Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a nice variety resp. a manifold over the complex numbers. One defines a connection on a vector bundle $V$ on over $X$ as a $\mathbb C-$linear sheaf homomorphism

$\nabla : V\rightarrow V\otimes \Omega^1$

which satisfies the Leibniz rule.

I have read that this is equivalent to giving for each local vector field $Y\in Der_{\mathbb C}(\mathcal O_X)$ a $\mathbb C-$ linear sheaf homomorphism

$\nabla_Y : V \rightarrow V$


(1) Leibniz rule

(2) $\nabla_{fY+gZ}=f\nabla_Y+g\nabla_Z$ for $f,g \in \mathcal O_X$ and $Y,Z$ local vector fields.

I can prove that each connection in the first sense implies a connection in the second sense. But I don't see how you get from the datum of thhe $\nabla_Y$ a connection in the first sense.

Remark: By a vector field I understand a linear derivation of the structure sheaf into itself.

share|cite|improve this question
up vote 3 down vote accepted

Using the fact that $\Omega^1$ is $\mathcal{O}_X$-locally free and dual to the sheaf of vector fields you should be able to prove $Hom_{\mathcal{O}_X}(\Theta_X,\mathcal{E}nd_\mathbb{C}(V)) = Hom_\mathbb{C}(V,V\otimes_{\mathcal{O}_X} \Omega^1)$. In local complex coordinates $(x_1,\ldots,x_n)$, $\{\nabla_Y\}$ is mapped to $\nabla: V \to V\otimes \Omega^1$ defined by $$ \nabla(v) = \sum_{i=1}^n \nabla_{\partial_i}(v) \otimes dx_i $$ (This the analogue of $df = \sum_i \frac{\partial f}{\partial x_i} dx_i$). Condition (2) ensures that this formula does not depend on the choice of coordinates.

share|cite|improve this answer
Hi YBL, thanks for your enlightening answer. But I think you mean "this formula does not depend of..."? – Veen Oct 23 '11 at 14:07
Of course. I edited it. – AFK Oct 23 '11 at 15:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.