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Let $M$ be a $m$-dimensional Riemannian manifold. I will follow the notation of the book "From calculus to cohomology - Madsen and Tornehave"

If $\xi $ is a $k$-dimensional complex vector bundle, what is the connection ?

If $\{e_i\}$ is a basis of $\Omega^0(\xi)$ then any section has the form $$ s=\sum_{i=1}^k [s_j(x_1, ... , x_m) + i t_j(x_1, ... ,x_m) ]e_j$$

Hence $$\nabla\ s = \{\ [ds_j + i\ dt_j ] + (s_l + it_l ) A_{lj}\ \}\ e_j $$

So $$\xi = \xi_0\otimes_{\mathbb R}\varepsilon^1_{\mathbb C}$$ and $A$ is the connection on $\xi_0$.

Further, $$ s=u+iv,\ F^{\nabla^\xi} (s) = F^{\nabla^{\xi_0}} (u) +i F^{\nabla^{\xi_0}} (v)$$

Am I right ? Thank you in advance.

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2 Answers 2

up vote 3 down vote accepted

There's a few issues. Firstly, asking "what is the connection" doesn't usually make sense: an arbitrary vector bundle does not have a canonical connection.

Secondly, to be able to decompose $s$ into it's real and imaginary parts as you did may not always be possible. Indeed, this happens if and only if the complex bundle is the complexification of a real vector bundle (which doesn't always happen and a necessary condition is that the Chern classes vanish in degrees $4k+2$).

Unless are you given that $\xi = \xi_0 \otimes \mathbb C$?

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For this, consult P. Deligne, LNM 163, Equations Differentielles a Points Singuliers Reguliers, Chap II.

That definition is good for analytic spaces(in particular for complex manifolds) and also for algebraic geometry. You can see that it is similar to the usual definition.

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