In this part we shall consider the Riemann surfaces of algebraic functions.
Motivation
The prototype of vanishing algebraic functions should be the graph of a holomorphic function. Let D⊂C be a domain, g∈O(D). Define
X:={(z,g(z))∈C×C:z∈D}⊂C×C
endowed with subspace topology. There is a natural projection π:X→D, which is also a homeomorphism with continuous inverse π−1(z)=(z,g(z)). Thus π is a global complex chart on X, making X a Riemann surface.
Smooth irreducible affine plane curve
Def. Let D⊂C be a domain, g∈O(D). If ∃f∈C[z,w], f=0 ,such that
f(z,g(z))=0,∀z∈D,
then g is said to be algebraic.
We shall discuss the space of zeros of algebraic functions, which are Riemann surfaces. The simplest space of algebraic functions are affine plane curves.
Def. Let f∈C[z,,w], f=0. Define Z(f):=f−1(0) to be an affine plane curve. Write
f=i,j∑aijziwj.
Define the degree of f to be
deg(f):=max{i+j:aij=0}
and the multiplicity of f at 0 to be $
mult0(f):=min{i+j∣aij=0}
and for p=(a,b)∈C×C, define the multiplicity of f at p to be
multp(f):=mult0(f(z−a,w−b))
With the notation of multiplicity, we can determine whether a zero of f is regular or singular.
Def. Let f∈C[z,,w], f=0 and p∈Z(f).
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If multp(f)=1, then p is said to be regular.
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If multp(f)>1, then p is said to be singular.
Moreover, denote S={p∈X:multp(f)>1} to be the singular set of f. In particular, in the case that f is a polynomial with two variables, S can be written as
S=Z(f,∂z∂f,∂w∂f)
Def. If S=∅, then f is said to be smooth. (All zeros are regular)
Def. If f cannot be written as the product of two polynomials, then f is reducible.
Cor. If f is irreducible in C[z,w], then
TpX=Z((z−a)∂z∂f(p)+(w−b)∂w∂f(p)).
The following is the main result of this section, showing that IT IS WORTHY RESEARCHING RIEMANN SURFACES!!! However, the proof is a little bit long, and it requires another post to present.
Theorem. A smooth irreducible affine plane curve X is a Riemann surface.
Proof. See other articles.□
Nonsingular projective plane curves
The main object of this section is projective space.
Fact. The n-dimensional complex projective plane Pn is a compact complex manifold of dimension n.
Remark. There exists a biholomorphism Cˉ≅P1.
Remark. The projective plane can be regarded as a complex of inclusion
P1↪P2↪⋯
where
P1=C1∪∞
P2=C2∪P1=C2∪C1∪∞.
More generally,
Pn=k≤n⨆Ck.
Def. The homogenization of a function f∈C[x1,…,xn] of degree d is
f∗:=x0df(x0x1,x0x2,…,x0xn)∈C[x0,…,xn]d.
Similarly, the dehomogenization of a homogeneous function F∈C[x0,…,xn]d is defined to be
F∗:=F(1,x1,…,xn)∈C[x1,…,xn].
The arithmetic properties of homogenization and dehomogenization are clear.
Cor. Up to some power of x0, factoring F∈C[x0,…,xn]d is the same as factoring F∗∈C[x1,…,xn]. In particular, if F∈C[z,w]d, then
F(z,w)=i=1∏k(aiz+biw)ri
for some ai,bi∈C, ri∈Z, and ∑ri=d.
Def. Let F∈C[x0,…,xn]d be a homogeneous polynomial of degree d. If the only solution to
{F∂xi∂F=0=0,i=0,1,…,n
is
(x0,…,xn)=(0,…,0),
then F is said to be nonsingular.
Now the case are down to Riemann surface (dimension 1), and the number of variables z0,…,zn is 3.
Remark. As it is stated is Algebraic Geometry, F∈C[x,y,z]d is not a well-defined function on P2, while the zeros Z(F) is well-defined on P2.
Def. Let F∈C[z0,z1,z2]d be a nonconstant homogeneous polynomial. Define X=Z(F) to be a projective plane curve. Moreover, if F is nonsingular, then X is said to be smooth.
Similar to the case of AFFINE PLANE CURVE, the following theorem shows that smooth projective plane curves are Riemann surface.
Theorem. Let F∈C[z0,z1,z2]d be a nonsingular homogeneous polynomial of degree d. Then X=Z(F)⊂P2 is a Riemann surface.
Proof. See other articles. □
An additional result about projective plane curve is as follows.
Def. Let f∈C[z1,z2] and X=Z(f)⊂C2 be an affine plane curve. Define
Xˉ:=Z(f∗)⊂P2
be the projective closure of X.
Theorem. ∣Xˉ−X∣<∞.
Proof. Decompose
f=i=0∑dfi,deg(fi)=i.
Then the homogenization of f can be written as
F:=f∗=i=0∑dx0d−ifi.
Thus
{F=0,z0=0}={fd=0,z0=0}
⇒Xˉ∩{z0=0}={[0,z1,z2]∈P2∣fd(z1,z2)=0}
Since fd∈C[z1,z2]d, then fd=∏i=1k(z1−αiz2)ri, then
∣Xˉ∩{z0=0}∣=∣Z(fd)∣=k<∞
Therefore,
Xˉ={F=0,z0=0}∪{F=0,z0=0}=X∪finitey many points
This completes the proof. □