Einstein 3-manifold has constant sectional curvature A Riemannian Einstein 3-manifold has constant sectional curvature. I know a proof of a stronger theorem I read somewhere (it was iff) that makes use of the Weyl tensor but for just this implication it should be possible to use far less machinery, namely the $r$ endomorphism of the tangent space, defined as $<r(x),y>=Ric(x,y)$ and the isomorphism $\phi$ from the tangent space to the alternating 2-forms on it defined as $\phi e_i=e_j\wedge e_k$ for every $(ijk)$ cyclic permutation of $(123)$, where $\{e_i\}_{i=1,2,3}$ is a local orthonormal frame of $TM$. In particular these ingredients alone and the scalar curvature should determine algebraically the curvature operator $\mathcal{R}$ from the 2-forms on the tangent space to itself. I struggled to find any book that approaches this matter in such way, but I know from an exercise from an old course I took years ago. it should be possible to prove it this way. Anybody knows how? Any reference helpful towards this approach would also be helpful.
 A: Assume $(\mathcal{M}^3, g)$ is Einstein, i.e $\operatorname{Ric}_{\mathcal{M}} = \lambda g$. By the Bianchi identity $\mathrm{d}  \operatorname{Scal}  = 2 \operatorname{div}(\operatorname{Ric})$ where $\operatorname{Scal}: \mathcal{M} \to \mathbb{R}$ is the scaIar curvature. If $\mathcal{M}$ is Einstein we then have $\mathrm{d}(\lambda \operatorname{tr}(g)) = n \cdot \mathrm{d} \lambda = 2 \operatorname{div}( \lambda g)$, which in coordinates can be written as
$$
2 g^{i j} \nabla_{i}\left(\lambda g_{j k}\right) \mathrm{d} x^{k}=2 g^{i j} g_{j k} \nabla_{i} \lambda \mathrm{d} x^{k}=2 \mathrm{~d} \lambda = n \cdot \mathrm{d}\lambda
$$
where I'm using Einstein notation. Hence if $n \geq 3$ and $\mathcal{M}$ is connected $\mathrm{d} \lambda \equiv 0$, and therefore $\lambda$ is constant. Then for an arbitrary $p \in \mathcal{M}$ and a geodesic frame $\{e_1, e_2, e_3\}$ around $p$, we get $$\sum_{i = 1}^{3} R(X, e_i, Y, e_i) = 2 \lambda g(X, Y)$$ Making $X = Y = e_j$ with $j \in \{1, 2, 3\}$ and evaluating at $p$ we then get $$\lambda = R_{1212}(p) = R_{1313}(p) = R_{2323}(p)$$ It follows from Schur's lemma that $\mathcal{M}$ has constant sectional curvature (and in particular all it's curvatures are constant as well).
A: EDIT: Based your comments, here is how to map your notation to mine: Given an orthonormal frame $(e_1,e_2,e_3)$, denote the components of the curvature tensor by
$$ R_{ijkl} = R(e_i,e_j,e_k,e_l) = \langle R(e_i,e_j)e_l,e_k\rangle $$
Define the symmetric tensor $Q$ by
$$ Q_{ab} = R(e_i,e_j,e_k,e_l), $$
where $(a,i,j)$ and $(b,k,l)$ are cyclic permutations of $(1,2,3)$. This is essentially the same as what you denote by $\phi^{-1}\circ\mathcal{R}\circ\phi$.
Therefore,
\begin{align*}
Q_{11} &= R_{2323}\\
Q_{12} &= R_{2331}\\
Q_{13} &= R_{2312}\\
Q_{22} &= R_{3131}\\
Q_{23} &= R_{3112}\\
Q_{33} &= R_{1212}
\end{align*}
On the other hand, the 6 components of the Ricci tensor are
\begin{align*}
\newcommand{\Ric}{\mathrm{Ric}}
\newcommand{\Ric}{\mathrm{Ric}}
\Ric_{11} &= R_{1212} + R_{1313} = Q_{33} + Q_{22}\\
\Ric_{12} &= R_{1323} = -Q_{21}\\
\Ric_{13} &= R_{1232} = -Q_{31}\\
\Ric_{22} &= R_{2121} + R_{2323} = Q_{33} + Q_{11}\\
\Ric_{23} &= R_{2131} = -Q_{32}\\
\Ric_{33} &= R_{3131} + R_{3232} = Q_{22} + Q_{11}
\end{align*}
From this, you can see that
\begin{align*}
\newcommand{\tr}{\operatorname{tr}}
\Ric_{ij} &= -Q_{ij} + (\tr Q)\delta_{ij}\\
\end{align*}
Taking the trace of this, we get
$$
\tr \Ric = 2\tr Q
$$
Therefore,
\begin{align*}
Q_{ij} &= - \Ric_{ij}+ (\tr Q)\delta_{ij}\\
&= -\Ric_{ij} + \frac{1}{2}(\tr \Ric)\delta_{ij}.
\end{align*}
Therefore, the Riemann curvature tensor is uniquely determined by the Ricci tensor. Moreover, if the metric is Einstein, then
$$
\Ric = \frac{1}{3}(\tr \Ric)\delta = \lambda \delta
$$
and
$$
Q = \frac{1}{6}(\tr \Ric)\delta
$$
and, by the contracted first Bianchi identity, $\tr\Ric$ is constant. From this, the sectional curvatures satisfy, if $(i,j,k)$ is a cyclic permutation of $(123)$,
$$
K(e_i,e_j) = R_{ijij} = Q_{kk} = \frac{1}{6}\tr \Ric = \frac{1}{2}\lambda.
$$
$Q$ is an interesting tensor. It looks like minus the Einstein tensor. Unlike the Ricci tensor, it is not transversally elliptic, and therefore, the analog of the Ricci flow but using $Q$ or $-Q$ instead does not give a geometric heat flow.
