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Is there a Riemannian metric on $\mathbb{R}^3$ for which the corresponding curvature tensor $R$ satisfies $R(X,Y)Z=(X\wedge Y)\wedge Z$?

I have already discussed this question in the following post and I recived very interesting partial answer but I search for a complete answer containing a classification of all metrics with this property.

https://math.stackexchange.com/questions/2937107/realizing-the-wedge-product-of-mathbbr3-as-torsion-or-curvature-tensor

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  • $\begingroup$ By $\wedge$ here do you mean the Euclidean cross product, or the "cross product" induced by the Riemannian structure of the putative solution? In other words, using the triple product formula, are you looking for (writing $R_g$ for the Riemann curvature of a metric $g$) a solution $g$ of $$ R_g(X,Y)Z = g(Z,X) Y - g(Z,Y)X $$ on $\mathbb{R}^3$ or are you looking for a solution $g$ of $$R_g(X,Y)Z = e(Z,X)Y - e(Z,Y)X$$ where $e$ is the Euclidean metric? $\endgroup$
    – Willie Wong
    Commented Oct 8, 2018 at 14:00
  • $\begingroup$ In the latter case, second Bianchi implies that relative to the standard coordinates, the components of the metric tensor necessarily satisfies $\partial_i g_{jk} = \partial_j g_{ik}$. This implies that $g$ is the (Euclidean) Hessian of a scalar $\sigma$. And the Riemann curvature tensor is (up to sign) $R_{mij}{}^k = \frac14 (\partial_m g^{kl} \partial_i g_{jl} - \partial_i g^{kl} \partial_m g_{il})$. $\endgroup$
    – Willie Wong
    Commented Oct 8, 2018 at 14:59

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There are two interpretations of your question.

Metric cross product

Assuming that you are looking for a Riemannian metric $g$ such that (by the triple-product formula) $$ R_g(X,Y)Z = g(Z,X)Y - g(Z,Y)X $$ this implies immediately that your metric has constant sectional curvature.

Euclidean cross product

In the case you are looking for a Riemannian metric $g$ such that $$ R_g(X,Y)Z = e(Z,X)Y - e(Z,Y)X $$ where $e$ is the Euclidean inner product, we note the following:

  1. $R_g(X,Y)$ is not the zero map for any $X,Y$ such that $X\wedge Y \neq 0$. This makes the curvature tensor "regular" in the sense of Kowalski, and by his theorem in "On regular curvature structures" this means that the solution is unique up to conformal rescaling.
  2. The Riemann curvature tensor as given is invariant under Euclidean group of motions. This means that given $\phi:\mathbb{R}^3 \to \mathbb{R}^3$ an isometry of the Euclidean metric (in particular translations and rotations), $\phi^*g$ is another solution to the problem.

Combining the two points we have that

  1. At any fixed $p\in \mathbb{R}^3$, $g_p$ is pointwise rotationally symmetric: for not, there exists a distinguished largest eigenvalue of $g_p$ relative to $e$, and a corresponding eigenvector. But any rotation of $g_p$ is conformally equivalent to $g_p$.
  2. And therefore there exists a non-vanishing, positive function $\sigma:\mathbb{R}^3 \to \mathbb{R}$ such that $g = \sigma e$.

This implies that $$ \sigma R_g(X,Y)Z = g(Z,X)Y - g(Z,Y)X $$ Taking the trace this implies that $g$ is Einstein, and hence $\sigma$ is constant. But when $\sigma$ is constant we know that the Euclidean metric is flat. And hence we conclude there does not exist a solution for this formulation.

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  • $\begingroup$ Thank you very much for your answer. I effort to understand the details. But it is very interesting that there is no metric with this property. $\endgroup$
    – Ali Taghavi
    Commented Oct 9, 2018 at 18:39

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