I struggle with making sense of the claim in Chapter 16.2 of Arthur Besse's book "Einstein manifolds".
Let $\nabla$ be an affine connection on the manifold $\mathcal{M}$. We want to decompose the covariant derivative of the Ricci tensor $\nabla Ric$ into the irreducible representations of the orthogonal group.
Since we assume the connection to be Levi-Civita, the tensor $\nabla Ric$ is a section of a vector bundle $H \subset T^*\mathcal{M} \otimes S^2 \mathcal{M}$, the fibre of which consists of tensors symmetric in its last two indices and obeying the Bianchi identity: $$ 2 \sum \xi(X_i, X_i,X) = \sum \xi(X,X_i, X_i) $$
For the purposes of the classification, the author introduces a number of natural vector bundle homomorphisms, and only two of them are relevant to my question, the contraction on the first two indices $$ \gamma(\xi)(X) = \sum \xi(X_i, X_i, X) $$ and the construction of rank-3 tensor (a section of $H$) out of a covector $$ \varphi(\zeta)(X,Y,Z) = (X,Y) \zeta(Z) + (X,Z) \zeta(Y) + 2n (n-2)^{-1} (Y,Z) \zeta(X) $$ where $X,Y,Z\in T_x \mathcal{M}$, $\zeta\in T^*_x \mathcal{M}$, and $n = \dim \mathcal{M}$.
I cannot understand the author's argument that $Q=Im\,\varphi$ and $\Gamma = H \cap Ker\, \gamma$ are orthogonal complements in $H$: $Q \oplus \Gamma = H$.
First, consider $\gamma \circ \varphi = \frac{2 n}{n-2} Id_{T^*M}$. This clearly implies that $Q \cap \Gamma = 0$. Next, the author claims that since $$ (\varphi(\zeta),\xi) \, =^? \frac{7n-6}{n-2}(\gamma(\xi), \zeta) $$ it follows that $Q \cup \Gamma = H$.
I understand neither the logic of the last claim nor the coefficient on the RHS. Should not it be $\dfrac{6n-4}{n-2}$?
