Besse p134 Riemann tensor in dimension 4

Question

Does someone have a reference for the proof of 4.72 page 134 of Einstein Manifolds? It is said that $$\check{R}-\vert R\vert^2g/4=S/3 (Ric-S/4) +2\mathring{W}(Ric -S/4) $$ because we are in dimension 4, where$\check{R}_{ab}=R_{ajkl}R^{jkl}_b$ and $\mathring{W}$ is Weyl acting on symmetric tensor. Is there is a simple idea to get it, except to develop everything? In fact I am trying to get it from a moving frame point of view, and more generally, I am looking for any reference making computation about curvature functional, Einstein metric and Bach tensor from the moving frame point of view. Thanks in advance.

Answer 1

I don't have a reference, but there is a conceptual way to see that you only have to check a few constants to get this formula.

Notice that it's a statement about a quadratic mapping from the space of curvature tensors in 4D to the space of traceless quadratic forms in 4D (because both sides are clearly traceless quadratic forms). Also, this mapping has to be invariant under change of frame, i.e., under the action of $\mathrm{O}(4)$, and, in particular, $\mathrm{SO}(4)$, so you could expect representation theory to help.

Now, the space of curvature tensors in 4D, when considered as an $\mathrm{SO}(4)$ representation space, is actually a representation of $\mathrm{SO}(4)/\{\pm I_4\} = \mathrm{SO}(3)\times\mathrm{SO}(3)$, which is a product of two simple groups of rank $1$, so the irreducible representations are of the form $V^{p,q} = V^{p,0}\otimes V^{0,q}$, where $V^{p,0}$ is the irreducible representation of $\mathrm{SO}(3)\times\{1\}$ of dimension $2p{+}1$ and $V^{0,q}$ is the irreducible representation of $\{1\}\times\mathrm{SO}(3)$ of dimension $2q{+}1$. With this notation, a curvature tensor $R$ in dimension $4$ is uniquely expressed as a sum $$ R = R^{0,0} + R^{1,1} + R^{2,0} + R^{0,2} $$ where $R^{0,0}$, with values in $V^{0,0}$, is the scalar curvature piece; $R^{1,1}$, with values in $V^{1,1}$, is the traceless Ricci tensor; $R^{2,0}$, with values in $V^{2,0}$, is the self-dual Weyl curvature; and $R^{0,2}$, with values in $V^{0,2}$, is the anti-self dual Weyl curvature.

Now, an $\mathrm{SO}(4)$-equivariant quadratic expression in $R$ taking values in $V^{1,1}$, which is what the left hand side of your equation is, must come from an $\mathrm{SO}(4)$-equivariant map $$ \mathsf{S}^2\bigl(V^{0,0} \oplus V^{1,1} \oplus V^{2,0} \oplus V^{0,2}\bigr)\longrightarrow V^{1,1} $$
But, from the Clebsch-Gordan formula for tensor products of representations of $\mathrm{SO}(3)$, we see that the left hand representation has four $V^{1,1}$-components, which come from the terms $$ \begin{aligned} V^{1,1} &\simeq V^{0,0}\otimes V^{1,1}\\ V^{1,1} &\subset \mathsf{S}^2\bigl( V^{1,1}\bigr) \simeq V^{1,1} \oplus V^{2,2}\oplus V^{0,2}\oplus V^{2,0}\oplus V^{0,0}\\ V^{1,1} &\subset V^{1,1}\otimes V^{2,0} \simeq V^{3,1}\oplus V^{2,1}\oplus V^{1,1}\\ V^{1,1} &\subset V^{1,1}\otimes V^{0,2} \simeq V^{1,3}\oplus V^{1,2}\oplus V^{1,1} \end{aligned} $$ The first line gives you a bilinear pairing between the scalar curvature and the traceless Ricci, which, up to a constant factor, is the first term on the right hand side of your equation. The second line gives a quadratic map from the traceless Ricci tensors into traceless symmetric tensors (but that potential term turns out not to show up in the final formula). The third and fourth lines give you pairings between the self-dual and anti-self-dual parts of the Weyl curvature and the traceless Ricci, which, up to a constant factor, is the second term on the right hand side of your equation. (Because the mapping actually has to be equivariant under $\mathrm{O}(4)$, which exchanges the self-dual and anti-self dual parts of the Weyl curvature, the two constants for those two terms have to enter symmetrically, which is why there is only one term there.)

To determine the correct constant multiples, it suffices to simply check the formula on curvature tensors that are made up of very simple cases: Take a curvature tensor $R$ that has no scalar or Weyl curvature, just a diagonalized traceless Ricci tensor, plug it into the left hand side, and you'll get zero, so that means that the potential quadratic in traceless Ricci term (from the second line) must be zero. Next, take a curvature tensor $R$ that has no Weyl curvature part and plug it in to determine the constant multiple for the first term in the right hand side of the above formula . Finally, take a tensor $R$ with no scalar curvature term and as simple as possible Ricci to check the coefficient for the second term.

While this method does not avoid all calculation, it does reduce the calculation to a few simple checks of constants. Also, this method illustrates the usefulness of representation theory in these calculations, which are generally extremely useful in getting explicit formulas such as the one you are trying to understand.