Realizing the wedge product of $\mathbb{R}^3$ as torsion or curvature tensor Is there a connection on $\mathbb{R}^3$ whose corresponding  torsion tensor is the standard cross product  of $\mathbb{R}^3$, that is $T(X, Y)= X \wedge Y$?
Is there a metric connection with this property?
Is there a Riemannian metric on $\mathbb{R}^3$ whose corresponding curvature tensor satisfies  $R(X,Y,Z)=(X\wedge Y) \wedge Z$?
 A: The first question is true. Let $E_i = \frac{\partial}{\partial x_i}$, $i=1,2,3$, be the usual coordinate vectors of $\mathbb{R}^3$. Let $\nabla$ be an affine connection and $a_{ij}^k$ functions such that
$$\nabla_{E_i}E_j = a_{ij}^k E_k, $$
where we use Einstein notation. Since the $E_i$ are coordinate vector fields, their Lie bracket is zero, so $T(E_i,E_j) = (a_{ij}^k-a_{ji}^k) E_k$. The condition $T(X,Y) = X\wedge Y$ gives only 9 non-trivial equations:
$$ 
\begin{align*}
&a_{ij}^j = a_{ji}^i, \text{ for $i\neq j$} \label{eq1}\tag{1}\\
&a_{ij}^k -a_{ji}^k=1, \text{ when $(ijk)$ is a positive permutation of (123)} \label{eq2}\tag{2}
\end{align*}
$$
So there are plenty of connections that agree with the cross product.
If we want $\nabla$ to be compatible with the usual Euclidean metric, then we also have the condition $a_{ij}^k=-a_{ik}^j$. This in combination with \eqref{eq1} implies that only the $a_{ij}^k$ where $i$, $j$, $k$ are all different are non-zero. If we also use \eqref{eq2}, we get that
$a_{12}^3=a_{23}^1=a_{31}^2=\frac{1}{2}$.
So there is only one such connection compatible with the usual metric:
$$\nabla_{E_i}E_j = \frac{1}{2}\varepsilon_{ij}^k E_k, $$
where $\varepsilon_{ij}^k$ is the Levi-Civita symbol.
If you want to find all the connections, it involves more calculations, since it's involves 6 extra unknowns $g_{ij}$.
I am not sure about your second question (yet). The amount of unknowns is a bit too big for a quick calculation, so finding all such metrics is a big task.
A short argument starts by noting that
$$ X\wedge (Y\wedge Z) = (X\cdot Z) Y - (X\cdot Y) Z.$$
Also the Riemann curvature tensor for a manifold with constant curvature $\kappa$  is given by $R(X,Y)Z=\kappa( \langle Y,Z\rangle X - \langle X,Z\rangle Y)$.
This gives me the hunch that the Riemannian metric should have constant curvature, but again, this is just a hunch for the moment.

Edit: I found a short answer. Suppose $\langle\cdot,\cdot\rangle$ is a metric such that $R(X,Y)Z=(X\wedge Y)\wedge Z$. By the triple vector product identity
$$ 
 \langle R(X,Y)Z, W\rangle = (X\cdot Z)\langle Y, W\rangle -(Y\cdot Z)\langle X, W\rangle. 
$$
From the identity $\langle R(X,Y)Z,W \rangle=\langle R(Z,W)X,Y \rangle$ it follows that
$$ 
 (Y\cdot Z)\langle X, W\rangle = (W\cdot X)\langle Z, Y\rangle.
$$
Now take $X=E_i$, $W=E_j$ and $Y=Z=E_k$. This gives $\langle E_i,E_j\rangle=\delta_{ij}\langle E_k, E_k\rangle$. Therefore there exists a non-vanishing function $\lambda$ such that
$(X\cdot Y) = \lambda \langle X,Y\rangle$. Substituting this in the Riemann curvature tensor gives
$$ 
 \langle R(X,Y)Z, W\rangle = \lambda\bigl(\langle X, Z\rangle \langle Y, W\rangle -\langle Y, Z\rangle \langle X, W\rangle\bigr). 
$$
A well-known theorem of Schur implies that $\lambda$ is a constant function (this follows from the Bianchi identities) and hence the curvature is constant (and non-zero). On the other hand, $\langle X, Y \rangle=\lambda^{-1}(X\cdot Y)$ equals the Euclidean metric (up to a constant), which implies that the curvature is zero. We obtain a contradiction and conclude that there exists no metric such that $R$ is given by the triple vector product.
