Quadratic forms play a huge role in math. This leads one to wonder: Is there a theory of cubic forms, quartic forms, quintic forms and so on? I have failed to discover any. Is there any such theory? If not, is it because:
I once asked André Weil the same question.
When I was college, taking a course that discussed quadratic forms, Weil gave a guest lecture to the students about that topic. After the talk, I raised my hand and asked him why there was such a big deal in math about quadratic forms while it seemed there was nothing comparable for higher-degree forms. Weil gave an answer, but to my regret I could not understand it (difficulty hearing him) and I did not ask him later to repeat what he had said. Now many years later, I can offer an answer that I think my former student self would have found satisfactory.
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).
That the bijection between quadratic forms and symmetric bilinear forms can be extended to higher degrees suggests there might be general theory in higher degree that's just like the quadratic case, but it turns out there really are significant differences between quadratic forms and forms of higher degree. Here are two of them.
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.
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.
In retrospect, the label used for the second topic ("Group theory") really applies to both topics. For a field $F$, the group ${\rm GL}_n(F)$ acts on the forms of degree $d$ in $n$ variables with coefficients in $F$, and the first topic is about the orbit of $x_1^d + \cdots + x_n^d$ under this action while the second topic is about the stabilizer of $f(x_1,\ldots,x_n)$ under this action.
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.
This is an extended comment on KConrad's discussion of symmetry groups. We can think of $k$-forms on a vector space $V$ (homogeneous polynomials of degree $k$) abstractly as elements of the symmetric power $S^k(V^{\ast})$. If we want to classify such things up to change of coordinates, or equivalently if we want to understand the automorphism groups of such things, we look at the action of $GL(V)$, and at its orbits and its stabilizers respectively.
This action is close to faithful (the kernel contains at least the $k^\text{th}$ roots of unity) so heuristically we can expect the generic orbit to have dimension $\max(\dim GL(V), \dim S^k(V^\ast))$. Now if $n = \dim V$ we have that
from which we see that it's only for $k \le 2$ that $\dim S^k(V^{\ast}) \le \dim GL(V)$ in general, and starting with $k \ge 3$ we have $\dim S^k(V^{\ast}) \gg \dim GL(V)$. This heuristically suggests that orbits will have large stabilizers for $k \le 2$ but generically have finite stabilizers for $k \ge 3$, which is apparently what we find. It also suggests that we might have a zero-dimensional space of orbits for $k \le 2$ (which e.g. is the case over $\mathbb{R}$ or $\mathbb{C}$) but a positive-dimensional space of orbits varying in families over the base field for $k \ge 3$. So based off these dimension counts alone we expect cubic forms and higher to both be much harder to classify than quadratic forms and to have much fewer symmetries. This is before getting into the additional difficulties of solving, say, cubic and higher degree Diophantine equations relative to quadratic ones.
Now for some philosophical speculation: we're working here with an abstract finite-dimensional vector space over an abstract field but of course our concepts of fields, vector spaces, and even quadratic forms is ultimately based on our preconceptual direct experiences with moving around in ~Euclidean space, the $3$-dimensional real vector space $\mathbb{R}^3$ equipped with the Euclidean form $x^2 + y^2 + z^2$. One wonders if the whole concept of fields and vector spaces somehow has Euclideanness baked into it from the start, in such a way that it inevitably distinguishes quadratic forms, and if intelligent beings living in a substantially less Euclidean world would set up the fundamentals of mathematics quite differently….
Some mild evidence against this is the classification of finite simple groups, which seems to suggest that Lie theory, or at least the theory of algebraic groups, is in some sense inevitable once you have the notion of a symmetry at all. And there the orthogonal groups (together with the symplectic groups etc.) clearly play a distinguished role.
Another aspect of the privilegiated role of quadratic forms is duality. Roughly speaking, a form $f$ of degree $n$ defines a differential map $x\mapsto\xi=df(x)$. If $m=2$, then $x\mapsto\xi$ is linear, and there is a simple criterion that it be one-to-one (the discriminant being non-zero). If $m>2$, the situation is much more intricate.
Let me mention however that $m$-forms over $\mathbb R$ or $\mathbb C$ play an important role in the theory of linear differential operators. The class of hyperbolic forms, introduced by Gårding is particularly important.
The most important hyperbolic form is the determinant (degree $n$) over the space ${\bf Sym}_n$ of real symmetric $n\times n$ matrices. This one turns out to have a very large isotropy group, namely the transformations $S\mapsto P^TSP$ where $P\in{\bf SL}_n$. This group $O(Det)$ is all but finite, thus this example is really exceptional , in Conrad's point of view.
Considering the determinant over ${\bf M}_n(k)$, then $O(Det)$ is even wider. It contains the transformations $M\mapsto PMQ$ for $P,Q\in{\bf SL}_n(k)$, as well as $M\mapsto M^T$.
A variant of the example above is the Pfaffian, an $m$-form over the space ${\bf Alt}_{2m}(k)$ of alternate $(2m)\times(2m)$ matrices. Its isotropic group is again ${\bf SL}_{2m}(k)$.
This is perhaps more of a comment than a full answer, but even ternary cubic forms are not well understood, and results about them are at the cutting edge of current research. See for example Ternary cubic forms having bounded invariants, and the existence of a positive proportion of elliptic curves having rank 0 by Bhargava and Shankar, or the research into numbers expressible as the sum of three cubes.
Also over the rationals (as @KConrad might have mentioned!), the theory of quadratic forms is nicer than the theory of higher forms:
Hasse’s principle holds for quadratic forms: if a quadratic form has a zero in the reals and in all the p-adics, then it has a rational zero too. By contrast this fails even for ternary cubics, as in Selmer's example.
There are simple questions about length and area which have been open for over 300 years and are equivalent to solubility of quadratic forms over the rationals, with nothing so old and geometric for higher forms. E.g.:
Is there a perfect Euler brick, whose side lengths and diagonals on the faces and main diagonal all have rational lengths?
The congruent number problem: Which numbers can be the area of a right triangle whose sides all have rational lengths?
Cubic forms are much more complicated than quadratic forms, so it may not be possible to develop a theory to end it all.
One direction of cubic forms is cubic composition laws, similar to Gauss's quadratic composition laws.
Another direction is to study special higher-degree forms that yield unusual automorphism groups. For example:
The Lie group $G_2$ is the stabilizer subgroup of the canonical differential 3-form $⟨-,(−)×(−)⟩$ in $\mathbb R^7$;
The Lie group $F_4$ is the stabilizer subgroup of the Jordan algebra product $A\circ B=\frac12 (AB+BA)$ (an entrywise 3-form) defined for 3x3 octonionic matrices;
Cartan found a cubic form (on 27 dimensions) for $E_6$ and a quartic form (on 56 dimensions) for $E_7$;
I'm not very sure for $E_8$, but it seems to require octic polynomials.
The study of forms appears to grow "exponentially harder" with their degree as the following example seems to indicate.
Let us work over the field of complex numbers since the following statements become even more intricate over other fields!
A homogeneous form $F(x_0,\dots,x_n)$ of degree $d$ is said to be smooth if the ideal generated by its partial derivatives contains all monomials of sufficiently large degree.
A parametric solution of a form is a collection of polynomials $\mathbf{u}(\mathbf{y})=(u_i(y_1,\dots,y_n))_{i=0}^n$ such that "most" solutions of the form are values of the form $\mathbf{u}(\mathbf{a})$ for a suitable set of complex numbers $(a_1,\dots,a_n)$. By "most" solutions we mean all solutions that do not satisfy some independent other fixed form $G(x_0,\dots,x_n)$. (For those who know the terminology, this means that the associated variety is unirational.)
It was known classically that linear and quadratic forms have parametric solutions.
It is probably only less than 150 years ago or so that it was shown that smooth cubic forms have parametric solutions in 4 or more variables.
It is even more recently shown that smooth forms of any degree $d$ have parametric solutions in sufficiently many variables, where the number of variables grows more than exponentially with $d$. (It is worth finding out the exact number of variables required for $d=4$.) More recent efforts have strengthened these results somewhat, but exponential growth remains a "feature".
We can also ask whether there is a parametric solution where most solutions have a unique associated $(a_1,\dots,a_n)$. (This corresponds to asking whether the variety is rational.) Such a solution is called a rational solution.
It was known classically that such rational solutions are possible for linear and quadratic forms.
Smooth cubic forms in 4 or more variables that do have rational solutions have probably been known for a long time. It was only about 150 years ago that it was shown that no smooth cubic form in 3 variables has a rational solution and every smooth cubic form in 4 variables has a rational solution. It was only in the 70's that it was shown that there are smooth cubic forms in 5 variables that do not have a rational solution. For 6 or more variables the locus of smooth cubic forms that have a rational solution is not fully understood.
The situation for quartic forms (or, more generally, degree $\geq 4$) is more dire. There is not a single example known of a smooth quartic form that has a rational solution. However, it has not been proven that there is no such form either!
Ref: The following exposition appears to cover quite a bit of what is stated above. Details are in the references to this reference! "Unirationality of Hypersurfaces" by Robert Mijatovic
This is really more of a remark. The zero set of a homogeneous polynomial $f(x_0,\ldots, x_n)$ over a field, is a hypersurface in the projective space $\mathbb{P}^n_k$. These objects have been studied in algebraic geometry, and surrounding areas, for a long time.
Suppose that $k= \mathbb{C}$. When $d=\deg f=2$, then up to a linear change of variables, $f= x_0^2+\dots + x_m^2$. In particular, there is only one nonsingular quadric in a given dimension. Its geometry can be understood, by observing that it is homogenous under the action of the special orthogonal group. When $d> 2$, the story becomes much complicated. Except for plane cubics (elliptic curves) these are never homogeneous. All of the nonsingular hypersurfaces with $d,n$ fixed are diffeomorphic, however, as algebraic varieties or complex manifolds, they can be nonisomorphic. In fact, there are "continuous families" or moduli spaces of isomorphism classes of hypersurfaces. I'll stop with that.
Here is an additional comment. Every homogeneous polynomial can be regarded as a symmetric tensor. Quadrics correspond to symmetric matrices; cubics correspond to order $3$ tensors; in general, a homogeneous polynomial of degree $d$ corresponds to a symmetric tensor of order $d$. Matrices are certainly easier to work with, more familiar, and more deeply understood, than higher-order tensors.
There is a theory of tensors. In fact it's a highly active area. But even so, it's leagues behind the theory of matrices.
In a little bit more detail: The symmetric $n \times n$ matrix $A$ with entries $a_{i,j} = a_{j,i}$ corresponds to the quadratic form
$$ Q(x) = \sum_{i,j} a_{i,j} x_i x_j = a_{1,1} x_1^2 + 2a_{1,2} x_1 x_2 + \dotsb = \vec{x}^T A \vec{x} , $$
and this is closely related to the bilinear form $B(x,y) = x^T A y$, or $\langle x | A | y\rangle$ if you like. When we go to cubic forms, the symmetric $n \times n \times n$ array $T$ with entries $a_{i,j,k} = a_{i,k,j} = a_{j,k,i} = \dotsb$ corresponds to the cubic form (degree $3$ homogeneous polynomial)
$$ C(x) = \sum_{i,j,k} a_{i,j,k} x_i x_j x_k = a_{1,1,1} x_1^3 + 3a_{1,1,2} x_1^2 x_2 + \dotsb + 6 a_{1,2,3} x_1 x_2 x_3 + \dotsb. $$
And it corresponds to the trilinear form $F(x,y,z)$ given by triple contraction of $T$ by the vectors $x,y,z$ along the three "sides" of $T$. How often do we deal with trilinear forms? They exist, but it's a bit of a minor topic compared with bilinear forms. Worse yet, $d$-linear forms for larger $d$.
I'm not claiming that higher-degree forms are lacking theory (or harder, or less interesting) because of the distinction between tensors and matrices. The relation here is probably more like an analogy or parallel, rather than causation.
