Take the 2-minute tour ×
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.

I've recently read a paper that used a covariant derivative product rule for cross products. It was something like $\nabla_v (A \times B) = (\nabla_v A) \times B + A \times (\nabla_v B)$. Here, $A, B$ are vector fields on a regular surface $S$, and $\nabla$ is the Levi-Civita connection.

From my limited knowledge of covariant derivatives, this seems implausible. For instance, I believed that the output of the covariant derivative always lies in the tangent plane, which seems to contradict the above rule. For example, assume $A$ and $B$ are tangent vectors, then $\nabla_v A, B$ are tangent vectors, so $\nabla_v A \times B \notin T_p(S)$.

I'd like to read up on this, but none of my standard books on differential geometry cover the cross product. I understand that the cross product is a (1, 2) tensor, so it should follow the product rule associated with tensors, but I'm not sure if that results in the product rule above. Could any differential geometers please give me a reference? Thanks!

share|improve this question
The output of the covariant derivative always lies in the tangent plane only if you take the covariant derivative of the tangential fields w.r.t. to a tangential field. In your case you differentiate something ambient. In fact, $A \times B$ will be normal to the surface. –  Yuri Vyatkin Mar 21 '12 at 22:22
add comment

1 Answer

The Leibniz rule indeed holds even in this case.

One of the quick ways to see that is to recall that the cross product can be equivalently defined by $$ A \times B := (A^{\flat} \wedge B^{\flat})^{\sharp} $$

Assuming that the covariant derivative works as expected we may write the following lines $$ \begin{align} \nabla_{v}(A \times B) &= (\nabla_{v}A^{\flat} \wedge B^{\flat} + A^{\flat} \wedge \nabla_{v}B^{\flat})^{\sharp} \\ &= (\nabla_{v}A^{\flat} \wedge B^{\flat})^{\sharp} + A^{\flat} \wedge \nabla_{v}B^{\flat})^{\sharp} \\ &= \nabla_{v}A \times B + A \times \nabla_{v}B \end{align} $$

If we look at this calculation more carefully we will observe that there are different covariant derivatives that are involved there! The normal connection acts on the normal fileds such as $A \times B$ here, while the "intrinsic" covariant derivative acts on the tangential fields $A$ and $B$.

We should have adorned our $\nabla$-s with some marks to distinguish them with regards to the bundle they act on, but this is quite customary in differential geometry to use the same $\nabla$ for all bundles involved in calculations provided the reader knows where the sections are taken from.

In fact, if one diligently writes down all the definitions it will be clearly visible that the coordinate presentations of the operations strikingly differ.

share|improve this answer
add comment

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.