Explicit formula for codifferential I want to show that $$\delta\eta \overset{!}{=} \nabla^*\eta = -\sum\limits_{i=1}^{n} E_i \lrcorner \nabla_{E_i}\eta$$ where $\delta:\Omega^{k+1} \to \Omega^k$ is the adjoint of the exterior derivative $d:\Omega^k \to \Omega^{k+1}$ and $\nabla^*:\mathcal{T}^0_{k+1} \to \mathcal{T}^0_{k}$ is the adjoint of the covariant derivative $\nabla:\mathcal{T}^0_k \to \mathcal{T}^0_{k+1}$ of $(0,k)$-tensor fields, both w.r.t to the averaged inner product $(\cdot,\cdot) = \int_M g(\cdot,\cdot)$ (assume $M$ is compact for simplicity).
I think usually one proves this using the Hodge star operator (see e.g. [Poor], Differential Geometric Strucutres, p.152--154).
What I tried to do is this:
I know that
$$
 d{\omega}(X_0,\dots,X_k) = \sum\limits_{j=0}^k (-1)^j(\nabla\omega)(X_j,X_0,\dots,\widehat{X_j},\dots,X_k).
$$
Note that $\nabla\omega$ is not skew-symmetric, so we can not simplify this any further.
Let $\tau_{i,j} : \mathcal{T}^0_k \to \mathcal{T}^0_k$ denote the transposition of the $i$th and $j$th entry, i.e.,
$$
 (\tau_{i,j} \circ T)(X_1,\dots,X_i,\dots,X_j,\dots,X_k) := T(X_1,\dots,X_j,\dots,X_i,\dots,X_k)
 $$
if $T\in \mathcal{T}^0_k$ and $X_1,\dots,X_k \in \mathfrak{X}(M)$. The adjoint of $\tau_{i,j}$ with respect to the  inner product $g$ on $\mathcal{T}^0_k$ is $\tau_{i,j}$ itself. Hence the same holds for the averaged inner product $(\cdot,\cdot)$ on $\mathcal{T}^0_k$.  With this notation the above formula for $d$ becomes
$$
  d{\omega} = \sum\limits_{j=0}^k (-1)^j (\tau_{0,j} \circ \nabla\omega)
 $$
and it follows that for all $\omega \in \Omega^k$ and $\eta \in \Omega^{k+1}$
\begin{align*}
  (d{\omega},\eta) &= \sum\limits_{j=0}^k (-1)^j (\tau_{0,j} \circ \nabla\omega, \eta) \\
  &= \sum\limits_{j=0}^k (-1)^j (\nabla\omega, \tau_{0,j} \circ \eta) \\
  &= \sum\limits_{j=0}^k (-1)^j (\omega, \nabla^*(\tau_{0,j} \circ \eta)) \\
  &= (\omega,\nabla^*\eta) + \sum\limits_{j}^{k} (-1)^j (\omega, \nabla^*(\underbrace{\tau_{0,j} \circ \eta}_{=-\eta})) \\
  &= \left(\omega,\left(1+\sum\limits_{j}^{k} (-1)^{j+1}\right)\nabla^*\eta \right),
 \end{align*}
so $\delta\eta = 2\nabla^*\eta$ if $k$ is odd and $\delta\eta = \nabla^*\eta$ is $k$ is even.
This is of course the wrong result and it also feels really weird. After reading this I realized that the problem probably is that $\tau_{i,j}$ is not globally defined (and obviously not independent of the frame (not even of its order)).
EDIT: I don't think that this is the problem. $\tau_{i,j}$ is globally defined. Then I wonder even more: Where is this going wrong?
But my hope is that this could be fixed somehow. The usual proof with the Hodge star or, I think, another possibility is to define the 1-form $\alpha(X) = g(X \lrcorner \eta, \omega)$ and then go on from
$$
  \text{div } \alpha^\sharp = g(\underbrace{\sum\limits_{i=1}^{n} E_i \lrcorner\nabla_{E_i}\eta}_{=-\nabla^*\eta}, \omega) + \sum\limits_{i=1}^{n}g(E_i \lrcorner \eta, \nabla_{E_i} \omega),
 $$
both involve "hardcore" combinatorics. (In the latter case I need to show that $\sum\limits_{i=1}^{n}g(E_i \lrcorner \eta, \nabla_{E_i} \omega)$ is $g(d\omega,\eta)$ which feels like an index war.)
I guess the combinatorics must be somewhere and the above approach with the $\tau_{i,j}$ might just be "too simple".
Still I'd like to know if this approach can be realized somehow and if not what's the real reason for this to go wrong.
Also, please correct me if anything else is wrong.
 A: I think I solved it. First of all, the formula for $d$ with the $\tau_{0,j}$'s was of course wrong. It should be
$$
  d{\omega} = \sum\limits_{j=0}^k (-1)^j (\tau_j \circ \nabla\omega).
 $$
where $\tau_j := \tau_{0,1} \circ \dots \circ \tau_{j-1,j}$.
The other problem was the (hidden) interchange of inner products. Let me elaborate:
For $k$-forms $\omega,\eta \in \Omega^k$ we defined the inner product as
$$
  g^\Omega(\omega,\eta) = \sum\limits_{1 \le i_1 < \dots < i_k \le n} \omega_{ i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k}
 $$
if we write $\omega$ and $\eta$ locally in an orthonormal frame. On the hand, if we would interpret $\omega$ and $\eta$ as skew-symmetric tensor fields, we would have
\begin{align*}
  g^\mathcal{T}(\omega,\eta) &= \sum\limits_{1 \le i_1,\dots,i_k \le n} \omega_{i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k} \\
  &= \sum\limits_{1 \le i_1\neq\dots\neq i_k \le n} \omega_{i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k} \\
  &= k! \sum\limits_{1 \le i_1<\dots<i_k \le n} \omega_{i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k} \\
 \end{align*}
by skew-symmetricity of $\omega$ and $\eta$. This pushes through to the averaged inner product, i.e., we have
$$
  (\omega,\eta)^\mathcal{T} = k!(\omega,\eta)^\Omega.
 $$
Then everything works out:
Let $\omega \in \Omega^k$ and $\eta \in \Omega^{k+1}$. The claim now follows from the definition of $\delta$ and
\begin{align*}
  (d{\omega},\eta)^\Omega &= \tfrac{1}{(k+1)!}\sum\limits_{j=0}^k (-1)^j (\tau_j \circ \nabla\omega, \eta)^\mathcal{T} \\
  &= \tfrac{1}{(k+1)!} \sum\limits_{j=0}^k (-1)^j (\nabla\omega,  \underbrace{\tau_{j-1,j}\circ\dots\circ\tau_{0,1} \circ \eta}_{(-1)^j\eta})^\mathcal{T} \\
  &= \tfrac{k+1}{(k+1)!} (\nabla\omega, \eta)^\mathcal{T} \\
  &= \tfrac{1}{k!} (\omega, \nabla^*\eta)^\mathcal{T} = (\omega,\nabla^*\eta)^\Omega.
 \end{align*}
