notes on non-euclidean geometry jiayin panweb.math.ucsb.edu/~j_pan/113notes.pdf · 2020. 12....

Notes on Non-Euclidean Geometry

Jiayin Pan

Abstract. Lecture notes for Math 113 Non-Euclidean geometry. We loosely

follow the textbook Geometries and Groups by Nikulin and Shafarevich. Most

proofs have been rewritten and more content has been added. This note is self-contained.

Last updated: December 14, 2020

Contents

Chapter 1. Motivating examples 51.1. Sphere 51.2. Cylinder 71.3. More examples 10

Chapter 2. Basic Concepts 132.1. Geometries 132.2. Groups 162.3. Group actions on Geometries 18

Chapter 3. Free and discrete isometric actions on R2 or S2 233.1. Hello again, linear algebra 233.2. Elements of Isom(R2) without fixed points 253.3. Subgroups of Isom(R2) 263.4. Subgroups of Isom(S2) 30

Chapter 4. Coverings 334.1. Covering maps 334.2. Cover a 2-dimensional locally Euclidean space by R2 354.3. Covering groups 37

Chapter 5. Hyperbolic Plane 415.1. Introduction 415.2. Pure imaginary half line under the SL2(R)-action 425.3. Distance function on H 445.4. Projection 485.5. The isometry group of (H, d) 51

3

CHAPTER 1

Motivating examples

1.1. Sphere

We consider the 2-dimensional unit sphere S2 in R3 centered at the origin.We would like to measure the distance between two points, say P and Q, on thesphere. One way is using the distance from R3, but a more natural way (from theperspective of S2) is the length of a shortest path on S2 between P and Q. Wedenote dS(P,Q) as their distance coming from the length of shortest path.

Definition 1.1.1. A great circle on S2 is the intersection of S2 and a plane throughthe origin.

Note that great circle is a circle of radius 1. For P,Q ∈ S2 and a great circlethrough them, P and Q divide the circle into two arcs. When speaking of the arcbetween P and Q, we mean the shorter one unless otherwise noted. Note that thisarc is unique unless P = −Q.

Theorem 1.1.2. The shortest path on S2 between P,Q ∈ S2 is the arc of a greatcircle through P and Q.

The proof is a contradicting argument. Before presenting the detailed proof,here is a sketch:1. Suppose the contrary, then there is a path l̃ between P,Q that is shorter thanthe arc l.2. Choose a series of points on l̃, then connecting them successively by small arcs.In this way, we obtain a “broken arc” l′ between P,Q. As a technical part, we willshow that with suitable intermediate points chosen, l′ has length less than that ofl.3. Deleting the intermediate points one by one, while connecting the remainingpoints by arcs. In this process, the new broken arc has length no greater than thatof the old broken arc (Surely, we will need a lemma here).4. Eventually, after all the intermediate points are deleted, we end in a arc betweenP and Q, but its length is shorter than the arc that we start with. A contradiction.

Now we prove some lemmas to address the non-trivial part in the sketch.

Lemma 1.1.3. Let A,B ∈ S2. Let d be the Euclidean distance between them andθ = ∠AOB. Then

d = θ −O(θ3),where O(θ3) means a term comparable to θ3 as θ → 0.

Proof. By cosine law, we have

12 + 12 − 2 · 1 · cos θ = d2,

5

6 1. MOTIVATING EXAMPLES

or d2 = 2− 2 cos θ. When θ is very small, we can apply the Maclaurin series

d2 = 2− 2 cos θ = 2− 2(

1− θ2

2+O(θ4)

)= θ2 −O(θ4).

Thus

d = θ√

1−O(θ2) = θ(1−O(θ2)) = θ −O(θ3).(You may check the Maclaurin series of

√1 + x if you are not sure about the second

equality above.) �

Remarks:(1) θ = ∠AOB is equal to the length of the arc between A and B.(2) d ≤ dS(A,B) ≤ θ.

Lemma 1.1.4. Let P,Q ∈ S2 and let l̃ be a path between them. Let P0 =P, P1, ..., Pn = Q on l̃ such that they divide the curve l̃ evenly. Let ln be the brokenarc through P0, ..., Pn. Then

Length(ln)− Length(l̃)→ 0as n→∞.

Proof. By assumptions, we have

dS(Pi, Pi+1) = Length(l̃)/n =: δ(n)

for each i = 0, 1, ..., n− 1. Since the sphere is soooo symmetric, the angle between−−→OPi and

−−−−→OPi+1 are the same for all i; we denote this angle as θ(n). Then

Length(ln) = nθ(n), Length(l̃) = nδ(n) ≥ nd(n),where d(n) is the Euclidean distance between Pi and Pi+1. By Lemma 1.1.3, weestimate

Length(ln)− Length(l̃) = n(θ(n)− δ(n)) ≤ n(θ(n)− d(n))

= n ·O(θ(n)3) = Length(l̃)δ(n)

·O(θ(n)3)

≤ Length(l̃) · θ(n)d(n)

·O(θ(n)2).

Since θ(n)/d(n)→ 1 and θ(n)→ 0 as n→∞, we see that the error above → 0 asn→∞. �

Lemma 1.1.5. Let A,B,C ∈ S2. Then∠AOC ≤ ∠AOB + ∠BOC.

Proof. We project B to p(B) on plane AOC. It suffices to show that

∠AOp(B) ≤ ∠AOB, ∠p(B)OC ≤ ∠BOC.We prove the first inequality (the second one is the same). Without lose of

generality, we assume that A = (1, 0, 0) and plane AOC is the xy-plane. We write

~v = ~OA = (1, 0, 0), ~w = ~OB = (x, y, z). Then ~p(w) = ~Op(B) = (x, y, 0). We have

x = ~v · ~w = cos∠AOB, x = ~v · ~p(w) = ||p(w)|| cos∠AOp(B).Since ||p(w)|| ≤ 1, we conclude that ∠AOp(B) ≤ ∠AOB. �

1.2. CYLINDER 7

Proof of Theorem 1.1.2. We follow the sketch. Let l be an arc betweenP,Q. Suppose that there is a path l̃ on S2 between P,Q such that

length(l̃) < length(l).

Let P0 = P, P1, ..., Pn = Q be a series of points of l̃ that divide l̃ evenly. With thesepoints, we construct a broken arc, by connecting Pi to Pi+1 successively by an arc.We call this broken arc ln. By Lemma 1.1.4, we can assume that n is sufficientlylarge such that

length(ln) < length(l).

Next we delete the intermediate points P1, ..., Pn−1 one by one. First delete P1,that is, we replace two arcs, P0 to P1 and P1 to P2, by a new arc from P0 directlyto P2. By Lemma 1.1.5, the length of the new arc is no greater than the sum oftwo original orcs (recall that the angle equals the length of arc). We call the newbroken arc through P0, P2, ..., Pn as ln,1. Then

length(ln,1) ≤ length(ln) < length(l).We continue this process until all intermediate points are deleted. We end in anarc ln,n−1 between P,Q with

length(ln,n−1) < length(l).

This is a contradiction to our choice l as the shortest arc between P and Q. �

1.2. Cylinder

We look for shortest paths on the cylinder C = S1 × R, as a surface in R3.We write a point on C as (θ, z), which corresponds to the point (cos θ, sin θ, z)

in R3. Given two distinct points P = (θ1, z1) and Q = (θ2, z2) on C, we define anarc connecting them as below:(1) if θ1 = θ2, then the arc is the vertical segment connecting P and Q;(2) if z2 = z2, then the arc is the usual arc (as part of a circle) lying in the horizontalplane containing P and Q;(3) if θ1 6= θ1 and z2 6= z2, we define the arc as part of a helix:

(α(t), z1 + t(z2 − z1)), t ∈ [0, 1],where α(t) is the usual arc from θ1 to θ2 in the circle.

Our goal is showing that these arcs are exactly all the shortest curves on C. Wefirst calculate the length of the arcs. Let P and Q be two points on the cylinder.Using the symmetries of the cylinder, we can assume that P = (0, 0) and Q = (θ, z)with θ ∈ [0, π]. For cases (1) and (2) above, the length of the corresponding arc isclear. For case (3), the arc has parametric equation

γ(t) = (cos(θt), sin(θt), tz), t ∈ [0, 1].It has tangent vector

γ′(t) = (−θ sin(θt), θ cos(θt), z)with norm

||γ′(t)|| =√θ2 + z2.

Thus the arc has length

Length(γ) =

∫ 10

||γ′(t)||dt =√θ2 + z2.


Once we know that these arcs are indeed the shortest curves, the distancebetween P = (0, 0) and Q = (θ, z) will be

√θ2 + z2, where θ ∈ [0, π]. Note that

this formula is identical to the Pythagorean Theorem of the plane. For P = (0, 0)and Q = (θ, z) with θ ∈ (π, 2π), we have

d(P,Q) = d(P,Q′) =√

(2π − θ)2 + z2,

where Q′ = (2π − θ, z)

Theorem 1.2.1. The shortest path between P,Q on C is the arc between them.

To show that these arcs are indeed the shortest curves, we follow a similarstrategy used in the sphere case.

Exercise 1.2.2. Write a proof based on the sketch below:Step 1. Suppose that there is a secret path whose length is shorter than the arc,then we construct a path by joining small broken arcs; show that the length this newpath can be arbitrary close (, as the partition becomes finer,) to that of the secretpath.Step 2. Similar to the sphere case, for the broken arcs, show that after deleting amidpoint, the new broken arc is shorter than the old one.Step 3. Derive a contradiction.

You may have a try on your own before taking a look at the hints below.Hint on Step 1. You may encounter with an issue at some point. Though the

cylinder is so symmetric, it is not soooo symmetric as the sphere. More precisely,once we know the spherical distance between two points, their Euclidean distance isdetermined (independent of the actual positions of these two points). This propertyis no longer true on the cylinder. We write A(A,B) as the length of the arcbetween A and B; we also write dE(A,B) as the Euclidean distance between A,B.Inspecting the proof in the spherical case, here we would similarly estimate

Length(ln)− Length(l̃) =∑i

A(Pi, Pi+1)− nδ(n)

=∑i

(A(Pi, Pi+1)− δ(n))

≤∑i

(A(Pi, Pi+1)− dE(Pi, Pi+1))

In the sphere case, each arc has the same length θ(n) and the Euclidean distance arealso the same d(n). On the cylinder, these two terms may depend on the positionof points. Fortunately, it suffices to have a estimate: can you find a useful upperbound for A(Pi, Pi+1)− dE(Pi.Pi+1)?

Hint on Step 2. This is indeed the triangle inequality of the plane (Why?)

Besides treating cylinder as a surface in R3, we can define it by using a equiv-alence relation.

Definition 1.2.3. For two points A = (x1, y1) and B = (x2, y2), we write x ∼ y,if x1 − x2 = 2kπ for some k ∈ Z and y1 = y2.

It is straight-forward to check that ∼ is an equivalence relation (Have a try ifyou are not sure).

1.2. CYLINDER 9

Equivalently, we can introduce a map φ : R2 → R2, φ(x, y) = (x+ 2π, y). Thenx ∼ y if and only if y = φk(x) for some k ∈ Z.

Let E(A) be the set of all points that is equivalent to A, that is,

E(A) = {A′ ∈ R2|A′ = φk(A) for some k ∈ Z}.

We write Ā and B̄ as the corresponding points in the quotient set R2/ ∼. Wedefine a distance function d̄ on R2/ ∼ as follows:

d̄(Ā, B̄) = d(E(A), E(B)) = infA′∈E(A),B′∈E(B)

d(A′, B′).

Remark: If you never see inf before, just treat it as minimum.

Proposition 1.2.4. The above infimum(minimum) can be achieved. Also, there isB0 ∈ E(B) such that

d̄(Ā, B̄) = d(A,B0).

Proof. Note that for any A′ = φk(A) ∈ E(A),

d(A′, E(B)) = d(φk(A), E(B)) = d(φ−kφk(A), φ−k(E(B)))

= d(A, φ−k(E(B))) = d(A,E(B)).

The second equality holds because φ−k, as a translation, preserves the distancefunction. Thus

d(E(A), E(B)) = d(A,E(B)) = minB′∈E(B)

d(A,B′).

Choose B0 ∈ E(B) such that d(A,E(B)) = d(A,B′). The result follows. �

It is straight-forward to check that d̄ satisfies:(1) d̄(Ā, B̄) ≥ 0, = holds if and only if Ā = B̄;(2) d̄(Ā, B̄) = d̄(B̄, Ā);(3) d̄(Ā, B̄) ≤ d̄(Ā, C̄) + d̄(C̄, B̄).

We give details on (3). Applying Proposition 1.2.4, we can find A0, B0, C0 ∈ R2such that

d̄(Ā, C̄) = d(A0, C0), d̄(C̄, B̄) = d(C0, B0).

Then

d̄(Ā, B̄) ≤ d(A0, B0) ≤ d(A0, C0) + d(C0, B0) = d̄(Ā, C̄) + d̄(C̄, B̄).

The next proposition explains how the “strip with portals” model is related tothe “quotient space” model.

Proposition 1.2.5. There is a one-to-one correspondence between R2/ ∼ (thequotient space model) and the half-open-half-closed stripe [0, 2π)× R.

Proof. We define

F : [0, 2π)× R→ R2/ ∼ (x, y) 7→ (x, y),

where (x, y) is the corresponding point in the quotient set.

F is surjective (onto): For any (x, y) ∈ R2/ ∼, since x ∈ R, we write x asx = 2kπ + x′ for some k ∈ Z and x′ ∈ [0, 2π). Note that (x, y) ∼ (x′, y). Thus

F ((x′, y)) = (x′, y) = (x, y).


F is injective (one-to-one): Suppose that F (x1, y1) = F (x2, y2), where (x1, y1),

(x2, y2) ∈ [0, 2π)× R. By the definition of F , this means (x1, y1) = (x2, y2). Thus(x1, y1) ∼ (x2, y2), that is,

x1 − x2 = 2kπ, y1 = y2,for some k ∈ Z. Recall that x1, x2 ∈ [0, 2π). It follows that k must be zero. Hence(x1, y1) = (x2, y2). �

Let Ā ∈ R2/ ∼. We write Br(Ā) as the (open) r-ball around Ā, that is,Br(Ā) = {P̄ ∈ R2/ ∼ |d̄(Ā, P̄ ) < r}.

Proposition 1.2.6. Let A ∈ R2 and Ā ∈ R2/ ∼. Then Bπ/2(Ā) is “identical”to Bπ/2(A) in R2, in the sense that, for any B,C ∈ Bπ/2(A), we have d̄(B̄, C̄) =d(B,C).

Here “identical” means the distance functions are the same. The identificationis through the quotient map

p : R2 → R2/ ∼, A 7→ Ā.Later in the course, we will introduce a rigorous definition to describe this.

Proof. We first show that if d(A,B) < π, then d(A,B) = d̄(Ā, B̄). To provethis, we draw two vertical lines in R2: The first one l1 has equal distance between Aand φ(A); the second one l2 has equal distance between A and φ

−1(A). Since φ isa translation along x-axis by 2π, each line has distance π to A. Since d(A,B) < π,B must stay inside the strip formed by l1 and l2. For convenience, we call this stripS. It is clear that

d̄(Ā, B̄) = min{d(A,B), d(φ(A), B), d(φ−1(A), B)}since all the other points equivalent to A is further away from B. Because B staysin the strip S, the above minimum is realized at A, that is,

d(A,B) = d̄(Ā, B̄).

Now for any B,C ∈ Bπ/2(A), note thatd(B,C) ≤ d(B,A) + d(A,C) < π/2 + π/2 = π.

Thus the result follows from what we proved in the first paragraph. �

Remark: When B is in the strip S above, then d(A,B) = d̄(Ā, B̄).

1.3. More examples

Torus: We define an equivalence relation on R2 as follows: we write (x1, y1) ∼(x2, y2), if there are integers k and l such that

x1 − x2 = 2kπ, l1 − l2 = 2lπ.It is direct to check that ∼ is indeed an equivalence relation.

The other way to say this is the following. We define two motions on R2. Theyare φ : (x, y) 7→ (x+ 2π, y) and ψ : (x, y)→ (x, y + 2π). Then (x1, y1) ∼ (x2, y2) ifand only if there are integers k and l such that

(x1, y1) = (φk ◦ ψl)(x2.y2).

We define the quotient distance on R2/ ∼ as usual. We denote the quotientdistance as dT . Similar to the cylinder case, we can prove the propositions below.

1.3. MORE EXAMPLES 11

Proposition 1.3.1. There is a one-to-one correspondence between R2/ ∼ and thehalf-open-half-closed square [0, 2π)× [0, 2π).

This allows us to write any point in the torus as (α, β), where α, β ∈ [0, 2π).

Proposition 1.3.2. For any Ā ∈ R2/ ∼, Bπ/2(Ā) is “identical” to Bπ/2(A) in R2,in the sense that, for any B,C ∈ Bπ/2(A), we have dT (B̄, C̄) = d(B,C).

Exercise 1.3.3. Prove the above two propositions.

We have seen a similar property for cylinders and twisted cylinders: local ballsin the space are identical to local balls in R2. We will call this property locallyEuclidean.

Proposition 1.3.2 also provides a formula for the distance dT between (0, 0) and(α, β), given that α and β are small:

dT ((0, 0), (α, β)) =√α2 + β2.

Twisted cylinder (Mobius strip): A glide reflection is a motion of R2 definedas follows: reflect along a line l, then translate along l. If we set l as the x-axis,we can write a glide reflection as ψ : (x, y) → (x + L,−y), where L is the lengthof the translation. For convenience, we will use L = 2π. Since both reflection andtranslation preserve the distance function of R2, so does ψ.

We define an equivalence relation and a quotient distance similarly:

Definition 1.3.4. For A,B ∈ R2, we say that A ∼ B, if there is an integer k suchthat B = ψk(A).

Definition 1.3.5. Distance function on R2/ ∼:d̄(Ā, B̄) = min

A′∈E(A),B′∈E(B)d(A′, B′).

We call the quotient space (with the quotient distance) (R2/ ∼, d̄) a twistedcylinder. Proposition 1.1 and the triangle inequality of d̄ extends directly for thetwisted cylinder, since we only need the fact φ preserves the distance function inthe proof of the cylinder case.

Exercise 1.3.6. Similar to Proposition 1.2.5, show that there is a one-to-onecorrespondence between the twisted cylinder and the half-open-half-closed stripe[0, 2π)× R.

Proposition 1.3.7. For any Ā ∈ R2/ ∼, Bπ/2(Ā) is “identical” to Bπ/2(A) in R2,in the sense that, for any point B,C ∈ Bπ(A), we have d̄(B̄, C̄) = d(B,C).

Proof. The proof is similar to the cylinder case, with some modifications. Itsuffices to show that d(A,B) = d̄(Ā, B̄) if d(A,B) < π. For A ∈ R2, we draw fourlines in R2. l1 has equal distance between A and φ(A), l2 between A and φ−1(A),l3 between A and φ

2(A), l4 between A and φ−2(A). Let R be the region containing

A and bounded by these four lines. Note that d(A,B) < π implies that B ∈ R.Also, we have

d(A,B) = minA′∈E(A)

d(A′, B)

Thusd̄(Ā, B̄) = min

A′∈E(A)d(A′, B) = d(A,B).

�


We have seen many examples coming from the quotient of R2. Here we list twomore examples. In each example, we define an equivalence relation on some spacevia certain motion, then define the quotient distance as usual.

• Cone: let φ be the rotation by 90 degrees around the origin. Note that φpreserves the distance and has order 4, that is φ4 is the identity map. Wedefine A ∼ B if A = φk(B) for some integer k. Note that the quotientspace has a cone tip, which is not locally identical to a ball in R2. Also,this cone tip corresponds to the origin in the plane.

• Closed half-plane: We know this is a subset of R2, but it can also berealized as a quotient. Let φ be reflection about the x-axis. φ has order2. We define A ∼ B if A = φk(B) for some integer k (in fact, k = 0 or 1).The quotient space has a “boundary part”; any point on the boundary isnot locally identical to a ball in R2.

Besides constructing new geometries from R2, we can apply a similar idea tothe sphere S2.

Real projective plane RP 2: We define φ : S2 → S2 by sending (x, y, z) to(−x,−y,−z). φ has order 2 and does not have fixed points on the sphere. Forpoints on the sphere, we define A ∼ B if A = φk(B) for some k (in other words,A = B or A = φ(B)). The quotient space (S2/ ∼, d̄) is called the real projectivespace.

Proposition 1.3.8. RP 2 is locally spherical.

Sketch. For a point A in the sphere, we need to find a region R around Asuch that the following holds: for any B ∈ R, we have d̄(Ā, B̄) = dS(A,B). Notethat by definition,

d̄(Ā, B̄) = min{dS(A,B), dS(φ(A), B)}.Therefore, we only need to make sure that B is closer to A than to φ(A). Let Cbe the great circle with equal distance to A and φ(A). We can set R to be regioncontaining A bounded by this great circle C. Then

dS(A,B) = d̄(Ā, B̄)

for any B ∈ R. �

CHAPTER 2

Basic Concepts

2.1. Geometries

Definition 2.1.1. Let X be a nonempty set. A distance function (or a metric) onX is a function d : X ×X → [0,∞) satisfying(1) d(x, y) = 0 if and only if x = y;(2) d(x, y) = d(y, x) for all x, y ∈ X;(3) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X.We call (X, d) a metric space.

Examples 2.1.2.• (Euclidean metric) Standard distance function d0 on R, or Rn.• Sphere with the distance defined as the length of the shortest curve.• Quotient distance that we have seen in previous lectures.• For a metric space (X, d) and a subset S ⊂ X, we can restrict d on S to obtaina metric on S.• (Discrete metric) Define d on a set X by setting d(x, y) = 1 if x 6= y, d(x, y) = 0if x = y.• (Manhattan metric) d on R2 defined as

d((x1, y1), (x2, y2)) = |x1 − x2|+ |y1 − y2|.• (C0-metric) We denote C[0, 1] as the set of all continuous functions on [0, 1].Define d on C[0, 1] by

d(f, g) = maxx∈[0,1]

|f(x)− g(x)|.

• Define a non-Euclidean metric d on R+ = (0,∞) by settingd(x, y) = | ln(x/y)|.

As a practice, we show that the last d above is indeed a distance function:(1) | ln(x/y)| = 0 if and only if x/y = 1 if and only if x = y.(2) ln(x/y) and ln(y/x) only differs by a sign. Thus d(x, y) = d(y, x).(3) d(x, y) = | ln(x/y)| = | lnx− ln y| ≤ | lnx− ln z|+ | ln z − ln y|

= | ln(x/z)|+ | ln(z/y)| = d(x, z) + d(z, y).

Definition 2.1.3. A curve (or a path) on (X, d) is a continuous map γ : I → X,where I is an interval of R.

We do not need the definition of continuity anyway, so you can just use yourintuition. If you are curious, here is the rigorous definition.

Definition 2.1.4. A map γ : I → (X, d) is continuous, if for any a ∈ I and for any� > 0, there is δ > 0 such that

d(γ(a), γ(t)) ≤ �

13

14 2. BASIC CONCEPTS

for all t with |t− a| ≤ δ.

Whether a map is continuous depends on d. For example γ(t) = t, t ∈ [0, 1] isnot continuous in R with the discrete metric.

Definition 2.1.5. Let γ : [a, b]→ (X, d) be a curve on (X, d). Let π be a partitionof [a, b], that is, π = {t0 = a, t1, ..., tk, ...tn = b}. We define the length of γ as

lim||π||→0

n−1∑i=0

d(γ(ti), γ(ti+1)),

(if the limit exists), where ||π|| = maxi |ti+1 − ti|.

The length of a path could be infinity in general. As a direct consequence ofthe triangle inequality, the distance between two endpoints is always no greaterthan the length of γ (As an exercise, you can have a try to prove this).

Definition 2.1.6. (X, d) is called a length metric space, if for any pair of pointsx, y ∈ X, there is a curve γ connecting them with

d(x, y) = Length(γ).

We call this γ a minimal geodesic between x and y.

Some literature may have a different name for Definition 2.1.6, for example,geodesic space, intrinsic metric space.

Example 2.1.7.• Sphere with the metric defined as the length of shortest curves.• Any convex subset of (Rn, d0) is a length metric space.• R2 − {0} is not a length metric space.

As a practice, we show that the last d in Examples 2.1.2 is a length metricspace. For x, y ∈ R, without lose of generality, we may assume that x < y. Weconsider the path

γ(t) = x+ t(y − x), t ∈ [0, 1].Let π = {t0 = 0, t1, ..., tk, ...tn = 1} be a partition of [0, 1] and let pi = γ(ti). Notethat pi < pi+1. Then

n−1∑i=0

d(pi, pi+1) =

n−1∑i=0

| ln(pi/pi+1)| =n−1∑i=0

(ln pi+1 − ln pi) = ln pn − ln p0 = d(x, y).

Since this is true for any partition π, we conclude that Length(γ) = d(x, y).

Definitions 2.1.8. Let (X, d) be a length metric space and let I be an interval.A curve γ : I → X is a geodesic if it is locally minimizing, that is, for any t ∈ I,there is � > 0 such that Length(γ|[t−�,t+�]) = d(γ(t− �), γ(t+ �)).

Examples 2.1.9. The followings are some examples of geodesics.• A great circles in the sphere.• A helix in the cylinder.

Definition 2.1.10. Let (X, dX) and (Y, dY ) be two metric spaces. An isometrybetween X and Y is a bijection f : X → Y such that

dX(x1, x2) = dY (f(x1), f(x2))

for all x1, x2 ∈ X. If there is such a map f , we say that X and Y are isometric.

2.1. GEOMETRIES 15

If two metric spaces are isometric, we regard them as the same thing. Thisprinciple applies to length.

Proposition 2.1.11. Let f : X → Y be an isometry between two metric spaces.Let γ : [a, b]→ X be a curve in X. Then length(γ) = length(f ◦ γ).

Proof. We denote π = {t0 = a, t1, ..., tk, ...tn = b} as a partition of [a, b].

length(f ◦ γ) = lim||π||→0

n−1∑i=0

dY (f ◦ γ(ti), f ◦ γ(ti+1))

= lim||π||→0

n−1∑i=0

dX(γ(ti), γ(ti+1))

= length(γ).

�

Definition 2.1.12. (Open) Metric balls. For x ∈ X and r > 0, we define

Br(x) = {y ∈ X | d(x, y) < r}.

Definition 2.1.13. Let (X, d) be a length metric space. We say that (X, d) is2-dimensional locally Euclidean, if there is r > 0 such that for any x ∈ X, Br(x) isisometric to the r-ball in the Euclidean plane (R2, d0).

Similarly, we can define 2-dimensional locally spherical spaces.

Remark 2.1.14. r > 0 is uniform (works for all x ∈ X) in the above definition. Ifwe do not require this, then any open subset of R2 would satisfies the property.

Recall that locally Euclidean spaces can be constructed from an equivalencerelation in R2. In these cases, we have seen that the quotient map p : R2 → R2/ ∼gives an isometry from Br(A) to Br(p(A)) for r > 0 small and any A ∈ R2.

Proposition 2.1.15. Let (R2/ ∼, d̄) is as above. Let γ be a straight line in R2.Then(1) p ◦ γ is a geodesic in R2/ ∼.(2) length(γ|I) = length((p ◦ γ)|I) for any closed and bounded interval I ⊂ R.

Proof. (1) Reparametrize γ is necessary, we assume that it has unit speed,that is

d(γ(s), γ(t)) = |t− s|for all t, s ∈ R. (For example, you can write γ(t) = v + tw, where v ∈ R2 and w isa unit vector.)

Let r > 0 such that p : Br(A) → Br(p(A)) is an isometry for all A ∈ R2. Wepick � = r/2. For any t ∈ R, we have

length(p ◦ γ|[t−�,t+�]) = length(γ|[t−�,t+�])= d(γ(t− �), γ(t+ �))= d̄(p ◦ γ(t− �), p ◦ γ(t+ �)).

The second equality above holds because γ|[t− �, t+ �] has image in Br(γ(t)):

d(γ(s), γ(t)) = |t− s| ≤ � < r


for all t ∈ [t− �, t + �]. This shows that p ◦ γ|[t−�,t+�] is a minimal geodesic for allt. Thus p ◦ γ is a geodesic.

(2) We again assume that γ is of unit speed. Let π = {t0, ..., ti, ..., tn} be apartition of I. When ||π|| < r, we have

d(γ(ti), γ(ti+1)) = |ti − ti+1| < r.

Since p is an isometry from Br(γ(ti)) to its image, we see that

d(γ(ti), γ(ti+1)) = d(p ◦ γ(ti), p ◦ γ(ti+1)).

It follows that

length(p ◦ γ|I) = lim||π||→0

n−1∑i=0

d(p ◦ γ(ti), p ◦ γ(ti+1))

= lim||π||→0

n−1∑i=0

d(γ(ti), γ(ti+1))

= length(γ|I).

�

Corollary 2.1.16. Let (R2/ ∼, d̄) is as above. Then (R2/ ∼, d̄) is a length metricspace.

Proof. Let Ā, B̄ ∈ R2/ ∼. We know that we can find A′ ∈ E(A) and B′ ∈E(B) such that

d̄(Ā, B̄) = d(A′, B′).

Let γ be a segment connecting A′ and B′. Then

d(A′, B′) = length(γ) = length(p ◦ γ).

Note that p ◦ γ is a curve between Ā and B̄. Thus p ◦ γ is a minimal geodesic. Thisshows that (R2/ ∼, d̄) is a length metric space. �

Together with the exercise below, (R2./ ∼, d̄) is 2-dimensional locally Euclidean.

Exercise 2.1.17. Show that the quotient map p : Br(A) → Br(p(A)) is bijectivefor r > 0 small.

We can establish corresponding for the real projective plane (S2/ ∼, d̄) as well.

Question 2.1.18. Can we classify (have a list) of all 2-dimensional locally Eu-clidean / spherical spaces?

2.2. Groups

Definition 2.2.1. Let G be a set. A multiplication of G is a map (operation)· : G×G→ G, which we usually write ·(a, b) as a · b, satisfying(1) there is an element e ∈ G such that e · a = a · e = a for any a ∈ G;(2) for each a ∈ G, there is an element a−1 such that a · a−1 = a−1 · a = e;(3) (a · b) · c = a · (b · c) for all a, b, c ∈ G.We call e the identity element, a−1 the inverse of a, and (G, ·) a group.

2.2. GROUPS 17

Examples 2.2.2.• (Z,+). The identity element is 0. The inverse of a ∈ Z is −a.• (R+, ·), where · is the usual multiplication. The identity element is 1. The inverseof a ∈ R+ is 1/a.• (Zp,+), the so called modulus that you learned in Math 8. Zp = {0̄, 1̄, ..., p− 1}.+ is defined as follows: for ā, b̄ ∈ Zp, there is a unique element c ∈ {0, 1, ..., p− 1}such that a+ b ≡ c(mod p), then we define ā+ b̄ as c̄. For instance, in Z4, we have2̄ + 2̄ = 0̄, 2̄ + 3̄ = 1̄.• The set of all translations in R2 along x−axis by a unit 2kπ, where k ∈ Z. Thecomposition of motions gives the multiplication.• Let φ be the rotation around the origin by 90 degrees in R2. Denote id as theidentity map of R2. Then {id, φ, φ2, φ3} forms a group with multiplication as thecomposition. Note that φ4 = id.• Permutation group Sn is the set of all bijections from {1, 2, ..., n} to itself.• The circle group (S1, ·). We treat S1 as the unit circle in the complex plane.In this way, we can write each element as eiθ. · is the multiplication of complexnumbers, that is, eiθ1 · eiθ2 = ei(θ1+θ2).• General linear group GLn(R): the set of all n × n real matrices with non-zerodeterminant. The multiplication is the usual multiplication of two matrices.• Special linear group SLn(R): the set of all n×n real matrices A with det(A) = 1.This is a group: if det(A) = det(B) = 1, then det(AB) = det(A)det(B) = 1.• Orthogonal group O(n): the set of all n × n orthogonal matrices A (, that is,AAT = In, where A

T is the transport of A and In is the n×n identity matrix). Asan exercise, have a try to prove that O(n) is indeed a group.• Special linear group SO(n) = O(n) ∩ SLn(R).

Definition 2.2.3. Let S be a subset of a group G. We say that S generated G,if any element of G can be written as a product of elements in S or S−1, whereS−1 = {a−1|a ∈ S}.

Examples 2.2.4.• (Z,+) is generated by a single element {1}.• (Zp,+) is generated by {1̄}.• In the fourth example of Examples 2.2, the translation by 2π generates the group.• In the fifth example of Examples 2.2, φ generates the group.• Sn can be generated by transpositions.• The circle group can be generated by a small neighborhood around e.• As a (nontrivial) fact, the set of reflections generates O(n).

Definition 2.2.5. Let (G, ·) be a group. We say that a subset H of G is a subgroupif (H, ·) is a group, or equivalently, H is closed under multiplication and takinginverses. This is usually denoted as H ≤ G.

Examples 2.2.6.• (Z,+) ≤ (Q,+) ≤ (R,+).• (S1, ·) ≤ (C− {0}, ·).• SO(n) ≤ SLn(R) ≤ GLn(R).

Definition 2.2.7. Let G and H be two groups. A group homomorphism is a mapψ : G→ H such that

ψ(g1 · g2) = ψ(g1) · ψ(g2)


for all g1, g2 ∈ G.We say that a group homomorphism ψ : G → H is a group isomorphism if it

is bijective.

Examples 2.2.8.• The determinant det : GLn(R)→ (R−{0}, ·) is a group homomorphism because

det(AB) = det(A)det(B).

• In the fourth example of Examples 2.2.2, the group is isomorphic to Z. Theisomorphism can be realized by sending the translation by 2π to 1 ∈ Z.• In the fifth example of Examples 2.2.2, the group is isomorphic to Z4. Theisomorphism sends φ to {1̄} ∈ Z4.

Proposition 2.2.9. Let (X, d) be a metric space. Then the set of all isometriesof (X, d) forms a group with multiplication as composition of maps. This group isdenoted as Isom(X).

Proof. The identity element is the identity map of X. We show that Isom(X)is closed under taking inverses and multiplications.

Let F ∈ Isom(X). Since F is bijective, it has an inverse F−1. Clearly F−1 isalso a bijection. We show that F−1 preserves the metric:

d(F−1(x), F−1(y)) = d(F (F−1(x)), F (F−1(y))) = d(x, y).

Let F1, F2 ∈ Isom(X). We show that F1 ◦ F2 ∈ Isom(X):d((F1 ◦ F2)(x), (F1 ◦ F2)(y)) = d(F2(x), F2(y)) = d(x, y).

�

Remark 2.2.10. If two spaces are isometric, then their isometry groups are iso-morphic. In principle, this is true because isometry group comes from the metric,since the metrics are the same, then the isometry groups should be the same as well.To prove this formally, let F : (X, dX)→ (Y, dY ) be an isometry. For g ∈ Isom(X),its conjugation under F , that is, F ◦g◦F−1, is an isometry of (Y, dY ). This providesa map

CF : Isom(X)→ Isom(Y ), g 7→ F ◦ g ◦ F−1.

Exercise 2.2.11. Prove that CF is indeed a group isomorphism.

2.3. Group actions on Geometries

Definition 2.3.1. Let X be a set and G be a group. A (left) G-action on X is amap µ : G×X → X satisfying(1) µ(e, x) = x for any x ∈ X;(2) µ(gh, x) = µ(g, µ(h, x)) for all g, h ∈ G and x ∈ X.When the map µ is clear, we usually write µ(g, x) as g(x) or g · x. With thisnotation, condition (2) can be written as (gh) · x = g · (h · x).

Examples 2.3.2.• The permutation group Sn acts on the set {1, 2, ..., n} by, well, permutations.• Z acts on R2 by translations: µ(k, (x, y)) = (x+ 2kπ, y).• Z acts on R2 by glide reflections: µ(k, (x, y)) = (x+ 2kπ, (−1)ky).• S1 acts on R2 by rotations around the origin:

µ(θ, (x, y)) = (x cos θ + y sin θ,−x sin θ + y cos θ).

2.3. GROUP ACTIONS ON GEOMETRIES 19

• A group acts on itself by left translations: µ(g, h) = g · h.• A group acts on itself by conjugations: µ(g, h) = ghg−1. As an exercise, provethis is a group action.

Exercise 2.3.3. Let H be the upper half complex plane H = {z ∈ C | Im(z) > 0}.Let SL2(R) be the special linear group of R2. We define a map

µ : SL2(R)×H→ C,([a bc d

], z

)7→ az + b

cz + d.

Show that(1) µ has image in H.(2) µ defines a SL2(R)-action on H.

Definition 2.3.4. Let X be a G-space and let x ∈ X. The G−orbit at x is definedas

G · x = {g · x|g ∈ G}.

As an exercise, think about what is the G-orbit in Examples 2.3.2.We define an equivalence relation on X: x1 ∼ x2, if x1 and x2 belong to the

same G−orbit. In other words, there is g ∈ G such that g · x1 = x2.

Proposition 2.3.5. ∼ in an equivalence relation on X.

Proof. Transitivity is the only non-trivial part. Suppose that x1 ∼ x2 andx2 ∼ x3. By definition, this means that there is g1 and g2 in G such that

x2 = g1 · x1, x3 = g2 · x2.

Then

x3 = g2 · x2 = g2 · (g1 · x1) = (g2g1) · x1.�

The quotient space with respect the above equivalence relation is called theorbit space, and is denoted as X/G. We usually write x̄ as an element in X/G.

Definition 2.3.6. Let (X, d) be a metric space. We say a G-action on X is iso-metric (or G acts on X by isometries), if

Ag : X → X, x→ g · x

is an isometry for each g ∈ G.

Let (X, d) be a length metric space with an isometric G-action. We want todefine the quotient metric on X/G to be minimal between two orbits, that is,

d̄(x̄, ȳ) = ming1,g2∈G

d(g1 · x, g2 · y)

Since G acts by isometries, RHS equals to

ming1,g2∈G

d(x, g−11 g2 · y) = minh∈G

d(x, h · y).

Ideally, we wish that this minimal can be achieved. However, this is not alwaystrue. For example, consider Q-action by (R, d0) by translations: µ(q, x) = q + x.Since the orbit Q·0 is dense in R, for any x ∈ R, the minimal (infimum) of d(q ·0, x),which is 0, cannot be attained.

Motivated by the cylinder example, we define


Definition 2.3.7. Let (X, d) be a metric space. We say that a subset S ⊂ X isdiscrete in X, if for any s ∈ S, there is r > 0 such that Br(s) ∩ S = {s}.

Let G be an isometric action on (X, d). We say that G-action is discrete, if theorbit G · x is discrete for all x ∈ X.

Example 2.3.8. The set {1/n|n ∈ Z} is not discrete in (R, d0).

For an orbit G · x to be discrete, we can indeed use a weaker condition.

Lemma 2.3.9. Let (X, dX) be a G-space and let x ∈ X. If there is s ∈ G · x andr > 0 such that Br(s) ∩ (G · x) = {s}, then G · x is discrete.

Proof. By assumption, there is s ∈ G ·x and r > 0 such that Br(s)∩ (G ·x) ={s}. We show that this r > 0 also works for other points in the orbit, that is,Br(g · s) ∩ (G · x) = {g · s} for any g ∈ G.

Suppose the contrary, then there are a distinct orbit point h · s 6= g · s lying inBr(g · s). We consider the point (g−1h) · s. This point is not s; otherwise (g−1h) ·xwould give h · x = g · x. Also,

d((g−1h) · s, s) = d(h · s, g · s) < r,

thus (g−1h) · s ∈ Br(s). This contradicts our choice of r. �

Other than requiring G-action to be discrete, we need another reasonable con-dition to make sure the minimum can be obtained. The Definition 2.3.10 andProposition 2.3.11 below are not required for this course.

Definition 2.3.10. We say that a metric space (X, d) is locally compact, if foreach x ∈ X, there is r > 0 such that the local ball Br(x) satisfies that any sequencein Br(x) has a convergent subsequence with a limit in X.

Proposition 2.3.11. Let (X, d) be a locally compact metric space with a discreteisometric G-action. For two non-equivalent points x and y, minh∈Gd(x, h · y) canbe achieved and is positive.

Proof. We can always take the infimum. Let m = infh∈G d(x, h · y). There isa sequence of point pi = hi · y such that d(x, pi) → m as i → ∞. The tail of thissequence lies in B2m(x). By assumption, {pi} has a convergent subsequence. Thiscontradicts to the fact that G · y is discrete. �

For spaces like Rn, which is locally compact, there is a easier way to prove.

Lemma 2.3.12. Let G be an isometric action on (Rn, d0). Suppose that G-actionis discrete, then for any x, y ∈ X and any R > 0, BR(x) ∩ (G · y) is a finite set.

Proof. Let r > 0 such that points of Br(y)∩ (G · y) = {y}. By Lemma 2.3.9,Br(s) ∩ (G · y) = {s} for any s ∈ G · y. We claim that

Br/2(s1) ∩Br/2(s2) = ∅,

for all s1 6= s2 ∈ G · y. In fact, suppose that z ∈ Br/2(s1) ∩ Br/2(s2), then bytriangle inequality, we have

d(s1, s2) ≤ d(s1, z) + d(z, s2) < r/2 + r/2 = r.

Thus s2 ∈ Br(s1) ∩ (G · y), a contradiction. This verifies the claim.

2.3. GROUP ACTIONS ON GEOMETRIES 21

Note that if s ∈ BR(x), then Br/2(s) ⊂ BR+r/2(x) by triangle inequality.Thusfamily of balls {Br/2(s)}s∈BR(x)∩(G·y) are all contained in BR+r/2(x) and pairwisedisjoint. We have

vol(BR+r/2)(x) ≥∑

s∈BR(x)∩(G·y)

vol(Br/2(s)) = |BR(x) ∩ (G · y)| · vol(Br/2(0)).

Thus BR(x) ∩ (G · y) has cardinality

|BR(x) ∩ (G · y)| ≤vol(BR+r/2(0))

vol(Br/2(0))=

(R+ r/2)n

(r/2)n 0 guarantees than the first conditionin Definition 1.1 is fulfilled. The second condition clearly holds. We check thetriangle inequality, which is indeed similar to the cylinder case.

For x, y, z ∈ X, we choose z0 ∈ G · z such that d̄(x̄, z̄) = d(x, z0). Then wechoose y0 ∈ G · y such that d̄(z̄, ȳ) = d(z0, y0). We have

d̄(x̄, ȳ) ≤ d(x, y0) ≤ d(x, z0) + d(z0, y0) = d̄(x̄, z̄) + d̄(z̄, ȳ).We need a lemma before proving (2). �

Lemma 2.3.14. Let p : (X, d) → (X/G, d̄), x 7→ x̄ be the quotient map. Letγ : I → X be a path of finite length in X. Then Length(p ◦ γ) ≤ Length(γ).

Proof. This follows from the definition of d̄ that p is distance non-increasing:d̄(x̄, ȳ) ≤ d(x, y) for all x, y ∈ X. Let π = {t0, ..., tk, ...tn} be any partition of I.Then

Length(p ◦ γ) = lim||π||→0

n−1∑i=0

d̄(p ◦ γ(ti), p ◦ γ(ti+1))

≤ lim||π||→0

n−1∑i=0

d(γ(ti), γ(ti+1))

= Length(γ).

�

Proof of Proposition 2.3.13(2). Given x̄, ȳ ∈ X/G, we need to find a pathbetween them whose length equals to d̄(x̄, ȳ). Let x, y ∈ X such that d(x, y) =d̄(x̄, ȳ). Because (X, d) is a length metric space, there is a path γ from x to y withLength(γ) = d(x, y). By the previous lemma,

Length(p ◦ γ) ≤ Length(γ) = d(x, y) = d̄(x̄, ȳ).On the other hand, because p ◦γ connects x̄ and ȳ, we have Length(p ◦γ) ≥ d̄(x̄, ȳ)by triangle inequality. Hence Length(p ◦ γ) = d̄(x̄, ȳ). �


Definition 2.3.15. We say that G-action is free, if for any g 6= e and any x ∈ X,g · x 6= x holds.

In other words, a free G-action means any element g ∈ G− {e} does not havefixed points on X.

Lemma 2.3.16. Let G be a subgroup of Isom(Rn). If G-action on Rn is free anddiscrete, then there is D > 0 such that d(g · x, x) ≥ D for all g 6= e ∈ G and allx ∈ X.

We will prove this lemma later for R2 when we have a better understanding onisometries of R2. Assuming this at the moment, we prove

Theorem 2.3.17. Let G be a subgroup of Isom(Rn). If G-action on Rn is free anddiscrete, then the quotient space (X/G, d̄) is n-dimensional locally Euclidean.

Proof. Let D > 0 be the constant in Lemma 2.3.16. We claim that if d(x, y) <D/2, then d̄(x̄, ȳ) = d(x, y). In fact, for any x′ ∈ G · x that is not x, by triangleinequality, we have

d(x′, y) ≥ d(x′, x)− d(x, y) ≥ D −D/2 = D/2 > d(x, y).This shows that

d(x, y) = minx′∈G·x

d(x′, y) = d̄(x̄, ȳ).

For any y, z ∈ BD/4(x), by triangle inequalityd(y, z) ≤ d(y, x) + d(x, z) < D/4 +D/4 = D/2.

It follows from the claim that d̄(ȳ, z̄) = d(y, z). This shows that the quotient mapp : X → X/G, when restricting it to BD/4(x), preserves the metric.

Next we show that p|BD/4(x) is a bijection between BD/4(x) and BD/4(x̄); thiswould finish the proof that BD/4(x̄) is isometric to D/4-ball in Rn.

p|BD/4(x) is injective: Suppose that y1, y2 ∈ BD/4(x) maps to the same pointthrough p. This means that y1 and y2 lie in the same orbit with d(y1, y2) < D/2,which contradicts our choice of D.

p|BD/4(x) is onto BD/4(x̄): For any ȳ with d̄(x̄, ȳ) < D/4. Recall that d̄(x̄, ȳ) isthe minimal between x and any point equivalent to y. Therefore, there is a pointy′ such that p(y′) = ȳ and d(x, y′) = d̄(x̄, ȳ) < D/4, that is, y′ ∈ BD/4(x). �

CHAPTER 3

Free and discrete isometric actions on R2 or S2

3.1. Hello again, linear algebra

In this section, we make use of linear algebra to prove the following:

Theorem 3.1.1. Every element g of Isom(Rn) can be uniquely expressed in theform of g(x) = Ax+ v, where A ∈ O(n) and v ∈ Rn.

Lemma 3.1.2. Every element g of Isom(Rn) can be uniquely written as a compo-sition T ◦ h, where T is a translation and h is an isometry fixing the origin.

Proof. Let v = g(0) ∈ Rn. If v = 0, then g itself is an isometry fixing theorigin and the statement holds clearly. We assume that v 6= 0. We define isometries

T : Rn → Rn, x 7→ x+ v;

h : Rn → Rn, x 7→ g(x)− v.h is an isometry fixing the origin because it is the composition of two isometriesand

h(0) = g(0)− v = v − v = 0.Note that

T ◦ h(x) = T (g(x)− v) = g(x)− v + v = g(x).The uniqueness part is left as an exercise. �

Lemma 3.1.3. Let h : Rn → Rn be a function. The followings are equivalent.(1) h is an isometry fixing the origin.(2) h preserves the dot product: h(x) · h(y) = x · y for all x, y ∈ Rn.(3) h is linear and its matrix representation is orthogonal.

Proof. First recall that the dot product and the distance is related by

d(x, y)2 = (x− y) · (x− y).We write x · x = ||x||2 below for simplicity.

(1)⇒(2): Suppose that h is an isometry. This means that||h(x)− h(y)|| = d(h(x), h(y)) = d(x, y) = ||x− y||

for any x, y ∈ Rn. Squaring both sides, we have(h(x)− h(y)) · (h(x)− h(y)) = (x− y) · (x− y)

h(x) · h(x)− 2h(x) · h(y) + h(y) · h(y) = x · x− 2x · y + y · y||h(x)||2 − 2h(x) · h(y) + ||h(y)||2 = ||x||2 − 2x · y + ||y||2.

Together with

||h(x)|| = d(h(x), 0) = d(h(x), h(0)) = d(x, 0) = ||x||,

23

24 3. FREE AND DISCRETE ISOMETRIC ACTIONS ON R2 OR S2

it follows that

h(x) · h(y) = x · y.We also need to show that h(0) = 0:

||h(0)||2 = h(0) · h(0) = 0 · 0 = 0.

(2)⇒(1): Suppose that h preserves the dot product. Then for any x, y ∈ Rn,

d(h(x), h(y))2 = (h(x)− h(y)) · (h(x)− h(y))= h(x) · h(x)− 2h(x) · h(y) + h(y) · h(y)= x · x− 2x · y + y · y= (x− y) · (x− y)= d(x, y)2

(2)⇒(3): We first show that h is linear. This means that we need to show that

h(ax+ by) = ah(x) + bh(y)

for any x, y ∈ Rn and any a, b ∈ R. Let {e1, e2, ..., en} be a standard basis of Rn.Since h preserves the dot product, {h(e1), ..., h(en)} is an orthonormal basis of Rn.For each j = 1, ..., n, we have

h(ax+ by) · h(ej) = (ax+ by) · ej = a(x · ej) + b(y · ej)= a(h(x) · h(ej)) + b(h(y) · h(ej))= (ah(x) + bh(y)) · h(ej).

Then h(ax + by) = ah(x) + bh(y) follows from the fact that two vector in Rnare equal if and only if they have the same dot product with each element in anorthonormal basis.

Next we show that the matrix representation of h (under the standard basis)is orthogonal. Let A be its matrix representation, that is, h(x) = Ax, where Axis understood as matrix multiplication. For all x, y ∈ Rn as column vectors (underthe standard basis), we have

x · y = h(x) · h(y) = Ax ·Ay = (Ax)T (Ay) = xT (ATA)y.

For x = ei and y = ej , the term xT (ATA)y is the (i, j)-th entry of the matrix ATA.

Hence the (i, j)-th entry of ATA should be ei · ej = δi,j . This shows that ATA isthe identity matrix, that is, A is orthogonal.

(3)⇒(2): Let x, y ∈ Rn. We regard x, y as column vectors under the standardbasis. Then the dot product x · y equals multiplication xT y. Since h : x 7→ Ax,where A is orthogonal, we have

h(x) · h(y) = Ax ·Ay = (Ax)T (Ay) = xT (ATA)y = xT y = x · y.

�

Theorem 3.1.1 follows directly from Lemmas 3.1.2 and 3.1.3.

Proof of Theorem 3.1.1. Let g ∈ Isom(Rn). By Lemma 3.1.2, we knowthat g can be written as

g : x 7→ h(x) + v,

3.2. ELEMENTS OF Isom(R2) WITHOUT FIXED POINTS 25

where h is an isometry fixing the origin and v ∈ Rn. Applying Lemma 3.1.3, h islinear and its matrix representation A is orthogonal. Thus

g(x) = h(x) + v = Ax+ v.

�

From Theorem 3.1.1, we know that every element g of Isom(Rn) is in the formof g(x) = Ax+ v, where A ∈ O(n) and v ∈ Rn. We write this element g as (A, v).By calculating the composition of (A, v) and (B,w), it is easy to get a formula forgroup multiplication:

(A, v) · (B,w)(x) = (A, v)(Bx+ w) = A(Bx+ w) + v= ABx+Aw + v = (AB,Aw + v)(x).

Thus(A, v) · (B,w) = (AB,Aw + v).

Using this rule for multiplication, we can also find the inverse of (A, v):

(A, v)−1 = (A−1,−A−1v).In fact, we can check that

(A, v) · (A−1,−A−1v) = (AA−1, A(−A−1v) + v) = (I, 0),(A−1,−A−1v) · (A, v) = (A−1A,A−1v + (−A−1v)) = (I, 0).

3.2. Elements of Isom(R2) without fixed points

Now we focus on R2. Recall that we are looking for isometries that does nothave fixed points.

Proposition 3.2.1. If g ∈ Isom(R2) does not have any fixed points, then g is atranslation or a glide reflection.

Proof. We write g = (A, v). Since it has no fixed points, the equation Ax+v =x has no solutions, that is, (A− I)x = −v has no solution. By basic linear algebra,we conclude det(A − I) = 0. In other words, A has an eigenvalue 1. Recall thatA ∈ O(2) and thus det(A) = ±1. Hence the product of two eigenvalues should be1 or −1. There are only two cases:Case 1: The other eigenvalue is also 1. Because A ∈ O(2), A must be the identitymatrix in this case (this step is left as an exercise). Then g = (I, v) is a translation.Case 2: The other eigenvalue is −1. Let e1 and e2 be the unit length eigenvectorwith eigenvalue 1 and −1, respectively. By linear algebra, these two vectors areorthogonal. Under the orthonormal basis {e1, e2}, we can rewrite g = (F,w), whereF is the diagonal matrix diag{1,−1}. F is indeed the reflection about the line Re1.With this, the fact that (F − I)x = w has no solutions translates to that(

0 00 −2

)·(x1x2

)=

(w1w2

)has no solutions. We conclude that w1 6= 0. Note that

g(x) = Fx+ w = Fx+ w1e1 + w2e2 = (Fx+ w2e2) + w1e1 = h(x) + w1e1,

where h(x) = Fx + w2e2 Because F reflects about Re1 and e2 is perpendicularto e1, the map h is indeed the reflection about the line L = {y = w2/2}. Thetranslation left w1e1 6= 0 is parallel to L. We move the coordinate up (along e2) by


w2/2. Then under the new coordinate, we can write g as (F,w1e1) with w1 6= 0,which is a glide reflection. �

Remark 3.2.2. With the proof above, we also showed that any glide reflection ofR2 can be written as (F,w1e1) under a suitable coordinate {e1, e2}.

3.3. Subgroups of Isom(R2)

With Proposition 3.2.1, we proceed to classify subgroups of Isom(R2), whoseactions are free and discrete. Let Γ be such a subgroup. By Proposition 3.2.1, weknow that any non-identity element of Γ is either a translation or a glide reflection.We define T be the set of all translations in Γ plus the identity element. It is clearthat T is a subgroup of Γ. We classify the types of Γ by how many directions thetranslations can move:• Type I. T = {e};• Type II. T consists of translations in one direction;• Type III. T contains at least two translations along two linearly independentvectors;Depending on whether Γ contains glide reflections, we further classify type II andIII, respectively, into:• Type IIa and IIIa: Γ = T , that is, Γ does not contain glide reflections;• Type IIb and IIIb: Γ contains glide reflections.

For Type I above, T = {e} implies Γ = {e}. In fact, suppose that Γ contains aglide reflection g = (F,w1e1), then

g2 = (F 2, F (w1e1) + w1e1) = (I, 2w1e1)

would be a translation in Γ, which contradicts to T = {e}.The following Lemma will be used when we study Type IIb and Type IIIb.

Lemma 3.3.1. Suppose that Γ contains a glide reflection g, then Γ = T ∪ Tg,where Tg = {t · g | t ∈ T}.

Proof. By Remark 3.2.2, we can write g = (F,w) under a suitable coordinate,where w is a multiple of e1. Then g

−1 = (F−1,−F−1w) = (F,−w). For any otherglide reflection g′ = (A, v) ∈ Γ, we show that the element g′g−1 ∈ Γ is a translationor identity. In fact,

g′g−1 = (A, v) · (F,−w) · (A, a) = (AF,−Aw + v).Because A has determinant −1 (by the proof of Proposition 3.2.1), AF has deter-minant 1. When g′g−1 is not identity, (AF,−Aw+ v) is a non-trivial element in Γ;in particular, this element does not have fixed points. As a result, AF must be theidentity matrix (again by the proof of Proposition 3.2.1). Thus g−1g′ = (I,−Aw+v)is a translation we denote as t. This shows g′ = t · g. In other words, any glidereflection of Γ is an element of Tg. Because Γ only contains translations and glidereflections, we see that Γ = T ∪ Tg. �

3.3.1. Type II. Recall that type II means that T , all translations of Γ, are inthe same direction. Let e1 be the unit vector parallel to this direction, then anytranslation of Γ is in form of (I, ce1) for some c ∈ R. For type IIa, we have Γ = T .Using the fact that Γ-action is discrete, we prove the following:

Proposition 3.3.2. Suppose that Γ is of type IIa. Then under a suitable orthonor-mal coordinate {e1, e2}, there is a > 0 such that (I, ae1) generates Γ.

3.3. SUBGROUPS OF Isom(R2) 27

In general, if a set S generates a group Γ, we write Γ = 〈S〉. With this notation,Proposition 3.3.2 says

Γ = 〈(I, ae1)〉 = {(I, kae1)|k ∈ Z}.

Proof. We know that any element of Γ are in the form of (I, ce1) for somec ∈ R. We choose a > 0 so that (I, ae1) is the shortest translation in Γ. In otherwords,

a = min{|c| : (I, ce1) ∈ Γ− {e}}.The minimum exists because there are only finitely many points in Br(0) that areorbit points of 0.

We claim that any element of Γ is a multiple of (I, ae1). Suppose that (I, be1) ∈Γ such that b/a is not an integer. We write b/a = k+r, where k ∈ Z and r ∈ (0, 1).Since (I,−kae1) ∈ Γ and Γ is a group, we conclude that

(I,−kae1) · (I, be1) = (I, rae1) ∈ Γ.This contradicts to the our choice of (I, ae1) as the minimal translation. �

Proposition 3.3.3. Suppose that Γ is of type IIb. Then under a suitable orthonor-mal coordinate {e1, e2}, there is a > 0 such that

T = 〈(I, ae1)〉, Γ = 〈(F, ae1/2)〉.

Proof. Under a suitable coordinate, we can write g as (F, be1). Note thatg2 = (I, 2be1) is a translation. Since all translation of Γ are in the same direction,by Proposition 3.3.2, there is a > 0such that the subgroup T can be generated bya single element (I, ae1). Thus there is an integer k such that ka = 2b.

We claim that k is odd. Suppose that k = 2l for some l ∈ Z. Then la = b.Note that (I,−lae1) is an element of Γ, thus

(I,−lae1) · g = (I,−lae1) · (F, lae1) = (F, 0) ∈ Γ.We reached a contradiction because Γ-action should be free. This verifies the claim.

Because k is odd, we write k = 2l + 1 for some l ∈ Z. So (2l + 1)a = 2b, thatis, b = (l + 1/2)a. We consider

h = (I,−lae1) · g = (I,−lae1) · (F, (l + 1/2)ae1) = (F, ae1/2),which is a glide reflection in Γ. h2 = (I, ae1) generates T . Also, Proposition 3.3.1assures that Γ = T ∪ Th. As a result, h generates Γ.

We check that Γ-action is free and discrete. It is clear that Γ-action is freebecause Γ = T ∪ Th consists of translations and glide reflections. It remains toshow it is discrete. For any x ∈ R2 and any (I, kae1) ∈ T − {e}, we estimate:

d(x, (I, kae1)(x)) = ||kae1|| = |ka| ≥ |a|.Denote p1 : R2 → Re1 as the projection map to the first component. Then for any(F, ae1/2 + kae1) ∈ Tg and any x ∈ R2,

d(x, (F, ae1/2 + kae1)(x)) = d(x, Fx+ ae1/2 + kae1)

≥ d(p1(x), p1(Fx) + p1(ae1/2 + kae1))= d(p1(x), p1(x) + ae1/2 + kae1)

≥ d(0, ae1/2 + kae1)= |a/2 + ka| ≥ |a|/2.


�

3.3.2. Type III.

Proposition 3.3.4. Suppose that Γ is of type IIIa. Then under a suitable or-thonormal coordinate {e1, e2}, there are a, b > 0 and a unit vector v that is linearlyindependent of e1 such that

Γ = 〈(I, ae1), (I, bv)〉 = {(I, kae1 + lbv)|k, l ∈ Z}.

Proof of Proposition 3.3.4. We start with one translation in Γ. Changingthe coordinate if necessary, this translation moves along e1 direction. We consider asubgroup H of Γ, which consists of identity and all translations along e1 direction.H is clearly a subgroup. By Proposition 3.3.2, H is generated by a single element(I, ae1). We will to find a second element (I, bv) so that they together generate Γ.

Let L = Re1 be the e1-axis. The orbit H ·0 lies in L. Let 0, ae1 be the segmentfrom 0 to ae1. We choose g ∈ Γ−H such that it is the closest orbit point to 0, ae1,that is,

d(g(0), 0, ae1) = ming′∈Γ−H

d(g′(0), 0, ae1).

We claim that g(0) is also the closest orbit point in (Γ−H) · 0 to the entire line L:

d(g(0), L) = ming′∈Γ−H

d(g′(0), L);

in particular, the minimal on the right hand side above exists. We argue this bycontradiction. Suppose that there is g′ ∈ Γ−H such that d(g′(0), L) < d(g(0), L).Let pe1 be the point on L with d(g

′(0), L) = d(g′(0), pe1). We write p = ka + r,where k ∈ Z and r ∈ [0, a). We consider g′′ = (I,−kae1) · g′ ∈ Γ−H. Then

d(g′(0), L) = d(g′(0), p) = d((I,−kae1) · g′(0), (I,−kae1) · p) = d(g′′(0), re1).

Therefore, we find g′′ ∈ Γ−H such that

d(g′′(0), 0, ae1) = d(g′′(0), re1) = d(g

′(0), L) < d(g(0), L) = d(g(0), 0, ae1).

This contradicts to our choice of g. Hence the claim is verified.As a translation, g can be written as (I, bv). (I, ae1) and (I, bv) together

generates a subgroup

{(I, kae1 + lbv)|k, l ∈ Z}.Its orbit at 0 gives a lattice:

L = {kae1 + lbv|k, l ∈ Z}.

Consider the lines that are parallel to e1 or v and goes through L. These lines dividethe plane into many parallelogram of equal size. Suppose that Γ has an elementα that not generated by (I, ae1) and (I, bv). Then α(0) is not on the lattice L.Note that α(0) cannot lie on horizontal line; otherwise, let kae1 + lbv be the orbitpoint that is left to α(0) on this line, then (I,−kae1− lbv) ·α(0) is a orbit point inL, whose distance to 0 is shorter than a, a contradiction to the choice of (I, ae1).Now we pick a parallelogram containing α(0) and let kae1 + lbv be the left-bottomvertex of this parallelogram. Then (I,−kae1 − lbv) · α(0) is an orbit point in theparallelogram with left-bottom vertex 0. This orbit point has distance to L beingshorter than d(bv, L), which contradicts to our choice of g = (I, bv). �

3.3. SUBGROUPS OF Isom(R2) 29

Proposition 3.3.5. Suppose that Γ is of type IIIb. Then under a suitable or-thonormal coordinate {e1, e2}, there are a, b > 0 such that

T = 〈(I, ae1), (I, be2)〉, Γ = 〈(F,1

2ae1), (I, be2)〉.

Proof. For any glide reflection in Γ, note that its square is a translation,which moves along a unit vector we call e1. Let (I, ae1) ∈ Γ be the translationof minimal length along e1. By the proof of Proposition 3.3.3, the glide reflectiong = (F, ae1/2) must be an element of Γ. Let T be the subgroup of Γ consists ofall translations. By the proof of Proposition 3.3.4, besides (I, ae1), we can findanother element (I, bv) so that these two elements together generates T .

We will show that v = e2, the unit vector perpendicular to e1. We write

bv = b1e1 + b2e2

for some b1, b2 ∈ R. Compositing (I, bv) by some element (I, kae1) if necessary, wecan assume that b1 ∈ [0, a). We consider the element

(F,−ae1/2) · (I, bv) · (F, ae1/2) = (F, ae1/2) · (F, ae1/2 + bv)= (I,−ae1/2 + F (ae1/2 + bv))= (I, F (bv)),

which is a translation in Γ. Hence

(I, bv) · (I, F (bv)) = (I, 2b1e1) ∈ T.

Recall that b1 ∈ [0, a) and any translation along e1 is in the form of (I, kae1) forsome k ∈ Z. As a result, we conclude that b1 = 0 or b1 = a/2. We need to rule outthe case b1 = a/2. If (I, ae1/2 + b2e2) ∈ Γ, then

(I,−ae1/2− b2e2) · (F, ae1/2) = (F,−b2e2) ∈ Γ.

Observe that (F,−b2e2) is a reflection, which has fixed points; this contradicts tothe assumption that Γ-action is free.

We summarize what we got so far: b1 = 0, g = (F, ae1/2) ∈ Γ, T is generatedby two elements: (I, ae1) and (I, be2). By Lemma 3.3.1, Γ = T ∪ Tg.

Finally, we check that Γ-action is free and discrete (this step confirms theexistence of a group of type IIIb). It is clear that Γ-action is free because Γ = T∪Tgonly consists of translations and glide reflections. It remains to show it is discrete.For any x ∈ R2 and any (I, v) ∈ T , we estimate:

d(x, (I, v)(x)) = ||v|| = ||kae1 + lbe2|| ≥ min{|a|, |b|}.

Denote p1 : R2 → Re1 as the projection map to the first component. Then for any(F, ae1/2 + v) ∈ Tg and any x ∈ R2,

d(x, (F, ae1/2 + v)(x)) = d(x, Fx+ ae1/2 + v)

≥ d(p1(x), p1(Fx) + p1(ae1/2) + p1(v))= d(p1(x), p1(x) + ae1/2 + p1(v))

≥ d(0, ae1/2 + p1(v))≥ |a|/2.

�


3.3.3. A small final step. Now we have classified all free and discrete isometricaction on R2. Recall that in Chapter 2, we postponed the proof of Lemma 2.3.14.The Lemma says that there is a uniform D > 0 such that d(gx, x) ≥ D for allg ∈ Γ − {e} and all x ∈ R2. We need this Lemma to assure that the quotientspace R2/Γ is locally Euclidean (see proof of Theorem 2.3.15). Now we prove thisLemma.

Inspecting the last paragraph of the proof of Proposition 3.3.5 and Proposition3.3.3 respectively, we already proved that for type IIb and IIIb. For instance,in type IIIb, we have d(gx, x) ≥ D for all g ∈ Γ − {e} and all x ∈ R2, whereD = min{|a|/2, |b|}. The remaining type IIa and IIIa are easier since they are justtranslations. To sum up, we can choose D > 0 for each type as below:• Type IIa: D = |a|;• Type IIb: D = |a|/2;• Type IIIa: D = min{|a|, |b|};• Type IIIb: D = min{|a|/2, |b|}.This completes the proof of Lemma 2.3.14 that we left.

3.3.4. Summary. Now we have classified all 2-dimensional locally Euclidean spacesthat coming from quotient spaces of free and discrete isometric actions on R2. Welist them as below.

Theorem 3.3.6. Let Γ be a subgroup of Isom(R2). Suppose that Γ-action on R2is free and discrete, then Γ is one of the followings:

Type Generators QuotientI identity PlaneIIa (I, ae1) CylinderIIb (F, ae1/2) Twisted cylinderIIIa (I, ae1), (I, bv) TorusIIIb (F, ae1/2), (I, be2) Klein bottle

The generators above are written under a suitable orthonormal coordinate {e1, e2}.

In the next chapter, we will prove that any 2-dimensional locally Euclideanspace can be realized as a quotient of a free and discrete isometric action on R2.

3.4. Subgroups of Isom(S2)

This section will be in the form of exercises, assigned through the course.We regard S2 as the unit sphere in R3 centered at the origin. The distance on

S2 is given as the length of shortest path on S2.Here we recall some facts from linear algebra that will be helpful.

• Any eigenvalue (real or complex) of any orthogonal matrix has norm 1.• Let A ∈ O(n). If V is an A-invariant linear subspace of Rn, then V ⊥, the

orthogonal complement of V , is also A-invariant.

Exercise 3.4.1. Show that there is a natural isomorphism between Isom(S2) andthe orthogonal group O(3).

3.4. SUBGROUPS OF Isom(S2) 31

Exercise 3.4.2. Let g ∈ Isom(S2). Show that under a suitable orthonormal basis,the matrix representation of g has the form±1 0 00 cos θ sin θ

0 − sin θ cos θ

Hint: First show that it must have a real eigenvalue.

Exercise 3.4.3. Let H be a subgroup of Isom(S2) that acts freely on S2. Show thatH is either the identity subgroup {I3} or {±I3}.

These exercises explore all 2-dimensional locally spherical spaces that comesfrom quotients of S2: they are either the sphere itself or S2/{±I3}. As introducedin Section 2.6, the quotient S2/{±I3} is the real projective space RP 2.

Similar to the locally Euclidean case, any 2-dimensional locally spherical spacesindeed arises from a quotient of S2. Together with the result in this section, thiswould fully classify 2-dimensional locally spherical spaces: they are S2 and RP 2.

A similar classification result can be achieved for even dimensional locally spher-ical spaces: they are the sphere S2n and the real projective space S2n/{±I2n+1}.Odd dimensional ones have much more examples:

Example 3.4.4. We consider S3 as the unit sphere in R4. We can identity R4with C2, by corresponding (x, y, u, v) ∈ R4 to (x + iy, u + iv) ∈ C2. We define aS1-action on C2:

S1 × C2 → C2, (eiθ, (z, w)) 7→ (eiθz, eiθw).This induces a free isometric S1-action on S3. For any p ∈ Z, we take a subgroupZp ≤ S1, then we have an isometric Zp-action on S3 that is free and discrete. Thecorresponding quotient space S3/Zp is locally spherical. When p = 2, it is the realprojective space.

Exercise 3.4.5. Let p, q be two coprime positive integers. We define a Zp-actionon C2 as follows. Let g be a generator of Zp,

g · (z, w) = (ei(2π/p)z, ei(2πq/p)w).This induces an isometric Zp-action os S3. Show that this Zp-action on S3 is free.

The corresponding quotient spaces in the above exercise are called lens spaces,denoted as L(p, q). By construction, they are also locally spherical.

CHAPTER 4

Coverings

This chapter is optional and will not be covered in lectures.Here is the main goal of this chapter:

Theorem 4.0.1. Let (X, d) be a 2-dimensional locally Euclidean space. Then thereis a subgroup Γ of Isom(R2), whose action on R2 is free and discrete, such that(X, d) is isometric to the quotient metric space R2/Γ.

Together with the classification of free and discrete isometric actions on Rn inthe last chapter, Theorem 4.0.1 fully classifies all 2-dimensional locally Euclideanspaces.

In Section 4.1, we introduce the notion of covering maps and study their prop-erties. In Section 4.2, we show that any 2-dimensional locally Euclidean space Xadmits a covering space R2. In Section 4.3, we define the covering group associatedto a covering map and finish the proof of Theorem 4.0.1.

4.1. Covering maps

Recall that for a free and discrete isometric Γ-action on R2, we know that thereis r > 0 such that the quotient map q provides an isometry from Br(x) to Br(q(x))for all x ∈ X. We extract this property as a definition.

Definition 4.1.1. Let (X1, d1) and (X2, d2) be two length metric spaces. We saythat a map φ : X1 → X2 is a covering map, if(1) φ is surjective, and(2) there is s > 0 such that for any a ∈ X1, the map

φ|Bs(a) : Bs(a)→ φ(Bs(a))

is an isometry onto Bs(φ(a)).We say that (X1, d1) is a covering space of (X2, d2).

Examples 4.1.2.• The quotient map R2 → R2/Γ, where Γ acts on R2 freely and discretely byisometries.• Wrapping a circle of circumference 2π twice over a circle of circumference π.• Wrapping R over a circle.

We mention that there is a more general definition of “covering maps”, whichdoes not require any metrics. Here for our purposes, we will focus on length metricspaces, actually 2-dim locally Euclidean spaces.

Proposition 4.1.3. Let φ : (X1, d1)→ (X2, d2) be a covering map. Let γ : R→ X1be a geodesic in X1, then φ ◦ γ is a geodesic in X2.

33

34 4. COVERINGS

Proof. Let s > 0 be the constant in (2) of Definition 4.1.1. Fix t ∈ R. Wechoose � > 0 such that(1) Length(γ|[t−�,t+�]) = d(γ(t− �), γ(t+ �)), and(2) d(γ(t− �), γ(t+ �)) < s.The the curve γ|[t−�,t+�] is contained in Bs(γ(t − �)). Because φ is an isometrywhen restricted to Bs(γ(t− �)). We have

Length(γ|[t−�,t+�]) = Length(φ ◦ γ|[t−�,t+�]),d(γ(t− �), γ(t+ �)) = d(φ ◦ γ(t− �), φ ◦ γ(t+ �)).

ThusLength(φ ◦ γ|[t−�,t+�]) = d(φ ◦ γ(t− �), φ ◦ γ(t+ �)).

�

Lemma 4.1.4. Let φ : (X1, d1) → (X2, d2) be a covering map between two lengthmetric spaces. Then

d2(φ(a), φ(b)) ≤ d1(a, b)for all a, b ∈ X1.

Proof. Let γ : I → X1 be a minimal geodesic between a and b. φ◦γ providesa curve between φ(a) and φ(b). We have

d2(φ(a), φ(b)) ≤ Length(φ ◦ γ) = lim||π||→0

∑i

d2(φ(γ(ti)), φ(γ(ti+1))),

where π is a partition of I. When ||π|| is sufficiently small, we haved1(γ(ti), γ(ti+1)) ≤ s

for all i, where s > 0 as in condition (2) of Definition 4.1.1. Applying the condition(2), we have

d2(φ(γ(ti)), φ(γ(ti+1))) = d1(γ(ti), γ(ti+1)).

Thusd2(φ(a), φ(b)) ≤ lim

||π||→0

∑i

d2(γ(ti), γ(ti+1)) = d1(a, b).

�

Lemma 4.1.5. Let φ : (X1, d1) → (X2, d2) be a covering map between two lengthmetric spaces. Suppose that φ is injective, then φ is an isometry.

Proof. By assumptions, φ is a bijection. We need to show that φ preservesthe distance. By Lemma 4.1.4, we have one side inequality

d2(φ(a), φ(b)) ≤ d1(a, b).To prove the other direction, we claim that the inverse φ−1 : X2 → X1 is also acovering map. Assuming this claim, then for φ(a) = x and φ(b) = y, we obtain

d1(a, b) = d1(φ−1(x), φ−1(y)) ≤ d2(x, y) = d2(φ(a), φ(b)).

This proves that φ is an isometry.It remains to verify the claim. We already knew that φ−1 is surjective. We

check condition (2) in Definition 4.1.1. Because φ is a covering map, there is s > 0such that φ maps to φ(Bs(a)) = Bs(φ(a)) as an isometry. For a point y ∈ X2, picka ∈ X with φ(a) = y. It is clear that φ−1|Bs(y) preserves the distance. Compositingφ−1 to both sides of φ(Bs(a)) = Bs(φ(a)), we have Bs(a) = φ

−1(Bs(φ(a))), thatis, Bs(φ

−1(y)) = φ−1(Bs(y)). This verifies condition (2). �

4.2. COVER A 2-DIMENSIONAL LOCALLY EUCLIDEAN SPACE BY R2 35

4.2. Cover a 2-dimensional locally Euclidean space by R2

Now we consider 2-dimensional locally Euclidean spaces, though all results canbe generalized to n-dimensional ones after some mild modifications. Our goal ofthis section is the Proposition below.

Proposition 4.2.1. Let X be a 2-dimensional locally Euclidean space. Then thereis a covering map φ : R2 → X.

In this section, we always assume that (X, d) is a 2-dimensional locally Eu-clidean space without mentioning.

Recall that a covering map sends geodesics to geodesics (Proposition 4.1.3).Now we make use of geodesics to construct covering maps and prove Proposition4.2.1.

Lemma 4.2.2. Let γ : [a, b]→ X be a minimal geodesic. Then γ can be extendedto a geodesic γ̃ : R→ X. Moreover, this extension is unique.

Proof. First notice that for a small piece of minimal geodesic γ : [b− �, b]→B2r (0) with γ(b) = 0, we can slightly extend the domain of γ beyond b by length r,by simply extending the segment in B2r (0). For a general locally Euclidean space,let r > 0 such that Br(x) is isometric to B

2r (0) for all x ∈ X. We extend the

geodesic γ piece by piece, every time inside some Br(x), which is isometric B2r (0).

Suppose that γ has two extensions γ1 and γ2. They agree on [a, b]. Withoutlose of generality, we assume that γ1(t0) 6= γ2(t0) for some t0 > b. Let

s = max{t ≥ b|γ1(t) and γ2(t) agree on [b, t]}.We write y = γ1(s) = γ2(s). Then on the ball Br(y), we have two geodesicsbranching at y: γ1 and γ2 agree on some interval [s − �, s] but disagree beyond s.This cannot happen since Br(y) is isometric to B

2r (0). �

Corollary 4.2.3. Let φ and ψ be two covering maps from R2 to X. Suppose thatφ and ψ agree on a r-ball in R2, then φ = ψ everywhere.

Proof. We assume that φ = ψ on B2r (0), but there is a ∈ R2 such thatφ(a) 6= ψ(a). Let x = φ(0) = ψ(0). Let γ be the straight line through 0 and a inR2. By Lemma 4.1.3, both φ ◦ γ and ψ ◦ γ are geodesics through x in X; moreover,they agree on a small piece around x since φ = ψ on B2r (0). This contradicts theuniqueness part of Lemma 4.2.2. �

Lemma 4.2.4 (Extension Lemma). Let r > 0 such that every r-ball in X isisometric to B2r (0). Let x ∈ X and let φ : B2r (0) → Br(x) be an isometry, then φhas an extension φ̃ : R2 → X with the following properties:(1) φ̃ maps geodesics through 0 to geodesics through x;(2) for any a, b ∈ R2 on a geodesic through 0 with d(a, b) < r, then

d(a, b) = d(φ(a), φ(b)).

Proof. We extend φ by extending geodesics. For any y 6∈ B2r (0), we drawthe unique straight ray γ emanating from 0 through y in R2 with γ(0) = 0. Apiece of γ lies in B2r (0). Via the isometry φ, this piece is mapped to a minimalgeodesic starting from x in Br(x). By Lemma 4.2.2, we can extend this minimalgeodesic in Br(x) to a geodesic with domain [0,∞) in X. We call this geodesic σwith σ(0) = x. For any point z on γ with d(z, 0) = l, we define φ̃(z) be the unique

36 4. COVERINGS

point on σ such that the piece of σ from x to φ̃(z) has length l. In particular, this

defines φ̃(y). (Alternatively, you may use lines through 0 instead of rays to define

φ̃.)Property (1) follows directly from our construction. It remains to prove Prop-

erty (2). Let a, b on a geodesic through 0 with d(a, b) < r. If a, b ∈ B2r (0), thendesired result follows directly from the fact that φ is an isometry on B2r (0). Thuswithout lose of generality, we can assume that a, b belong to a ray γ emanatingfrom 0. Let l = d(a, 0). For convenience, we write σz1,z2 as the piece of σ from z1to z2. By definition, φ(a) is the point on σ with Length(σ0,φ(a)) = l, and φ(b) isthe point on σ with Length(σ0,φ(b)) = l + d(a, b). Thus

Length(σφ(a),φ(b)) = (l + d(a, b))− l = d(a, b) < r.Because X is locally Euclidean, every small piece of σ of length < r is a minimalgeodesic. Hence

d(φ(a), φ(b)) = Length(σφ(a),φ(b)) = d(a, b).

�

We will show that the φ̃ that we constructed in Lemma 4.2.4 is indeed a coveringmap. We need a lemma on verifying covering maps.

Lemma 4.2.5. Let φ : R2 → X be a map. Suppose that there is s > 0 such thatd(φ(a), φ(b)) = d(a, b)

for all a, b with d(a, b) < s, then φ is a covering map.

Proof. Let r > 0 such that every r-ball in X is isometric to B2r (0). We putt = min{r, s/2} > 0.

We first verify condition (2) of Definition 4.1.1, that is, φ|Bt(x) is an isometryonto Bt(φ(x)). Let a, b ∈ Bt(x), then

d(a, b) ≤ d(a, x) + d(x, b) < 2t ≤ s.By assumptions, we know that

d(φ(a), φ(b)) = d(a, b),

thus φ(Bt(x)) ⊆ Bt(φ(x)) and φ preserves the metric on Bt(x). Hence we onlyneed to show that φ maps Bt(x) onto Bt(φ(x)). Suppose that there is a pointz ∈ Bt(φ(x)) but not in φ(Bt(x)). Because φ is an injection when restricted toBs(x), the image φ(Br(x)) contains at least one point other than φ(x); we picksuch a point φ(y). Let lx = d(z, φ(x)) and ly = (z, φ(y)). Recall that both Br(x)and Br(φ(x)) are isometric to B

2r (0). By planar geometry, there are two points (or

exactly one point) w1 and w2 in Br(x) such that

d(wj , x) = lx, d(wj , y) = ly, j = 1, 2.

A similar statement also holds on Br(φ(x)), and z is one of the two points. Becauseφ preserves the distance from Br(x) to φ(Br(x)), we have

d(φ(wj), φ(x)) = lx, d(φ(wj), φ(y)) = ly, j = 1, 2.

Therefore, one of the φ(wi) must be z. A contradiction.Next we prove that φ is surjective. For any point y ∈ X, we need to find

a point w ∈ R2 with φ(w) = y. Let γ : I → X be a minimal geodesic fromφ(0) to y. We choose a partition {t0, ..., tn} of I so that d(γ(ti), γ(ti+1)) < r for

4.3. COVERING GROUPS 37

all i = 0, ..., n − 1. We will find points zi ∈ R2 successively with φ(zi) = γ(ti).We start with φ(0) = γ(t0). For γ(t1) ∈ Br(γ(t0)), since φ|B2r(0) provides anisometry between B2r (0) and Br(φ(0)) = Br(γ(t0)), there is z1 ∈ B2r (0) such thatφ(z1) = γ(t1). Repeating this process over balls Br(γ(ti)) successively (or byinduction), we can find zi+1 with φ(zi+1) = γ(ti+1). Eventually we obtain zn withφ(zn) = γ(tn) = y. �

Proof of Proposition 4.2.1. Lemma 4.2.4 provides a map φ : R2 → X.We show that φ is a covering map. By Lemma 4.2.5, it suffices to find s > 0 so that

d(φ(a), φ(b)) = d(a, b)

for all a, b with d(a, b) < s. We pick s = r/2, where r > 0 as in Lemma 4.2.4. Ifboth a and b belong to a geodesic through 0, then Property (2) of Lemma 4.2.2 canbe directly applied.

For the remaining case, we draw two rays, γa and γb, from 0 to a and b re-spectively. On each ray, we evenly divide the segment from 0 to a (or to b) intok many short segments so that each segment has length < r/2, we call the in-termediate points m1,m2, ...,mk = a and n1, n2, ..., nk = b. By planar geometry,d(mi, ni) < r/2 for all i. Put m0 = n0 = a. By Property (2) of Lemma 4.2.4, wehave

d(mi,mi+1) = d(φ(mi), φ(mi+1)), d(ni, nn+1) = d(φ(ni), φ(ni+1)).

Note that 4φ(mi)φ(ni)φ(ni+1) is contained in Br(φ(mi)); since Br(φ(mi)) is iso-metric to B2r (0), we can view the triangle as one in R2.

Sketch of the remaining proof: Using cosine law, successively show that

4minini+1 ∼= 4φ(mi)φ(ni)φ(ni+1), 4mimi+1ni+1 ∼= 4φ(mi)φ(mi+1)φ(ni+1)

for all i. �

Remark 4.2.6. Here we proved that the map constructed in Lemma 4.2.4 is indeeda covering map. Recall that in Lemma 4.2.4, φ : R2 → X is extended from a localisometry φ : B2r (0)→ Br(x).

A special case is when X is R2. For a local isometry B2r (0) → B2r (x) ⊂ R2,through the same procedure, we obtain a covering map φ : R2 → R2.

Exercise 4.2.7. Prove that φ in the above remark is an isometry.Hint: By Lemma 4.1.5, it suffices to show that φ is injective.

4.3. Covering groups

Definition 4.3.1. Let φ : (X1, d1) → (X2, d2) be a covering map. We define thecovering group Cov(φ) to be

{g ∈ Isom(X1) | φ(g · x) = φ(x) for all x ∈ X1}.

Lemma 4.3.2. Cov(φ) is a group.

Proof. Clearly the identity map of X1 belongs to Cov(φ).Let g ∈ Cov(φ). Since g is an isometry, it has an inverse g−1. We check that

g−1 ∈ Cov(φ). This is true because for any x ∈ X1,

φ(x) = φ(g(g−1x)) = φ(g−1x).

38 4. COVERINGS

Let g1, g2 ∈ Cov(φ). For any x ∈ X1,φ((g1g2) · x) = φ(g1(g2x)) = φ(g2(x)) = φ(x).

This shows g1g2 ∈ Cov(φ). �

Lemma 4.3.3. Let φ : R2 → X be a covering map. Let s > 0 as in Definition4.1.1. Then for any a ∈ R2 and any non-identity element g ∈ Cov(φ), we haved(g · a, a) ≥ s. In particular, Cov(φ)-action on R2 is free and discrete.

Proof. We first prove that the statement when ga 6= a. By Definition 4.3.1,we have φ(ga) = φ(a). If we write x = φ(a), then both g · a and a are points in thepre-image φ−1(x). Since φ|Bs(a) is injective, we conclude d(g · a, a) ≥ s.

Now we rule out the situation ga = a. Suppose that ga = a. Since g is not theidentity map, there is a point b ∈ R2 with gb 6= b. Let γ be the segment between aand b. Through the isometry g, g ◦ γ is a segment between a and g · b. By planargeometry, there is a point b′ on γ that is close to a such that d(gb′, b′) < s. On theother hand, we know this cannot happen from the last paragraph. �

Lemma 4.3.4. Let φ : R2 → X be a covering map with covering group Γ = Cov(φ).For any a ∈ R2 and x = φ(a), the Γ-orbit at a and the pre-image φ−1(x) agrees.

Proof. By the Definition 4.3.1, it is clear that Γ · a ⊆ φ−1(x). Only the otherside inclusion is nontrivial. Let a′ ∈ φ−1(x). We need to find g ∈ Γ with g(a) = a′.Since φ is a covering map, there is s > 0 and isometries

f = φ|Bs(a) : Bs(a)→ Bs(x), f′ = φ|Bs(a′) : Bs(a

′)→ Bs(x).Then (f ′)−1 ◦ f : Bs(a) → Bs(a′) is an isometry between two s-balls of R2. ByRemark 4.2.6, we can extend this local isometry to an isometry of R2. We callthis extension g. By construction, g(a) = a′. We show that g ∈ Γ. According toDefinition 4.3.1, it suffices to show that φ(gy) = φ(y) for all y ∈ X. Note that bothφ ◦ g and φ are covering maps from R2 to X. Moreover, by our construction, forany y ∈ Bs(a),

φ ◦ g(y) = φ ◦ ((φ|Bs(a′))−1 ◦ φ|Bs(a))(y) = φ|Bs(a)(y) = φ(y).

Hence φ ◦ g and φ agree on Bs(a). By Corollary 4.2.3, φ ◦ g = φ everywhere. Thisshows g ∈ Γ and thus φ−1(x) ⊆ Γ · a. �

Proof of Theorem 4.0.1. Let (X, d) be a 2-dimensional locally Euclideanspace. By Proposition 4.2.1, there is a covering map φ : R2 → X. Let Γ be thecorresponding covering group. We prove that X is isometric to R2/Γ.

We construct a map F : X → R2/Γ as follows. Let x ∈ X and let a ∈ φ−1(x).We define F (x) as ā, where ā ∈ R2/Γ is the equivalent class of a. By Lemma 4.3.4,we know that Γ · a = φ−1(x). Hence F is well-defined. One way to interpret Fis using the quotient map q : R2 → R2/Γ. By the construction above, F (x) =q(φ−1(x)) (φ−1(x) can be either understood as the pre-image, or the inverse of alocal isometry around x).

We prove that F is an isometry between X and R2/Γ.• F is onto. For any equivalent class ā ∈ R2/Γ, by construction, F (φ(a)) has

image q(a) = ā.• F is one-to-one. Suppose that we have x, y ∈ X with F (x) = F (y). This

means that φ−1(x) and φ−1(y) lie in the same Γ-orbit. By Lemma 4.3.4, we seethat x = y.

4.3. COVERING GROUPS 39

• F is a covering map. (φ|Bs(a))−1 is an isometry from Bs(x) to Bs(a). Recallthat Γ-action satisfies d(g · a, a) ≥ s for all a ∈ R2 and all g ∈ Γ − {e}. ByTheorem 2.3.17, there is s′ > 0 such that q|Bs′ (a) : Bs′(a)→ Bs′(ā) is an isometry.Compositing these two isometries together from Bs′(x) to Bs′(ā), we obtain anisometry

(q|Bs′ (a)) ◦ (φ|Bs′ (a))−1 : Bs′(x)→ Bs′(ā).

This map is indeed F |Bs′ (x) by our construction. This shows that F is a coveringmap.

By Lemma 4.1.5, we conclude that F is an isometry. �

To end this chapter, we combine Theorem 4.0.1 with Theorem 3.1 in Note Freeand isometric actions on R2 together, as the classification result below.

Theorem 4.3.5 (Classification of 2-dimensional locally Euclidean spaces). Let Xbe a 2-dimensional locally Euclidean space. Then X is isometric to a quotient met-ric space R2/Γ, where Γ acts on R2 freely and discretely by isometries. Moreover,Γ is one of the following groups with the corresponding quotient as X.

Type Generators XI identity PlaneIIa (I, ae1) CylinderIIb (F, ae1/2) Twisted cylinderIIIa (I, ae1), (I, bv) TorusIIIb (F, ae1/2), (I, be2) Klein bottle

where a, b > 0, {e1, e2} a suitable orthonormal coordinate, v a unit vector indepen-dent of e1, F = diag{1,−1}.

One can follow a similar approach in Sections 4.2 and 4.3, with some mildmodifications, to prove that any 2-dimensional locally spherical space can be coveredby the standard sphere S2. Then we can conclude the classification of these spacesas well.

Theorem 4.3.6 (Classification of 2-dimensional locally spherical spaces). Let Xbe a 2-dimensional locally spherical space. Then X is isometric to either S2 or thereal projective plane RP 2 = S2/Z2, where Z2 acts on S2 by sending a point to itsantipodal point.

CHAPTER 5

Hyperbolic Plane

5.1. Introduction

We look for a geometry that satisfies the first four axioms for the Euclideanplane geometry and a different 5th axiom (parallel postulate):• Given a point not on a given line, there are at least two lines through the givenpoint that does not intersect the given line.

We also wish geometry to have a large symmetry group.

Definition 5.1.1. A metric space (X, d) is homogeneous, if for any x, y ∈ X, thereis g ∈ Isom(X) with g · x = y.

A different name that commonly used is saying that Isom(X)-action is transitiveon X.

For example, sphere and the Euclidean plane are homogeneous.Recall that we have seen a non-Euclidean metric on R+ = (0,∞) defined by

d(x, y) = | ln(x/y)|.

Proposition 5.1.2. (R+, d) is homogeneous.

Proof. We consider the R+-action on R+ by multiplication:µ : R+ × R+ → R+, (a, x)→ ax.

This action is by isometries since

d(ax, ay) = | ln((ax)/ay)| = | ln(x/y)| = d(x, y).Then (R+, d) is clearly homogeneous since for any x, y ∈ R+, we have

µ(y/x, x) = (y/x) · x = y.�

Also recall that in Exercise 2.3.3, we have seen a SL2(R)-action on H, whereH = {z ∈ C|Im(z) > 0}:

µ : SL2(R)×H→ H,((

a bc d

), z

)7→ az + b

cz + d.

Note that for the vertical half line {iy|y > 0} ⊂ H, the action by diag{a, 1/a} leavesthis half line invariant: (

a 00 1/a

)· (iy) = a(iy)

1/a= a2(iy).

We will find a distance function d on H so that• (H, d) is a length metric space;• the imaginary half line {iy|y > 0} is a geodesic;• SL2(R)-action on H is isometric;

41

42 5. HYPERBOLIC PLANE

• (H, d) is homogeneous;• (H, d) satisfies the modified 5th axiom.

5.2. Pure imaginary half line under the SL2(R)-action

We study the SL2(R)-action on H by tracking the pure imaginary half lineL = {iy | y > 0} under the action.

We first point out a caveat here: different matrices could correspond to thesame map of H. For example, both I2 and −I2 correspond to the identity mapz 7→ z. In general, A and −I ·A also corresponds to the same map. Thus when wewrite the composition of maps into matrix multiplication, we may have to add anadditional −I (or equivalently, change the signs of all entries in a matrix) to get anequality.

We check all the elements in SL2(R) that leaves L invariant.

Lemma 5.2.1. Suppose that A ∈ SL2(R) maps L to L. Then

A =

(a 00 1/a

)or

(0 −b

1/b 0

).

Proof. By direct calculation, A · (iy) has real partbd+ acy2

d2 + c2y2.

By assumption, this real part should be zero. Note that this is impossible if ac 6= 0since y could be any positive real number. If ac = 0, then bd = 0.

Case 1: a = 0. Because det(A) = 1 and bd = 0, we have d = 0 and c = −1/b.Case 2: c = 0. By the same reasons, we end in c = 0 and d = 1/a. �

Now we keep track of (L, d) under SL2(R)-action. For A =(a bc d

), if c = 0

(then ad = 1),

A · z = az + bd

=a

dz +

b

d= a2z +

b

d.

This is can be realized as a composition a multiplication (by a real number) z 7→ a2·zand a translation (by a real number) z 7→ z + b/d and . The multiplication and

the translation correspond to the matrix

(a 00 1/a

)and

(1 b/d0 1

), respectively.

In terms of matrix multiplication, the composition says(a b0 d

)=

(1 b/d0 1

)·(a 00 1/a

).

Under these two elements (and their compositions), it’s clear that L moves to othervertical half lines. If c 6= 0

A · z = az + bcz + d

=ac (cz + d) + b− ad/c

cz + d

=a

c+ (b− ad/c) · 1

az + d

=a

c+ (bc− ad) · 1

c2z + cd

=a

c+

−1c2z + cd

.

5.2. PURE IMAGINARY HALF LINE UNDER THE SL2(R)-ACTION 43

We write the above as a composition of maps:

zc2·7−→ c2z +cd7−→ c2z + cd 7−→ −1

c2z + cd

+a/c7−→ ac

+−1

c2z + cd.

Besides the translation and multiplication (by real numbers) we have seen in thecase c = 0 above, it has a new map involved in the composition: z 7→ −1/z, which

corresponds to the matrix C =

(0 −11 0

). (Here we cannot use z 7→ 1/z because

1/z would be outside H.) In terms of matrix multiplication, we can check(a bc d

)=

(1 a/c0 1

)·(

0 −11 0

)·(

1 cd0 1

)·(c 00 1/c

).

This proves an algebraic result on SL2(R) by writing z 7→ A · z into a compositionof maps.

Proposition 5.2.2. SL2(R) is generated by the matrices of the forms below:(1 a0 1

),

(a 00 1/a

),

(0 −11 0

),

where a ∈ R.

We note that C switches the first and second quadrant in H: for z = reiθ,

C · z = −1z

= −1re−iθ =

1

rei(π−θ).

Lemma 5.2.3. C maps the vertical half line La = {a+ iy | y > 0}, where a 6= 0,to the upper semi-circle centered at −1/(2a) ∈ C with radius 1/|2a|.

Proof. For z = a+ iy, we calculate

C(a+ iy) = − 1a+ iy

= − a− iya2 + y2

.

Then we calculate its distance square to the point −1/(2a):

dR2(C(z),−1/(2a)) =(− aa2 + y2

+1

2a

)2+

(y

a2 + y2

)2=

1

4a2.

Thus all point in C(La) has equal distance 1/|2a| to the point −1/(2a). Since wealready we know that C(La) is contained in H, we see that C(La) is in the describedsemi-circle. To see that C(La) equals this semi-circle, not

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