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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(f g) = f dg + 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 oc- currences 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 = dx 2 y
dy 3 x^2 + 1
Is it really OK to have fractions? According to the definition we should have
ω = f dx + gdy with f, g ∈ k[X].
Here 2 yω = 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, 3 x^2 + 1) = ∅, and this is true: the curve is nonsingular. Of course we can also compute explicitly: take
f (x, y) = −
xy and g(x, y) =
x^2 + 1.
Then 2yf + (3x^2 + 1)g = − 92 x(x^3 + x) + ( 32 x^2 + 1)(3x^2 + 1) = 1. Hence
ω =
dx 2 y
dy 3 x^2 + 1
xydx + (
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( (^) X^1 )d( (^) X^1 ) = g( (^) X^1 ). (^) X−^12 .dX but that is impossible: there are no polynomials f (X) and g(X) such that f (X) = g( (^) X^1 ). − X^12. 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 6 = 0 and coordinates (X, Y, 1), A 2 is the part with Y 6 = 0 and coordinates (U, 1 , V ), A 3 is the part with X 6 = 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 = dX 2 Y
dY 3 X^2 + 1
dU 2 U V − 1
−dV 3 U 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 dX 2 Y =^
dY 3 X^2 +1 is a regular differential form on that affine piece. In^ A^2 we have the equation V = U 3 + U V 2 and in the same way we see that (^2) U VdU − 1 = (^3) U− 2 dV+V 2 is a regular differential form on that affine piece. Finally, in the intersection A 1 ∩ A 2 we have
ω = dY 3 X^2 + 1
3( UV )^2 + 1
V 2
.dV = −dV 3 U 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 dimk 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.
