# Non-degenerate multilinear forms

Is there a standard notion of non-degeneracy for multilinear forms?

My motivation is simple curiosity, by the way!

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For alternating forms one definition is that it's symplectic -- wedging it with itself enough times produces a volume form. –  Ryan Budney Mar 22 '10 at 22:03
@WW: The quadratic form F(x,y) = xy is nondegenerate over any field, even though f(0,y) = f(x,0) = 0. Your parenthetical definition is correct, but it is not equivalent to what you or fpqc said. –  Pete L. Clark Mar 22 '10 at 22:31
(Deleted some of my earlier comments.) So for bilinear forms F:V^2 to R, it is said to be non-degenerate if the map V -> V^* induced by F is an isomorphism. For multilinear forms V^k to R, asking that the induced map V -> (V^*)^k is isomorphic is obviously too strong. Is injectivity not good enough? –  Willie Wong Mar 22 '10 at 23:19
That map should be into the $n-1$st tensor power, and no, I don't think that it is correct. I have an inductive definition for you: Say an n-form is nondegenerate if all of the maps induced by substitution into the jth variable $f_j:V\to (V^*)^{\otimes(n-1)}$ map nonzero vectors of $V$ to nondegenerate $n-1$-forms. We define a 1-form to be nondegenerate if it is not identically zero. This recovers the bilinear case. –  Harry Gindi Mar 23 '10 at 0:34

I'm not sure if this notion is "standard", but there is one such notion, used for example in Nigel Hitchin's paper Stable forms and special metrics (arXiv:math/0107101) for alternating multilinear forms. The idea is that symplectic structures on a vector space $V$ can be characterised by the fact that they lie in an open orbit of $\mathrm{GL}(V)$ on $\Lambda^2V^*$. Hitchin calls these stable forms and shows that apart from the the case of symplectic forms, there are stable $3$-forms in dimensions 6,7 and 8; the $G_2$-invariant $3$-form in a seven-dimensional vector space (i.e., the imaginary component of the multiplication of imaginary octonions) being one such example.

Hitchin's notion is very fruitful, as it provides a variational approach to 7-dimensional riemannian metrics of weak $G_2$ holonomy, for example.

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Very interesting. Thanks! –  Mariano Suárez-Alvarez Mar 23 '10 at 0:13

Basic idea: a bilinear form $B$ is degenerate if there are two nonzero vectors $v$ and $w$ so that not only is $$B(v,w)=0,$$ but also either vector is enough to kill $B$ without help from the other: $$B(v,-) = 0 \mbox{ and } B(-,w)=0.$$

Another way to say it: even if we perturb $v$ and $w$, $B(v \otimes w)=0$ to first order. From this point of view, we see that a bilinear form is degenerate iff it is an element of the variety dual to the Segre embedding of $\mathbb{P}V \times \mathbb{P}W$ in $\mathbb{P}(V \otimes W)$.

This motivation generalizes gracefully to the following definition from Gelfand, Kapranov, and Zelevinsky's book Discriminants, resultants, and multidimensional determinants:

A $p$-linear form $T$ is said to be degenerate if either of the following equivalent conditions holds:

• there exist nonzero vectors $\beta_i$ so that, for any $1 \leq j \leq p$, $$T \left( \beta_1, \ldots , \beta_{j-1} , x_j ,\beta_{j+1}, \ldots , \beta_{p} \right) = 0 \mbox{ for all x_{j};}$$

• there exist nonzero vectors $\beta_{i}$ so that $T$ vanishes at $\otimes \beta_{i}$ along with every partial derivative with respect to an entry of some $\beta_{i}$: $$T \mbox{ and } \frac{\partial T}{\partial \beta^{(j)}_{i}} \mbox{ vanish at \otimes \beta_{i}.}$$

In certain favorable cases (when the dimensions of the vector spaces $V_i$ are not too different) the dual to the Segre is a hypersurface; in other words, there is a single polynomial---the hyperdeterminant---which vanishes exactly at the degenerate multilinear forms. This polynomial possesses many magical properties and is much subtler than determinants of bilinear forms.

I can attest that this definition is at least useful, if not standard, since it came up in a substantial way in an elementary question about coin flipping: http://arxiv.org/abs/1009.4188 .

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