It may be a very simple question,but I can't figure it out.

Let $(M,I)$ be an almost complex manifold where $I$ is an almost complex structure.Then $-I$ is also an almost complex structure.Then the identity map $id:(M,I)\rightarrow (M,-I)$ is (pseduo)-holomorphic.

Here,enter image description here

Here is what I think:if $z_i\in T^{1,0}(M)$,then $I(z_i)=iz_i$,$-I(z_i)=-i(z_i)$,however,$id_*=id$,I can't get $id_*\circ I=(-I)\circ id_*$.I wanna know where I did wrong.Can anyone give me some advice?

By the way,any notes including the explicit examples about (pseduo)-holomorphic map are appreciated.Thanks a lot!

  • $\begingroup$ Please do not use $i$ as both a complex number and an index. This is very confusing in complex geometry. ... You should be computing on tangent vectors, not on coordinates, but, nevertheless what is true is that the image of the tangent space is a complex subspace under the new almost-complex structure, but the equation sure seems to fail vector by vector. (Compute $I(\partial/\partial x_j) = \partial/\partial y_j$ and $I(\partial/\partial y_j) = -\partial/\partial x_j$, etc.) $\endgroup$ Sep 28, 2019 at 17:59
  • $\begingroup$ Dear @TedShifrin,thanks for your comment.So do you mean the claim is not right?In fact,I came across this problem math.stackexchange.com/questions/3367602/… ,so I ask this question.Can you give me some advice for my another question math.stackexchange.com/questions/3367602/… ?Thanks a lot. $\endgroup$
    – Invariance
    Sep 29, 2019 at 5:18
  • 2
    $\begingroup$ The identity is definitely not $(I,-I)$-holomorphic. Changing $I$ by $-I$ essentially says that all holomorphic objects according to the first structure become anti-holomorphic with respect to the second structure, but the two concepts are not the same. $\endgroup$
    – Ivo Terek
    Sep 29, 2019 at 20:08
  • $\begingroup$ Dear @IvoTerek,thank you! $\endgroup$
    – Invariance
    Sep 30, 2019 at 2:28
  • $\begingroup$ @Steve: I commented on your other question. $\endgroup$ Sep 30, 2019 at 23:33

1 Answer 1


The claim is wrong. Take for example the complex manifold $\mathbb C$ which is an alomost complex manifold as well. $\operatorname{id}:(\mathbb C,i)\to (\mathbb C,-i)$ being (pseduo)-holomorphic is equivalent to $\mathbb C \to \mathbb C$, $z \mapsto \overline z$ being holomorphic which is of course not the case.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .