TITLE: Curvature tensors induced by multilinear forms
QUESTION [1 upvotes]: Suppose I have a finite-dimensional vector space $V$ and two $4$-linear forms $b_1, b_2$ on it (that is, elements of $(V^*)^{\otimes 4}$) that satisfy the symmetries
$$
\begin{align*}
b_j(w,y,z,x) &= b_j(x,y,z,w) \\
b_j(x,z,y,w) &= b_j(x,y,z,w) \\
b_j(y,x,w,z) &= b_j(x,y,z,w)
\end{align*}
$$
for all $x,y,z,w \in V$ and $j = 1,2$. Then I can define two curvature-like tensors
$$
R_j(x,y,z,w) = b_j(x,w,y,z) - b_j(x,z,y,w).
$$
If we know that $R_1 = R_2$, can we conclude that $b_1 = b_2$?
For motivation, I have a problem where I assume that a certain metric is of constant sectional curvature. Then one of the forms above is $b(x,y,z,w) = g(x,w)g(y,z)$. I can also write the curvature tensor as being induced by a very complicated multilinear form as above, and I'd like to prove that this complicated form is actually the one I get from the underlying metric.

REPLY [1 votes]: Not necessarily. 
Let $b_1$ be a linear form satisfying the desired symmetries, and let $a$ be a symmetric linear form. Then $b_2 = b_1 + a$ also satisfies the desired symmetries.
Note that 
\begin{align*}
R_2(x, y, z, w) &= b_2(x, w, y, z) - b_2(x, z, y, w)\\
&= b_1(x, w, y, z) + a(x, w, y, z) - b_1(x, z, y, w) + a(x, z, y, w)\\
&= b_1(x, w, y, z) + a(x, y, z, w) - b_1(x, z, y, w) + a(x, y, z, w)\\
&= b_1(x, w, y, z) - b_1(x, z, y, w)\\
&= R_1(x, y, z, w).
\end{align*}
This is the only possible problem. That is, if $b_1$ and $b_2$ are linear forms satisfying the desired symmetries such that $R_1 = R_2$, then $a:= b_2 - b_1$ is a symmetric linear form. To see this, note that $a$ satisfies the symmetries that $b_1$ and $b_2$ do, as well as the additional symmetry $a(x, w, y, z) = a(x, z, y, w)$ which follows from the fact that $R_1 = R_2$. So the form $a$ is invariant under the subgroup of $S_4$ generated by $\{(2\ 4), (1\ 4), (2\ 3), (1\ 2)(3\ 4)\}$. This subgroup is in fact $S_4$.
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