Realizing the cross product of $\mathbb{R}^3$ as the curvature tensor of a Riemannian metric on $\mathbb{R}^3$ 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
 A: 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:


*

*$R_g(X,Y)$ 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. 

*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 


*

*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 $g_p$. 

*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. 
