0

Let $E\longrightarrow X$ be a rank $n$ vector bundle equipped with an euclidean metric and let $\nabla$ be a connection which is compatible with the metric. We extend $\nabla$ to the exterior algebra bundle obtained from $E$. If $e_1, \cdots, e_n$ are local orthonormal sections forming a local basis do we have $$\nabla \left(e_1 \wedge \cdots \wedge e_n\right)=0$$

?

I do not see why it has to be true but I think I need it to be true.

flag
Yes. With a metric you are working on an $O(n)$ bundle. Extending the structure group to the top-order $\wedge^nE$, you see that its structure group is $O(1)$. – Willie Wong Dec 2 2010 at 22:44
Another way to see it is that if you take any curve $\gamma$ and compare the frame $e_1, \ldots, e_n$ at $\gamma(1)$ with the parallel transport from $\gamma(0)$ to $\gamma(1)$ of $e_1,\ldots, e_n$, by definition they differ by an element $A$ of $O(n)$. Therefore a simple computation shows that the difference between $e_1\wedge \cdots \wedge e_n$ at $\gamma(1)$ and the one parallel transported from $\gamma(0)$ is precisely $\det(A) = \pm 1$. My continuity only $+1$ is allowed, so the two top $n$-vectors are in fact identical. – Willie Wong Dec 2 2010 at 22:56
3 
 I don't consider this question appropriate for MathOverflow. The answer is a straightforward consequence of the definition of the connection induced on the exterior product and the compatibility of the connection with the metric. – Deane Yang Dec 2 2010 at 23:21

closed as too localized by José Figueroa-O'Farrill, Deane Yang, Willie Wong, Sergei Ivanov, Daniel Moskovich Dec 3 2010 at 13:05

Browse other questions tagged or ask your own question.