284
X – The Riemann Surface of an Algebraic Function
has exactly one root ζ ∈ W . We need to show that this unique root ζ is a
holomorphic function of z.
Regard the map
P : (z, ζ) −→ P (z, ζ)
from C
2 to C as a map from R
4 to R
2 . At each point, it has a tangent linear
map P
(z, ζ) which is C-linear since P is holomorphic at z and ζ, namely
(Chap. IX, formula (2.24))
(h, k) −→ D 1 P (z, ζ)h + D 2 P (z, ζ)k ,
(2.4)
where h, k are vector variables in C and where the derivatives are taken to
be in the complex sense; (4) shows that P
is surjective at every point of
the open subset Ω of C
2 , where D 1 P and D 2 P are not both zero, so that
P : Ω −→ R
2 is a submersion. Hence, every equation P (z, ζ) = c defines a
closed submanifold of Ω having dimension 4 − 2 = 2 [Chap. IX, n
◦ 13, (ii)],
hence a submanifold of R
4 = C
2 ; this is in particular the case of E. The
tangent vector space to E at a point (a, α) ∈ E where D 2 P = 0 is the set of
(h, k) such that
D 1 P (a, α)h + D 2 P (a, α)k = 0 ,
i.e. such that
k = −D 1 P (a, α)h/D 2 P (a, α) ;
hence the manifold E is the graph of a map ζ = f (z) in the neighbourhood
of (a, α), where f is C
∞ , with a tangent linear map given by
f
(a)h = −D 1 P (a, α)h/D 2 P (a, α)
[Chap. IX, n
◦ 13, (iv)]. This formula shows that f
(a) is C-linear, and so f
is holomorphic, qed.
Similarly, E could be shown to be the graph of a holomorphic function
z = g(ζ) in the neighbourhood of a point (a, α) where D 1 P = 0.
Returning to the Riemann surface X, i.e. the graph of F over B, consider
some a ∈ B = p(X). Denoting by α k the n simple roots of the equation
P (a, ζ) = 0, there are holomorphic functions f k (z), 1 ≤ k ≤ n in the neighbourhood of a, satisfying
P [z, f k (z)] = 0 , f k (a) = α k .
(2.5)
If D is a sufficiently small disc centered at a, these n local uniform branches
f k at a are all defined on D and, being continuous, are pairwise distinct at
all z ∈ D since so are the f k (a) = α k . Denoting by D k ⊂ X the image of
D under z → (z, f k (z)), the D k are seen to be pairwise disjoint and to have
X – The Riemann Surface of an Algebraic Function
has exactly one root ζ ∈ W . We need to show that this unique root ζ is a
holomorphic function of z.
Regard the map
P : (z, ζ) −→ P (z, ζ)
from C
2 to C as a map from R
4 to R
2 . At each point, it has a tangent linear
map P
(z, ζ) which is C-linear since P is holomorphic at z and ζ, namely
(Chap. IX, formula (2.24))
(h, k) −→ D 1 P (z, ζ)h + D 2 P (z, ζ)k ,
(2.4)
where h, k are vector variables in C and where the derivatives are taken to
be in the complex sense; (4) shows that P
is surjective at every point of
the open subset Ω of C
2 , where D 1 P and D 2 P are not both zero, so that
P : Ω −→ R
2 is a submersion. Hence, every equation P (z, ζ) = c defines a
closed submanifold of Ω having dimension 4 − 2 = 2 [Chap. IX, n
◦ 13, (ii)],
hence a submanifold of R
4 = C
2 ; this is in particular the case of E. The
tangent vector space to E at a point (a, α) ∈ E where D 2 P = 0 is the set of
(h, k) such that
D 1 P (a, α)h + D 2 P (a, α)k = 0 ,
i.e. such that
k = −D 1 P (a, α)h/D 2 P (a, α) ;
hence the manifold E is the graph of a map ζ = f (z) in the neighbourhood
of (a, α), where f is C
∞ , with a tangent linear map given by
f
(a)h = −D 1 P (a, α)h/D 2 P (a, α)
[Chap. IX, n
◦ 13, (iv)]. This formula shows that f
(a) is C-linear, and so f
is holomorphic, qed.
Similarly, E could be shown to be the graph of a holomorphic function
z = g(ζ) in the neighbourhood of a point (a, α) where D 1 P = 0.
Returning to the Riemann surface X, i.e. the graph of F over B, consider
some a ∈ B = p(X). Denoting by α k the n simple roots of the equation
P (a, ζ) = 0, there are holomorphic functions f k (z), 1 ≤ k ≤ n in the neighbourhood of a, satisfying
P [z, f k (z)] = 0 , f k (a) = α k .
(2.5)
If D is a sufficiently small disc centered at a, these n local uniform branches
f k at a are all defined on D and, being continuous, are pairwise distinct at
all z ∈ D since so are the f k (a) = α k . Denoting by D k ⊂ X the image of
D under z → (z, f k (z)), the D k are seen to be pairwise disjoint and to have