Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

If I have a contraction of a vector field with a 1-form valued 2-form, what would be the appropiate product rule?

$$d_{\left[a\right.} \left(P_{[bc]i} v^i \right)_{\left. \right]} = \, ?$$

This expression should be torn appart to have some kind of $d v$ and $d P$.

The underlying 3D manifold is a metric one.

As far as I understand one will need to define an exterior derivative for vector-valued differential forms. In turn, that will require a connection, but which one?

share|cite|improve this question
up vote 2 down vote accepted

As I can see, you have a 2-form $P$ with values in a bundle $E^i$, and a 0-form $v$ with values in the mentioned bundle. We can write the contraction as $$ C(P \wedge v)_{ab} = g_{i j} P_{[ab]}{}^{i} v^j $$ and then use the Levi-Civita connection $\nabla$ of the metric given in your manifold. This connection extends to the covariant exterior derivative on forms: $d = \wedge \circ \nabla$

(Of course, $E^i$ is the tangent bundle of your manifold)

Applying this to our contraction we compute $$ (d\,C(P \wedge v))_{abc} = g_{i j} (3 \, \nabla_{[a} P_{bc]}{}^i) v^j + g_{ij} \, 3 \, P_{[bc}{}^i \nabla_{a]}v^j $$

Invariantly, this means $$ d(C(P \wedge v)) = C(d^{\nabla}P \wedge v) + (-1)^2 C(P \wedge d^{\nabla}v) $$ where by $C$ we denote the appropriate contraction operator (with which the derivatives commute).

Compare with the calculation I added to my answer here.

share|cite|improve this answer

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.