# Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold?

Let $M$ be a smooth Riemannian manifold, let $R$ be the Riemannian curvature operator, and let $p$ be a point in the manifold. With respect to any orthonormal basis of the tangent bundle at the point $p$, the operator $R$ is a skew-symmetric matrix with entries that are two forms. Thus, $R = [R_{ij}]$ where the transpose of $R$ is $-R$. It's well-known that for a skew-symmetric matrix with REAL entries, we can choose a basis such that the matrix is similar to a block diagonal matrix with $2 \times 2$ blocks of the form $[a_{ij}]$ where $a_{11} = a_{22} = 0$, $a_{12} = - a_{21}$. Can the same be done for a matrix of 2-forms, specifically for the curvature matrix $R = [R_{ij}]$, where the entries of the $2 \times 2$ blocks are 2-forms?

Precisely, my question is:

Can we choose a basis of the tangent space at $p$ such that $R = [R_{ij}]$ is similar to a block diagonal matrix with $2 \times 2$ blocks of the form $[a_{ij}]$ where $a_{11} = a_{22} = 0$, $a_{12} = - a_{21}$, and all other entries are zero, and $a_{12}= - a_{21}$ are 2-forms?

On page 84 and 86 of Professor Freed's notes,

www.ma.utexas.edu/users/dafr/DiracNotes.pdf‎,

he states that this is indeed possible. However, because the 2-forms (of course) do not form a field, one cannot apply any standard proofs in the case the matrix has REAL entries to prove Professor Freed's claim. Moreover, people that I talk with seem to doubt the claim, exactly because the 2-forms don't form a field. (Note that the entries $a_{12}$, in the case the matrix $R$ has REAL entries, are the eigenvalues of the complex matrix $iR$.)

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Let us start it from the other end. Assume you can block-diagonalize the curvature tensor. Then in the most of coordinate sectional directions the curvature is zero.

In fact your curvature tensor equals to a curvature of product of few surfaces and maybe the real line --- this is a very special case.

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P.S. For example, if sectional curvature has definite sign then the curvature tensor can not be block-diagonalized. Further, the rank of generic curvature tensor is $n\cdot(n-1)/2$, but any block-diagonalized has rank at most $\lfloor\tfrac n2\rfloor$ –  Anton Petrunin May 17 '13 at 17:41
@Anton: Yes, but more than just having rank at most $\lfloor \frac{n}{2}\rfloor$, the curvature must take values in an abelian subspace of the endomorphism algebra. By the way, just because the curvature 'looks like' that of a product, that doesn't mean that the metric is a product. For example, the space of Riemannian metrics in dimension $3$ whose curvature operators have rank at most $1$ everywhere depends (up to diffeomorphism) on $3$ functions of $2$ variables, while the products of surfaces with lines depend (up to diffeomorphism) only on $1$ function of $2$ variables. –  Robert Bryant May 17 '13 at 22:24
@Robert, You are right, but note that I state "curvature tensor equals to a curvature of product"; it does not mean that space is a product. –  Anton Petrunin May 18 '13 at 1:36
Thank you Professor Petrunin and Professor Bryant. –  Jorge May 18 '13 at 22:19

If you are asking whether there always exists an orthonormal basis where $R_{ijkl} = 0$ unless $\{i,j\} = \{k,l\}$, the answer is yes in dimension 2 or 3 and no in higher dimensions. In dimension 3 it's the same as diagonalizing the Ricci tensor. As for higher dimensions, in a paper with Dennis DeTurck on diagonal coordinates we show that those conditions hold if and only if the Weyl tensor vanishes.

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@Deane: I think the OP is asking a different question, which is whether the curvature operator at each point must take values in a maximal torus in the Lie algebra of skew-symmetric endomorphisms of the tangent space. This is not true generally even in dimension $3$, where the condition would be that the curvature operator has rank $1$ at each point. In fact, for the generic metric at the generic point, the curvature operator is surjective onto the skew-symmetric endomorphisms of the tangent space, so this doesn't work for $n>2$. –  Robert Bryant May 17 '13 at 12:28