What, if anything, makes homogeneous polynomials so great?

Question

It should be obvious from the question that I am not any kind of algebraic geometer, so if there are definitions of hom-polys as comonoidal dyadic functors or whatnot, let's leave that to one side for the purposes of this question. I really mean hom-polys in the most pedestrian sense possible.

From the outside, it seems that homogeneous polynomials get a lot of attention in alg-geom. (Perhaps in other areas as well?) I know that they are, well, homogeneous with respect to dilations, and that this allows one to look at their zeros in projective space in a natural way.

I usually like to keep a safe distance between myself and projective space, and have always looked at hom-polys as "merely" a technical tool. But, just the other day, I was able to quickly solve a small, elementary number-theoretic problem by converting it into "homogeneous" form. (Some vague memories of homogeneous problem-solving heuristics prompted this.) To be precise, the problem was that of computing how many solutions there are to m^2 + n^2 = 1 in Z_p, where p is a prime congruent to 3 mod 4. (Of course this is trivial by a change of variables when p = 1 mod 4.) The problem seemed unfriendly at first, but passing to the related question of the solutions to m^2 + n^2 - t^2 = 0 revealed a lot of symmetry and made it quite trivial. (Of course, solving small cases of the first problem showed a lot of symmetry, but it wasn't obvious how to get a handle on it.)

My question is: do hom-polys help to solve a lot of seemingly unrelated problems via some process of homogeneization of the problem? Do you have examples? Is this one reason for their popularity? I'm looking for heuristics mostly, but if you think an actual theorem makes precise or helps formulate a heuristic, go nuts.

Answer 1

I would rephrase the question as "what is so great about projective space (as compared to affine space)?" I would give two answers:

1. Projective space has a larger symmetry group: dimension $n^2+2n$ rather than $n^2+n$.
2. Projective space is compact.

The first is what you are using when you turn $x^2+y^2=1$ into $x^2+y^2=z^2$ and then $(z+y)(z-y)=1$; the corresponding change of coordinates is a symmetry of projective space which doesn't pass to affine space. The second is why homogenous polynomials work better for intersection theory: intersections can't run off to infinity. It is also why point counting over finite fields gives nicer answers in projective space: point counting is related to cohomology, and cohomology for smooth compact spaces obeys Poincare duality.

Answer 2

From a practical perspective, putting a grading on an algebra usually organizes the algebra into a collection of finite-dimensional vector spaces, each indexed by a natural number. This opens the door to induction arguments which at each step, only have to deal with a finite-dimensional vector space. Its this crude idea which seems to motivate most of the computational techniques in the theory of commutative algebras (see, anything with Gr\"obner bases).

More generally, graded algebras/projective spaces allow finite-dimensional-type techniques to be used in the study of infinite-dimensional algebras and modules. As an example, if $A$ is a polynomial ring, and $M$ is a f.g. graded $A$ module, then the double dual $$ Hom_\mathbb{C}(Hom_{\mathbb{C}}(M,\mathbb{C}),\mathbb{C}) $$ is a monstrosity: infinitely-generated and non-graded. However, its graded double dual $$\underline{Hom}_\mathbb{C} (\underline{Hom}_\mathbb{C} (M,\mathbb{C}),\mathbb{C})$$ is isomorphic to $M$.

This finite-dimensionality mantra is even more prominent in the study of projective schemes. A coherent sheaf of modules $\mathcal{M}$ on $\mathbb{P}^n$ will not only have finite-dimensional global sections, but all higher cohomologies of $\mathcal{M}$ will be finite dimensional. This means you can talk about things like the dimension of these cohomologies, which allows definitions of things like 'genus' and 'Euler characteristic'. These concepts have to be heavily modified to make any sense in the affine cases.

Answer 3

Well, if you are interested in counting solutions mod p, you could note that the "good" formulae are indeed related to the homogeneous approach/projective space. It is not just a question of restoring the symmetry between variables, though that can be part of it: there are more projective transformations than affine transformations. The general theory of local-zeta functions would explain that counting points mod p works best in the homogeneous setting, and making things inhomogeneous is going to cut out some points, in a way that is relatively random.

Try http://en.wikipedia.org/wiki/Pl%C3%BCcker_coordinates for example, for geometry.

Answer 4

Several of the other answers make the point that

projective varieties are compact

in one way or another. Let me say the same thing, in a sort of low-concept way, which you might find appealing from an algebraic point of view:

Homogeneous polynomials allow you to clear denominators.

The formal way of saying these two things are the same is called the valuative criterion for properness (combined with the fact that projective space is proper). The analogous statement for complex manifolds is as follows--a complex manifold $X$ is compact if and only if every meromorphic map from a punctured disc to $X$ extends across the puncture. The proof that this is true for $X=\mathbb{CP}^n$ is precisely: clear denominators.

In general, the valuative criterion for properness says that a variety $X$ is compact (read: a Noetherian scheme is proper) if and only if, for every dvr $R$ with fraction field $K$, every solution to the equations defining $X$ in $K$ comes from a unique solution in $R$ (for the actual statement, click the Wikipedia link). Again, the proof that this is true for projective varieties is precisely: clear denominators. You should think of $K$ as being analogous to a punctured disc, and $R$ to the whole disc.

Answer 5

One thing you can do with them is to solve the following problem: given $f_1,\ldots,f_k$ polynomials in $x_1,\ldots,x_n$, is there a polynomial over a field $K$, call if $g$, such that $g=0$ if and only if $f_1=\ldots=f_k=0$? The solution that I know of involves taking an irreducible polynomial over $K$ (so $K$ can't be algebraically closed), homogenizing it, and then substituting the original polynomial in for the new variable, and iterating the process. The solution may not be homogeneous, but homogeneous polynomials are useful for solving some problems stated entirely in terms of arbitrary polynomials.

Answer 6

As you already noticed, thinking in terms of homogeneous coordinates in a complex projective space $\mathbb{P}^n$, in order to define a function (any old complex-valued function, not necessarily holomorphic or even continuous) on $\mathbb{P}^n$ we must require that $f$ is homogeneous of order zero, i.e., $f(\lambda x)=f(x)$ for any $\lambda \in \mathbb{C}\setminus 0$ and any $x \in \mathbb{P}^n$. (I am cutting some corners with this notation.) But in order to define a zero set of a function on $\mathbb{P}^n$ , we have a little more leeway: requiring that the the ``function" be homogeneous of order $k \geq 0$, i.e., $f(\lambda x)=\lambda^kf(x)$ for any $\lambda \in \mathbb{C}\setminus 0$ and any $x \in \mathbb{P}^n$, will do. A more profound statement (=Chow's theorem) is true (sort of converse to the above): every analytic set in $\mathbb{P}^n$ is the (common) zero set of finitely many homogeneous polynomials.

Aside 1: This point of view generalizes to toric varieties, since they admit an analog of homogeneous coordinates (first introduced in the smooth case in MR1106194 Audin, Michèle: The topology of torus actions on symplectic manifolds. Translated from the French by the author. Progress in Mathematics, 93. Birkhäuser Verlag, Basel, 1991. 181 pp. ISBN: 3-7643-2602-6, then developed for the general case in MR1299003 Cox, David A.: The homogeneous coordinate ring of a toric variety. J. Algebraic Geom. 4 (1995), no. 1, 17–50)

Aside 2: the interpretation of sections of a line bundle over a complex manifold as homogeneous functions has nothing to do with projectivity and/or Serre's theorem (as an analyst, I'll stick to the language of complex manifolds and line bundles instead of schemes and sheaves). It is more general and follows from the relation between the $k$-th tensor power of the line bundle $\pi: L \to X$ over an $n$-dimensional complex manifold $X$ and the dual bundle $\pi': L' \to X$. Namely (as can be checked using respective definitions, taking into account relations between trivializations), every holomorphic function $f$ on $L'$ which is homogeneous of degree $k$ on the fibers is a section in $L^k$ (not just a ''stand-in"). In the specific case of $X=\mathbb{P}^n$ and $L=\mathcal{O}(1)$, the space of sections $\Gamma(\mathbb{P}^n, \mathcal{O}(k))$ is isomorphic to the space of homogeneous polynomials of degree $k$ in the variables $z_0,...,z_n$.