Regular differential forms

1 Regular differential forms - the affine case

Let X be an affine algebraic variety with coordinate ring k[X]. The k[X]-module Ω[X] of regular differential forms is generated by elements df (f ∈ k[X]) with relations

d(f + g) = df + dg
d(fg) = fdg + gdf
dα = 0 (α ∈ k)

So, elements of Ω[X] are sums of terms gdf with f, g ∈ k[X].

Example The affine parabola y = x^{2} has dy = 2xdx and, using this, all occurrences of y and dy can be eliminated. The regular differential forms are ω = g(x)dx with g ∈ k[x].

Example The affine cubic curve y^{2} = x^{3} + x has 2ydy = (3x^{2} + 1)dx. An example of a regular differential form is

ω = \frac{dx}{2y} = \frac{dy}{3x^{2} + 1}.

Is it really OK to have fractions? According to the definition we should have

ω = fdx + gdy with f, g ∈ k[X].

Here

2yω = dx
(3x^{2} + 1)ω = dy

so we need 2yf + (3x^{2} + 1)g = 1 for certain f, g ∈ k[X]. By the Nullstellensatz that means X ∩ V (2y, 3x^{2} + 1) = ∅, and this is true: the curve is nonsingular. Of course we can also compute explicitly: take

f(x, y) = −\frac{9}{4}xy and g(x, y) = \frac{3}{2}x^{2} + 1.

Then 2yf + (3x^{2} + 1)g = −\frac{9}{2}x(x^{3} + x) + (\frac{3}{2}x^{2} + 1)(3x^{2} + 1) = 1. Hence

ω = \frac{dx}{2y} = \frac{dy}{3x^{2} + 1} = −\frac{9}{4}xydx + (\frac{3}{2}x^{2} + 1)dy.

2 Regular differential forms - the projective case

If X is a projective variety, it has a covering with affine pieces. Now a regular differential form is one that is regular in each piece.

Example Take the projective line P^{1}. It has projective coordinates (X, Y). It is covered by the two affine pieces A_{1} = P^{1} \ {(1, 0)} and A_{2} = P^{1} \ {(0, 1)}. In A_{1} the projective coordinates can be chosen as (X, 1), and in A_{2} the projective coordinates can be chosen as (1, Y). In A_{1} ∩ A_{2} the projective point (X, Y) corresponds to (X/Y, 1) in A_{1} and to (1, Y/X) in A_{2}, so the Y of A_{2} is the 1/X of A_{1}.

Suppose we have a regular differential form on P^{1}. Restricted to A_{1} it looks like f(X)dX. Restricted to A_{2} it looks like g(Y)dY. And both forms agree on A_{1} ∩ A_{2}. That is, f(X)dX = g(Y)dY = g(\frac{1}{X})d(\frac{1}{X}) = g(\frac{1}{X}).\frac{−1}{X^{2}} .dX but that is impossible: there are no polynomials f(X) and g(X) such that f(X) = g(\frac{1}{X}).\frac{−1}{X^{2}} . It follows that there are no regular differential forms on P^{1}.

Example Take the projective curve Y^{2}Z = X^{3} + XZ^{2}. The projective plane P^{2} is covered by three affine pieces: A_{1} is the part with Z ≠ 0 and coordinates (X, Y, 1), A_{2} is the part with Y ≠ 0 and coordinates (U, 1, V), A_{3} is the part with X ≠ 0 and coordinates (1, S, T), where on A_{1} ∩ A_{2} we have U = X/Y, V = 1/Y, and on A_{1} ∩ A_{3} we have S = Y/X, T = 1/X, and on A_{2} ∩ A_{3} we have S = 1/U, T = V/U. In our case (where Y^{2}Z = X^{3} + XZ^{2}) the part A_{3} is superfluous, since already A_{1} and A_{2} cover the curve. (In the projective plane the only point not covered by A_{1} ∪ A_{2} is (1, 0, 0), but that does not lie on our curve.)

Claim:

ω = \frac{dX}{2Y} = \frac{dY}{3X^{2} + 1} = \frac{dU}{2UV − 1} = \frac{−dV}{3U^{2} + V^{2}}

is a regular differential form.

Check: In A_{1} we have the equation Y^{2} = X^{3} + X and we already saw that \frac{dX}{2Y} = \frac{dY}{3X^{2}+1} is a regular differential form on that affine piece. In A_{2} we have the equation V = U^{3} + UV^{2} and in the same way we see that \frac{dU}{2UV −1} = \frac{−dV}{3U^{2}+V^{2}} is a regular differential form on that affine piece. Finally, in the intersection A_{1} ∩ A_{2} we have

ω = \frac{dY}{3X^{2} + 1} = \frac{1}{3(\frac{U}{V})^{2} + 1}.\frac{−1}{V^{2}}.dV = \frac{−dV}{3U^{2} + V^{2}} .

Thus ω is a regular differential form as claimed.

3 An algebraic definition of the genus

So far we saw that there are no regular differential forms on the projective line P^{1} and we found one such form for the elliptic curve Y^{2} = X^{3} + X.

Theorem 3.1 Let X be a nonsingular projective curve. Then dim_{k} Ω[X] = g.

Earlier the genus g was defined as the number of holes in the two-dimensional real surface that is the one-dimensional complex curve. This theorem can be taken as definition when k is not the field of complex numbers.

Description of the regular differential forms

If the nonsingular projective curve X is given by the equation f(X, Y) = 0 for some polynomial f of degree d, then f_{X}dX + f_{Y} dY = 0 (where f_{X} and f_{Y} are the derivatives of f w.r.t. X and Y), and the regular differential forms look like

ω = \frac{gdX}{f_{Y}} = \frac{−gdY}{f_{X}}

for some polynomial g(X, Y) of degree at most d − 3.

(Nonsingularity implies that f_{X} and f_{Y} do not vanish simultaneously on the curve, so that ω can be written without fractions.)

(Why d − 3? Consider the change from (X, Y, 1) to (U, 1, V), with Y = \frac{1}{V} and dY = −\frac{1}{V^{2}} dV. If f(X, Y) has degree d and f_{X}(X, Y) has degree d − 1 and g(X, Y) has degree e, then

\frac{−g(X, Y)dY}{f_{X}} = \frac{g(\frac{U}{V} , \frac{1}{V})}{f_{X}(\frac{U}{V} , \frac{1}{V})}.\frac{1}{V^{2}} .dV

behaves like V^{−e}.V^{−2}.V^{d−1} near V = 0, and since the value must be well-defined for V = 0 we must have e ≤ d − 3.)

Since a polynomial of degree at most d has (d+1)+d+...+1 = \frac{1}{2}(d+2)(d+1) coefficients, we find g = \frac{1}{2}(d − 1)(d − 2) for a nonsingular curve of degree d.

4 Rational differential forms

Rational differential forms are sums of terms gdf where now f, g ∈ k(X). For a nonsingular curve X, the set Ω(X) of rational differential forms is a 1-dimensional vector space over k(X).