Let $E \to M$ be a complex vector bundle over an almost complex manifold.

Is there always an almost complex structure on $E$? I.e. does there always exist some $J : E \to E$ such that $J^2 = -Id$?

My question is motivated by two cases. First, if $M$ is a complex manifold, we know this is always true. This is because on a complex manifold, almost complex structures on $E$ (where a Hermitian metric $h$ is given) are in correspondence with Hermitian/unitary connections (see here). And one can easily show that any complex bundle $(E,h)$ over $M$ admits a Hermitian connection.

Second, one can further show that any complex vector bundle over any (compact?) smooth manifold admits a Hermitian connection. This is achieved by taking the corresponding classifying map into a large enough conplex Grassmannian and pulling back the Chern connection of the tautological bundle with respect to a given metric on it. Since pullbacks of metric compatible connections are also metric compatible with respect to the pullback metric, the pullback bundle has a metric compatible connection, i.e. a Hermitian connection.

So one can take, for example, the tangent bundle of $S^4$ which we know does not admit an almost complex structure, yet it admits a Hermitian connection. (EDIT: this is not a complex vector bundle) So the correspondence between Hermitian connections and almost complex structures fails for a manifold that is not almost complex, let alone complex.

I'm wondering if this correspondence works for almost complex manifolds? I.e.:

Let $E \to M$ be a complex vector bundle over an almost complex manifold, with a Hermitian metric $h$. Are Hermitian connections on $E$ in correspondence with almost complex structures on $E$, as is the case for when $M$ is a complex manifold?

  • $\begingroup$ $S^4$ doesn't admit a hermitian connection because its tangent bundle is not a complex vector bundle. $\endgroup$ Commented Oct 31, 2022 at 13:57
  • $\begingroup$ @MichaelAlbanese right, silly mistake of mine. $\endgroup$ Commented Oct 31, 2022 at 14:10

1 Answer 1


Let $E$ be a complex vector bundle. Consider the bundle endomorphism $J : E \to E$ given by $J(v) = iv$. As $J \circ J = -\operatorname{id}_E$, we see that $J$ is an almost complex structure on $E$.

Conversely, if $E$ is a real vector bundle with almost complex structure $J$, then $E$ obtains the structure of a complex vector bundle by defining $(a + ib)v := av + bJ(v)$.

In particular, when applied to $TM$, we see that $M$ admits an almost complex structure if and only if $TM$ admits the structure of a complex vector bundle.

  • $\begingroup$ I see, thanks. So does this mean the correspondence in my second question still makes sense? $\endgroup$ Commented Oct 31, 2022 at 14:34
  • $\begingroup$ I think you meant "[...] with almost complex structure $J$ [...]" and not "$E$". $\endgroup$
    – Didier
    Commented Feb 23, 2023 at 23:42
  • $\begingroup$ @Didier: Indeed I did. Thanks! $\endgroup$ Commented Feb 24, 2023 at 0:52

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