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.