Riemann Surfaces
Lectured by Prof Gabriel Paternain
1 0 Brief Review of material from IB Complex Analysis
A smooth function f on an open subset U ∈ C is called a holomorphic function or analytic if equivalently:
(i) f is complex differentiable. i.e. For each z0 ∈ U the derivative 0 f(z)−f(z0) f (z0) = lim exists z−z0 (ii) f admits a power expansion, if we let D(a, r) = z : |z − a| < r then P∞ n f(z) = n=0 cn(z − a) valued for all z ∈ D(a, r)
r P∞ n−r (ii)⇒ We can write f(z) = g(z)(z − a) locally with g(z) = r cn(z − a) analytic on D(a, r) and g(a) = cr 6= 0 This gives the ”principle of isolated zeroes”. If f 6= 0 on a domain U ⊂ C the zeroes are isolated. Each point z has an open neighbourhood ∆ ⊂ U for which either f|∆ ≡ 0 or f is nowhere zero on ∆ \{z}. This gives the identity pricniple. If two analytic functions f, g on a domain U ⊂ C agree on a non-discrete subset of U then f ≡ g on U.
Definition. Domain=open and connected subset of C Suppose f is analytic on a punctured disc D∗(a, r) = {z ∈ C : 0 < |z−a| < r} for some r > 0. i.e. f has an isolated singularity at a. There is a Laurent expansion of f at D ∗ (a, r)
∞ X n f(z) = cn(z − a) −∞
The singularity at a is:
• Removable if cn = 0 for all n < 0. i.e. f analytic at a.
• A Pole if ∃N > 0 : c−N 6= 0 but cn = 0 for all n < −N. N is the order of the pole. And we can write f(z) = (z − a)−N g(z) with g analytic on D(a, r), g(a) = c−N 6= 0
• Essential if cn 6= 0 for infinitely many n < 0. Removable Singularity theorem(RST): Singularity is removable ⇔ f is bounded on some punctured neighbourhood of a Casaroti-Weierstrass Theorem(CWT): If f has an essential singularity at a, then ∗ ∀w ∈ C∞ = C ∪ {∞} ∃ a sequence zn ∈ D (a, r) with zn → a and f(zn) → w Together these imply: f has a pole at a ⇔ f(z) → ∞, z → a
1 Example. (i) ez −1 has a pole of order 1 at z = 0 while z ez −1 has a removable singularity.
2 1 (ii) e z has an essential singularity at z = 0.
Recall that a function f on an open set U ⊂ C is called meromorphic if it is analytic apart from isolated singularities and they are poles. 1 ∗ For example 1 is meromorphic in C = C \{0} but not meromorphic in C. e z −1 1 Poles are z = 2πin , n 6= 0, n ∈ Z
0.1 The Complex Logarithm We will deal with ”multivalued functions” which are locally analytic but it cannot be defined globally. e.g. The complex logarithm For a given z = reiθ 6= 0 possible values of log z are log r + i(2πn + θ) where n ∈ Z (possible w ∈ C such that ew = z Let U1 = C \ R≥0. For each n ∈ Z, we can define log analytically on U by log(reiθ) = log r + i(2πn + θ) where 0 < θ < 2π. It is analytic because it is an inverse of the holomorphic map exp : {x+iy : x ∈ R, 2πn < y < 2(n+1)π} → U and invoke the inverse function theorem. R z dw Alternatively, we can say h(z) = −1 w + (2n + 1)πi and prove it is holomor- 0 1 phic with h (z) = z ln(z+τ)−ln(z) = 1 R z+τ dw = R 1 dt = 1 → 1 τ τ z w 0 z+tτ z+t0τ z Claim:h(z) is the required inverse for ez such that g(z) = ez
eh(z) 0 h(z) Proof. g(z) = z on U and g (z) = 0 ⇒ g = constant g(−1) = 1 ⇒ e = z
3 1 Analytic Continuation in the Plane
Definition. A function element is an ordered pair (f, U) where U is a domain and f is a function defined on it. Definition. A function element in domain U is a pair (f, D) where D is a subdomain of U and f is holomorphic on D Definition. Let (f, E), (g, E) be function elements in U. (i)( g, E) is a direct analytic continuation of (f, D) written (f, D) ∼ (g, E) if D ∩ E 6= φ and f = g on D ∩ E (ii)( g, E) is an analytic continuation of (f, D) along a path γ : [0, 1] → U (Written (f, D) ≈γ (g, E)) if there are function elements n (fi,Di)i=1 and o = t0 < t1 < ... < tn = 1 such that γ([tj−1, tj]) ∈ Dj j = 1, ..., n and (f, D) = (f0,D0) ∼ (f1,D1) ∼ ... ∼ (fn,Dn) = (g, E) (g, E) is an analytic continuation of (f, D) if ∃ such a path. Then (f, D) ≈ (g, E)
Remark. (i) If (f, D) ≈γ (g, E) and (f, D) ≈γ (h, E) for some γ ⇒ (g = h) We will prove this later when it is less messy and when we have more machinery. (ii) ∼ is not an equivalence relation. (Even if intersection is not empty.) However ≈ is Definition. An equivalence class F under ≈ is called a complete analytic func- tion. Example. We consider log from last time. U1 = C \ R≥0 U = C Let J = (α, β) = {θ : R, α < θ < β} iθ DJ = {z = re 6= 0, θ ∈ J} fJ (z) = log r + iθ LJ = (fJ ,DJ ) is a function element(last time we saw DJ = U : J = (0, 2π) −pi π Consider the intervals A = ( 2 , 2 ) π 7π B = ( 6 , 6 ) 5π 11π C = ( 6 , 6 and consider DA,DB,DC . We have a curve
γ : [0, 1] → C t 7→ e2πit
1 1 5 0 = t < 6 < 2 < 6 < 1 = t4 LA ∼ LB,LB ∼ LC But LA 6∼ LC Points in i π π DA ∩ DC are of the form re θ where θ ∈ (− 2 , − 6 ) giving fA(z) = log z + iθ i(θ+2π) 3π 11π And it is also of the form re with θ + 2π ∈ ( 2 , 6 ) ⊂ C
4 So fC (z) = log r + i(θ + 2π) = fA(z) + 2πi This analytic continuation of LC on DA produces (DA, fA + 2πi) = LA+2π so LA+2πγγ LA, so we can go around γ n times in both directions to produce all possible values of log This gives a F for example, the F determined by the equivalence class of (fA,DA)
1.1 The Riemann surface of log
Consider U = C \ R+ fn(z) = log r + i(2πi + θ) where n ∈ Z, θ ∈ (0, 2π) This glueing construction gives the ”Riemann Surface” R of log and is now a univalued function on R. i fn(re θ) = log r + i(2πn + θ) 0 < θ < 2π, n ∈ Z More precisely let a ∗ R = (C × {k}) k∈Z (the disjoint union) Topology: Open sets are unions of sets of the form
• D((η, k), r) where D((η, k), r) = {(z, k): |z − η| < r} × if η ∈ C \ R≥0(r < dist(η, R≥0) • A((η, b), r) := {(z, b), |z − η| < r, Im(z) ≥ 0} `{(z, b − 1), |z − η| < r, Im(z) < 0} if η ∈ R≥0 (r < |η|) Check this makes R into a Hausdorff connected(also path-connected topological space with π : R → C∗,(z, b) 7→ z continuous. Obviously, π|D((η,b),r) : D((η, b), r) → D(η, r) is a homeomorphism and the same holds for π|A((η,b),r) : A((η, b), r) → D(η, r) We capture this in the following definition. Definition. A covering map of a topological space is a continuous map π : X˜ → X where X,X˜ Hausdorff path-connectedd topological spaces and π is a local homeomorphism. i.e. For eachx ˜ ∈ X˜ then there is an open neighbourhood N˜ ofx ˜ such that ˜ π|N˜ : N → N is a homeomorphism where N is an open neighbourhood of x = π(˜x) X is called the base space. X˜ is called the covering space. Warning: Terminology, don’t confuse with the covering maps in Algebraic Topology. This notion is weaker. The map π : R → C∗ from the example above has a property that for each z ∈ C∗, there is a disc D ⊂ C∗ such that z ∈ D
5 −1 ` π (D) = Di where π|D : Di → D is a homeomorphism. These are called i∈Z i regular and match the ones in Algebraic Topology. We will come back to this. Define f : R → C f(eiθ, k) = log r +i(2πk +θ) where 0 ≤ θ < 2π
f R C π exp C∗ To be upgraded to a holomorphic diagram Additional Example: We could do 1 n 1 ∗ something quite similar with z = exp( n log z) on C , n ≥ 0(integers) 1 iθ 2iπk Possible branches are r n e n e n where k = 0, 1, ..., n − 1 This time we only need to glue n copies of C \ R≥0 and again we obtain a diagram
f R C π zn C∗
1.2 Power Series
Recall that a power series about a ∈ C is uniformly and absolutely convergent on any closed disc inside its radius of convergence. If r < ∞, what can we say about analytic continuations (f, D) where
∞ X k f = ak(z − a) k=0 where D = D(a, r) Without loss of generality, we will assume that a = 0, r = 1 so we work on ∆ = D(0, 1) and Π := ∂D(0, 1) So we consider
∞ X k f = akz k=0 Definition. A point z ∈ Π is said to be regular if ∃g analytic on an open neighbourhood N of Z such that f = g on N ∩ ∆ ( f on D Defining ˜(f) = g on N we have a direct analytic continuation to ∆ ∪ N If z is not regular, it is called singular. Note: Regular points form open sets. Singular points form a closed set. P k Example/Warning. 1. z regular 6⇒ akz converges. P∞ k 1 e.g. f(z) = 0 z = 1−z z ∈ ∆, Every point z ∈ Π \{1} is regular but P zk diverges for all z ∈ Π
6 P k 2. akz converges 6⇒ z regular P∞ z f(z) = 2 k(k−1) Absolutely convergent if all z ∈ ∆. 1 is a singular point. 00 P∞ k 1 If f extends analytically to a neighbourhood of 1 then f = 0 z = 1−z extends as well. Absurd
P∞ k Theorem 1.1. Suppose f(z) = 0 akz has radius of convergence 1 then exists at least one singular point.
Proof. ∃ an open disc Dz such that f extends to a function fz on ∆ ∪ Dz
If Dz1 ∩ Dz2 6= φ
But fz1 = f = fz2 on the open subset Dz1 ∩ Dz2 ∩ ∆ 6= φ of Dz1 ∩ Dz2 .
By the identity principle fz1 = fz2 on Dz1 ∩ Dz2 . Since Π is compact we can cover it with finitely many discs Dz1 , ..., Dzn and f has an analytic extension to
∆ ∩ Dz1 ∩ ... ∩ Dzn
However ∃δ > 0: D(0, 1 + δ) ⊂ ∆ ∪ Dz1 ∪ ... ∪ Dzn . And f has an analytic extension to D(0, 1 + δ). This is absurd since 1 is the radius of convergence. Recall from last time that ∆ = D(0, r) and Π = ∂D(0, r) Π is called the natural boundary if every point is singular. Example. ∞ X f(z) = zk 0 has Π as a natural boundary. 2πip 2πip Consider ω = e q where p, q ∈ Z. We note that the set S = {ω = e q : p, q ∈ Z} is dense in Π. Since the singular points form a closed subset, it suffices to show that every point in S is singular. Take r = 1. q−1 ∞ X X f(r, ω) = rk!ωk! + rk! → ∞ as r → 1 0 q ⇒ Every point is singular.
7 ∞ X zk! Variations. 1. f(z) = also has Π as a natural boundary because f 0 k! 1 does.
2. We can extend the notion of natural boundary to other curves(rather than Π). For example, the imaginary axis is a natural boundary for the analytic P∞ k!z function. f(z) = 0 e defined in Re(z) < 0
8 2 Riemann Surfaces
Definition. A Riemann surface R is a connected Hausdorff topological space,together with a collection of homeomorphisms φα : Uα → Dα ⊂ C with Uα open and φα(Uα) = Dα open in C such that 1. [ Uα = R α
−1 2. If Uα ∩Uβ 6= φ then φβ ◦φα : φα(Uα ∩Uβ) → φβ(Uα ∩Uβ) is holomorphic. R
Uα Uβ
φα φβ
Dα Dβ
−1 φβ ◦ φα
φα(Uα ∩ Uβ) φβ(Uα ∩ Uβ)
The pair (Uα, φα) are called charts. The collection of all charts is called an −1 atlas. The functions φβ ◦ φα are called the transition functions. (They are conformal equivalence.)
Remark. Each point z ∈ R has a neighbourhood holomorphic to a disc in C. Hence R assumed to be connected automatically gives path-connected. (R is −1 connected and locally path-connected as φα is continuous.) Example. 1. R = C single chart φ(z) = z 2. R = C single chart φ(z) = z + 1
9 Definition. Two atlases a1, a2 on a toplogical space R are equivalent if a1 ∩ a2 is also an atlas. −1 i.e. For any charts (Uα, φα) ∈ a1 and (Vβ, ψβ) ∈ a2 we have ψβφα : φα(Uα ∩ Vβ) → ψα(Uα ∩ Vβ) holomorphic. Example. (See above) Atlases (1) and (2) are clearly equivalent but not equiv- alent to an atlas with the single chart φ(z) =z ¯ Check: In definitions above equivalence of atlases, check that it is indeed an equivalence class. An equivalence class is called a conformal structure.
Remark. If R is a Riemann Surface with atlas {(Uα, φα)} and S is a connected open subset of R then S is a Riemann Surface with atlases {(Uα ∩ S, φα|Uα∩S)} Example. Every domain of C is a Riemann surface. Note: Whenever we refer to C as a Riemann Surface, we mean C with the standard conformal structure determined by (C, φ(z) = z)
Definition. Let R and S be Riemann surfaces with atlases {Uα, φα} and {(Vβ, ψβ)} respectively. A continuous map f : R → S is analytic/holomorphic if ∀α, β such that
−1 • Uα ∩ f (Vβ) 6= φ −1 −1 • ψβfφα is analytic on φα(Uα ∩ f (Vβ)) f is called a conformal equivalence(analytic isomorphism/holomorphism) if ∃ analytic inverse g : S → R
Example. (C, φ(z) = z) 7→ (C, ψ(z) =z ¯) Remark. f : R → S holomorphic iff given z ∈ R, ∃(U, φ) with z ∈ U and (V, ψ) with f(z) ∈ V such that ψfφ−1 : φ(U ∩ f −1(V )) → ψ(V ) is holomorphic. This is because if for example, (U 0, φ0) is another chart around z, then ψf(φ0)−1 = ψfφ−1φ(φ0)−1
f g Lemma 2.1. The composition of analytic R → S → G is analytic.
Proof. Write h = gf. Suppose charts are {(Uα, φα)}, {(Vβ, ψβ)}{(Wδ, θδ)} Uα ∩ −1 −1 −1 h (Wδ) is covered by Uα ∪ (f (Vβ) ∩ h (Wδ)). For β s.t. intersection is non-empty, we have −1 −1 −1 −1 ψβfφα : φα(Uα ∩ f (Vβ) ∩ h (Wδ)) → ψβ(Vβ ∩ g (Wδ)) −1 −1 θδgψβ : ψβ(Vβ ∩ g (Wδ)) → θδ(Wδ) both analytic.
Uα Vβ Wδ f g
φα ψβ θδ
10 So composition is analytic. −1 −1 −1 ⇒ θδgfφα : θδhφα is analytic on each φα(Uα ∩ f (Vβ) ∩ h (Wδ)) and these −1 cover φα(Uα ∩ h (Wδ))
Tautology: If a1, a2 are atlases for R and τ :(R, a1) → (Ra2) holomoprhic⇔ a1, a2 equivalent. 0 0 0 If f :(R, α1) → (S, α2) analytic then ai ∼ ai then f :(R, ai) → (S, ai) analytic. τ −1 0 0 τR f S 0 f :(R, a1) → (R, a1) → (S, a2) → (S, a2) Can move between equivalent atlases, equivalence of atlas is an equivalence relation. Remark. Sufficient to check ∀z ∈ R, ∃ chart about z ∈ U, chart (U, ϕ) and chart (V, ψ) s.t. V 3 f(z), ψfϕ−1 holomorphic. ϕ(U ∩ f −1(N)) → ψ(V ). Then if (U 0, ϕ0) another chart at z,(V 0, ψ0) at f(z) then ψ0f(φ0)−1 = (ψ0ψ−1)(ψfϕ−1)(ϕ(ϕ0)−1) is holomorphic.
Lemma 2.2. If R is a Riemann surface and π : R˜ → R is a covering map of topological spaces, then there exists a unique conformal structure on R˜ such that π : R˜ → R is analytic. ˜ ˜ ˜ ˜ ˜ Proof. Given a point z ∈ R, choose N s.t. z ∈ N, pi|N˜ : N → π(N) is a homeomorphism. Let (V, ψ) be any chart of R with π(z) ∈ V and take chart φ = ψπ : U → ψ(N ∪ V ). This chart (U, φ) form an atlas a˜ of R˜ transition functions are just restriction of transitition fucntions of R. Let us check π analytic.
ψπφ−1 : ψπ(ψπ)−1 = ψππ−1ψ−1 = id holomorphic on some appropriate domain. ∗ Uniqueness: Suppose a = {(Wα, θα} is an atlas on R˜ s.t. pi :(˜(R), a∗) → (R, a) is analytic. We wish to show that a∗ is equivalent to a˜. Given (U, φ) ∈ a˜ as above. Assume that ˜ ∗ π|N˜ :(N, a |N˜ ) → (N, a|N ) is holomorphic. ∗ −1 Given (Wα, θα) ∈ a s.t. Wα ∩ N˜ ∩ π (V ) 6= φ have a diagram
π Wα ∩ U N ∩ V φ θα ψ
θα(Wα ∩ U) ψ(N ∩ V )
−1 By(*) ψπ θα is holomorphic. |{z} φ −1 ∗ This means φθα is holomorphic. i.e. a is equivalent to a˜.
11 Example. Let us revisit log. Two lectures ago we constructed,
f R C π exp C∗
a ∗ R = C × {k} By Lemma 2.2, ∃ unique conformal structure on R making π k∈Z holomorphic. ∗ (C , ψ(z) = z). Charts on R are making (U, φ) where φ = πN˜ ˜ −1 To check that f is holomorphic we only need to check that f ◦ (π|N˜ ) is holomorphic. But this is just a branch of log. In fact, f is a bijection and its inverse is also holomorphic. −1 Since exp(f(z)) = π(z), locally (π|N ) ◦ exp and this is holomorphic. Similarly, with zn we get a n-folded cover.
f R C π zn C∗ and using Lemma 2.2, we can upgrade the diagram to a diagram of holomorphic maps between Riemann Surfaces. Important Examples: Riemann sphere, C∞ = C ∪ {∞} Topology: Open subsets are either open subset of C or of the form {∞} ∩ (C \ compact set ) 2 3 With this topology, C∞ is homeomorphic to S = {x ∈ R : ||x|| = 1} 2 Homeomorphism is given by the stereographic projection π : S → C∞. In particular, C∞ is compact, connected and Hausdorff. 1 Atlas: (C, φ1), φ1(z) = z,(C \{0}, φ2), φ2(z) = z Transition Function:
∗ ∗ C → C 1 z 7→ z Alternatively, we can define charts in S2 directly. 2 U1 = S \{(0, 0, 1)}, π1 projection from (0, 0, 1). 2 U2 = S \{(0, 0, −1)}, π2 projection from (0, 0, −11). Recall π1(x)π2(x) = 1 x + iy x − iy π = , π = 1 1 − z 2 1 + z Consider atlas {(U1, π1), (U2, π2)}. The harmonic function is again. Complex Tori C Consider a lattice
12 Λ = τ1Z + τ2Z τ1, τ2 6= 0, τ1/τ2 6∈ R π : C → C/Λ We endow C/Λ with the quotient topology. This makes π continuous. C/Λ is a compact connected Hausdorff topological space. For any a ∈ T , choose w ∈ C s.t. π(U) 3 a and open neighbourhood N 3 W s.t. π|N is a homeomorphism. −1 Take as chart:φ = (π|N ) : U → N Two such charts φ1 : U1 → N1 φ2 : U2 → N2 with U1 ∩ U2 6= φ −1 give rise to a transition function φ2φ : φ1(U1 ∩ U2) → φ(U1 ∩ U2) which is locally just a translation by some w ∈ Λ hence analytic. (Example sheet 1). Theorem 2.3 (Inverse Function Theorem). Given any analytic function g on an open set V ⊂ C and a ∈ V such that g0(a) 6= 0, there exists an open neighbourhood N of a such that g|N : N → g(N) is a conformal equivlance of N onto the open neighbourhood g(N) of g(a). Proof. By considering g(z) − g(a), we may assume without loss of generality that g has a simple zero at a Take a small disc D = D(a, ) with D¯ ⊂ V such that • g is nowehre zero in D¯ \{a}
• g is nowhere zero in D Principle of Argument ⇒ n(g ◦ γ, 0) = 1 = zeroes of g in D where γ a closed loop around a. Note that g ◦ γ([0, 1]) compact hence closed. Therefore, C \ g(γ([0, 1])) is open and choose ∆ disc centred at 0 contained in the connected component containing g(z). Hence n(g ◦ γ, w) = 1 for all w ∈ ∆ By the principle of a argument again, N = (g|D)−1(∆) is an open neighbour- hood of a and g|N : N → ∆ is bijective. −1 Then, the opening mapping theorem, (IB Complex Analysis)⇒ h = (g|N ) ) is continuous and hence g|N is a homeomorphism. Pick w0 ∈ ∆ and w 6= w0 and let z0 = h(w0)andz = h(w) and consider h(w0)−h(w) z0−z 1 = → 0 w0−w g(z0)−g(z) g (z) Since w → w0 iff z → z0 as w → w0. (g|N is a homeomorphism.)
13 2.1 Local Form of Holomorphic Function
Suppose we have a ∈ U ⊂ C and f =6∼ 0 is analytic in the domain U with f(a) = 0. Then we know we can write locally around a
f(z) = (z − a)rh(z) where h is analytic, h0(a) 6= 0 and r > 0. We can find a disc D containing a iα such that h(D) ⊂ C \ R≥0e So on D we can choose an analytic function 1 r 1 r l(z) = [h(z)] = exp( r log(h(z)) Hence f(z) = [g(z)] where g(z) = (z − a)l(z) By Theorem 2.3, g is locally a conformal equivalence which means if we siad w = g(z) then f(g−1(w)) = wr More generally, if f : R → C locally around a point p0 ∈ R we can choose a chart φ : U → C and by replacing f by f(p)−f(p0) −1 we may assume f(p0) = 0. If a = φ(p0) we can locally write the function fφ as [g(z)]n as above. Locally around p0 we have a chart ψ = gφ and a diagram. r i.e. f(p) − f(p0) = ψ Finally, if f : R → S when R and S are both Riemann Surfaces take a chart (v, θ) around f(p0) with θ(f(p0)) = 0 and apply the above to θf so locally we have that f is of the form z 7→ zr Theorem 2.4 (Open Mapping Theorem). Let f : R → S be a non-constant analytic map of Riemann Surfaces. Then f is an open map.
Proof. Suppose W is open in R, z ∈ W . We are required to prove that f(W ) contains an open neigbourhood of f(z). Choose charts (U, φ) in R with z ∈ U (V, ψ) in S with f(z) ∈ V Let D be an open disc with φ(D) s.t. φ−1(D) ⊂ U ∩ W ∩ f −1(V ) ψfφ−1 is non-constant. (If not by the identity principle of the Riemann Surfaces on Example Sheet 1 Q 11, f is constant on R.) By the opne mapping theorem for plane domains(IB)⇒ ψfφ−1(D) is open. Have fφ−1(D) is open in S. Also, f(z) ∈ f(φ−1(D)) ⊂ f(W ) in S. Hence fφ−1(D) is the required open neighbourhood.
Corollary 2.5. Let f : R → S be analytic and non-constant. If R is compact then f(R) = S and S is also compact. Proof. By (2.4) and f(R) is open. Since R is compact, f(R) is compact and since S is Hausdorff. f(R) is closed. Since S is connected and f(R) 6= φ we have f(R) = S.
Corollary 2.6. Any analytic function f : R → S with R compact must be constant.
Proof. Since C is not compact, thus it follows from (2.5) Remark/Application: Suppose f : R \{a} → C is analytic and a ∈ R. Taking local charts around a, we see f has a removable singularity ⇔ f is
14 bounded in a neighbourhood of a. For instance, f : C → C holomorphic and bounded.Then f extends to a holo- morphic map f : C∞ → C and hence is constant.(i.e.Liouville’s theorem)
15 2.2 Harmonic Functions
Definition. Let D be a plane domain and u : D → R is harmonic in u ∈ C2(D) ∂2u ∂2u ∂x2 + ∂y2 = 0 Laplace’s equation. Remark. If f = u + iv is analytic in D then u, v harmonic. (just use Cauchy- Riemann equations.) If D is a disc and u is harmonic then there is a holomorphic function, f : D → C s.t. u = Re(f).(Example sheet 1, Q12) Definition. A harmonic function h or a Riemann surface R is a continuous function f : R → R s.t. for any chart φ : U → φ(U) ⊂ C on R the function hφ−1 is harmonic on φ(U). The remark above implies that this definition is independent for the choice of charts. Moreover, a continuous function h : R → R is harmonic iff each z ∈ R has a neigbourhood N s.t.h|N = Re(g) for some analytic function g on N.
Proposition 2.7. Non-constant harmonic functions h : R → R on a Riemann surface are open maps. Hence if R is compact all harmonic functions are con- stant. Proof. Suppose that U ⊂ R is open. For each z ∈ U, there exists open neighbourhood N ⊂ U s.t. z ∈ N and h = Re(g) where g : N → C holomorphic. g is non-constant. If g is constant ⇒ h is constant on open set. Example sheet 1 Q 13⇒ h globally constant. Open mapping theorem for g ⇒ ∃ neighbourhood g(z) = a + ib of the form. (a − , a + ) × (b − , b + ) ⊂ g(N). Thus (a − , a + ) ⊂ h(N) ⊂ h(U) ⇒ h(U) open. If R is compact and h is non-constant then h(R) is both open and closed. ⇒ h(R) = R is compact.#
16 2.3 Meromorphic functions Definition. A meromorphic function f on a Riemann Surface R is an analytic map f : R → C∞ Lemma 2.8. If U ⊂ C is a plane domain, then this definition coincides with the usual one of isolated poles.
Proof. Suppose f : U → C∞ analytic. If a ∈ U has f(a) ∈ C then on each neighbourhood N of a, f is an analytic function on N. If f(a) = ∞ need to take a chart 1/w at ∞ 1 f analytic at a ⇒ g(z) = f(z) is analytic in the neighbourhood of a with g(a) = 0. Writing g(z) = (z − a)rh(z) h analytic around a and h(a) 6= 0 r > 0 1 1 f(z) = (z − a)−r with holomorphic around a. h(z) h(z) ⇒ f has an isolated pole at a. Conversely if f s meromorphic in the usual sense with a pole at a ∈ U we can write f(z) = (z − a)−rl(z) locally around a for some holomorphic function l with l(a) 6= 0. 1 r 1 ⇒ g(z) = f(z) = (z − a) l(z) is analytic around a. f : U → C∞ holomorphic. Example. (Related to Q 15 in Example Sheet 1) 3 f(z) = √z − z g(z) = z3 − z 2 { (z, w) : w = f(z) ⊂ C} | {z } | {z } algebraic cutting and curve pasting U = C \ [−1, 0] ∩ [1, ∞) Locally around every point in C \{0, ±1}, ∃ precisely two functions g such that [g(z)]2 = f(z) Claim: ∃g holomorphic on U s.t. g2(z) = f(z). We can do this by analytic continuation along paths. Principle of Argument: n(f ◦ γ, 0) = n(γ, −1) + n(γ, 0) + n(γ, 1) | {z } | {z } same →0 n(f ◦ γ, 0) = 2n(γ, −1) ∈ 2Z n(f ◦ γ, 0) even implies that as we analytically continue along a closed curve γ we return exactly where we started so such a g exists. R: Riemann surface obatined from glueing. R is topologically a torus with four points removed.
17 3 The space of Germs and the Monodromy The- orem
First clear some topological preliminaries. Definition. Suppose π : X˜ → X is a covering map of topological spaces and γ : [0, 1] → X is a path. A lift of γ is a pathγ ˜ : [0, 1] → X˜ such that π ◦ γ˜ = γ
Proposition 3.1. Uniqueness of Lifts Supposeγ ˜1 andγ ˜2 are lifts of γ with γ˜1(0) =γ ˜2(0) thenγ ˜1 =γ ˜2.
Proof. Consider I = {t ∈ [0, 1] :γ ˜1(t) =γ ˜2(t)}
• I is open If τ ∈ I choose open neighbourhood N˜ ofγ ˜1(τ) =γ ˜2(τ) such that π|N˜ is homeorphic. By continuity there exists δ > 0: for t ∈ (τ − δ, τ + δ) ⇒ γ˜1(t), γ˜2(t) ∈ N But π(γ ˜1(t) = π(γ ˜2(t)) = γ(t) ˜ ˜ ⇒ γ1(t) = γ2(t) for all t ∈ (τ − δ, τ + δ)
C • I is closed. If τ ∈ I thenγ ˜1τ 6=γ ˜2τ X Hausdorff⇒ ∃ open sets U1,U2 ⊂ X˜ such thatγ ˜1(τ) ∈ U1, γ˜2(τ) ∈ U2,U1 ∩ U2 = φ Continuity ⇒ ∃δ > 0 : |t − τ| < δ ⇒ γ˜1(t) ∈ U1, γ˜2(t) ∈ U2
Definition. A covering π : X˜ → X is called regular if every x ∈ X has an open neigbourhood U such that −1 a π (U) = U˜ν ν where U˜ are disjoint open subsets of X˜ such that π| : U˜ → U is a homeo- ν U˜ν ν morphism for each ν. Example. 1. Covering map
∗ C → C z 7→ exp(z)
is regular but the restriction to −2π < =(z) < 2π is not.
2. π : C → C/Λ, Λ lattice Proposition 3.2. Let π : X˜ → X be a regular covering and γ a path in X, z ∈ X˜ such that π(z) = γ(0). Then ∃ a liftγ ˜ such thatγ ˜(0) = z. (Uniquely by (3.1)) Proof. Set I = {t ∈ [0, 1] : ∃ a liftγ ˜ : [0, t] → X˜ withγ ˜(0) = z} Clearly 0 ∈ I so I 6= φ. Set τ := sup I
18 Claim: ∃δ > 0 such that [0, τ + δ] ∩ [0, 1] ⊂ I. This implies τ = 1, 1 ∈ I and hence the result. Proof of claim: Choose an open neigbourhood of γ(τ) −1 a with π (U) = U˜ν as above ν Continuity⇒ ∃δ > 0: t ∈ [τ − δ, τ + δ] ∩ [0, 1] ⇒ γ(t) ∈ U By definition of τ, ∃τ1 ∈ [τ − δ, τ]: γ has a liftgamma ˜ (0) = z −1 a Nowγ ˜(τ1) ∈ π (U) = U˜ν . ν ∃!ν s.t.γ ˜(τ1) ⊂ U˜ν Definingγ ˜ := (π| )−1 ◦ γ on [τ, τ + δ] ∩ [0, 1] extendsγ ˜ continuously to a lift U˜ν on [0, τ + δ] ∩ [0, 1]
3.1 Homotopy of Lifts Definition. Let X be a topological space and α, β : [0, 1] → X be paths with same initial and final points. i.e. α(0) = β(0), α(1) = β(1) We say that α, β are homotpic in X if there exists a family (γs)s∈[0,1] of paths in X s.t.
(i) γ0 = α, γ1 = β
(ii) γs(0) = α(0) = β(0), γs(1) = α(1) = β(1) (iii)
(t, s) 7→ γs(t) [0, 1] × [0, 1] → X
is continuous Definition. X is called simply connected if X is path-connected and every closed path γ : [0, 1] → X is homotopic to the constant path γ(0). Theorem. Monodrony Theorem Let π : X˜ → X be a covering map and α, β paths in X. Assume 1. α, β homotopic in X
2. α, β have liftsα, ˜ β˜ withα ˜(0) = β˜(0) 3. Every path γ in X with γ(0) = α(0) = β(0) have a liftγ ˜ withγ ˜(0) = α˜(0) = β˜(0). Thenα, ˜ β˜ homotopic in X and in particularα ˜(1) = β˜(1)
Proof. (Non-examinable) See Beardon p.106-107
19 Note: Regular⇒ (2) and (3) hold. Let G ⊂ C be a domain. Definition. Let z ∈ G,(f, D), (g, E) be function elements in G. Write Write (f, D) ∼z (g, E) if z ∈ D ∩ E,f = g on some open neighbourhood of z in D ∩ E This defines an equivalence relation and the equivalence classes containing (f, D) is called the germ of f at z denoted by [f]z
[f1]z1 = [f2]z2 ⇔ z1 = z2 = z, f = g on a neighbourhood of z. G = G (G) := {all germs [f]z over all z ∈ G} Let (f, D) be a function element [f]D := {[f]z : z ∈ D} ⊂ G
1. Topology of G Open sets are unions of set of the form [f]D Required to prove U1,U2 are open in G ⇒ U1 ∩ U2 is open
Proof. Suppose [f]z ∈ U1∩U2 since U1,U2 are open, ∃(f1,D1) and (f2,D2) s.t.
[f]z ∈ [f1]D1
[f]z ∈ [f2]D2 Choose representatives (f, D) of the germ [f]z such that z ∈ D ⊂ D1 ∩ D2 [f]z = [f1]z = [f2]z ⇒ ∃ a neigbourhood of z: f = f1 = f2 on it. By the identity principle f = f1 = f2 on D Hence, [f]z ∈ [f]D ⊂ U1 ∩ U2
2. Topology is Hausdorff Suppose [f]z 6= [g]w. Choose domains D,E such that z ∈ D,w ∈ E f analytic on D, g analytic on E. If z 6= w choose D,E disjoint and [f]D ∩ [g]E = φ If z = w choose D and E to coincide and claim [f]D ∩ [g]E = φ. Otherwise [h]s ∈ [f]D ∩ [g]E for s ∈ D. ⇒ h = f = g is a neighbourhood of S. ⇒ f = g on D by the identity principle. ⇒ [f]z = [g]z #Absurd 3. Projection map
π : G → G
[f]z 7→ z
• π is continuous Given V ⊂ G open, note π−1(V ) =
• π|[f]D :[f]D → D is a homeomorphism.
Certainly bijective and π|[f]D is continuous.
π|[f]D is an open map. Take U open in [f]D.
20 U open in [f]D [ ⇔ U = [f]Dα where Dα are open in D α [ ⇔ π(U) = Dα α ⇔ π(U) open in D.
4. Atlas on G On each connected component π is a covering map so we use Lemma 2.2 to give each connected component the structure of a Riemann surface for which π is analytic.
Charts are simply (U, φ), U = [f]D
φ = π|[f]D
5. Evaluation Map
E : G → C [f]z 7→ f(z) This is analytic
This is analytic. −1 Eφ (z) = E([f]z) = f(z) holomorphic. (f, D), z ∈ D
Theorem 3.3. Let (f, D) and (g, E) be function elements in G and γ : [0, 1] → G be a path with γ(0) ∈ D, γ(1) ∈ E. Then (g, E) is an analytic continuation of (f, D) along γ ⇔ ∃ a pathγ ˜ : [0, 1] → G such that πγ˜ = γ and γ˜(0) = [f]γ(0) γ˜(1) = [g]γ(1)
Proof. Assume (f, D) ∼γ (g, E) ⇒ ∃(fj,Dj)j=1 and 0 = t0 < t1 < ... < tn = 1 with (f, D) = (f1,D1) ∼ (f2,D2) ∼ ... ∼ (fn,Dn) = (g, E) and γ([tj−1, tj]) ⊂ Dj for j = 1, ..., n. Defineγ ˜(t) := [fj]γ(t) for tj−1 ≤ t ≤ tj Note this is well-defined since fj = fj−1 on Dj−1 ∩ Dj (Thus take care of endpoints.) Obviously πγ˜ = γ Required to proveγ ˜ is continuous.γ ˜ : [0, 1] → G Consider a basic open set [h]∆ and supposeγ ˜(τ) ∈ [h]∆ Without loss of generality, we may assume that τ ∈ [tj−1, tj] so γ˜(τ) = [fj]γ(τ) ∈ [h]∆ i.e. γ(τ) ⊂ ∆ and fj = h is some open neighbourhood N of γ(τ) in ∆ ∩ Dj By continuity of γ, there is a δ > 0 s.t. if |t − τ|z < δ ⇒ γ(t) ∈ N Thenγ ˜(t) = [fj]γ = [h]γ(t) ∈ [h]∆
21 ⇒ γ˜ is continuous. (⇐) Assume ∃γ˜ → G is a path s.t.γ ˜(0) = [f]γ(0),γ ˜(1) = [g]γ(0) and πγ˜ = γ
Claim: We can find basic open sets [f]D1 , ..., [f]Dn in G for some n and 0 = t0 < t1 < ... < tn = 1 s.t.γ ˜([tj−1, tj]) ⊂ [fj])Dj for each j. Proof of claim: Note that by continuity ofγ ˜ for each t ∈ [0, 1] we can find a basic open set [ft]Dt 3 γ˜(t) in G and an open interval I(t) 3 t such that γ˜(I(t) ∩ [0, 1]) ⊂ [ft]D The intervals Ik cover [0, 1] hence by compactness we can find a finite subcover, which we can reorder to obtain intervals Ik = [ak, bk] with ak+1 < bk for all k = 1, ..., n.
Choosing tk ∈ (ak+1, bk) we can arrange fk,γ ˜([tj−1, tj]) ⊂ [fj]Dj Adjustments: WLOG may assume that
1. Each Dj to be a disc.
2. Also, sinceγ ˜(0) = [f]γ(0) and˜γ(1) = [g]γ(1) We may assume that D1 ⊂ D, Dn ⊂ E with f = f1 in D1 and g = fn in Dn.
For each 1 ≤ j ≤ n − 1,γ ˜(tj) ∈ [fj]Dj ∩ [fj+1]Dj+1 hence fj = fj + 1 on a range of γ(tj) and hence in all of Dj ∩ Dj+1 (Identity Principle on intersection Dj ∩ Dj+1 is connected since Di are disc(s). Thus, (f, D) ∼ (f, D1) = (f1,D1) ∼ ... ∼ (fn,Dn) = (g, Dn) ∼ (g, E)
And we have γ([tj−1, tj]) = πγ˜([tj−1, tj]) ⊂ π([fj]Dj ) = Dj (We can always enlarge the dissection t1 tn−1+1 0 < 2 < t1 < ... < tn−1 < 2 < 1 to account for his first and last continuation.)
Corollary 3.4. Let F be a complete analytic function in G and put GF = S (f,D)∈F [f]D. Then GF is a connected component of G .
Proof. (3.3)⇒ GF is path-connected, hence connected. GF is open by definition. T Also GF = G F 06=F GF 0. Hence GF is a connected component.
Summary: G ⊂ C, F complete analytic function then we have constructed a Riemann Surface GF and a diagram. π : GF → G projection map which is analytic covering. GF is the Riemann Surface associated with F.
22 GF
[f]D
G C f D
Theorem 3.5 (Uniqueness of Continuation along paths). If (g, E) and (h, E) are analytic continuations of (f, D) along some path γ in G then g = h in E.
∗ ∗ Proof. (3.3)⇒ ∃γ,˜ γ˜ s.t.γ ˜0 =γ ˜ (0) = [f]γ(0) inGF and γ˜(1) = [g]γ(1) ∗ γ˜ (1) = [h]γ(1) Uniquness of lifts (3.1)⇒ γ˜(1) = γ˜∗(1) ⇒ g = h in a neighbourhood of γ(1) ⇒ g = h on E by the identity principle. Theorem 3.6 (Classical Monodromy Theorem). Suppose (f, D) can be contin- ued along all paths in G which start in D. If (g, E) and (h, E) are continuations of (f, D) along paths α, β which are homotopic in G, then g = h on E.
Proof. (3.3)⇒ ∃ liftsα ˜ and β˜ s.t. ˜ α˜(0) = [f]α(0) , β(0) = [f]β(0) ˜ α˜(1) = [f]α(1) , β(1) = [f]β(1) Since α(0) = β(0) we haveα ˜(0) = β˜(0). So by Monodromy Theorem(Topological) now liftsα, ˜ β˜ are homotopic and in particular,α ˜(1) = β˜(1) Hence as before g = h on E.
Corollary 3.7. Suppose G ⊂ C is simply-connected and (f, D) is a function element in G which can be analytically continued along all paths in G which start in D. Then, ∃ a single-valued analytic continuation f˜ of (f, D) to all G. Proof. If (g, E) and (h, E) are two analytic continuations of (f, D) then the two analytic continuations of each other along some closed path γ in G. Since G is simply-connected, γ is homotopic to a point, hence (3.6)⇒ g = h on E. Deforming f˜ = g on E(for all (g, E) ∈ F) gives an analytic extension of (f, D) to all of G.
23 4 Multiplicities and Degrees f : R → S with R,S compact Rimeann surfaces, f non-constant. Given z ∈ U ⊂ C(domain) and a non-constant holomorphic function f, we can write locally m (z) f(w) = f(z) + (w − z) f f1(w) where f1(z) 6= 0 We call mf the multiplicity or valency of f at z.
Lemma 4.1. Suppose g and h are analytic and non-constant on domains in C with image(h) ⊂ image(g) then,
mg◦h(z) = mg(h(z))mh(z)
m (z) Proof. 1. h(w) = h(z) + (w − z) h h1(w) locally with h1(z) 6= 0
mg (η) 2. g(η) = g(η0) + (η − η0) g1(η) locally with g1(η) 6= 0 and η0 = h(z). Substitute (1) into (2). m (z)mg (η0) mg (η0) g(h(w)) = g(h(z)) + (w − z) h (h1(w)) g1(h(w)) locally with g1(h(z)) = g(η0) 6= 0 and h1(z) 6= 0 Consider f : R → S analytic and non-constant and z ∈ R. Take charts (U, φ) around z and (V, ψ) around f(z) and define mf (z) = mψfφ−1 (φ(z)) This does not depend on the choice of charts. If we choose (U,˜ φ˜), (V,˜ ψ˜) instead then,
ψf˜ φ˜−1 = ψψ˜ −1ψfφ−1φφφ˜ −1 | {z }| {z }| {z } local conformal equivalence
Since mψψ˜ −1 (ψ(f(z))) = 1 and mφφ˜ −1 (φ(f(z))) = 1 ˜ (4.1)⇒ mψf˜ φ˜−1 (φ(z)) = mψfφ−1 (φ(z)) Given any chart θ : V → C in S, where f(z) ∈ V ⊂ S with θ(f(z)) = 0, we saw in §2 that there is a chart ψ : N → C s.t. ψ(z) = 0, z ∈ N ⊂ f −1(V ) and −1 r θfψ : ψ(N) → C is just z 7→ z where r = mf (z)
f N ⊂ R S
ψ θ zr C C
Remark. 1. mf (z) > 1 only at isolated points of R.(f non-constant.) Clear from the plane case given by vanishing of f 0.
2. Points z ∈ R s.t. mf (z) > 1 are called ”Ramification Points” and mf (z) is called the ramification index. The image of a ramification is called a branch point.
24 d X k Example. p(z) = akz polynomial of degree d, ad 6= 0. 0 Define p(∞) = ∞. Then p becomes a holomorphic map C∞ → C∞. 1 1 Take charts φ(w) = ψ(w) = w , w on C∞ \{0}. 1 zd ψpφ−1(z) = = = zdg(z) with g(0) 6= 0 p(1/z) Pd d−k 0 akz ⇒ mp(∞) = d Theorem 4.2 (Valency Theorem). Suppose f : R → S is a non-constant holo- morphic map with R compact(⇒ f(R) = S and S compact.) Then ∃ an integer n ≥ 1 s.t. f is an n to 1 map of R onto S counting multiplic- ities. X n = mf (z) z∈f −1(w)
∀w ∈ S
Definition. n is called the degree of f, also denoted n(f), deg(f).
Proof. f −1(w) is finite. Note f −1(w) is compact and discrete.(Identity Princi- ple) P Define n(w) := z∈f −1(w) mf (z) for w ∈ S. Our aim is to prove that w 7→ n(w) is constant. −1 Take w0 ∈ S and write f (w0) = {z1, ..., zq} For any point z0 ∈ R, we know that ∃ charts
ψ : N 7→ C θ : N 7→ C
with f(n) ⊂ V and θfψ−1 is just z 7→ zmp(z0) Hence ∃ disjoint neighbourhoods N1, ..., Nq of z1, ..., zq s.t. f is an mf (zj) to 1 map. (Counting multiplicities) of Nj onto f(Nj) for each j − 1, ...q. Include image q Claim.:There exists neighbourhood M of w0 s.t. f(R \ ∪1Nj) ∩ M = φ q [ Proof of claim.: R \ Nj closed in R. 1 q [ Hence compact ⇒ f(R \ Nj) is compact hence closed. 1 q j [ [ Then M = S \ f(R \ Nj) is open containing w0 and M ∩ f(R \ Nj) = φ 1 1
Corollary 4.3 (Fundamental Theorem of Algebra, one more time!). Let P be a polynomial of degree d, then P has d zeroes counted with multiplicities.
25 Proof 1. We have seen that P gives rise to P : C∞ → C∞ holomorphic with X P (∞) = ∞ and mP (∞) = d ⇒ # of zeroes of P = mP (z) = d (4.2) z∈P −1(0) Proof 2. By (2.5) p is onto. Hence there exists a zero of p. Use remainder theorem to get d zeroes.
4.1 Rational Maps
Proposition 4.4. f : C∞ → C∞ holomorphic and non-constant c(z − α1)...(z − αn) ⇔ f(z) = a non-constant rational function in C. (z − β1)...(z − β2)
1 Proof. (⇐)f(z) and f( z ) are meromorphic and so define a holomorphic map of C∞. 1 −1 (⇒) W.l.o.g assume f(∞) ∈ C. Otherwise consider f . Then f (∞) is a finite subset of C say {z1, ..., zq} and f is meromorphic in C as each zj is a pole. Near zj, f has the form,
∞ X i f(z) = ai(z − zj)
−kj
−1 X i and set Qj = ai(z − zj)
−kj | {z } rational function ˜ ˜ Consider f = f − (Q1 + ... + Qq). f has now only removable singularity at ˜ ˜ ˜ z1, ..., zq. Hence f gives rise to a holomorphic map f : C∞ → C∞. By (2.6) f is constant and thus f = Q1 + ... + Qq+constant.
Remark. f(∞) ∈ C ⇔ n ≤ m and in this case deg(f) = m = # of poles counting multiplicities. (Example Sheet 2) deg(f) = max{m, n} If f is an analytic isomorphism, deg(f) = 1 ⇒ m = n = 1. az + b Hence f(z) = with ad − bc 6= 0, a M¨obiusmap. cz + d So Aut(C∞) =M¨obvious group.
4.2 Quick review of triangulations and Euler characteris- tic(from IB Geometry) S compact surface. Topological triangle on S=homeomorphic image of a closed triangle T ⊂ R2. Definition. A topological triangulation consists of a finite collection of topo- logical triangles whose union is all of S s.t. (i) Two triangles are either disjoint or their intersection is a common vertex or common edge.
26 F = # of triangles V = # of vertices E = # of edges e := F − E + V Euler-number of triangulation Fact. Fact 1 One can always find a triangulation. Fact. Fact 2 e does not depend on the chosen triangulation; it is a topological invariant χ(S).
Example. 1. S2 F = 8 E = 12 V = 6 χ(S2) = 2
2. Torus T 2 F = 18,E = 27,V = 9 χ(T 2) = 0 Fact. A compact Riemann Surface R is topologically a sphere with g handles. e.g.g=0. χ(R) = 2 − 2g g genus of R. f : R → S be a non-constant holomorphic map between compact Riemann Surfaces. Given p ∈ R, let ep be the ramification index mf (p). Recall ep > 1 ⇔ p is a ramification point and the set of these points i actually finite. In particular, the following sum X ep − 1 is finite. p∈R
Theorem 4.5 (Riemann Hurwitz). If f has degree n, then X χ(R) = nχ(S) − (ep − 1) p∈R
Proof. Let Q1, ..., Qr denote the branch points of f. Suppose we have a trian- gulation on S by subdividing it we can assume that the Qi are vertices. The proof of the valency theorem theorem (4.2) gives us an open cover U1, ..., Ur,Ur+1, ..., Um of S such that
−1 (i) If j > r, f (Uj) has a disjoint union of open subset of V for which f|V : V → U is a conformal equivalence so U contains no branch points.
−1 (ii) If j ≤ r Qj ∈ Uj and for each point p ∈ f (Qj) there exists a unique −1 component V of f (Uj) so that f|V : V → Uj is an eP to 1 map over ep Uj \{Qj} given in terms of suitable charts by z 7→ z Now we ???? our trinagulation by means of Euclidean triangles as follows.
27 ...to ensure that each topological triangle is centred in some Uj. T then lifts to a triangle on R if no vertex of T is a branch point, these triangles are disjoint; either T ⊂ Uj if j > r(when it is obvious) or T ⊂ Uj if j < r and still okay. If T has a vertex at a branch point Qi,it still lifts to n triangles of R −1 but there will be ep vertices in common at each ramification point p ∈ f (Qj). We therefore get at triangulation on R corresponding to the one in S s.t. 0 0 0 F,F ,V,V ,E,E denoted the # of faces, vertices and edges respectively on S and R and we see that 0 0 0 X F = uF, E = uE, V = nV − ep − 1. p∈R X Hence χ(R) = nχ(S) − ep − 1. p∈R Trivial Remark. X 2 − 2g(R) = n(2 − 2g(S)) − ep − 1 p∈R
⇒ g(R) ≥ g(S)
Example. R ⊂ C × C R = {(z, w)|w2 = f(z)} where f(z) = z3 − z
−1 π : R → C (0, 0) = π (0) (z, w) 7→ z (1, 0) = π−1(1) (−1, 0) = π−1(−1)
g : R → C (z, w) 7→ w g around these 3 points is an inverse. g−1(w) = (f −1(w2), w) Look at π around (0, 0) and (±1, 0)
πg−1(w) = f −1(w2)
−1 dπg 0 = (f −1) (w2) · 2w dw Assume R can be compactified. We can add a point c s.t. R¯ = R ∪ {c} is a ¯ compact Riemann Surface and π(c) = ∞ so that π : R → C∞ is holomorphic. deg(π) = 2, branch points 0, ±1 ramification points (0, 0), (±1, 0) with ep = 2. −1 Since π (∞) = c then ∞ is also a branch point and ec = 2. ¯ ¯ Riemann Hurwitz χ(R) = 2χ(C∞) − 4 = 2 × 2 − 4 = 0 ⇒ g(R) = 1 Compactificaiton R¯ is a torus. One could push this further and consider f a polynomial with 2g + 1 simple roots to obtain a Riemann Surface R¯ with genus g.
28 5 Meromorphic Periodic Functions f : C → C∞ analytic and non-constant. We say w is a period for f if f(z +w) = f(z) for all z ∈ C. The set of period Ω satisfies
1. Ω is an additive subgroup of C. 2. Ω contains only isolated points. Example sheet #3 Q1 ⇒ Only 3 possibilities for Ω.
1.Ω= {0} non-periodic
2.Ω= Zw1 w1 6= 0 simply periodic
w1 3.Ω= w1 + w2 w1, w2 6= 0, 6∈ Doubly periodic or Elliptic. Z Z w2 R
5.1 Simply Periodic Functions
Without loss of generality, assume Ω = Z. i.e. f(z + 1) = f(z) for all z ∈ C Proposition 5.1. Define p : C → C∗ by p(z) = e2πiz. If f is meromorphic in C with periods Z then there exists a unique f˜ meromorphic on C∗ s.t. f = f˜◦ p
f C C∞ P f˜ C∗
Proof. Define f˜(e2πiz) = f(z). 2πiz1 2πiz2 ˜ Well defined: If e = e ⇔ z1 −z2 ∈ Z ⇔ f(z1) = f(z2) Clearly f = f ◦p and f˜ is unique. −1 ˜−1 −1 If V is open in C∞ then p f V = f V is open in C | {z } open ⇒ f −1V open ˜ ˜ log(w) f meromorphic locally with f(w) = f 2πi
Corollary 5.2. Suppose f has no poles on the strip S = {z : α < Im(z) < β} P∞ 2πikz Then f(z) = −∞ ake is locally uniformly convergent on S. Proof. f˜ is analytic on the annulus, p(S) = {w : e−2πβ < |w| < e−2πα} Lau- rent’s theorem ∞ ˜ X k P∞ 2πzk ⇒ f(w) = akw locally uniformly convergent on p(S). ⇒ f(z) = −∞ ake −∞ (Fourier expansion of simply periodic functions)
29 5.2 Doubly Periodic Functions
Let Λ = Zw1 + Zw2 be a lattice in C. A meromorphic function f on C with period Λ is called elliptic. These functions form a field. An elliptic function is determined by its value on any parallelogram P = {z + t1w1 + t2w2 : t1, t2 ∈ [0, 1)} If it has no poles ⇒ bounded in C. ⇒ constant by Liouville’s theorem. Now we look at the complex torus π : C → C/Λ. Proposition 5.3. If fis an elliptic function with period Ω then there exists a unique meromorphic map f˜ on C/Λ s.t. f = f˜◦ π
f C C∞ π f˜ C/Λ
Proof. Same as Proposition 5.1(Exercise). Corollary 5.4. The degree of f˜ is n ≥ 2. ˜ ˜ Proof 1. If deg(f) = 1 then f : C/Λ → C∞ is a conformal equivalence. But this is absurd since they have different genus. (C/Λ and C∞) Proof 2. Consider a period parallelogram P such that ∂P contains no poles of f. X 1 Z Res f = f(z)dz = 0 z 2πi z∈P ∂P ⇒ f cannot just have a simple pole.
5.3 Weierstrass ℘ function 1 P 1 Definition. For a given lattice Λ define ℘(z) = ℘λ(z) = z2 + z∈Λ\{0}{ (z−w)2 }− 1 { w2 } Convergence? X 1 Lemma. < ∞ ⇔ t > 2 |w|t w∈Λ\{0}
Proof. The function (t1, t2) 7→ t1w1 + t2w2 is continuous and non-zero on the 2 ∈R compact set {(t1, t2): |t1| + |t2| = 1} And hence its modulus is bound above and below by constants C, c > 0. k l Set t1 = |k|+|l| and t2 = |k|+|l| (k, l) ∈ Z2 \{0, 0} ⇒ c(|k| + |l|) ≤ |kw1 + lw2| ≤ C(|k| + |l|)
30 X Hence, < ∞ (k,l)∈ 2\{0,0} 1 t Z |kw1+lw2| ∞ X X 1 ⇔ < ∞ qt q=1 |k|+|l|=1 ∞ X 4 ⇔ < ∞ ⇔ t > 2 qt−1 q=1
Theorem 5.5. ℘ is elliptic with periods Λ Moreover, ℘ is an even function. ℘ is 2 to 1 on a fundamental/period parallelogram. Proof. Let |z| < r and |w| > 2r.
2 1 1 2wz − z − = (z − w)2 w2 (w − z)2w2 2r|w| + r2 ≤ |w|4/4 | {z } |w| |z|
℘(z + w) − ℘(z) = c = ℘(−z) = −℘(−z − w) = −c
⇒ c = 0 So Λ is contained in the periods of ℘ and since ℘ only has roots on Λ, these are all the periods. In a fundamental parallelogram, ℘ has a double pole. So by the valency theorem, this implies that ℘ defines a 2 to 1 map C/Λ → C∞
31 Remark. ℘ is characterised by having periods Λ, poles only on Λ and ℘(z) − 1 → 0 as z → 0 z2 Why? Let f be another meromorphic function. Then f − ℘ is periodic with at worst simple poles...It’s constant(deg ≥ 2) Now considering 1 1 f(z) − ℘(z) = f(z) − − ℘(z) − z2 z2 We see f − ℘ ≡ 0. The function ℘0 has triple poles on Λ and valency 3. It is clearly odd and w −w w ℘0 1 = ℘0 1 = −℘0 1 z ↑ z z periodicity
w1 0 ⇒ z is a zero of ℘ / w2 w1+w2 0 Similarly, 2 and 2 are zeroes of ℘ (z). w1 w2 w1+w2 Therefore the analytic map ℘ : C/Λ → C∞ is ramified at the points 0, 2 , 2 , 2 and it has branch points w w w + w ∞, ℘ 1 ℘ 2 ℘ 1 2 2 2 2 =e1 =e3 =e2 Check: 4 branch points is compatible with degree-genus formula Riemann-Hurwitz Note: The points e1, e2 and e3 are all distinct because otherwise ℘ would have valency at least 4. Proposition 5.6. There is an algebraic relation
0 2 3 (℘ (z)) = 4 (℘(z)) − g2℘(z) − g3 for constant g2, g3 ∈ C depending only on Λ. Proof. Using 1 X 1 1 ℘(z) = + − z2 (z − w)2 w2 w∈Λ\{0} , the Laurent expansion of ℘(z) over 0 has 1 ℘(z) = + az2 + O(z4) z2 So −2 ℘0(z) = + 2az + O(z3) z3 −g (℘0(z))2 − 4(℘(z))3 = 2 + (locally holomorphic function) z2 0 ⇒ ℘ (z) − 4℘(z) + g2℘(z) is constant by periodicity. (Liouville’s theorem)
32 w w w + w Remark. Since ℘0(z) = 0 if z = 1 , 2 , 1 2 , 2 2 2 we find that the cubic equation,
3 4x − g2x − g3 = 4(x − e1)(x − e2)(x − e3)
0 2 (℘ (z)) = 4(℘(z) − e1)(℘(z) − e2)(℘(z) − e3)
In particular, note that e1 + e2 + e3 = 0 Remark. Another interpretation of Proposition 5.6. Consider the map (℘(z), ℘0(z)) as defining an analytic map.
2 F :(C/Λ) \{0} → C 2 3 The image of F lies on the cubic curve, y = 4x − g2x − g3. This can be compactified (cf. Example Sheet 1 Q15) to a curve in P3(C) defined by the homogeneous equation
2 3 2 3 y z = 4x − g2xz − g3z
[x : y : z] coordinate on P3(z) Last Task: We want to understand all elliptic functions on C/Λ Theorem 5.7. Let f be an elliptic function with period Λ. Then there are rational functions on Q1,Q2.
0 f = Q1(℘) + ℘ Q2(℘)
In particular, if f is even then you can take Q2 ≡ 0 Proof. We first treat the case when f is assumed to be even. 1 Let E = t ∈ |z ∈ Λ or f 0(z) = 0 C 2 Then f(E) is finite, so we can choose distinct , d ∈ C \ f(E). f(z) − c Let g(z) = so at poles of f, g(z) → 1 f(z) − d Then g is elliptic and its zeroes,(where f(z) = c) and poles (f(z) = d) are all simple as f 0(z) = 0 at each point and g is even since f is. Let us list the zeroes and poles of g as {a1, ..., am, −a1, ..., −am} has zeroes 1 (since f(aj) = c ⇒ aj 6∈ 2 Λ,aj = −aj) Similarly list the poles of g as {b1, ..., bn, −b1, ..., −bn}. Let (℘(z) − ℘(a ))(℘(z) − ℘(a ))...(℘(z) − ℘(a )) h(z) = 1 2 m (℘(z) − ℘(b1))...(℘(z) − ℘(bn)) Then h is elliptic with the same zeroes and poles as g, all of which are simple. g So is an elliptic function without poles so is constant A. h g f(z) − c = A ⇒ Ah = h f(z) − d
33 ⇒ f = Q1(℘) Algebraic Rearrangement 0 0 If f is odd then f/℘ (z) is an elliptic function s.t. f/℘ (z) = Q2(z) and hence 0 f = ℘ · Q2 In general we can write f(z) + f(−z) f(z) − f(−z) f(z) = + 2 2 and apply the previous cases. Remark. The upshot of the previous results is that the field of meromorphic functions on C/Λ is just the field of fractions
2 3 C[x, y]/(y = 4x − g2x − g3)
34 6 The Uniformization Theorem
Definition. A group of homeomorphism of a topological group space X is said to act properly discontinuously if every x ∈ X has an open neighbourhood U s.t. all g(U) for g ∈ G are pairwise disjoint. Consider the quotient map π : X → X/G. Continuous wrt the quotient topology. −1 ` Given x ∈ X and a neighbourhood U as above π (πU) = g∈G gU ⇒ πU is a bijection and hence a homeomorphism. Thus if X is path-connected, the π is a regular covering. Remark. If X is a Riemann Surface and G ⊂ Aut(X) is a subgroup of ana- lytic automorphisms automorphisms acting discontinuously, then X/G has the structure of a Riemann Surface; we may assume that (U, φ) is a chart with U −1 as above then φ ◦ (π|U ) is the region contained on X/G This is a conformal strucutre: Transition functions are analytic since G ⊂ Aut(X).
Example. Λ ⊂ C is a lattice by translations ⇒ C/Λ is a Riemann Surface. Note: If X˜ → X is a regular covering, the group G of ”deck transformations” acts properly and discontinuously and X = X/G˜
6.1 Uniformization Theorem 1. (Hard) Any simply connected Riemann Surface is conformally equivalent to one of C∞, C and ∆. Note that C homeomorphic to ∆ but not conformally equivalent by Liou- ville. 2. Easier Any Riemann Surface R is conformally equivalent to R/G˜ for R˜ to one of the Riemann Surface in (1) and G ⊂ Aut(R˜), a group of analytic automorphism acting properly and discontinuously. We say that R is uniformized by R˜ Example. S2 has only one confromal structure; if R is homeomorphic to S2 it is simply connected and hence by (1) is conformally equivalent to C∞ Now consider the 3 possibilities for R˜ in turn ˜ az+b (i) R = C∞ Aut(C∞) consists of M¨obiustransformations z 7→ cz+d ad−bc 6= 0 and they have 1 or 2 fixed points ⇒ if G acts properly discontinuously ⇒ G = id ⇒ R = C∞ (ii) R˜ = C , Aut(C) consists of maps of the form az + b, a ∈ C∗. If a 6= 1, ∃ a fixed point so G consists of translations and so G,→ C g 7→ g(0). Moreover G consists only of isolated points(Check!). Example Sheet 3 Q1⇒ G = {id}. Zw1 on λ lattice. No R is conformally equivalent to C, C∗ on C/Λ(complex torus)
35 (iii) R˜ = ∆, R is not conformally equivalent to R, C, C∗. Indeed suppose it is. Get f˜ and using the Monodromy theorem and Liouville, f is constant. ∗ Hence any Riemann Surface which is not C∞, C, C or C/Λ is uniformized by ∆ z − a Recall: Aut(∆) consists of M¨obiustransformations z 7→ eiθ where a ∈ 1 − az¯ ∆ Large literature concerning G ⊂ Aut(∆) acting properly discontinuously Fuschian Groups Consequences: (a) Compact Riemann Surfaces of genus 1 are uniformized by ∆
(b) Riemann Mapping Theorem: If U 6⊂ C domain and simply connected, then it is conformally equivalent to ∆. Just need to show 6 ∃ conformal equivalence f : C → U. This is the standard arugement saying that f can’t have an essential singularity. If it did the Casorati-Weierstrass contradicts the injectivity of f.
Hence f extends to an analytic map f : C∞ → f(C∞) ⊂ C∞. This is however not surjective.
(c) If a Riemann Surface R is uniformized by ∆ ,∃ a hyperbolic metric on R since Aut(∆) consists of isometries of the hyperbolic metric on ∆. Hence if R is compact, Guass-Bonnet⇒ χ(R) < 0 so a compact surface of genus 1 must be uniformized by C. ⇒ It is a complex torus C/Λ!
(d) Picard Theorem If f : C → C \{0, 1} analytic, f is constant.
Proof. Example Sheet 3 Q9⇒ C\{0, 1} is uniformized by ∆. Monodromy Argument f lifts up to f˜ (cf. Example Sheet 2 Q3) Hence by Liouville, f˜ is constant.⇒ f is constant.
Any abstract Reimann Surface can be realised as an algebraic object somewhere.
36