Decomposition of the Curvature operator and Matrix representation I'm trying do this question from Peter Petersen's Book and I can't do some parts.
I know that

$$R=\frac{scal}{2n(n-1)}g\circ g+\left(Ric-\frac{scal}{n}g\right)\circ g+W$$

Where, $R$ is the Riemannian curvature tensor, $scal$ is the scalar curvature, $W$ is the Weyl tensor, and we define the Kulkarni-Nomizu product of two (0,2)-tensor as a (0,4)-tensor:

$$h\circ k(v_1,v_2,v_3,v_4)=h(v_1,v_3)\cdot k(v_2,v_4)+h(v_2,v_4)\cdot k(v_1,v_3)-h(v_1,v_4)\cdot k(v_2,v_3)-h(v_2,v_3)\cdot k(v_1,v_4)$$

I can't do this parts:
A) Show that these decompositions are orthogonal, in particular:
$$|R|^2=\left|\frac{scal}{2n(n-1)}g\circ g\right|^2+\left|\left(Ric-\frac{scal}{n}g\right)\circ g\right|^2+|W|^2$$
B) (Look this question to understanding the doubt:Block Decomposition of a linear map on $\Lambda^2TM$) 

(I KNOW DO THIS) The curvature operator $\mathcal{R}:\Lambda^2TM\to\Lambda^2TM$ has the following block decompostion:
  $$\mathcal{R}=\begin{bmatrix}
 A&B \\ 
 C&D 
\end{bmatrix}$$
  Since $\mathcal{R}$ is symmetric, we get $A,C$ are symmetric and that $D=B^*$ is the adjoint of $B$.

(I DON'T KNOW THIS) One can furthermore show that 
$$A=W^++\frac{scal}{12}I$$
$$C=W^-+\frac{scal}{12}I$$
Where the Weyl tensor can be written 
$$\begin{bmatrix}
W^+&0\\
0&W^-
\end{bmatrix}$$
Someone can help me with this two questions?
 A: Answer to (A): The inner product on $(0,4)$-tensors is just the full contraction with the metric $\langle S,T \rangle = S_{ijkl} T^{ijkl}$. First compute (hopefully you can fill in the gaps here)
$$
\begin{eqnarray*}
\left\langle g\circ g,g\circ g\right\rangle  & = & 4\left(g_{ik}g_{jl}-g_{il}g_{jk}\right)\left(g^{ik}g^{jl}-g^{il}g^{jk}\right)\\
 & = & 4\left(2\left(\delta_{i}^{i}\right)^{2}-2\delta_{i}^{i}\right)=8\left(n^{2}-n\right).
\end{eqnarray*}
$$
A similarly boring computation shows
$$ 
\begin{eqnarray*}
\left\langle \mbox{Rc}\circ g,g\circ g\right\rangle  & = & \left(R_{ik}g_{jl}+R_{jl}g_{ik}-R_{il}g_{jk}-R_{jk}g_{il}\right)\left(2g_{ik}g_{jl}-2g_{il}g_{jk}\right)\\
 & = & 8\left(n-1\right)\mbox{Scal}.
\end{eqnarray*}
$$
We can now see the orthogonality of the Scalar and Traceless Ricci parts: $$\left\langle \left(\mbox{Rc}-\frac{\mbox{Scal}}{n}g\right)\circ g,g\circ g\right\rangle =\left\langle \mbox{Rc}\circ g,g\circ g\right\rangle -\frac{\mbox{Scal}}{n}\left\langle g\circ g,g\circ g\right\rangle =0.$$
As for the Weyl part, the key fact is that all traces/contractions of the Weyl tensor are zero. When we form the inner products $\langle W, g \circ g\rangle$ and $\langle W, \mbox{Rc} \circ g \rangle$ as above, we will find that every term contracts the metric with $W$ and therefore vanishes - so the three components are mutually orthogonal. (In fact in retrospect my first computation was unnecessarily involved; we could just show $(\mbox{Rc} - \mbox{Scal}/n\ \ g )\circ g$ is traceless and apply a similar argument).
From here the bilinearity of the inner product and the orthogonality of the components give the desired expression for $|R|^2 = \langle R , R \rangle$.
Answer for (B): Note that this question amounts to showing that $R = W + \frac{\rm Scal}{12} I$ when restricted to $\Lambda^+ \otimes \Lambda^+$ or $\Lambda^- \otimes \Lambda^-$. Since $I$ is just some multiple of $g \circ g$, we know from the decomposition of $R$ that this is equivalent to showing $\mathring{ \textrm{Rc}}\circ g= \left(\textrm{Rc}- \frac{\rm Scal}{n} g\right)\circ g$ vanishes on $\Lambda^+ \otimes \Lambda^+$ and $\Lambda^- \otimes \Lambda^-$. This is fairly easy to check by working in the basis
$$\varphi_{1}^{\pm}=e_{1}\wedge e_{2}\pm e_{3}\wedge e_{4}\quad\varphi_{2}^{\pm}=e_{1}\wedge e_{3}\pm e_{4}\wedge e_{2}\quad\varphi_{3}^{\pm}=e_{1}\wedge e_{4}\pm e_{2}\wedge e_{3}$$
and just checking that $\mathring{\textrm{Rc}} \circ g (\varphi_i^\pm, \varphi_j^\pm) = 0$. For $i \ne j$ this is just a consequence of symmetry; for $i=j$ you also need that the fact that $\mathring{\rm Rc}$ is trace-free.
