Invariant description of the Weitzenböck curvature operator by Bourguignon I recently came across the paper Les variétés de dimension 4 à signature non nulle dont
la courbure est harmonique sont d’Einstein by Jean Pierre Bourguignon. What he shows in §8 is that the Weitzenböck curvature operator $\mathfrak{Ric}_\text{R}$ on $p$-forms is given by $$\mathfrak{Ric}_\text{R}(\omega)(X_1,\dots,X_p) = (\hat{\omega} \circ \hat{R_p})(X_1,\dots,X_p)$$
where
$$R_p = \left(\tfrac{1}{2(p-1)}\text{Ric} \mathbin{\bigcirc\mspace{-19mu}\wedge\mspace{3mu}} \text{g} - \text{Rm}\right) \mathbin{\bigcirc\mspace{-19mu}\wedge\mspace{3mu}} \text{g}^{p-2}.$$
Here, I just can't figure out what he means by $\hat{\omega} \circ \hat{R_p}$.
Earlier in the paper he defines (in (2.11)) for a $C \in S^2\Lambda^{2}V$, interpreted as a self-adjoint map $\Lambda^{2}V \to \Lambda^{2}V$, and a form $\eta \in \Lambda^2(V)$, that $$\hat{C}(\eta) = \sum\limits_{i,j = 1}^n \eta(e_i,e_j)C(e_{i},e_{j})$$ for an orthonormal basis $(e_i)_{1 \le i \le n}$ of $V$.
I see how this can be generalized into:
For a $C \in S^2\Lambda^{p}V$, interpreted as a self-adjoint map $\Lambda^{p}V \to \Lambda^{p}V$, and a form $\eta \in \Lambda^p(V)$, define $$\hat{C}(\eta) = \sum\limits_{i_1, \dots, i_p = 1}^n \eta(e_{i_1},\dots,e_{i_p})C(e_{i_1},\dots,e_{i_p})$$ for an orthonormal basis $(e_i)_{1 \le i \le n}$ of $V$.
And I am thinking that this is what he means here for $C = R_k$. But this still doesn't answer what $\hat{\omega}$ would be.
I am thinking that $(\hat{\omega} \circ \hat{R_p})$ may just mean $\hat{R_p}(\omega)$; but why wouldn't he have written it like this then? Probably I am missing something central here..
The paper can be found here for free:
Les variétés de dimension 4 à signature non nulle dont la courbure est harmonique sont d’Einstein by Jean Pierre Bourguignon
 A: I think the idea is to think of $\hat{R}_p$ as a mapping from $\Lambda^pM$ to itself, and $\hat{\omega}$ as a mapping from $\Lambda^pM$ to $E$ (the vector bundle in which $\omega$ takes values), and then just compose these two operators (to get something $E$ valued in the end).
In the case where $E$ is the trivial (scalar) bundle, it works out to be $\hat{R}_p(\omega)$ since $R_p$ is symmetric.
A: This answer basically just writes out Willie Wong's.
In particular, the definition of $\hat{C}$ works for $\omega$ as well; if $\omega$ is a form with values in $E$, then $\hat{\omega}(\eta) = \sum\limits_{i_1, \dots, i_p = 1}^n \eta_{i_1,\dots,i_p}\omega_{i_1,\dots,i_p} \in E$. But then
\begin{align} 
(\hat{\omega} \circ \hat{R_p})(X_1,\dots,X_p) &= \hat{\omega}(\hat{R_p}(X_1,\dots,X_p)) \\
&= \sum\limits_{i_1, \dots, i_p = 1}^n \omega_{i_1,\dots,i_p}R_p(X_1,\dots,X_p)(e_{i_1},\dots,e_{i_p}) \\
&= \sum\limits_{i_1, \dots, i_p = 1}^n \omega_{i_1,\dots,i_p}R_p(e_{i_1},\dots,e_{i_p})(X_1,\dots,X_p) \\
&= \hat{R_p}(\omega)(X_1,\dots,X_p).
\end{align}
At the second equality, we used that $R_p$ is symmetric; this is the case because the (generalized Kulkarni-Nomizu) product on the graded algebra $KV$ is commutative.
EDIT: Actually, I am not at all satisfied with this answer anymore. What did I do at the second equality yesterday? That doesn't make sense.
I will try again: Working directly with the definitions we have with $\eta = e_{j_1}\wedge\dots\wedge e_{j_p}$ that
\begin{align*}
(\hat{\omega} \circ \hat{R_p})(\eta) &= \hat{\omega}(\hat{R_p}(\eta)) \\
&= \hat{\omega}\left(\sum\limits_{i_1, \dots, i_p = 1}^n  (e^{j_1} \wedge \dots \wedge e^{j_p})(e_{i_1}\wedge\dots\wedge e_{i_p}) \cdot (R_p)(e_{i_1},\dots,e_{i_p})\right) \\
&=\hat{\omega}\left(\sum\limits_{\tau \in S_p} (R_p)(e_{j_{\tau(1)}},\dots,e_{j_{\tau(p)}}) \right) \\
&= \sum\limits_{i_1, \dots, i_p = 1}^n \sum\limits_{\tau \in S_p} \omega_{i_1,\dots,i_p} (R_p)(e_{j_{\tau(1)}},\dots,e_{j_{\tau(p)}})(e_{i_1},\dots ,e_{i_p}).
\end{align*}
On the other hand, using the symmetry of $R_p$, we get
\begin{align*}
\hat{R}_p(\omega)(\eta) &= \left(\sum\limits_{i_1, \dots, i_p = 1}^n \omega_{i_1,\dots,i_p} \cdot (R_p)(e_{i_1},\dots ,e_{i_p}) \right)(\eta) \\
&= \sum\limits_{i_1, \dots, i_p = 1}^n \omega_{i_1,\dots,i_p} \cdot (R_p)(e_{i_1},\dots ,e_{i_p})(\eta) \\
&= \sum\limits_{i_1, \dots, i_p = 1}^n \omega_{i_1,\dots,i_p} \cdot (R_p)(\eta)(e_{i_1},\dots,e_{i_p}).
\end{align*}
Actually, typing this out now I see that, using that $R_p$ is skew-symmetric in the first $p$ arguments, that the two terms are just off by a factor $\vert S_p \vert = p!$.
