# Can one define an almost complex structure on any complex vector bundle over an almost complex manifold?

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?

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

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.
• I think you meant "[...] with almost complex structure $J$ [...]" and not "$E$". Commented Feb 23, 2023 at 23:42