$d\langle \alpha,\beta\rangle$ for two $k$ forms $\alpha,\beta$ on a manifold

Question

I want to write an expression for $d\langle\alpha,\beta\rangle$ where $\alpha,\beta$ are two $k$ forms on a closed Riemannian manifold say $M$. Now I would want to write something like \begin{align*} d\langle\alpha,\beta\rangle=\langle d\alpha,\beta\rangle+\langle\alpha,d\beta\rangle \end{align*} But it won't be right as $d\alpha$ and $d\beta$ would become $k+1$ forms and also the end product should be a $1$-form. What should be the correct expression and the interpretation for that? Here the inner product is the usual inner product on forms using Hodge-star operator.

Answer 1

Here's a direct calculation.

$$\langle \alpha, \beta\rangle = g^{i_1j_1} \cdots g^{i_kj_k} \alpha_{i_1\cdots i_k} \beta_{j_1\cdots j_k}$$

Assuming that the calculation is the center of a normal coordinates,

\begin{align} \partial_i \langle \alpha, \beta\rangle &= \partial_i \alpha_{i_1\cdots i_k} \ \beta_{i_1\cdots i_k} + \alpha_{i_1\cdots i_k}\ \partial_i \beta_{i_1\cdots i_k}\\ &= \nabla_i \alpha_{i_1\cdots i_k} \ \beta_{i_1\cdots i_k} + \alpha_{i_1\cdots i_k}\ \nabla_i \beta_{i_1\cdots i_k} \end{align}

where $\nabla \alpha$ is the covariant differentiation of $\alpha$, which is a $(0,k+1)$-tensor (which might not be a $(k+1)$-form). Formally one can still write

$$d\langle \alpha, \beta\rangle = \langle \nabla \alpha, \beta \rangle+ \langle \nabla \beta, \alpha\rangle$$

if by the inner product one refers to

$$\langle \nabla \alpha, \beta \rangle = C ( \nabla \alpha \otimes \beta^\sharp),$$

where $(\beta^\sharp)^{j_1\cdots j_k} = g^{i_1j_1}\cdots g^{i_kj_k} \beta_{i_1\cdots i_k}$ and $C$ is some contraction.

I guess it is not possible that $d\langle \alpha, \beta\rangle$ depends only on $\alpha, d\alpha, \beta , d\beta$ but not the full derivative $\nabla \alpha, \nabla \beta$: For example, if $\alpha$ is closed, there are cases where $\|\alpha\|^2$ is non-constant, so $d\langle \alpha, \alpha\rangle$ should be a non-zero one form.