Skip to main content
6 of 11
added 125 characters in body
KConrad
  • 50.6k
  • 9
  • 196
  • 277

This is a great question.

Have you seen Manin’s book “Cubic forms: algebra, geometry, arithmetic”?

There are important forms of degree greater than $2$, such as the determinant form of degree $n$ in $n$ variables for all $n$. But forms of higher degree can have differences from the quadratic case. Here are a two basic comparisons between degree $2$ and higher degree.

  1. Diagonalizability.

Outside characteristic 2, a quadratic form can be diagonalized after a linear change of variables, but for $n \geq 3$, a form of degree $n$ might not be diagonalizable after any linear change of variables. While any binary cubic form over $\mathbf C$ can be diagonalized (edit: when it is nondegenerate, i.e., has a nonzero discriminant - see the start of the proof of Lemma 1.7 here) , in more than two variables cubic forms over $\mathbf C$ need not be diagonalizable. For example, the three-variable cubic form $x^3 - y^2z - xz^2$ can't be diagonalized over $\mathbf C$ for a reason related to elliptic curves: see my comments on the MO pages here and here. For each $d \geq 3$ and $n \geq 2$ except for $d=3$ and $n = 2$, there are nondiagonalizable forms of degree $d$ in $n$ variables over $\mathbf C$ that are smooth away from $(0,0,\ldots,0)$. Note the diagonal form $x_1^d + \cdots + x_n^d$ is smooth away from the origin.

  1. Group theory.

For a form $f(x_1,\ldots,x_n)$ over a field $F$, its orthogonal group is the linear changes of variables on $F^n$ that preserve it: $$ O(f) = \{A \in {\rm GL}_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}. $$ Quadratic forms have a rich orthogonal group (many reflections). But for a "nondegenerate" form $f$ of degree $d \geq 3$ over an algebraically closed field of characteristic $0$, $O(f)$ is a finite group. The version of this over $\mathbf C$ goes back to Jordan. I believe it also holds in characteristic $p$ when $p > d$. When $f = x_1^d + \cdots + x_n^d$ for $d \geq 3$, its orthogonal group over $\mathbf C$ is a finite group: the compositions of $n!$ coordinate permutations and scaling of the $n$ coordinates by $d$th roots of unity: a finite group of order $d^n n!$. That's a big difference between $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$.

KConrad
  • 50.6k
  • 9
  • 196
  • 277