C$$, then $$f(z)=\bar z$$ is a pseudo-holomorphic map from $$(M,I)$$ to $$(M,-I)$$. More generally, if you have a complex submanifold of $$\Bbb C^n$$ or of $$\Bbb CP^n$$ that is given by real equations (homogeneous equations, of course, in the projective case), then this construction may still work (the mapping preserves the submanifold, but you have work to do to check what happens to tangent vectors).

• Dear @Ted,thanks a lot!I now figure it out.In fact,this problem comes from Huybrechts's Complex Geometry exercise 2.6.8. Commented Oct 1, 2019 at 4:50
• Yes, context helps. He of course meant $\Bbb P^3 = \Bbb CP^3$, and what I sketched here is doubtless what he had in mind. Commented Oct 1, 2019 at 4:56