Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
8  
This is a bit too short for an answer, so I'll just leave it here: What makes the projective plane so great? The fact that it doesn't have many of the annoying limitations that Euclidean geometry has. In Euclidean geometry, two lines may intersect at one point, but in some rare but still existent cases will be parallel, and thus you have two cases to consider - which in fact means $2^n$ cases if you have more than just two lines. In projective geometry, the second case is almost nonexistent - the only bad thing that can happen is that the two lines coincide, which is much more seldom than... –  darij grinberg Jun 1 '10 at 21:03
3  
... parallelism. So the two cases are reduced to one (well, except of the coinciding case, but as I said, it is much more seldom and easier to rule out usually). That makes projective geometry easier to handle than Euclidean/affine geometry (when I say Euclidean/affine, I mean $\mathbb R^2$ as opposed to $\mathbb P\mathbb R^2$; I am not talking about distances, angles etc.). Now, the functions of interest on the Euclidean plane are polynomials, while those on the projective plane are homogeneous polynomials... –  darij grinberg Jun 1 '10 at 21:05
4  
Another reason homogeneous polynomials are popular in introductory discussions of algebraic geometry is that they are stand-ins for global sections of line bundles. If $X$ is a projective scheme, and $L$ a very ample line bundle on $X$, then $L$ gives an embedding into some $P^n$. By Serre vanishing, for large enough powers of $d$, $H^{0}(X,L^d)$ can be identified with the quotient of the degree $d$ homogeneous polynomials (on $P^n$) by the ideal of $X$ in degree $d$. This equivalence allows us to talk around global sections of line bundles without actually introducing line bundles.... –  Mike Roth Jun 1 '10 at 21:16
4  
Another interesting fact is this: in projective space, varieties of complementary dimensions always intersect. Thus, it is much easier to find (in the nonconstructive sense) nontrivial solutions to homogeneous systems of polynomials, than solutions to inhomogeneous systems. (The portion of Darij's remark above about parallel lines is a special case of this.) –  Charles Staats Jun 1 '10 at 21:35
1  
Thanks to everyone for the comments! I was aware of the fact pointed out by Darij, but to me it was always something a bit abstract, which I had never really used to solve any problems. I had never even made the obvious extension to varieties pointed out by Charles! That helps a lot. Unfortunately, though I'm sure Mike's comment will be helpful to many, I have only the vaguest sense of what he's saying. :( –  Pietro KC Jun 3 '10 at 5:53
show 1 more comment

6 Answers 6

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.

share|improve this answer
5  
I gather that there is also some deeper sense in which projective varieties are somehow tamer than other complete varieties. I could make a stab at a summary of the reasons, but I might be interested in an insider's explanation of why (or whether) it is so. –  Greg Kuperberg Jun 2 '10 at 6:11
    
Hi David, thanks for your answer! Phrasing the success of the construction as a "change of coordinates which doesn't pass to affine space" is a very intuitive and elegant way of putting it. That was very helpful. I'm familiar with the "running off to infinity" problem in analysis, but I guess I'll have to think about it more in the arithmetic setting. Thanks again! –  Pietro KC Jun 3 '10 at 6:08
    
Greg, if you can find the time, I would be delighted to hear what you have to say on tameness of varieties! –  Pietro KC Jun 3 '10 at 6:09
add comment

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.

share|improve this answer
    
Hi Greg, thanks for your answer! I was aware of the inductive uses of grading an algebra; due to my background I guess the example that struck me the most was Dvir's proof of the finite field Kakeya conjecture. I don't fully appreciate your last paragraph, since "coherent sheaf of modules on P^n" and "global sections" are not things on which I have much intuition. The most familiar interpretation for me would be the sheaf on P^n which assigns to each open set U the $\mathbb{R}$-module of continuous (real) functions on U. Would a global section just be a continuous function on all of P^n? –  Pietro KC Jun 3 '10 at 5:37
    
I don't understand what finite-dimensionality of global sections would be. If P^n is real projective space it seems like the global continuous functions would be an infinite-dimensional space, so I guess I have the wrong idea. Please forgive the elementary questions! If this would take too long to explain then just a reference would be great. –  Pietro KC Jun 3 '10 at 5:46
    
I'm thinking about complex projective space when I say P^n. Here, you want the sheaf of algebraic functions. Since algebraic functions are always complex analytic, they satisfy the maximum modulus principle, which states that they attain their maximum norm on the boundary of the domain they are defined on. The function must then be constant, since for any point p, the norm of the function at p must be greater than or equal to the norm on all of P^n\p. –  Greg Muller Jun 3 '10 at 15:57
    
The reason for algebraic functions and not continuous ones can be traced back to the kind of ring we care about. the scheme P^n is closely related to C^{n+1}, and the ring of algebraic functions on C^{n+1} is C[x_0,...x_n], while the ring of continuous functions is horrible from an algebraic perspective (infinitely-generated, non-Noetherian, etc). In general in algebraic geometry, you want algebraically closed fields and algebraic functions, unless you have a very good reason otherwise. –  Greg Muller Jun 3 '10 at 15:59
add comment

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.

share|improve this answer
    
Your remark about the random cutting out of points is very helpful, thanks! I hadn't looked at it this way. Coincidentally, just today I listened to a talk on local zeta functions and it was also very clarifying. –  Pietro KC Jun 3 '10 at 5:20
    
In view of the offence caused by a comment I left when rolling this back to Charles Matthews's original text, I am removing the comment - the original text has been quoted on meta.mathoverflow.net/a/415/763 should you wish to see how I erred –  Yemon Choi Jul 9 '13 at 3:36
add comment

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.

share|improve this answer
    
Why can we think of $R$ as a disc and $K$ as a punctured disc? I see that it fits to the picture etc., but is there any explanation which is more rigorous? Perhaps within rigid geometry? –  Martin Brandenburg Jul 9 '13 at 7:33
2  
I would say rather that a dvr is a tiny bit of a smooth curve (e.g. it's a tiny, smooth $1$-dim'l scheme), hence a disc, and its fraction field is the "generic" part of this disc, hence a punctured disc. –  Daniel Litt Jul 9 '13 at 16:24
add comment

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.

share|improve this answer
    
Hi Charles, thanks for your answer! It seems like your remark is spot-on (using hom-polys to solve another problem), but I don't quite understand the construction. Are the f_i given over K, and then you want to find a g, also over K, the zeros of which are exactly the intersection of the zeros of all f_i? Or are the f_i given over some F and you can then choose your K? Even granting the 1st option, I'm not sure I understand the construction: what is the "original polynomial"? How are the f_i coming in? –  Pietro KC Jun 3 '10 at 6:02
    
There's only one field, $K$. Take $f(x)$ to be an irreducible poly in one var over $K$, then take $f'(x,y)$ the homogenization. Then look at $f'(x,f(y))$ and iterate this construction, because $f'(x,y)=0$ if and only if $x=y=0$ in the field, because it was irreducible. –  Charles Siegel Jun 3 '10 at 12:04
add comment

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 $z_0,...,z_n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.