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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.

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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
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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
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@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
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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
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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
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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 $c\in R\setminus 2\pi Z.$

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