$\newcommand{\sig}{\sigma}$ $\newcommand{\tr}{\operatorname{tr}_{\eta}}$ $\newcommand{\al}{\alpha}$ $\newcommand{\be}{\beta}$ $\newcommand{\til}{\tilde}$

Let $E$ be a smooth vector bundle over a manifold $M$. Suppose $E$ is equipped with a metric $\eta$ and with a compatible metric connection $\nabla$. Note that $\nabla$ induces a connection on $E \otimes E$.

Let $\sig \in \Omega^k(M,E \otimes E)$ be an $E \otimes E$-valued differential form of degree $k$.

We denote by $d_{\nabla}$ the corresponding *covariant exterior derviative* on $E \otimes E$-valued forms:

$$ d_{\nabla}:\Omega^k(M,E \otimes E) \to \Omega^{k+1}(M,E \otimes E) $$

**Now we can apply contraction and exterior differentiation in two orders:**

(1) Let $\sigma \in \Omega^k(M,E \otimes E)$; $\, \,d_{\nabla} \sig \in \Omega^{k+1}(M,E \otimes E)$, so $\tr(d_{\nabla} \sig) \in \Omega^{k+1}(M)$ (a real valued form).

(2) $\tr(\sig) \in \Omega^{k}(M) \Rightarrow d\left(\tr(\sig)\right) \in \Omega^{k+1}(M)$ ($d$ is the standard exterior derivative of course)

**Question:** Is it true that $\tr(d_{\nabla} \sig) =d\left(\tr(\sig)\right)$ ?

I tried to prove this via induction on the degree of the form, but I am unable to finish the proof (see details below).

I suspect there is a more general commuting property hiding in the shadows, i.e commutation between exterior derivative and contraction for $E^* \otimes E$-valued forms.

I would be happy to find a reference (or a self-contained proof, of course).

**Here is an attempt to prove this by induction on the degree:**

$k=0$: $\sig \in \Omega^0(M,E \otimes E)=\Gamma(E \otimes E)$. Let $X \in \Gamma(TM)$.

$$ d_{\nabla} \sig \,(X)=\nabla_x^{E \otimes E} \sig \in \Gamma(E \otimes E)$$

Since the assertion is local, we can assume $\sig=\al \otimes \be$, where $\al ,\be \in \Gamma(E)$. So, on the one hand

$$ (*) \tr(d_{\nabla} \sig) (X) = \tr \left(d_{\nabla} \sig \left(X\right )\right) = \tr(\nabla_x^{E \otimes E} \left (\al \otimes \be\right))=\tr(\nabla_x^E \al \otimes \be) + \tr( \al \otimes \nabla_x^E \be)$$

$$ =\widetilde{\nabla_x^E} \al \,(\be) + \til \al (\nabla_x^E \be) $$

where given a section $\gamma \in \Gamma(E)$, we denote by $ \til \gamma \in \Gamma(E^*)$ its corresponding section obtained using the metric $\eta$.

On the other hand,

$$ \tr(\sig) =\tr(\al \otimes \be)= \til \al (\be)= \langle \al , \be \rangle_{\eta} \in \Omega^0(M)=C^{\infty}(M),$$ hence

$$ (**) \, d\left(\tr(\sig)\right) (X)= X \cdot \langle \al , \be \rangle_{\eta} = \langle \nabla_X^E \al , \be \rangle_{\eta} + \langle \al , \nabla_X^E \be \rangle_{\eta} =\widetilde{\nabla_x^E} \al \,(\be) + \til \al (\nabla_x^E \be)$$

Now $(*), (**)$ implies the desired equality.

**Now, let us assume the assertion holds for all $j$-forms when $j < k$.**
(I am not sure how to proceed from here)