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Before I begin, let me point out that we all know one important higher-degree form: the determinant form of degree $n$. So it is reasonable to ask what kind of general theory there could be for higher degree forms.

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 nondegenerate binary cubic form over $\mathbf C$ can be diagonalized (see the start of the proof of Lemma 1.7 here; in the binary case, nondegeneracy of a cubic form is equivalent to the dehomogenization being a cubic polynomial with nonzero discriminant), nondegenerate cubic forms over $\mathbf C$ in more than two variables need not be diagonalizable. For example, the three-variable cubic form $x^3 - y^2z - xz^2$ is nondegenerate and 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 nondegenerate forms of degree $d$ in $n$ variables over $\mathbf C$ that are smooth away from $(0,0,\ldots,0)$ and are not diagonalizable. Note the diagonal form $x_1^d + \cdots + x_n^d$ is smooth away from the origin.

Concerning papers and books, I'll just mention one of each. There is Harrison's paper "A Grothendieck ring of higher degree forms" in J. Algebra 35 (1978), 123-138 here and Manin’s book Cubic forms: algebra, geometry, arithmetic. Manin mentioned a recurring nightmare he had about this book, soon after he finished it, in an interview with Eisenbud here.

Before I begin, let me point out that we all know one important higher-degree form: the determinant form of degree $n$. So it is reasonable to ask what kind of general theory there could be for higher-degree forms.

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 nondegenerate binary cubic form over $\mathbf C$ can be diagonalized (see the start of the proof of Lemma 1.7 here; in the binary case, nondegeneracy of a cubic form is equivalent to the dehomogenization being a cubic polynomial with nonzero discriminant), nondegenerate cubic forms over $\mathbf C$ in more than two variables need not be diagonalizable. For example, the three-variable cubic form $x^3 - y^2z - xz^2$ is nondegenerate and 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 nondegenerate forms of degree $d$ in $n$ variables over $\mathbf C$ that are smooth away from $(0,0,\ldots,0)$ and are not diagonalizable. Note the diagonal form $x_1^d + \cdots + x_n^d$ is smooth away from the origin.

Concerning papers and books, I'll just mention one of each. There is Harrison's paper "A Grothendieck ring of higher degree forms" in J. Algebra 35 (1978), 123–138 here and Manin’s book Cubic forms: algebra, geometry, arithmetic. Manin mentioned a recurring nightmare he had about this book, soon after he finished it, in an interview with Eisenbud here.

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: $$ O(f) = \{A \in {\rm GL}_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}. $$ Nondegenerate quadratic forms have a rich orthogonal group (many reflections) and some higher-degree forms have a large orthogonal group: if $f$ is the determinant form of degree $n$ then its orthogonal group contains ${\rm SL}_n(F)$. But for a form $f$ of degree $d \geq 3$ over an algebraically closed field of characteristic $0$, $O(f)$ is sometimes a finite group. This happens if the corresponding symmetric $d$-multilinear form $\Phi$ satisfies $\Phi(\mathbf x, \ldots, \mathbf x,\mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = 0$. When $d \geq 3$ this condition is different from nondegeneracy as defined above. Let's say such $f$ and $\Phi$ are nonsingular. That nonsingular forms of degree $d$ have a finite orthogonal group over $\mathbf C$ is due to Jordan. It also holds over algebraically closed fields of characteristic $p$ when $p > d$ (so $d! \not= 0$ in the field). As an example, the orthogonal group of $x_1^d + \cdots + x_n^d$ over $\mathbf C$ when $d \geq 3$ has order $d^n n!$: it contains only the compositions of $n!$ coordinate permutations and scaling of each of the $n$ coordinates by $d$th roots of unity. Taking $n = 2$, this reveals a basic difference between the concrete binary forms $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$ that you can tell anyone who asks you in the future how higher degree forms are different from quadratic forms.

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: $$ O(f) = \{A \in {\rm GL}_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}. $$ Nondegenerate quadratic forms have a rich orthogonal group (many reflections) and some higher-degree forms have a large orthogonal group: if $f$ is the determinant form of degree $n$ then its orthogonal group is ${\rm SL}_n(F)$. But for a form $f$ of degree $d \geq 3$ over an algebraically closed field of characteristic $0$, $O(f)$ is sometimes a finite group. This happens if the corresponding symmetric $d$-multilinear form $\Phi$ satisfies $\Phi(\mathbf x, \ldots, \mathbf x,\mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = 0$. When $d \geq 3$ this condition is different from nondegeneracy as defined above. Let's say such $f$ and $\Phi$ are nonsingular. That nonsingular forms of degree $d$ have a finite orthogonal group over $\mathbf C$ is due to Jordan. It also holds over algebraically closed fields of characteristic $p$ when $p > d$ (so $d! \not= 0$ in the field). As an example, the orthogonal group of $x_1^d + \cdots + x_n^d$ over $\mathbf C$ when $d \geq 3$ has order $d^n n!$: it contains only the compositions of $n!$ coordinate permutations and scaling of each of the $n$ coordinates by $d$th roots of unity. Taking $n = 2$, this reveals a basic difference between the concrete binary forms $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$ that you can tell anyone who asks you in the future how higher degree forms are different from quadratic forms.

Before I begin, let me point out that we all know one important higher-degree form: the determinant form of degree $n$ in $n$ variables for all $n$. So it is reasonable to ask what kind of general theory there could be for higher degree forms.

First let's see that the bijection between quadratic forms and symmetric bilinear forms generalizes to higher degree. Recall for a field $F$ not of characteristic $2$, there is a bijection between quadratic forms $Q : F^n \to F$ and symmetric bilinear forms $B : F^n \times F^n \to F$ by $Q(\mathbf x) = B(\mathbf x,\mathbf x)$ and $$ B(\mathbf x,\mathbf y) = \frac{1}{2}(Q(\mathbf x + \mathbf y) - Q(\mathbf x) - Q(\mathbf y)). $$ Replacing $2$ by a degree $d \geq 1$, for a field $F$ where $d! \not= 0$ (meaning $F$ has characteristic $0$ or characteristic $p$ for $p > d$) there is a bijection between forms $f : F^n \to F$ of degree $d$ and symmetric $d$-multilinear maps $\Phi : \underbrace{F^n \times \cdots \times F^n}_{d \ {\sf copies}} \to F$ where $f(\mathbf x) = \Phi(\mathbf x,\ldots,\mathbf x)$ and $$ \Phi(\mathbf x_1,\ldots,\mathbf x_n) = \frac{1}{d!}\sum_{\substack{J \subset \{1,\ldots, d\} \\ J \not= \emptyset}} (-1)^{d - |J|}f\left(\sum_{j \in J} \mathbf x_j\right), $$ (You could include $J = \emptyset$ in the sum by the usual convention that an empty sum is $\mathbf 0$, since $f(\mathbf 0) = 0$.) For example, when $d = 3$ $$ \Phi(\mathbf x,\mathbf y,\mathbf z) = \frac{1}{6}(f(\mathbf x + \mathbf y + \mathbf z) - f(\mathbf x + \mathbf y) - f(\mathbf x + \mathbf z) - f(\mathbf y + \mathbf z) + f(\mathbf x) + f(\mathbf y) + f(\mathbf z)). $$ For example, if $f : F^3 \to F$ by $f(x_1,x_2,x_3) = x_1^3+x_2^3+x_3^3$ then $\Phi(\mathbf x,\mathbf y,\mathbf z) = x_1y_1z_1 + x_2y_2z_2+x_3y_3z_3$. The general formula for $\Phi$ in terms of $f$ shows why we want $d! \not= 0$ in $F$. Over fields of characteristic $0$, I think this bijection is due to Weyl.

Using this bijection, we call a form $f$ of degree $d$ nondegenerate if, for the corresponding symmetric multilinear form $\Phi$, we have $\Phi(\mathbf x,\mathbf y, \ldots, \mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = \mathbf 0$. In degree $2$, this is the usual notion of a nondegenerate quadratic form (or nondegenerate symmetric bilinear form).

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: $$ O(f) = \{A \in {\rm GL}_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}. $$ Nondegenerate 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. It also holds in characteristic $p$ when $p > d$ (so that $d! \not= 0$ in the field). For $d \geq 3$, the orthogonal group of $x_1^d + \cdots + x_n^d$ over $\mathbf C$ has order $d^n n!$: the compositions of $n!$ coordinate permutations and scaling of each of the $n$ coordinates by $d$th roots of unity. Taking $n = 2$, this reveals a basic difference between the concrete binary forms $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$ that you can tell anyone who asks you in the future how higher degree forms are different from quadratic forms.

Before I begin, let me point out that we all know one important higher-degree form: the determinant form of degree $n$. So it is reasonable to ask what kind of general theory there could be for higher degree forms.

First let's see that the bijection between quadratic forms and symmetric bilinear forms generalizes to higher degree. Recall for a field $F$ not of characteristic $2$, there is a bijection between quadratic forms $Q : F^n \to F$ and symmetric bilinear forms $B : F^n \times F^n \to F$ by $Q(\mathbf x) = B(\mathbf x,\mathbf x)$ and $$ B(\mathbf x,\mathbf y) = \frac{1}{2}(Q(\mathbf x + \mathbf y) - Q(\mathbf x) - Q(\mathbf y)). $$ Replacing $2$ by a degree $d \geq 1$, for a field $F$ where $d! \not= 0$ (meaning $F$ has characteristic $0$ or characteristic $p$ for $p > d$) there is a bijection between forms $f : F^n \to F$ of degree $d$ and symmetric $d$-multilinear maps $\Phi : \underbrace{F^n \times \cdots \times F^n}_{d \ {\sf copies}} \to F$ where $f(\mathbf x) = \Phi(\mathbf x,\ldots,\mathbf x)$ and $$ \Phi(\mathbf x_1,\ldots,\mathbf x_d) = \frac{1}{d!}\sum_{\substack{J \subset \{1,\ldots, d\} \\ J \not= \emptyset}} (-1)^{d - |J|}f\left(\sum_{j \in J} \mathbf x_j\right), $$ (You could include $J = \emptyset$ in the sum by the usual convention that an empty sum is $\mathbf 0$, since $f(\mathbf 0) = 0$.) For example, when $f$ is a cubic form ($d = 3$, $n$ arbitrary), the associated symmetric trilinear form is $$ \Phi(\mathbf x,\mathbf y,\mathbf z) = \frac{1}{6}(f(\mathbf x + \mathbf y + \mathbf z) - f(\mathbf x + \mathbf y) - f(\mathbf x + \mathbf z) - f(\mathbf y + \mathbf z) + f(\mathbf x) + f(\mathbf y) + f(\mathbf z)). $$ For example, if $f : F^3 \to F$ by $f(x_1,x_2,x_3) = x_1^3+x_2^3+x_3^3$ then $\Phi(\mathbf x,\mathbf y,\mathbf z) = x_1y_1z_1 + x_2y_2z_2+x_3y_3z_3$. The general formula for $\Phi$ in terms of $f$ shows why we want $d! \not= 0$ in $F$. Over fields of characteristic $0$, I think this bijection is due to Weyl.

Using this bijection, we call a form $f$ of degree $d$ nondegenerate if, for the corresponding symmetric multilinear form $\Phi$, we have $\Phi(\mathbf x,\mathbf y, \ldots, \mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = \mathbf 0$. (Equivalently, we have $\Phi(\mathbf x,\mathbf x_2, \ldots, \mathbf x_d) = 0$ for all $\mathbf x_2, \ldots, \mathbf x_d$ in $F^n$ only when $\mathbf x = \mathbf 0$.) When $d = 2$ (the case of quadratic forms), this is the usual notion of a nondegenerate quadratic form (or nondegenerate symmetric bilinear form).

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: $$ O(f) = \{A \in {\rm GL}_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}. $$ Nondegenerate quadratic forms have a rich orthogonal group (many reflections) and some higher-degree forms have a large orthogonal group: if $f$ is the determinant form of degree $n$ then its orthogonal group contains ${\rm SL}_n(F)$. But for a form $f$ of degree $d \geq 3$ over an algebraically closed field of characteristic $0$, $O(f)$ is sometimes a finite group. This happens if the corresponding symmetric $d$-multilinear form $\Phi$ satisfies $\Phi(\mathbf x, \ldots, \mathbf x,\mathbf y) = 0$ for all $\mathbf y$ in $F^n$ only when $\mathbf x = 0$. When $d \geq 3$ this condition is different from nondegeneracy as defined above. Let's say such $f$ and $\Phi$ are nonsingular. That nonsingular forms of degree $d$ have a finite orthogonal group over $\mathbf C$ is due to Jordan. It also holds over algebraically closed fields of characteristic $p$ when $p > d$ (so $d! \not= 0$ in the field). As an example, the orthogonal group of $x_1^d + \cdots + x_n^d$ over $\mathbf C$ when $d \geq 3$ has order $d^n n!$: it contains only the compositions of $n!$ coordinate permutations and scaling of each of the $n$ coordinates by $d$th roots of unity. Taking $n = 2$, this reveals a basic difference between the concrete binary forms $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$ that you can tell anyone who asks you in the future how higher degree forms are different from quadratic forms.