300
X – The Riemann Surface of an Algebraic Function
The map p from ˆ
X onto ˆ
C is obtained by setting p(η Y ), for each open
set ˆ
Y , to be the point a ∈ ˆ
C from which Y was obtained. If D(a) is the disc
centered at a chosen to get the connected components Y over a neighbourhood of a, the inverse image p
−1 (D(a)) is the union of open sets ˆ
Y , and so is
open in ˆ
X, obviously so is also p
−1 (D
) for any disc D
⊂ D(a). As a result,
p is continuous and, even better, maps every open subset of ˆ
X onto an open
subset of ˆ
C. But ˆ
X is not, strictly speaking, a covering of ˆ
C.
Let us show that ˆ
X is compact. For each a ∈ ˆ
C−B, consider the open subset p
−1 (D(a)) of ˆ
X. It is the union of open subsets of type ˆ
Y . Let D a ⊂ D(a)
be a closed disc, hence compact in ˆ
C, centered at a ; for all Y ⊂ p
−1 (D(a)),
the chart (Y, ϕ Y ) of ˆ
X transforms p
−1 (D a ) ∩ ˆ
Y homeomorphically into a
closed, hence compact, disc centered at 0. As a result, p
−1 (D a ) is the finite
union of compact sets, and so is compact. Similarly, for all a ∈ B, there is a
closed disc D a centered at a such that p
−1 (D a ) is compact. Since the interiors
of the D a cover the compact space ˆ
C, ˆ
C can be covered by finitely many discs
D a . Thus ˆ
X is the union of finitely many compact sets, qed.
(iii) The algebraic function F(z) as a meromorphic function on ˆ
X. Let
us now show that the functions (z, ζ) → z and F : (z, ζ) → ζ, defined
and holomorphic on X have meromorphic extensions on ˆ
X. It is enough to
prove this on each open set ˆ
Y , using the the chart ( ˆ
Y , ϕ Y ) to verify it. If ˆ
Y
corresponds to a point a ∈ ˆ
C−B and if Y is of order k, then ϕ Y (z, ζ)
k = z −a
or 1/z, whence the result related to (z, ζ) → z ; in particular, (z, ζ) → z has
a pole of order k at η Y if this point projects onto the point ∞ of ˆ
C. As an
aside, note that the point ∞ having been excluded during the construction
of X, ˆ
X may very well be a genuine covering space of a neighbourhood of
∞, in other words that k = 1 for each η Y projecting onto ∞ ; the Riemann
surface of ζ(ζ − 1)z = 1 has two points over ∞, the function ζ taking values
0 and 1 at these points.
The case of the algebraic function F : (z, ζ) → ζ is less obvious. We will
suppose that η Y projects onto a point a = ∞, the other case being similarly
dealt with. As ( ˆ
Y , ϕ Y ) is a holomorphic chart of ˆ
Y and as F is holomorphic
on Y = ˆ
Y − η Y , there is a Laurent series expansion
F (z, ζ) = ζ =
Z
c n q
n = h(q) , q = ϕ y (z, ζ)
(4.3)
in Y , and it all amounts to showing that c n = 0 for sufficiently large n < 0.
Since P (z, ζ) = 0 and q
k = z − a,
P 0
q
k + a
h(q)
n + . . . + P n
q
k
= 0
and so
h(q)
n + s 1 (q)h(q)
n−1 + . . . + s n (q) = 0 for q = 0 .
(4.4)
X – The Riemann Surface of an Algebraic Function
The map p from ˆ
X onto ˆ
C is obtained by setting p(η Y ), for each open
set ˆ
Y , to be the point a ∈ ˆ
C from which Y was obtained. If D(a) is the disc
centered at a chosen to get the connected components Y over a neighbourhood of a, the inverse image p
−1 (D(a)) is the union of open sets ˆ
Y , and so is
open in ˆ
X, obviously so is also p
−1 (D
) for any disc D
⊂ D(a). As a result,
p is continuous and, even better, maps every open subset of ˆ
X onto an open
subset of ˆ
C. But ˆ
X is not, strictly speaking, a covering of ˆ
C.
Let us show that ˆ
X is compact. For each a ∈ ˆ
C−B, consider the open subset p
−1 (D(a)) of ˆ
X. It is the union of open subsets of type ˆ
Y . Let D a ⊂ D(a)
be a closed disc, hence compact in ˆ
C, centered at a ; for all Y ⊂ p
−1 (D(a)),
the chart (Y, ϕ Y ) of ˆ
X transforms p
−1 (D a ) ∩ ˆ
Y homeomorphically into a
closed, hence compact, disc centered at 0. As a result, p
−1 (D a ) is the finite
union of compact sets, and so is compact. Similarly, for all a ∈ B, there is a
closed disc D a centered at a such that p
−1 (D a ) is compact. Since the interiors
of the D a cover the compact space ˆ
C, ˆ
C can be covered by finitely many discs
D a . Thus ˆ
X is the union of finitely many compact sets, qed.
(iii) The algebraic function F(z) as a meromorphic function on ˆ
X. Let
us now show that the functions (z, ζ) → z and F : (z, ζ) → ζ, defined
and holomorphic on X have meromorphic extensions on ˆ
X. It is enough to
prove this on each open set ˆ
Y , using the the chart ( ˆ
Y , ϕ Y ) to verify it. If ˆ
Y
corresponds to a point a ∈ ˆ
C−B and if Y is of order k, then ϕ Y (z, ζ)
k = z −a
or 1/z, whence the result related to (z, ζ) → z ; in particular, (z, ζ) → z has
a pole of order k at η Y if this point projects onto the point ∞ of ˆ
C. As an
aside, note that the point ∞ having been excluded during the construction
of X, ˆ
X may very well be a genuine covering space of a neighbourhood of
∞, in other words that k = 1 for each η Y projecting onto ∞ ; the Riemann
surface of ζ(ζ − 1)z = 1 has two points over ∞, the function ζ taking values
0 and 1 at these points.
The case of the algebraic function F : (z, ζ) → ζ is less obvious. We will
suppose that η Y projects onto a point a = ∞, the other case being similarly
dealt with. As ( ˆ
Y , ϕ Y ) is a holomorphic chart of ˆ
Y and as F is holomorphic
on Y = ˆ
Y − η Y , there is a Laurent series expansion
F (z, ζ) = ζ =
Z
c n q
n = h(q) , q = ϕ y (z, ζ)
(4.3)
in Y , and it all amounts to showing that c n = 0 for sufficiently large n < 0.
Since P (z, ζ) = 0 and q
k = z − a,
P 0
q
k + a
h(q)
n + . . . + P n
q
k
= 0
and so
h(q)
n + s 1 (q)h(q)
n−1 + . . . + s n (q) = 0 for q = 0 .
(4.4)