[Math] Leibniz rule for exterior derivative of a contraction

differential-formsdifferential-geometry

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?

Best Answer

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.