Let $(M,g)$ be a $4$-dimensional Riemannian manifold. The Riemann curvature tensor can be viewed as an operator $\mathcal{R}:\Lambda^2(T^{\star}M)\longrightarrow \Lambda^2(T^{\star}M)$ defined in this way (I'm using Einstein's notation):
$$(\mathcal{R}(\omega))_{ij}=\frac{1}{2}R_{klij}\omega_{kl}$$
where $\omega_{ij}$ are the components of the $2$-form $\omega$ with respect to an orthonormal basis $\{e^i \wedge e^j\}_{i,j=1,..,4}$ and $R_{ijkt}$ are the components of the Riemann curvature tensor.
Because of the splitting $\Lambda^2 (T^{\star}M)=\Lambda_{+}\oplus \Lambda_{-}$, the operator $\mathcal{R}$ can be written in a block form
$$\begin{pmatrix} A & B \\ ^{t}B & C \end{pmatrix}$$
where $^{t}A=A$, $^{t}C=C$ and $trA=trC=\frac{S}{4}$, with $S$ the scalar curvature of $M$.
So basically, if we write a $2$-form $\omega=\omega_{+} + \omega_{-}$ according to the splitting, we have that we can also write $\mathcal{R}\omega=(\mathcal{R}\omega)_{+}+(\mathcal{R}\omega)_{-}$: in particular, $\mathcal{R}(\omega_{\pm})=\mathcal{R}(\omega_{\pm})_{+}+\mathcal{R}(\omega_{\pm})_{-}$ and, for example, $A$ "sends" $\omega_+$ in $\mathcal{R}(\omega_{+})_{+}$.
Now, I read in a few papers that, if we call $A_{ij}$ the components of the matrix $A$, with $i,j=1,2,3$, we have that $A_{12}=A_{13}=0$ if and only if $A$ is a multiple of the identity matrix. How is that possible?
I tried to work directly with the components of the Weyl tensor $W$ of the manifold, since its self-dual part is strictly related to the matrix $A$, but I couldn't get anything. I tried also to exploit the fact that a matrix is a multiple of the identity matrix if and only if it commutes with any other matrix, but I don't know if it's useful.
EDIT: In these days I read also that this fact may be related to the property of the curvature form $\Omega$, i.e. $R_g ^{\star}\Omega = g^{-1}\Omega g$, where $R_g$ denotes the right multiplication by $g\in SO(4)$ in the bundle of orthonormal frames of $N$. How could it help?
Best Answer
I think you may be over complicating things, just quickly looking this is what I saw.
The operator $\mathcal{R}$ can be written in a block form
$$\begin{pmatrix} A & B \\ ^{t}B & C \end{pmatrix}$$
where $^{t}A=A$, $^{t}C=C$ and $trA=trC=\frac{S}{4}$, with $S$ the scalar curvature of $M$.
Since $^{t}A=A$, we know the $3\times 3$ matrix will be symmetric about it's diagonal. The reverse direction is easy to show, since if we assume the contrapositive, (i.e. $A$ is not a multiple of the identity matrix) then $A_{1,2}\neq A_{1,3}\neq 0$ follows naturally.
Now, assume $^{t}A=A$ and $A_{1,2}=A_{1,3}=0$. Write out $A$ with all of the zeros in place (all positions such that $i\neq j$). This could be a multiple of the identity matrix. What's left to show is the diagonals $\alpha$, $\beta$, $\gamma$ are in fact all equal. So use the fact that $trA = \frac{S}{4}$, which I'm assuming applies to each entry of the diagonal. Since $\frac{S}{4}$ is constant, then $\alpha=\beta=\gamma$, and can be written as some scaled version of the identity matrix.
EDIT
I didn't explicitly mention this, it was inherent to my line of reasoning above.
If $A_{2,3} \neq 0$ then the operator $\mathcal{R}$ has for matrix $A$ in block form a free variable when reduced to echelon form. If the free-variable is present, "$A$ sending $\omega_+$ in $\mathcal{R}(\omega_+)_+$" is not unique. I'm assuming the spectral properties of the operator are commutative, if not, the entire argument falls apart.