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>



Realizing the cross product of $\mathbb{R}^3$ as the curvature tensor of a Riemannian metric on $\mathbb{R}^3$
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"](https://link.springer.com/article/10.1007%2FBF01110924) 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**. 


