# An expression with an alternating trilinear form, written in terms of the determinant and a symmetric bilinear form

I am trying to understand a line from MacLachlan/Reid's The Arithmetic of Hyperbolic 3-Manifolds (it's in 3.4 if you have the book), that seems it should be elementary but I can't seem to find where it's coming from.

Suppose I have a vector space $V$ over a field $F$, and an alternating trilinear form $f:V^3\rightarrow F$. Suppose I also have a symmetric bilinear form $\sigma:V^2\rightarrow F$. Is it true in general that there must exist a constant $c\in F$ such that:

$f(x_1,x_2,x_3)f(y_1,y_2,y_3)=c$ det$\Big(\big(\sigma(x_i,y_j)\big)_{i,j}\Big)$?

I believe this is the property the authors quote to get their result, though it's possible I have overgeneralized it. The statement it's used for in the book is for the case when $V=M_2(\mathbb{C})$, $F=\mathbb{C}$, $f$ is $(x_1,x_2,x_3)\mapsto tr(x_1x_2x_3)$, and $\sigma$ is $(x_1,x_2)\mapsto tr(x_1x_2)$, with the added condition that we require the $x_i$ and $y_i$ to have trace $0$ (obviously products of these need not have trace zero), and in this case $c$ turns out to be $-\frac{1}{2}$. Obviously the equation in this case can be verified by a simple but tedious calculation, but this is not how the author derives it. He quotes the property above (or something close to it). If the statement is true, what is an elegant way to see it, perhaps from differential geometry, or just multilinear algebra? If I have overgeneralized the property, what is the correct one?

• There would have to be some relation between $f$ and $\sigma$. As it stands, if $f$ is nonzero and $\sigma$ is zero there is no hope. – Jack Huizenga Apr 25 '13 at 4:28
• Both sides are invariant under a lot of group actions - $GL_2$ acts by conjugation on $M_2(\mathbb C)$, and two copies of $GL_3$ act on the vectors $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$. You can count the dimension of the space of invariants of these group actions - if it's $1$, you get the result. – Will Sawin Apr 25 '13 at 4:54

It's certainly not true in general. One can see how far this can go by considering the following: If $f$ is not identically zero, then choosing three elements $y_i\in V$ such that $f(y_1,y_2,y_3)\not=0$, one sees that $c$ cannot be zero and that the determinant of the $3$-by-$3$ symmetric matrix whose entries are $\sigma(y_i,y_j)$ cannot be zero, so $\sigma$ has to be nondegenerate on the $3$-plane spanned by the $y_i$. Now, freezing such $y_i$, the above equation implies that $$f(x_1,x_2,x_3) = c\ \frac{\det\bigl(\sigma(x_i,y_j)\bigr)}{f(y_1,y_2,y_3)}$$ for all $x_i\in V$. In particular, if $W\subset V$ is the codimension $3$ subspace that is the $\sigma$-orthogonal of the span of the $y_i$, then $f$ is actually the pullback of a volume form on the $3$-dimensional quotient $\bar V = V/W$ under the natural quotient projection $V\to \bar V$. Moreover, if $\sigma$ were nondegenerate on any $3$-plane $E\subset V$ such that $E\cap W\not= (0)$, then $f$ would be nonzero when restricted to $E$, which is impossible, so it follows that $W$ is the null space of $\sigma$, so that $\sigma$, too, is the pullback of a nondegenerate quadratic form on $\bar V$.
Thus, the conclusion is that, except when $f\equiv0$, then $f$ (respectively, $\sigma$) must be the pullback of a volume form (respectively, nondegenerate quadratic form) on a $3$-dimensional vector space $\bar V$ under a surjection $V\to \bar V$. Conversely, in this case, there is always a nonzero $c$ for which the above equation holds.
• Reading your answer made me wonder: is there any natural projection $\Lambda^3(S^2 V^*)\to S^2(\Lambda^3 V^*)$? (By $\Lambda$ and $S$ I mean skew-symmetric and symmetric tensors, respectively.) If yes, then we are dealing with something like $f^2=c\sigma^3$: when $V$ is 3D, then (the square of) a nonzero $f$ spans the whole $S^2(\Lambda^3 V^*)$ and then the positive answer follows from the surjectivity of the above map. But when $V$ is larger, not all the $\sigma^3$ will project on $f^2$, and one must impose the condition you've explained. – Giovanni Moreno Aug 23 '13 at 10:06
• @G_infinity: For vector spaces $V$ of dimension greater than $3$, there is no nonzero $\mathrm{GL}(V)$-equivariant linear mapping between $\Lambda^3(S^2V^\ast)$ and $S^2(\Lambda^3V^\ast)$ because these representations of $\mathrm{GL}(V)$ have no common constituents. – Robert Bryant Aug 23 '13 at 11:33
I think this is specific to the $M_2(\mathbb{C})$ situation, or rather $\mathfrak{s}\mathfrak{l}_2(\mathbb{C})$. The identity boils down to $det(A)det(B)=det(AB)$ for $3\times 3$ matrices. It's a simple calculation using Pauli matrices.