Sunday, 18 September 2011

Commutative Algebra and Algebraic Geometry


Last time we saw how to create variables for use in polynomial arithmetic. This time I want to look at some of the things we can do next.

First, let's define the variables we are going to use:

> :l Math.CommutativeAlgebra.GroebnerBasis
> let [t,u,v,x,y,z,x',y',z'] = map glexvar ["t","u","v","x","y","z","x'","y'","z'"]


So now we can do arithmetic in the polynomial ring Q[t,u,v,x,y,z,x',y',z']. For example:

> (x+y)^2
x^2+2xy+y^2


The branch of mathematics dealing with the theory of polynomial rings is called commutative algebra, and it was "invented" mainly in support of algebraic geometry. Algebraic geometry is roughly the study of the curves, surfaces, etc that arise as the solution sets of polynomial equations. For example, the solution-set of the equation x^2+y^2=1 is the unit circle.

If we are given any polynomial equation f = g, then we can rewrite it more conveniently as f-g = 0. In other words, we only need to track individual polynomials, rather than pairs of polynomials. Call the solution set of f = 0 the zero-set of f.

Sometimes we're interested in the intersection of two or more curves, surfaces, etc. For example, it is well known that the hyperbola, parabola and ellipse all arise as "conic sections" - that is, as the intersection of a cone with a plane. So define the zero-set of a collection (or system) of polynomials to be the set of points which are zeros of all the polynomials simultaneously. For example, the zero-set of the system [x^2+y^2-z^2, z-1] is the unit circle x^2+y^2=1 situated on the plane z=1.


Okay, so how can commutative algebra help us to investigate curves and surfaces? Is there a way for us to "do geometry by doing algebra"? Well, first, what does "doing geometry" consist of? Well, at least some of the following:
- Looking at the shapes of curves and surfaces
- Looking at intersections, unions, differences and products of curves and surfaces
- Looking at when curves or surfaces can be mapped into or onto other curves or surfaces
- Looking at when two different curves or surfaces are equivalent, in some sense (for example, topologically equivalent)

(That phrase "curves and surfaces" is not only clumsy but also inaccurate, so from now on I'll use the proper term, "variety", for the zero-set of a system of polynomials, whether it's a set of isolated points, a curve, a surface, some higher dimensional thing, or a combination of some of the preceding.)

So can we do all those things using algebra? Well, let's have a go.

Let's start by looking at intersections and unions of varieties (remember, that's just the fancy name for curves, surfaces, etc.).

Well, we've already seen how to do intersections. If a variety V1 is defined by a system of polynomials [f1...fm], and a variety V2 is defined by [g1...gn], then their intersection is defined by the system [f1...fm,g1...gn] - the zero-set of both sets of polynomials simultaneously. We'll call this the "sum" of the systems of polynomials. (Note to the cognoscenti: yes, I'm really talking about ideals here.)

sumI fs gs = gb (fs ++ gs)

Don't worry too much about what that "gb" (Groebner basis) call is doing. Let's just say that it's choosing the best way to represent the system of polynomials. For example:

> sumI [x^2+y^2-z^2] [z-1]
[x^2+y^2-1,z-1]


Notice how the gb call has caused the first polynomial to be simplified slightly. The same variety might arise as the zero-set of many different systems of polynomials. That's something that we're going to need to look into - but later.

Okay, so what about unions of varieties. So suppose V1 is defined by [f1...fm], V2 is defined by [g1...gn]. A point in their union is in either V1 or V2, so it is in the zero-set of either [f1...fm] or [g1...gn]. So how about multiplying the polynomials together in pairs. That is, let's look at the system [fi*gj | fi <- fs, gj <- gs]. Call the zero-set of this system V. Then clearly, any point in either V1 or V2 is in V, since we then know that either all the fs or all the gs vanish at that point, and hence so do all the products. Conversely, suppose that some point is not in the union of V1 and V2. Then there must exist some fi, and some gj, which are non-zero at that point. Hence there is an fi*gj which is non-zero, so the point is not in V.

This construction is called, naturally enough, the product of the systems of polynomials.

productI fs gs = gb [f * g | f <- fs, g <- gs]


> productI [x^2+y^2-z^2] [z-1]
[x^2z+y^2z-z^3-x^2-y^2+z^2]


Just in case you're still a little unsure, let's confirm that a few arbitrary points in the union are in the zero-set of this polynomial:

> eval (x^2*z+y^2*z-z^3-x^2-y^2+z^2) [(x,100),(y,-100),(z,1)]
0
> eval (x^2*z+y^2*z-z^3-x^2-y^2+z^2) [(x,3),(y,4),(z,5)]
0


The first expression evaluates the polynomial at the point (100,-100,1), an arbitrary point on the plane z=1. The second evaluates at (3,4,5), an arbitrary point on the cone x^2+y^2=z^2. Both points are in the zero-set of our product polynomial.

Since we're in the neighbourhood, let's have a look at the other conic sections. First, let's rotate our coordinate system by 45 degrees, using the substitution x'=x+z, z'=z-x. (Okay, so this also scales - to save us having to handle a sqrt 2 factor.)



> let cone' = subst (x^2+y^2-z^2) [(x,(x'-z')/2),(y,y'),(z,(x'+z')/2)]
> cone'
-x'z'+y'^2


In these coordinates, the intersection of the cone with the plane z'=1 is the parabola x'=y'^2:

> sumI [cone'] [z'-1]
[y'^2-x',z'-1]


Alternatively, the intersection with the plane y'=1 is the hyperbola x'z'=1:

> sumI [cone'] [y'-1]
[x'z'-1,y'-1]


Okay, so we've made a start on seeing how to do geometry by doing algebra, by looking at unions and intersections of varieties. There's still plenty more to do. We mustn't forget that we have some unfinished business: we need to understand when different polynomial systems can define the same variety, and in what sense the gb (Groebner basis) function finds the "best" representation. That will have to wait for another time.

Incidentally, for the eval and subst functions that I used above, you will need to take the new release HaskellForMaths v0.4.0. In this release I also removed the older commutative algebra modules, so I revved the minor version number.

No comments:

Post a Comment

Followers