Formulae for covariant derivative of vector valued form

Question

Suppose $E$ is a vector bundle over $M, d^E$ a covariant derivative, $\sigma\in\Omega^p(E)$ and $\mu$ a q-form.

I have seen the following pair of formulae for wedge products:

$d^E(\mu\wedge \sigma)=d\mu\wedge\sigma+(-1)^q\mu\wedge d^E\sigma$

$d^E(\sigma\wedge\mu)=d^E\sigma\wedge\mu+(-1)^p\sigma\wedge d\mu$

I am quite happy with the first one but I can only get the second one if $q$ is even.

Here is my calculation when we write $\sigma=\omega\otimes s$:

$d^E((\omega\wedge\mu)\otimes s)=(d\omega\wedge\mu+(-1)^p\omega\wedge d\mu)\otimes s+(-1)^{p+q}(\omega\wedge\mu)d^E s$

$=(-1)^p\sigma\wedge d\mu+(d\omega+(-1)^{p+q}\omega d^Es)\wedge\mu$

I think I must have something wrong as there is a similar formula involving the connection on the endomorphism bundle.

Answer 1

You need an extra step of permuting forms to get $\mu$ all the way to the right in your second term, permuting past the one-form $d^Es$ changes the sign by a factor $(-1)^q$, leaving the correct sign $(-1)^p$ in your expression.

Answer 2

Just to add to the discussion, I'd like to write the derivation I went through using abstract index notation. It also allows one to work without writing $\sigma = \omega\otimes s$.

I introduce a bit of notation

$$\begin{align} \sigma &\equiv {\sigma^{c}}_{a_1\dots a_p} & \sigma&\in\Omega^{p}\otimes E\\ \mu &\equiv \mu_{b_1\dots b_q} & \mu&\in\Omega^{q}\\ (d\omega)_{a_0\dots a_k} &= (k+1)\nabla_{[a_0}\omega_{a_1\dots a_k]} &\omega&\in\Omega^{k}\\ {(d^{E}\omega)^{c}}_{a_0\dots a_k} &= (k+1)\nabla_{[a_0}{\omega^{c}}_{a_1\dots a_k]} &\omega&\in\Omega^{k}\otimes E\\ (\alpha\wedge\beta)_{a_1\dots a_{p+q}} &= \frac{(p+q)!}{p!q!} (\alpha\otimes\beta)_{[a_1\dots a_{p+q}]} & \alpha&\in\Omega^{p}, \beta\in\Omega^{q} \end{align}$$

Now, in our case: $${(\sigma\wedge\mu)^{c}}_{a_1\dots a_p b_1\dots b_q} = \frac{(p+q)!}{p!q!}{\sigma^{c}}_{[a_1\dots a_p}\mu_{b_1\dots b_{q}]}$$ Hence $$\begin{align} {[d^{E}(\sigma\wedge\mu)]^{c}}_{da_1\dots a_pb_1\dots b_q} &= (p+q+1)\frac{(p+q)!}{p!q!}\nabla_{[d}({\sigma^{c}}_{a_1\dots a_p}\mu_{b_1\dots b_{q}]}) \\ &= \frac{(p+q+1)!}{p!q!}\left( \nabla_{[d}{\sigma^{c}}_{a_1\dots a_p}\mu_{b_1\dots b_{q}]} + {\sigma^{c}}_{[a_1\dots a_p}\nabla_{d}\mu_{b_1\dots b_{q}]} \right) \\ &= \left( \frac{((p+1)+q)!}{(p+1)!q!}(p+1)\nabla_{[d}{\sigma^{c}}_{a_1\dots a_p}\mu_{b_1\dots b_{q}]} + \frac{(p+(q+1))!}{p!(q+1)!}{\sigma^{c}}_{[a_1\dots a_p}(q+1)\nabla_{d}\mu_{b_1\dots b_{q}]} \right) \\ &= \left( \frac{((p+1)+q)!}{(p+1)!q!}{(d^{E}\sigma)^{c}}_{[da_1\dots a_p}\mu_{b_1\dots b_{q}]} + \frac{(p+(q+1))!}{p!(q+1)!}{\sigma^{c}}_{[a_1\dots a_p}(d\mu)_{db_1\dots b_{q}]} \right) \\ &= {(d^{E}\sigma\wedge\mu)^{c}}_{da_1\dots a_pb_1\dots b_q} + {(\sigma \wedge d\mu)^{c}}_{a_1\dots a_pdb_1\dots b_q} \\ &= {(d^{E}\sigma\wedge\mu)^{c}}_{da_1\dots a_pb_1\dots b_q} + (-1)^{p}{(\sigma \wedge d\mu)^{c}}_{da_1\dots a_pb_1\dots b_q} \end{align}$$

I.e

$${[d^{E}(\sigma\wedge\mu)]^{c}}_{da_1\dots a_pb_1\dots b_q} = {(d^{E}\sigma\wedge\mu)^{c}}_{da_1\dots a_pb_1\dots b_q} + (-1)^{p}{(\sigma \wedge d\mu)^{c}}_{da_1\dots a_pb_1\dots b_q}$$

or, without all the indices:

$$d^{E}(\sigma\wedge\mu) = d^{E}\sigma\wedge\mu + (-1)^{p}\sigma \wedge d\mu$$

I know it is quite messy, but for me it makes things so much clearer. Hope it is helpful.