# Hilbert spaces are induced by a bilinear form. How about n-linear forms?

A Hilbert space is a complete vector space equipped with scalar product, i.e. a symmetric positive definite bilinear form.

What if we replace 'bilinear' by 'n-linear'? One might wonder, whether the $l^3$-norm might be induced by a trilinear form in a similar fashion like the $l^2$-norm by a bilinear form is.

Is there any interesting theory on this?

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The scalar product isn't bilinear; it's sesquilinear (unless the base field is the reals, in which case the two concepts are equivalent). –  Darsh Ranjan Dec 5 '09 at 2:29
As long as the form is positive definite and the unit ball is convex, you get a perfectly good Banach space using any symmetric $n$-linear form on a real vector space $V$. The degree $n$ is necessarily even. It is equivalent to defining the norm as the $n$th root of a homogeneous degree $n$ polynomial. $\ell^p$ is an example for any even integer $p$. There are many other examples. I found a paper, Banach spaces with polynomial norms, by Bruce Reznick, that studies these norms. He obtains various results; the most appealing one to me at a glance is that these Banach spaces are all reflexive.
Off-hand I can't think of any simple way to recover positive definiteness starting with odd polynomials. The cube of the norm on $\ell^3$ is a polynomial in the absolute values of the coordinates rather than the coordinates themselves.
Addendum: To address Darsh's comment, what you would look at in the complex case is self-conjugate polynomials of degree $(n,n)$. Equivalently, as with all complex Banach norms, the realification is a real Banach norm which is invariant under complex scalar rotation.