# Kulkarni-Nomizu square root of the Riemann tensor

Given a Riemann tensor $$Riem$$, what are conditions such that $$Riem=B\star B$$ for some bilinear symmetric form $$B$$, where $$\star$$ is the Kulkarni-Nomizu product? It follows from the proof of Proposition 15 (Chapter 4, Section 2, page 99) in the book of Petersen "Riemannian geometry" - Second Edition, that positivity of the curvature operator $${\mathscr{R}}$$ on $$\Lambda^2T_pM$$ is sufficient in dimension 3. Moreover, when $$n=3$$, $$\det{\mathscr{R}}>0$$ is sufficient.
Anyone have a reference for $$n>3$$, please?

This is how to know if there exists a square matrix $$n\times n$$ with assigned all its $$2\times2$$ subdeterminants (and possibly how to recover it). In the following paper of Greenhill http://web.maths.unsw.edu.au/~csg/papers/ext-matrix.pdf, an algorithm to find such a matrix is showed, but apparently it does not state a theoretical ("easy") criterium for its existence.

My question is related to my other question 2x2 subdeterminants of a matrix which was related to uniqueness.

Note that Petersen's Prop 15 does not directly address your question. The assumption that there already exists an embedding from the $$n$$ dimensional $$M$$ to $$(n+1)$$ dimensional Euclidean space is crucial. Petersen's Proposition 15 concerns not solvability of $$\mathrm{Riem} = S\star S$$ but whether the solution $$S$$ is unique.

What you are looking for is a theorem due to Jaak Vilms. He described it in two different places:

1. Local isometric imbedding of Riemannian $$n$$-manifolds into Euclidean $$(n+1)$$-space, J. Differential Geom. 12(2): 197-202 (1977). DOI: 10.4310/jdg/1214433981
2. Factorization of Curvature Operators, Transactions of the American Mathematical Society Vol. 260, No. 2 (Aug., 1980), pp. 595-605. https://www.jstor.org/stable/1998025

Note that this discussion presumes that you have a given Riemannian metric (or I guess in your case a positive definite inner product on $$T_pM$$).

Essentially: if one looks at the Gauss-Codazzi equations, one sees that when $$M$$ has an isometric immersion into Euclidean space of $$p$$ dimensions higher, then one has a decomposition of $$\mathrm{Riem} = \sum_{i = 1}^p S_i \star S_i$$ And so the question you are asking for is identical the (local) isometric immersivity of $$M^n$$ into $$E^{n+1}$$.

### Notation

Denote by $$\mathscr{R}: \Lambda^2 T_pM \to \Lambda^2 T_pM$$ the curvature operator acting on two forms. In indices its components are $$R^{ij}_{kl}$$ antisymmetric in $$i,j$$ and in $$k,l$$ separately.

Define $$\phi(\mathscr{R}) = R^{ij}_{kl} R^{kp}_{iq} R^{lq}_{jp}$$ and $$\psi(\mathscr{R}) = R^{ij}_{kl} R^{kl}_{pq} R^{pq}_{ij}$$

The curvature operator $$\mathscr{R}$$ is assumed to satisfy Bianchi identity.

The curvature operator $$\mathscr{R}$$ is said to "preserve decomposability" if for arbitrary 2 forms $$\alpha, \beta$$: $$\mathscr{R}\alpha \wedge \mathscr{R}\beta = 0 \iff \alpha \wedge \beta = 0$$ Note that this condition is trivial if $$n = 3$$.

Given a curvature operator $$\mathscr{R}$$, its rank subspace is the smallest subspace $$W$$ of $$T_pM$$ such that the image of $$\mathscr{R}$$ lies in $$\Lambda^2 W$$; below when we refer to rank of $$\mathscr{R}$$ we mean the dimension of $$W$$.

### Theorem

A curvature operator with rank $$> 1$$ decomposes as $$S\star S$$ if and only if

1. $$\mathcal{R}$$ preserves decomposability
2. $$\phi + \frac14 \psi > 0$$; in the case $$n = 3 \pmod 4$$ this condition can be replaced by positivity of $$\det \mathscr{R}$$ when restricted to the complement of its kernel.
3. a technical "non singularity condition".

The non-singularity condition is a bit annoying to state; see the second paper for details. It is trivially satisfied when $$\mathscr{R}$$ is positive (or negative) definite, and $$n$$ is either 3 or $$>4$$. (I think it may also be satisfied when $$n = 4$$ with definite curvature operator, but the computations are less obvious.)

• Thanks, you are absolutely right about Petersen's book, I edited my question. Thanks for the references and the Theorem, they are very useful. – Carlo Mantegazza Apr 6 at 23:28