Let $M$ be a real manifold with complex structure $J$, making $M$ into an almost complex manifold. I know that the complexification $T_{\textbf{C}}M = TM\otimes \textbf{C}$ of the tangent bundle $TM$ may be decomposed as $T^{1,0}M\oplus T^{0,1}M$, the first term of which is called the holomorphic tangent bundle and the second the antiholomorphic tangent bundle (defined by $J$ being either $i\cdot \text{id}$ or $(-i)\cdot \text{id}$ on the summands). I'm trying to understand what happens when we move from the tangent bundle to arbitrary vector bundles. My question is:

What does it mean for $E$ to be a holomorphic vector bundle on $M$?

Does this question even make sense? I'm trying to understand what Higgs bundles are, and the paper I'm reading starts off by taking a holomorphic vector bundle on a Kähler--Einstein manifold. As far as I understand, being Kähler only needs an almost complex manifold and being Einstein is a condition on the Ricci curvature (and doesn't seem to imply the manifold is complex).

It seems that $E$ should certainly be a complex vector bundle, but I can't figure out where the notion of something being holomorphic comes in here. If $M$ were complex, we would need the projection map $\pi:E\to M$ to be holomorphic, but in this case $\pi$ isn't even complex-valued. Maybe if we first project from $E$ onto $T_{\textbf{C}}M$ and then onto $M$ we could get something? I have no idea. I'm certain I'm missing some key component here. Maybe we have to have $M$ be complex to have a Higgs bundle? That would definitely solve things.


2 Answers 2


As far as I understand, being Kähler only needs an almost complex manifold

This is false.

A Kähler manifold is a complex manifold equipped with a hermitian metric such that the associated two-form is closed. The corresponding notion for an almost complex manifold as opposed to a complex manifold is called an almost Kähler manifold.

Your confusion has little to do with the definition of Kähler, but rather with the definition of a holomorphic vector bundle.

By definition, a holomorphic vector bundle is a complex vector bundle over a complex manifold $X$ such that the total space $E$ is a complex manifold and the projection map $\pi : E \to X$ is holomorphic. If $X$ is not complex, no bundle $\pi : E \to X$ is holomorphic.

In particular, while an almost complex structure on $M$ gives rise to a splitting $TM\otimes\mathbb{C} = T^{1,0}M\oplus T^{0,1}M$, the complex bundle $T^{1,0}M \to M$ is a holomorphic vector bundle only when the almost complex structure is integrable, in which case $M$ is complex.

Note, calling $T^{0,1}M$ the antiholomorphic tangent bundle is potentially misleading as there is no such thing as an antiholomorphic vector bundle. A holomorphic vector bundle has holomorphic transition functions, but a vector bundle cannot have antiholomorphic transition functions as the composition of antiholomorphic functions is holomorphic.

  • $\begingroup$ My comment about Kahler manifolds was only in reference to the base manifold not necessarily being complex (which would make the question trivial). If I understand correctly, you're saying a holomorphic vector bundle may only be considered over a complex manifold (so the question is indeed trivial). $\endgroup$ Jul 24, 2016 at 17:25
  • $\begingroup$ Yes, part of the definition of a holomorphic vector bundle is that the base is complex. $\endgroup$ Jul 24, 2016 at 17:36

Take two smooth manifolds $N$ and $M$ with almost-complex structures $I$ and $J$, respectively. Consider a smooth map $f \colon N \to M$, with derivative $f_* \colon TN \to TM$.

When $I$ and $J$ are complex structures, $f$ is holomorphic if and only if $J f_* = f_* I$. If we use this characterization as the definition of "holomorphic," we get a definition that makes sense even when $I$ and $J$ are merely almost-complex structures.

Now, consider a complex vector bundle $\pi \colon E \to M$. From the definition of a complex vector bundle, $E$ comes with a complex structure $i \colon E_m \to E_m$ on every fiber. Pick an almost-complex structure $I$ on the total space $E$ which is compatible with the fiber-wise complex structure, in the sense that it agrees with the action of $i$ on vertical tangent vectors. (A "vertical tangent vector" is a vector in the kernel of $\pi_* \colon TE \to TM$. I'll leave it as an exercise to guess how $i$ is supposed to act on vertical tangent vectors.)

Now we can say $E$ is "holomorphic" if $\pi$ is holomorphic, using our earlier definition of holomorphicity for maps between almost-complex manifolds.

This answer comes with the standard caveat that even if you can generalize a definition, it doesn't mean you should. This definition of a holomorphic vector bundle may only turn out to be useful when one or both of $I$ and $J$ are complex structures. At the very least, however, this definition should help you understand which results about holomorphic vector bundles fail, and why, when the base space or the total space is merely almost-complex.

  • 3
    $\begingroup$ I would be tempted to call this a pseudoholomorphic vector bundle. $\endgroup$ Jul 24, 2016 at 21:41
  • $\begingroup$ Aha! I knew there was a name I liked better than "$J$-holomorphic." @MichaelAlbanese, thanks for the tip. When I get a chance, I'll revise my answer to change the terminology. $\endgroup$
    – Vectornaut
    Jul 25, 2016 at 19:11

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