Let $(M,g)$ be an oriented riemannian manifold, $(E,h)$ be an hermitian fiber bundle over $M$ and $\nabla$ a connection compatible with respect to the metric. Is it possible to define a wedge product that trasfmors two $E$-valued differential forms onto an usual differential form?
I think so. If $\xi$ and $\eta$ are $E$-valued differential forms, the wedge product $\xi\wedge \eta$ is an $E\otimes E$-valued differential form (cfr https://en.wikipedia.org/wiki/Vector-valued_differential_form). Nevertheless, the metric $h$ provides a canonical isomorphism $E\cong E^\ast$ so i can define
$$
(\alpha\otimes s)\wedge (\beta\otimes t):=(\alpha\wedge \beta)\,h(s,t)
$$
for every couple of usual differential forms $\alpha,\beta$ over $M$ and sections $s,t$ of $E$.
Is this a good definition of a wedge product? If the answer to the previous question is positive, then is there a formula for
$$
\mathrm d(\xi\wedge \eta)?
$$
I imagine something like this
$$
d(\xi\wedge \eta)=\nabla \xi\wedge \eta+(-1)^{\text{deg}(\xi)} \xi\wedge \nabla \eta
$$
but i wasn't able to prove it. Can you help me, please?
Best Answer
This is almost the definition that leads to your desired Leibniz's rule, you only need to accomodate sesquilinearity instead of complex bilinearity :)
In short, define a pointwise pairing $$ (E\otimes \Lambda^r T^*_\mathbb{C}M) \times (E\otimes \Lambda^s T^*_\mathbb{C}M)\rightarrow \Lambda^{r+s}T^*_\mathbb{C}M $$ by your formula on basic elements
$$ (\alpha\otimes s)\wedge (\beta\otimes t):=(\alpha\wedge \bar{\beta})\,h(s,t) $$ and then extend by linearity. In this way this pairing is complex antilinear in the second component, which is the notion that makes sense so as to extend a Hermitian product to a Hermitian wedge product. A word of caution: I am referring to the complexification of the cotangent bundle, by $T^*_\mathbb{C}M$ I actually mean $T^* M \otimes_{\mathbb{R}} \mathbb{C}$ the complex vector bundle of differential 1-forms over $M$ with complex coefficients!
I suggest you check the following references, if you haven't already, even though usually authors in complex differential geometry tend to skip over this (in my opinion!) very important and confusing notational aspects!
If I recall correctly, all of the above introduce the notion of a metric connection $\nabla$ on a Hermitian vector bundle $(E,h)$ as those satisfying
$$ d(s,t)= \nabla s \wedge_h t + s \wedge_h \nabla t $$
for every pair of sections $s,t\in \Gamma_{C^{\infty}}(X,E)$. This immediately implies that for $E$-valued forms, the induced covariant derivative satisfies the graded Leibniz's rule above
$$d(\xi\wedge_h \eta) = \nabla\xi\wedge_h \eta + (-1)^{\deg(\xi)} \xi \wedge_h \nabla \eta$$
for $E$-valued forms $\xi,\eta$. Let us actually prove what you wanted, but for basic forms. $$ d[ (\alpha\otimes s) \wedge_h (\beta \otimes t)] = d[ (\alpha\wedge\bar{\beta}) h(s,t)] = d\alpha \wedge \bar{\beta} h(s,t) + (-1)^{\deg{\alpha}} \alpha\wedge \bar{d \beta} h(s,t) + (-1)^{\deg \alpha + \deg\beta}\alpha\wedge\bar{\beta}\wedge dh(s,t) = (d\alpha \otimes s)\wedge_h (\beta \otimes t) + (-1)^{\deg \alpha} (\alpha\otimes s)\wedge_h (d\beta\otimes t) + (-1)^{\deg \alpha + \deg \beta} (\alpha\otimes \nabla s)\wedge_h (\beta\otimes t) + (-1)^{\deg \alpha + \deg \beta} (\alpha\otimes s)\wedge_h (\beta\otimes \nabla t) $$ Now recalling that the connection should satisfy $$ \nabla(\alpha\otimes s) = d\alpha \otimes s + (-1)^{\deg\alpha} \alpha\otimes \nabla s $$ then the above yields precisely
$$ \dots = (d\alpha \otimes s + (-1)^{\deg\alpha} \alpha \otimes \nabla s) \wedge_h (\beta \otimes t) +(-1)^{\deg\alpha} (\alpha \otimes s)\wedge_h (d\beta \otimes t + \beta \otimes (-1)^{\deg\beta} \beta \otimes \nabla t) = \nabla(\alpha\otimes s) \wedge_h (\beta\otimes t) +(-1)^{\deg \alpha} (\alpha\otimes s) \wedge_h \nabla(\beta\otimes t) $$
DISCLAIMER: for me, Hermitian metrics are linear in the first and antilinear in the second arguments. Some authors (mostly physicists) would choose the other convention. Not a big deal.
BONUS: Note that you have not used the Riemannian metric g yet, and you can (after extending it sesquilinearly to $T^*_\mathbb{C}M=T^*M\otimes \mathbb{C}$) defined yeat another pairing, this times giving an honest Hermitian product in each fiber $$ (E\otimes \Lambda^r T^*M) \times (E\otimes \Lambda^r T^*M)\rightarrow \mathbb{C} $$ The latter $\mathbb{C}$ should really be read as the trivial complex line bundle $M\times\mathbb{C}$. This pairing should be defined as
$$ \langle(\alpha\otimes s),(\beta\otimes t)\rangle:=h_g(\alpha, {\beta})\,h(s,t) $$ where $h_g$ is the Hermitian metric on $T^*_\mathbb{C}M$, which is the sesquilinear extension of the Riemannian metric $g$ on $T^*M$, $$ h_g(\alpha\otimes z,\beta\otimes w) := z\bar{w}g(\alpha,\beta) $$ for evert $\alpha,\beta$ elements of $T^*M$ and $z,w$ complex numbers.