Why isn't the dimension of the space of curvature-like tensors equal to the dimension of the Grassmannian? It's a well known fact that if $V$ is a vector space of dimension $n$, then the vector space consisting of all curvature-like tensors (see this text for a precise definition and a proof) has dimension $\frac{n^2(n^2-1)}{12}$. It's also well known that any curvature-like tensor $R$ is determined by its sectional curvatures (i.e, the values $R(x, y, x, y)$ with $x, y$ going through all of $V$). The sectional curvature can be thought of as a function $K: \mathrm{Gr}_2(V) \to \mathbb{R}$. In my mind these two separate facts appear to be in contradiction: if the sectional curvatures determine $R$, shouldn't the vector space consisting of all curvature-like tensors have the same dimension as $\mathrm{Gr}_2(V)$ (which has dimension $2(n-2)$)? Can someone explain what I'm missing here? Thanks in advance!
 A: The issue here is that while the map $$\mathcal{R}(V) \ni R \mapsto K^R\in {\rm C}^\infty({\rm Gr}_2(V), \mathbb{R})$$is linear and injective, it is not surjective. The Bianchi identity cuts down the dimension of the range for what it has to be or, in other words, it is not true that every smooth map ${\rm Gr}_2(V) \to \mathbb{R}$ arises as the sectional curvature function of some curvaturelike tensor on $V$.
To elaborate more on the proof of this dimension formula, writing $S^2(V)$ for the space of symmetric bilinear forms on $V$, one considers the Bianchi map ${\sf b} \colon S^2(V^{\wedge 2}) \to (V^*)^{\wedge 4}$ given by $${\sf b}(R)(x,y,z,w) = \frac{1}{3}(R(x,y,z,w)+R(y,z,x,w)+R(z,x,y,w)),$$and note that ${\sf b}$ is surjective with kernel $\ker {\sf b} = \mathcal{R}(V)$, so $$\dim S^2(V^{\wedge 2}) = \dim \mathcal{R}(V) + \dim ((V^*)^{\wedge 4})$$reads $$\frac{1}{2}\frac{n(n-1)}{2}\left(\frac{n(n-1)}{2}+1\right) = \dim\mathcal{R}(V) + {n \choose 4}$$and directly gives $\dim \mathcal{R}(V) = n^2(n^2-1)/12$. To see why surjectivity of ${\sf b}$ holds, observe that the symmetry group $S_4$ acts on all rank $4$ tensor spaces over $V$ simply by permuting the arguments, and that ${\sf b}(\sigma\cdot R) = {\rm sgn}(\sigma) {\sf b}(R)$ holds, as well as ${\sf b}(\zeta)=\zeta$ for every $\zeta \in (V^*)^{\wedge 4}$.
Another consequence of this is the formula for the dimension of the space $\mathcal{W}(V)$ of Weyl curvature tensors in $V$, i.e., the kernel of the abstract Ricci contraction map ${\rm Ric}\colon \mathcal{R}(V) \to S^2(V)$, once one has shown that $$\mathcal{R}(V) = (g\wedge S^2(V))\oplus \mathcal{W}(V), $$where $\wedge$ stands for the Kulkarni-Nomizu product of symmetric bilinear forms. Then $$\frac{n^2(n^2-1)}{12} = \frac{n(n+1)}{2} + \dim\mathcal{W}(V)$$directly gives us that $$\dim \mathcal{W}(V) = \frac{n(n+1)(n+2)(n-3)}{12}.$$
