what is the geometry picture of Riemann tensor identity $R(X\wedge Y,V\wedge W) = R(V\wedge W, X\wedge Y)$ For symmetries in Riemann tensor of $R(X,Y)V:=\nabla_X\nabla_YV-\nabla_Y\nabla_XV-\nabla_{[X,Y]}V$, there are excellent explanations on the intuition behind it like this relevant question. If we think $R(X,Y)V$ measures the change of tangent vector $V$ along $X$ and $Y$ direction, it is easily understood that $R(X,Y) = -R(Y,X)$. But along this line, how can we see through the skew symmetry of the latter pair $ R(.,.,V,W) = -R(.,.,W,V) $ and the symmetry of exchange pair of $X,Y$ and $V,W$ beyond algebraic proof?  
$$R(X\wedge Y,V\wedge W) = R(V\wedge W, X\wedge Y)$$ 
 A: Skew-symmetry in last two indices. By definition the Levi-Civita connection $\nabla$ of a Riemannian metric $g$ preserves that connection, so that $\nabla g = 0$. Thus, if we view the curvature as a section $R \in \bigwedge^2 T^*M \otimes \operatorname{End}(TM)$, the induced action of $R$ on $g$ is $R \# g = 0$. Expanding gives
$$0 = (R\#g)(Z, W) = -g(R\#W, Z) - g(R\#Z, W) .$$
But rearranging gives exactly that $g(R\#W, Z)$ is skew in $W, Z$ as claimed.
Equivalently, since $\nabla$ preserves $g$, so does $R$, in the sense that it takes values in $\bigwedge^2 T^*M \otimes \mathfrak{o}(g)$, and lowering an index with $g$ gives an isomorphism $\mathfrak{o}(g) \cong \bigwedge^2 T^*M$. In particular, this does not hold for general linear connections (which generically do not preserve metrics).
Pair-swapping symmetry. The pair-swapping identity for a Levi-Civita connection $\nabla$ is generated by the two transposition symmetries and the Bianchi identity, $$\mathfrak{S}[R(X, Y) W)] = 0 ,$$ where $\mathfrak{S}[\cdots]$ denotes the sum over cyclic permutations in $Y, Z, W$. But the Bianchi Identity can be proved by writing the expression $\mathfrak{S}[R(X, Y) W)]$ using the definition of curvature, rewriting everything in terms of Lie brackets (which in particular requires torsion-freeness of $\nabla$), and applying the Jacobi identity.
