Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a Riemann surface $S$, e.g. $\mathbb{P}^1(\mathbb{C})$, with complex conjugation on the coordinates and a holomorphic vector bundle $E$ over $S$.

The complex conjugation $f$ is not holomorphic, but $C^{\infty}$. So we can look at $f^*E$ as a smooth real vector bundle. Since $f$ is an involution $E$ and $f^*E$ are isomorphic as smooth real bundles (this is most likely wrong). But somehow there should be more structure on $f^*E$. I mean $f$ is antiholomorphic, that should be useful somehow.

Is there a way to define a (canonical?) complex structure on $f^{*}E$, such that for any point $p$ we have an antilinear isomorphism $E_{p} \rightarrow f^{*}E_{p}=E_{\overline{p}}$?

I think we don't get a linear iso, since the complex conjugation itself is only antilinear.

share|improve this question
I'm a little confused by the statement that $E$ and $f^*E$ are necessarily isomorphic as smooth bundles. An anti-holomorphic $f$ involution of a Riemann surface reverses its orientation. It follows that given a line bundle $L$ of degree $d$, the bundle $f^*L$ has degree $-d$ and so is rather isomorphic to $L^*$. –  Joel Fine Mar 9 '10 at 21:06
Given the third paragraph, I'd guess the questioner means "smooth real bundles" but it would be useful to have this confirmed. –  Andrew Stacey Mar 9 '10 at 22:25
@Andrew Stacey, ah that would make more sense since, of course, $\bar E \cong E^*$ as smooth bundles. What confused me is that $f^*E$ is already naturally a complex vector bundle. Reading the last sentence of the question again, it seems that what you suggest is what TonyS actually means. –  Joel Fine Mar 9 '10 at 22:37
I don't know much about the subtleties of complex geometry so this might be obvious, but I don't see why $E$ and $f^*E$ are isomorphic as smooth bundles. Is complex conjugation on $S$ always homotopic to the identity? Thinking about the real situation, it's not true that if I have a diffeomorphism $\alpha : M \to M$ and a vector bundle $E \to M$ that $\alpha^*E$ and $E$ are isomorphic: take $M$ to be a coproduct of two identical manifolds and $E$ trivial over one factor, non-trivial over the other, and $\alpha$ the swap. What am I missing? –  Andrew Stacey Mar 10 '10 at 8:10
To return to the actual question, you certainly get a complex structure on $f^*E$ - simply pull-back the complex structure on $E$! Then the morphism $f^*E \to E$ is a morphism of complex bundles (note that this covers $f : S \to S$ rather than the identity). But this identifies $(f^*E)_p$ with $E_{f(p)}$ as complex vector spaces so to get the identity that TonyS wants, one should take the pull-back of $\overline{E}$. I wouldn't be surprised to learn that $f^*\overline{E}$ was then a holomorphic vector bundle over $S$: the two antiholomorphic factors should combine to a holomorphic one. –  Andrew Stacey Mar 10 '10 at 8:14
show 2 more comments

1 Answer

up vote 3 down vote accepted

For any antiholomorphic Diffeomorphism $f\colon S\to S$ we get a canonical identification $f^\star\bar K=K,$ $ K $ and $\bar K$ being the canonical and anticanonical bundle of the Riemann surface. A holomorphic structure on a complex vectorbundle $E$ is the same as an complex operator $D\colon\Gamma(E)\to\Gamma(\bar KE)$ satisfying the (Cauchy Riemann) Leibnitz rule. (holomorphic sections are exactly the one in the kernel of D). (For higher dimensions this is not true.)

Now, $f^* E$ has a natural complex structure (it's just i). Therefore one gets an anti-holomorphic structure $\bar D\colon\Gamma(f^\star E)\to\Gamma(Kf^* E)$ satisfying the antiholomorphic Cauchy Riemann equation. But the complex conjugate bundle $\bar E$ also has a anti-holomorphic structure, since $\overline{\bar K E}=K\bar E.$ Therefore, $f^* \bar E$ has a natural holomorphic structure.

These two holomorphic structures are not isomorphic in general: In the case of a line bundle $L=E$ of degree $0$ one might see this as follows: every holomorphic structure $D$ gives rise to an unique unitary flat connection $\nabla$ such that $D=1/2(\nabla+i*\nabla).$ Then the anti-holomorphic structure on $\bar L$ is given by $1/2(\nabla-i*\nabla)$ and, the unitary flat connection corresponding to the holomorphic structure on $f^* L$ is the connection $f^* \nabla.$ But this connection is not gauge equivalent to $\nabla$ in general: For example, on a square torus with f given by $z\mapsto \bar z$ the connection $d+c idx$ is not gauge equivalent to $d+ci dy$ for $c\in R\setminus 2\pi Z.$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.