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A sufficient condition is if the Riemannian manifold is conformally flat, this implies that the Weyl curvature vanishes, and the Riemann curvature tensor is a linear combination of the identity operator on two forms and the operator formed by the Kulkarni-Nomizu product of the Ricci curvature and the metric. Using that the Ricci curvature is a symmetric bilinear form, you can diagonalize it relative to the metric, and explicitly show (as in the 3 dimensional case) that the Kulkarni-Nomizu product of Ricci and the metric can be diagonalized over a basis formed by ${e_i\wedge e_j}$.

On the other hand, there are also large classes of manifolds for which it is impossible to satisfy your requirement. For example, consider the four dimensional (anti)-self-dual Einstein manifolds with nonvanishing Weyl curvature. The Einstein equation $Ric = \lambda g$ means that the Ricci and scalar parts of the curvature are just multiplies of the identity. But the self-duality of the Weyl part means any eigen-twoform of the curvature operator must be either self-dual or anti-self-dual, which rules them out from being rank two.

Here are also some possibly relevant papers.

• Vilms considered in this paper conditions related to the curvature operator having bounded rank.
• In this paper the same author studied curvature operators of the form $R = b\wedge b$, where $b$ is symmetric bilinear. In general one sees that a necessary and sufficient condition for curvature operators to be diagonalisable in your sense is that $R = \sum_{i = 1}^{M} b_i\wedge b_i$, where the $b_i$'s are symmetric bilinear forms that can all be simultaneously diagonalised.
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A sufficient condition is if the Riemannian manifold is conformally flat, this implies that the Weyl curvature vanishes, and the Riemann curvature tensor is a linear combination of the identity operator on two forms and the operator formed by the Kulkarni-Nomizu product of the Ricci curvature and the metric. Using that the Ricci curvature is a symmetric bilinear form, you can diagonalize it relative to the metric, and explicitly show (as in the 3 dimensional case) that the Kulkarni-Nomizu product of Ricci and the metric can be diagonalized over a basis formed by ${e_i\wedge e_j}$.

On the other hand, there are also large classes of manifolds for which it is impossible to satisfy your requirement. For example, consider the four dimensional (anti)-self-dual Einstein manifolds. The Einstein equation $Ric = \lambda g$ means that the Ricci and scalar parts of the curvature are just multiplies of the identity. But the self-duality of the Weyl part means any eigen-twoform of the curvature operator must be either self-dual or anti-self-dual, which rules them out from being rank two.

Here are also some possibly relevant papers.

• Vilms considered in this paper conditions related to the curvature operator having bounded rank.
• In this paper the same author studied curvature operators of the form $R = b\wedge b$, where $b$ is symmetric bilinear. In general one sees that a necessary and sufficient condition for curvature operators to be diagonalisable in your sense is that $R = \sum_{i = 1}^{M} b_i\wedge b_i$, where the $b_i$'s are symmetric bilinear forms that can all be simultaneously diagonalised.