UNIVERSITY OF CHICAGO REU 2007: GROUPS OF 2 × 2 MATRICES

COURSE BY URI BADER JULY 16-27, 2007 NOTES TYPED BY SPENCER DOWDALL

1. Linear Transformations

Let K be a field (such as ℝ, ℂ, …), and let V = K^{2} = \{( \begin{matrix}s \\ t \end{matrix}) : s, t ∈ K\} =

\{se_{1} + te_{2} : s, t ∈ K\} (where e_{1} = ( \begin{matrix}1 \\ 0 \end{matrix} ) , e_{2} = ( \begin{matrix}0 \\ 1 \end{matrix} )).

Definition 1.1. A linear transformation is a map T : V → V such that

T (αx + βy) = αT (x) + βT (y)

for all x, y ∈ V and all α, β ∈ K.

Claim 1.2. T is determined by the four scalars ( \begin{matrix}a & b \\ c & d \end{matrix} ) where T (e_{1}) = ae_{1} + ce_{2}, T (e_{2}) = be_{1} + de_{2}

Proof. For the vector ( \begin{matrix}s \\ t \end{matrix}) = se_{1} + te_{2} ∈ V, we have

T ( \begin{matrix}s \\ t \end{matrix}) =T (se_{1} + te_{2}) = sT (e_{1}) + tT (e_{2}) = s(ae_{1} + ce_{2}) + t(be_{1} + de_{2})
=(as + bt)e_{1} + (cs + dt)e_{2} = ( \begin{matrix}as+bt \\ cs+dt \end{matrix} ) =: ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}s \\ t \end{matrix})
□

Matrix mutliplication is given by the composition rule: If T : V → V and T′ : V → V are linear transformations corresponding to ( \begin{matrix}a & b \\ c & d \end{matrix} ) and ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ), then T ∘ T′ is the linear transformation corresponding to

( \begin{matrix}aa′+bc′ & ab′+bd′ \\ ca′+dc′ & cb′+dd′ \end{matrix} ) =: ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} )

In particular, matrix multiplication is associative and non-commutative. (Why? Because we know this for linear transformations). Also, if 0 = ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) is the zero matrix and I = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) is the identity matrix, then

( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) and ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

Definition 1.3. A matrix ( \begin{matrix}a & b \\ c & d \end{matrix} ) is invertible if there exists a matrix ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) such

that ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) = I.

Claim 1.4. The matrix ( \begin{matrix}a & b \\ c & d \end{matrix} ) is invertible if and only if det ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ad − bc ≠ 0

Proof. Firstly, if ad − bc ≠ 0 then one easily checks that ( \begin{matrix}a & b \\ c & d \end{matrix} )^{−1} = \frac{1}{ad−bd} ( \begin{matrix}d & −b \\ −c & a \end{matrix} ) so that ( \begin{matrix}a & b \\ c & d \end{matrix} ) is indeed invertible. Conversely, if ad−bc = 0 then ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}d \\ −c \end{matrix} ) = ( \begin{matrix}0 \\ 0 \end{matrix} ).

Thus, if ( \begin{matrix}a & b \\ c & d \end{matrix} )^{−1} = ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ), then

( \begin{matrix}d \\ −c \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}d \\ −c \end{matrix} ) = ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}d \\ −c \end{matrix} ) = ( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) ( \begin{matrix}0 \\ 0 \end{matrix} ) = ( \begin{matrix}0 \\ 0 \end{matrix} ) ⟹ c, d = 0

Similarly, ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}−b \\ a \end{matrix}) = ( \begin{matrix}0 \\ 0 \end{matrix} ) ⟹ a, b = 0. This is a contradiction since ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) is not invertible. □

Exercise. Show that ( \begin{matrix}a & b \\ c & d \end{matrix} ) is invertible as a matrix if and only if the corresponding linear transformation T : V → V is one-to-one and onto.

Definition 1.5. The above shows that the set of invertible matrices over K forms a group. This group is called the General Linear Group and is denoted by GL_{2}(K).

Remark. GL_{2}(ℝ) ⊂ ℳ_{2×2}(ℝ) ≃ ℝ^{4} (ℳ_{2×2}(K) is the set of 2 × 2 matrices over K). The compliment of this subset is the inverse image of {0} under the determinant map det : ℳ_{2×2}(ℝ) → ℝ; so GL_{2}(ℝ) = det^{−1}(ℝ − {0}) which is an open set. Thus GL_{2}(ℝ) is not only an algebraic object (group) but is also a topological and geometrical object.

2. The Projective Line

Definition 2.1. Let ℙ(V) denote the set of all linear lines in V. (A linear line in V is a straight line through the origin, i.e. a linear subspace of dimension 1; such lines should not be confused with affine lines which need not pass through the origin). Thus

ℙ(V) = {all linear lines in V } = {all 1-dimensional linear subspaces of V }

To each nonzero vector we may associate the unique line containing it and the origin, thus we have a map V − {0} → ℙ(V), x ↦ {λx : λ ∈ K}. In this way, two nonzero vectors define the same line iff they are scalar multiplies of eachother; whence ℙ(V) ≃ V − {0}/ ∼ where u ∼ v if there exists a scalar λ ∈ K such that λu = v (this is obviously an equivalence relation).

Definition 2.2. ℙ^{1}(K) := ℙ(K^{2}) is called the projective line over K. (It is called a line because it is (in some sense) a one-dimensional object. In higher deminsions we set ℙ^{n}(K) = ℙ(K^{n+1})).

If v ∈ V − {0}, then let [v] = {u ∈ V − {0} : u ∼ v} be the equivalence class of all nonzero vectors equivalent to v (i.e., all nonzero vectors on the same linear line as v). Then

ℙ^{1}(K) = \{[ \begin{matrix}s \\ t \end{matrix}] : s, t ∈ K\} = \{[ \begin{matrix}1 \\ t \end{matrix}] : t ∈ K\} ∪ \{[ \begin{matrix}0 \\ 1 \end{matrix} ]\} ≅ K ∪ \{∞\} = K̂

(think “one-point compactification”).

Exercise. Convince yourself that ℙ^{1}(ℂ) = ℂ̂ is homeomorphic (that is, topologically equivalent) to the two-sphere 𝕊^{2}.

Note that the projective line is homogeneous: it looks the same in every direction (the point {∞} is not special). Observe that the action of GL_{2}(K) on K^{2} preserves lines (i.e., it sends lines to lines). Therefore this action decsends to an action of GL_{2}(K) on ℙ^{1}(K).

Claim 2.3. The action of GL_{2}(K) on ℙ^{1}(K) is three-transitive (but not fourtransitive), that is, every (ordered) triple of distinct lines can be mapped to any other (ordered) triple of disinct lines by some element of GL_{2}(K).

The proof of this is given below, but try to prove it yourself first as an exercise. (Hint: consider the triple 0, 1, ∞ ∈ K̂ = ℙ^{1}(K). Show that every triple of distinct lines can be mapped to it.)

Remark. The action of GL_{2}(K) on ℙ^{1}(K) has a kernel. (If a group G acts on the set X, then the kernel of the action is the set of those group elements g ∈ G which act trivially on X.) The kernel of this action is exactly the subgroup of scalar matrices:

kernel = \{( \begin{matrix}λ & 0 \\ 0 & λ \end{matrix} ) : λ ∈ K\} = Scalars

(note that this subgroup is exactly the center of GL_{2}(K)).

Since the scalar matrices are the kernel of the action of GL_{2}(K) on ℙ^{1}(K), the action factors through an action of the quotient group P GL_{2}(K) := GL_{2}(K)/Scalars on ℙ^{1}(K) (note that scalar matrices are a normal subgroup of GL_{2}(K)). In fact, P GL_{2}(K) acts simply transitively on the set of distinct ordered triples in ℙ^{1}(K), that is, for every two ordered triples of distinct points in ℙ^{1}(K) there is a unique element in P GL_{2}(K) that maps the one to the other.

Exercise. These remarks are justified below, but you should try to prove them yourself first.

3. Möbius transformations: another point of view

Recall that ℙ^{1}(K) = K ∪ \{∞\}. In coordinates (z ↔ [ \begin{matrix}z \\ 1 \end{matrix} ], ∞ ↔ [ \begin{matrix}1 \\ 0 \end{matrix} ]) the action of GL_{2}(K) on ℙ^{1}(K) is given by

( \begin{matrix}a & b \\ c & d \end{matrix} ) [ \begin{matrix}z \\ 1 \end{matrix} ] = [ \begin{matrix}az+b \\ cz+d \end{matrix} ] = [ 1^{\frac{az+b}{cz+d}} ]

On K ⊂ ℙ^{1}(K) this is the mapping z ↦ \frac{az+b}{cz+d} (such mappings are called Möbius transformations). More precisely, ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{2}(K) acts on ℙ^{1}(K) by

z ↦ \begin{cases}∞ & cz + d = 0 \\ \frac{a}{c} & z = ∞, c ≠ 0 \\ ∞ & z = ∞, c = 0 \\ \frac{az+b}{cz+d} & else \end{cases}

Exercise. Verify that composition of Möbius transformations corresponds to matrix multiplication.

Remark. If Möb(K) is the group of Möbious transformations of K, then Möb(K) ≃ P GL_{2}(K) ≃ Gal(K(z), K)

We now give a proof the of Claim 2.3:

Proof. (of 2.3) We will show that for every triple of lines L_{1}, L_{2}, L_{3} ∈ ℙ^{1}(K), we can find a g ∈ GL_{2}(K) such that g ( \begin{matrix}1 \\ 0 \end{matrix} ) ∈ L_{1}, g ( \begin{matrix}0 \\ 1 \end{matrix} ) ∈ L_{2}, and g ( \begin{matrix}1 \\ 1 \end{matrix} ) ∈ L_{3}. Pick u_{i} ∈ L_{i}. {u_{1}, u_{2}} is a basis, hence u_{3} = αu_{1} + βu_{2} for some α, β ∈ K. Set v_{1} = αu_{1}, v_{2} = βu_{2}, v_{3} = u_{3} = αu_{1} + βu_{2} = v_{1} + v_{2}. Let g ∈ GL_{2}(K) be the unique element sending ( \begin{matrix}1 \\ 0 \end{matrix} ) ↦ v_{1} and ( \begin{matrix}0 \\ 1 \end{matrix} ) ↦ v_{2} (if in coordinates v_{i} = ( \begin{matrix}v_{ix} \\ v_{iy} \end{matrix}), then take g = ( \begin{matrix}v_{1x} & v_{2x} \\ v_{1y} & v_{2y} \end{matrix})). Then g satisfies ( \begin{matrix}1 \\ 1 \end{matrix} ) ↦ v_{1} + v_{2} = v_{3}. Hence it will satisfy 0 ↦ L_{1}, ∞ ↦ L_{2}, and 1 ↦ L_{3}. □

We now justify the remark made earlier that the action of P GL_{2}(K) on the set of ordered triples of distinct points in ℙ^{1}(K) is simply transitive.

If a group G acts on a set X then the stabilizer of a point x ∈ X is the set Stab(x) = {g ∈ G : gx = x}.

Claim 3.1. Suppose that x, y ∈ X are such that y = gx for some g ∈ G (such x and y are said to lie in the same orbit), Then

Stab(y) = {ghg^{−1} : h ∈ Stab(x)} =: Stab(x)^{g}

Proof. If h ∈ Stab(x) then (ghg^{−1})y = ghg^{−1}gx = gx = y so that ghg^{−1} ∈ Stab(y). Conversely, if γ ∈ Stab(y) then g^{−1}γg ∈ Stab(x) so that γ = g(g^{−1}γg)g^{−1} ∈ Stab(x)^{g}. □

In the action of GL_{2}(K) on ordered triples of distinct points in ℙ^{1}(K), the stabilizer of (0, 1, ∞) is the subgroup of scalar matrices. If (L_{1}, L_{2}, L_{3}) is any ordered triple, then (L_{1}, L_{2}, L_{3}) = g(0, 1, ∞) for some g ∈ GL_{2}(K). Whence 3.1 implies that Stab(L_{1}, L_{2}, L_{3}) = {ghg^{−1} : h ∈ Scalars} = Scalars. Since the stabilzer of every triple is exactly the subgroup of scalar matrices, the quotient group P GL_{2}(K) indeed acts simply transitively. In fact, there is a bijection of sets: P GL_{2}(K) ≃ {(distint, ordered) triples in ℙ^{1}(K)}.

4. Cross-Ratio

Observe that the mapping z ↦ \frac{(z−z_{1})(z_{2}−z_{3})}{(z−z_{3})(z_{2}−z_{1})} is a Möbius transformation that sends (z_{1}, z_{2}, z_{3}) ↦ (0, 1, ∞) (in our coordinates z ↔ [ \begin{matrix}z \\ 1 \end{matrix} ] and ∞ ↔ [ \begin{matrix}1 \\ 0 \end{matrix} ]).

Exercise. Write this in matrix form

Definition 4.1. (first definition) For distinct points w, x, y, z ∈ ℙ^{1}(K), let g ∈ GL_{2}(K) be such that (gw, gx, gy) = (0, 1, ∞). The cross-ratio of w, x, y, z is defined to be the element (w, x, y, z) := gz ∈ ℙ^{1}(K). Thus the corss-ratio is a mapping CR : {distinct 4-tuples in ℙ^{1}(K)} → ℙ^{1}(K). Observe that the cross-ratio is GL_{2}(K)- invariant: for h ∈ GL_{2}(K), (hw, hx, hy, hz) = (w, x, y, z).

Definition 4.2. (second definition: in coordinates) Representing elements of ℙ^{1}(K) by elements in K (via z ↔ [ \begin{matrix}z \\ 1 \end{matrix} ]), we define a cross-ratio CR : \{distinct 4-tuples in K\} → K by (z_{1}, z_{2}, z_{3}, z_{4}) ↦ z, where ([ \begin{matrix}z_{1} \\ 1 \end{matrix} ] , [ \begin{matrix}z_{2} \\ 1 \end{matrix} ] , [ \begin{matrix}z_{3} \\ 1 \end{matrix} ] , [ \begin{matrix}z_{4} \\ 1 \end{matrix} ]) = [ \begin{matrix}z \\ 1 \end{matrix} ]. This definition leads to an explicit formula:

(z_{1}, z_{2}, z_{3}, z_{4}) = \frac{(z_{4} − z_{1})(z_{2} − z_{3})}{(z_{4} − z_{3})(z_{2} − z_{1})}

Corollary 4.3. The formula is invariant under Möbius transformations.

Proof. Exercise. □

Theorem 4.4. A self-map of ℙ^{1}(K) that preserves the cross-ratio is from P GL_{2}(K) (i.e. a Möbius transformation).

Proof. Exercise. □

Claim 4.5. Every Möbius transformation can be written as a composition of transformations of the following forms:

z ↦ az,  z ↦ z + b,  z ↦ \frac{−1}{z}

Proof. Consdier z ↦ \frac{az+b}{cz+d}. If c = 0, then the map is

z ↦ \frac{az+b}{d} = \frac{a}{d}z + \frac{b}{d} = (z ↦ z + \frac{b}{d}) ∘ (z ↦ \frac{a}{d}z)

If c ≠ 0, then

\frac{az+b}{cz+d} = ^{\frac{\frac{a}{c} (cz+d)+(b− \frac{ad}{c})}{cz+d}} = (b − \frac{ad}{c})\frac{1}{cz+d} + \frac{a}{c}

Therefore

z ↦ \frac{az+b}{cz+d} = (z ↦ z + \frac{a}{c}) ∘ (z ↦ (b − \frac{ad}{c})z) ∘ (z ↦ \frac{1}{z}) ∘ (z ↦ z + d) ∘ (z ↦ cz)

□

Corollary 4.6. This decomposistion of P GL_{2}(K) can be used to give a decomposition of GL_{2}(K), known as the Bruhat decomposition of matrices.

When K = ℂ, the complex projective line ℙ^{1}(ℂ) = ℂ̂ = ℂ ∪ {∞} ≃ 𝕊^{2} is called the Riemann Sphere. In this case the transformation z ↦ z + b is a translation by b ∈ ℂ, the transformation z ↦ λz is scaling by |λ| and a rotation by α where λ = |λ| e^{iα}, and the transformation z ↦ \frac{−1}{z} is an inversion and reflection in the unit circle.

Definition 4.7. A generalized circle in ℂ̂ is a Euclidean circle or line in ℂ. (Lines in ℂ are thought of as circles through {∞} ∈ ℂ̂.) Therefore, any three distinct points in ℂ̂ determine a unique generalized circle.

Theorem 4.8. Let w, x, y, z ∈ ℙ^{1}(ℂ) be distinct. Then for their corss-ratio we have (w, x, y, z) ∈ ℝ iff the four points w, x, y, z all lie on a generalized circle (i.e., iff z lies on the generalized circle determined by w, x, y).

Proof. Exercise. (Hint: first prove the case w, x, y ∈ ℙ^{1}(ℝ), then generalize by 3-transitivity.) □

Corollary 4.9. Möbius transfromations take generalized circles to generalized circles. Moreover, Möb(ℂ) acts transitively on the set of generalized circles.

Proof. This follows from 3-transitivity and Thm 4.8 since the cross-ratio is preserved by Möbius transformations. □

Question. What is the stabilizer of a circle? (recall that since the action is transitive, the stabilizers of different circles will be conjugate subgroups in Mob(ℂ); thus it is enough to know the stabilizer of one circle).

Claim 4.10. The stabilizer of ℙ^{1}(ℝ) ⊂ ℙ^{1}(ℂ) is exactly P GL_{2}(ℝ).

Proof. It is clear that P GL_{2}(ℝ) stabilizes the circle ℙ^{1}(ℝ) ⊂ ℙ^{1}(ℂ). The fact that this is the whole stabilizer follows from 3-transitivity: if g ∈ P GL_{2}(ℂ) stabilizes ℙ^{1}(ℝ), then g(0), g(1), g(∞) ∈ ℙ^{1}(ℝ). By 3-transitivity of the action of P GL_{2}(ℝ) on ℙ^{1}(ℝ), there is a unque h ∈ P GL_{2}(ℝ) with h(0, 1, ∞) = (g(0), g(1), g(∞)). Since P GL_{2}(ℂ) acts simply 3-transitively on ℙ^{1}(C), the the fact that g and h ∈ P GL_{2}(ℝ) ⊂ P GL_{2}(ℂ) agree on the three points 0, 1, and ∞ implies that they agree everywhere, whence g = h ∈ P GL_{2}(ℝ). □

Considering the circle ℙ^{1}(ℝ) ⊂ ℙ^{1}(ℂ) = Ĉ ≅ 𝕊^{2} as the equator, the upper and lower hemispheres are the sets:

upper hemisphere =U = {z ∈ ℂ : Imz > 0}
lower hemisphere =L = {z ∈ ℂ : Imz < 0}

(so the points {i} and {−i} are the north and south poles).

Corollary 4.11. P GL_{2}(ℝ) preserves the decomposition ℙ^{1}(ℂ) = ℙ^{1}(ℝ) ∪ U ∪ L.

Question. What is the stabilizer of the upper-hemisphere in P GL_{2}(ℝ)?

Answer. By the corollary, it suffices to consider whether the image of i ∈ U lies in

U or L. If ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ P GL_{2}(ℝ), then

Im (( \begin{matrix}a & b \\ c & d \end{matrix} ) i) = Im (\frac{ai+b}{ci+d}) = Im (\frac{(ai+b)(d−ci)}{c^{2}+d^{2}}) = ^{\frac{det( \begin{matrix}a & b \\ c & d \end{matrix} )}{c^{2}+d^{2}}} ∈ U ⟺ det ( \begin{matrix}a & b \\ c & d \end{matrix} ) > 0

Whence ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ P GL_{2}(ℝ) stabilizes the upper-hemisphere iff det ( \begin{matrix}a & b \\ c & d \end{matrix} ) > 0.

Definition 4.12. Let GL_{2}^{+}(ℝ) := {g ∈ GL_{2}(ℝ) : det g > 0} be the subgroup of invertible matrices with positve determinant. This is an index-2 subgroup of GL_{2}(ℝ). The Special Linear group (over ℝ) is the subgroup SL_{2}(ℝ) := \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{2}(ℝ) : det ( \begin{matrix}a & b \\ c & d \end{matrix} ) = 1\} of invertible matrices with determinant equal to 1. Whence we have the inclusion of subgroups SL_{2}(ℝ) ⩽ GL_{2}^{+}(ℝ) ⩽ GL_{2}(ℝ). The images of these subgroups under the natural quotient map GL_{2}(ℝ) → P GL_{2}(ℝ) are denoted by P SL_{2}(ℝ) and P GL_{2}^{+}(ℝ).

Exercise. Show that P SL_{2}(ℝ) = P GL_{2}^{+}(ℝ) and that P GL_{2}^{+}(ℝ) is an index-2 subgroup of P GL_{2}(ℝ). Whence we have the diagram

\xymatrix{SL_{2}(ℝ) \ar[d]^{q} & ⩽ & GL_{2}^{+}(ℝ) \ar[d]^{q} & ⩽^{index-2} & GL_{2}(ℝ) \ar[d]^{q} \\ P SL_{2}(ℝ) & = & P GL_{2}^{+}(ℝ) & ⩽^{index-2} & P GL_{2}(ℝ)}

Observations about the action of P SL_{2}(ℝ):

(i) P SL_{2}(ℝ) acts on the upper hemisphere {z : Im(z) > 0} and on the circle ℙ^{1}(ℝ).

(ii) The action of P SL_{2}(ℝ) is 2-transitive on ℙ^{1}(ℝ). (For two lines L_{1}, L_{2} ∈ ℙ^{1}(ℝ), choose two nonzero vectors v_{i} ∈ L_{i}. Rescale v_{1} so that the matrix having v_{1} and v_{2} as its columns has determinant 1. Then this matrix sends e_{i} to v_{i} and hence ∞ ↦ L_{1}, 0 ↦ L_{2}.

(iii) The action of P SL_{2}(ℝ) is not 3-transitive on ℙ^{1}(ℝ). Rather, the action on (distinct, ordered) triples has two orbits. (Given three points x, y, z ∈ ℙ^{1}(ℝ), one may map x and y to 0 and ∞ by 2-transitivity. An easy calculation shows that the stabilizer of the pair (0, ∞) is diagonal matrices, i.e. matrices of the form ( \begin{matrix}λ & 0 \\ 0 & λ^{−1} \end{matrix} ). Such matrices act on the third point z by ( \begin{matrix}λ & 0 \\ 0 & λ^{−1} \end{matrix} ) [ \begin{matrix}z \\ 1 \end{matrix} ] = [ \begin{matrix}λz \\ λ^{−1} \end{matrix} ] = [ \begin{matrix}λ^{2}z \\ 1 \end{matrix} ]. Whence the third point z can be scaled by any positive real number, but its sign cannot be changed. Therefore the orbits of the tirples (0, ∞, 1) and (0, ∞, −1) are distinct.)

(iv) The action of P SL_{2}(ℝ) on the upper hemisphere U = {z : Im(z) > 0} is transitive. (By choosing the appropriate a, b ∈ ℝ, the matrix ( \begin{matrix}a & ba^{−1} \\ 0 & a^{−1} \end{matrix} ) will map the point [ \begin{matrix}i \\ 1 \end{matrix} ] ↔ i ∈ U to any desired point in U.)

(v) The stabilizer of i is \{( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) : θ ∈ ℝ\} = SO_{2}(ℝ) (exercise: verify this by a calculation). Identifying i ∈ ℂ̂ ≃ 𝕊^{2} with the north pole and ℙ^{1}(ℝ) ⊂ ℂ̂ with the equator, such matrices rotate the sphere an angle of θ about the axis through i and −i. Be careful! This action of SO_{2}(ℝ) on ℂ̂ by Möbius transformations is not linear and should not be confused with the usual (linear) action of SO_{2}(ℝ) on ℝ^{2} ≃ ℂ by rotation.

Observations about the action of Möb(ℂ):

(i) The subgroup of Möb(ℂ) that fixes the origin and the unit circle is {z ↦ e^{2πiα}z : α ∈ [0, 1]}. (Clearly such maps fix the origin and unit circle. Use 3-transitivity to show that all such maps are of this form.)

(ii) The action of the Möbius group is conformal (i.e. it preserves angles). (This follows from the complex-differentiability of Möbius transformations: the action of the differential is by scalar multiplication, which is conformal).

5. Hyperbolic Space

Definition 5.1. Consider the unit disk 𝔻 = {z ∈ ℂ̂ : |z| < 1}. A hyperbolic line in 𝔻 is a (generalized) circle that intersects the unit circle 𝕊^{1} = ∂𝔻 in right angles.

By 3-transitivity, there is an element g ∈ Möb(ℂ) sending the circle ℙ^{1}(ℝ) to the unit circle ∂𝔻 and the upper-hemisphere {Imz > 0} to the unit disk 𝔻. It follows that the conjugate subgroup 𝒟 = g(P SL_{2}(ℝ))g^{−1} is the subgroup of Möb(ℂ) which preserves 𝔻. The action of 𝒟 on 𝔻 is therefore conformal, 2-transitive on ∂𝔻, 1transitive on 𝔻, and sends hyperbolic-lines to hyperbolic lines.

Claim 5.2. For every two (distinct) points in \overline{𝔻}, there is a unique hyperbolic line containing them.

Proof. This is immediate for a pair of antipodal points on ∂𝔻 as diameters are clearly hyperbolic lines. The claim thus follows for arbitrary pairs on ∂𝔻 by the 2-transitivity of the confomal action of 𝒟 on ∂𝔻. Similarly, the claim is clear for pairs of the form (0, y) with y ∈ \overline{𝔻} as the diameter containing y again suffices. Therefore, the result holds for abitrary pairs by the 1-transitivity of the action of 𝒟 on 𝔻. □

Claim 5.3. The stabilizer of ∂𝔻 in Möb(ℂ) acts transitively on the set of hyperbolic lines.

Proof. This follows from the 3-transitivity of the action of Möb(ℂ) on the unit circle and the bijective correspondence between hyperbolic lines and pairs of points on the unit circle. □

We now adjust our ordering convention for the cross ratio: For z_{0}, z_{1}, z_{2}, z_{3} ∈ ℂ̂, set their cross ratio to be (z_{0}, z_{1}, z_{2}, z_{3}) = \frac{(z_{2}−z_{0})(z_{3}−z_{1})}{(z_{1}−z_{0})(z_{3}−z_{2})}. (Note: this is the standard convention for the cross-ratio; it is a permutation of the definition given earlier).

Definition 5.4. For x, y ∈ 𝔻, let L be the unique hyberbolic line through x and y, and let s, t be the intersections of L with the unit circle ∂𝔻 (labeled so that the points appear in the order (s, x, y, t) on L). Define a metric on 𝔻 by setting

d(x, y) = \frac{1}{2} |log(s, x, y, t)|

(recall that by Thm 4.8 the cross-ratio of four points on the same circle is a real number).

Note that there are some symetries in the cross-ratio (s, x, y, t) obtained by permuting the vaiables. Exercise: Find all of the possibilities for the cross-ratio of four distinct points.

Remark. d is invariant under the subgroup 𝒟 of Möb(ℂ) preserving 𝔻.

Observation. Let z_{0}, z_{1}, …, z_{4} be any 5 points on the same circle. Then

(z_{0}, z_{1}, z_{3}, z_{4}) =\frac{(z_{3} − z_{0})(z_{4} − z_{1})}{(z_{1} − z_{0})(z_{4} − z_{3})} = \frac{(z_{2} − z_{0})(z_{4} − z_{1})}{(z_{1} − z_{0})(z_{4} − z_{2})} · \frac{(z_{3} − z_{0})(z_{4} − z_{2})}{(z_{2} − z_{0})(z_{4} − z_{3})}
=(z_{0}, z_{1}, z_{2}, z_{4}) · (z_{0}, z_{2}, z_{3}, z_{4})

Corollary 5.5. Hyperbolic lines are geodesics in the metric space (𝔻, d). (A geodesic is an isometric embedding (copy) of ℝ into a metric space.)

Theorem 5.6. d is a metric on 𝔻.

Proof. The only non-trivial thing to check is the triangle inequality. We sketch a proof of this: Let x, y, z ∈ 𝔻 and consider the (hyperbolic) triangle Δ ⊂ 𝔻 having these points as vertices. Let a, b, c denote the lengths of these sides so that d(x, z) = a, d(x, y) = b, and d(y, z) = c, and let A, B, C be the cooresponding hyperbolic lines comprising the sides of Δ. We must show a ≤ b + c.

Let P be the hyperbolic line through y and perpendicular to A. and let w be the intersection point of A and P. If a′ = d(x, w) and a′′ = d(w, z), then, by the above observation, we have a = a′ + a′′. Thus it suffices to prove a′ ≤ b and a′′ ≤ c, so we have reduced to the case of a right triangle.

By applying a Möbius transformation, we may assume that x = 0 so that the hyperbolic lines A and B connconnecting x to w and y are diameters (i.e. straight Euclidian lines). Therefore, if ã and b̃ are the Euclidean distances from x to w and y, it suffices to check that ã ≤ b̃. The hyperbolic line P from w to y is now a circular arc perpendicular to the diameter A. Draw the striaght Euclidean line E through w perpendicular to A (so E is tangent to P at w). Since (looking at things Euclideanly) the line P is curving away from the origin x, we see that E intersects B closer to x than does P; call this Euclidean distnace b̃′ so that b̃′ ≤ b̃. Finally, the lines A, B, E form a Euclidean right triangle with hypotenuse of length b̃′. As ã is the length of one leg of this triangle, Euclidean geometry implies that ã ≤ b̃′ ≤ b̃ as desired. □

Corollary 5.7. (𝔻, d) is a metric space and the subgroup 𝒟 ⩽ P GL_{2}(ℂ) preserving 𝔻 acts on it by isometries. Indeed, the same holds for any other circle in ℂ̂, in particular, for the upper half plane {z : Im(z) > 0} on which P SL_{2}(ℝ) acts by isometries.

Exercise. Show that P SL_{2}(ℝ) ⩽ Isom(Upper half-plane) is an index-2 subgroup.

Definition 5.8. The metric space (𝔻, d) is called the hyperbolic plane, and is denoted by ℍ^{2} (sometimes ℍ^{2}(ℝ)).

6. The group P SL_{2}(ℤ)

The goal in this section is to look at images of P SL_{2}(ℤ) in P SL_{2}(ℝ) and understand the corresponding actions on ℍ^{2}. To this end, we study the space of unimodular lattices.

What are all of the basis of ℂ? Each such is given by a map (of abelian groups) ℤ^{2} →^{φ} ℂ such that φ ( \begin{matrix}1 \\ 0 \end{matrix} ) and φ ( \begin{matrix}0 \\ 1 \end{matrix} ) are not on the same line. Considering ℂ as a 2-dimensional vector space, all such maps are linear, so we may set Map^{+}(ℤ^{2}, ℂ) = {φ : ℤ^{2} → ℂ : det(φ) > 0} to be the space of all such maps with positive determinant.

We will now mod out by some obvious equivalences on this space: There is an obvious left-action of ℂ^{∗} on Map^{+}(ℤ^{2}, ℂ) by post-multiplication by a scalar: λ·φ = λφ. Similarly, there is a right-action of Aut^{+}(ℤ^{2}) = SL_{2}(ℤ) on Map^{+}(ℤ^{2}, ℂ) by pre-composition: φ · α = φ ∘ α. Since these two actions correspond to post- and pre-composition respectively (λ · φ · α = ℤ^{2} →^{α} ℤ^{2} →^{φ} ℂ →^{λ} ℂ), the associativity of composition of maps implies that the two actions commute with each other

(Note: For R a ring,

GL_{2}(R) =\{2 × 2 invertible matrices over R\} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) : ad − bc ∈ R^{∗}\}

(where R^{∗} denotes the set of units of R). Therefore GL_{2}(ℤ) ≠ ℳ_{2×2} ∩ GL_{2}(ℝ); but rather GL_{2}(ℤ) = ℳ_{2×2} ∩ {det = ±1}. Whence SL_{2}(ℤ) is an index-2 subgroup of GL_{2}(ℤ).)

There is a natural correspondence: Map^{+}(ℤ^{2}, ℂ)/C^{∗} → ℍ^{2} (in upper half-plan model) given by φ ↦ φ ( \begin{matrix}0 \\ 1 \end{matrix} ) φ ( \begin{matrix}1 \\ 0 \end{matrix} )^{−1}; this is clearly a bijection becaues for every z ∈ ℍ^{2}, there is a φ ∈ Map^{+}(ℤ^{2}, ℂ) with φ ( \begin{matrix}0 \\ 1 \end{matrix} ) = z and φ ( \begin{matrix}1 \\ 0 \end{matrix} ) = 1. Since the actions of SL_{2}(ℤ and ℂ^{∗} on Map^{+}(ℤ^{2}, ℂ) commute, we get an induced action of SL_{2}(ℤ) on the quotient Map^{+}(ℤ^{2}, C)/C^{∗} ≅ ℍ^{2}. We also have the natural action of SL_{2}(ℤ) on ℍ^{2} by Möbius transformations. These two actions are related very simply:

Claim 6.1. The action of SL_{2}(ℤ) on the quotient Map^{+}(ℤ^{2}, ℂ)/C^{∗} ≅ ℍ^{2} agrees with the usual action on ℍ^{2} by Möbius transformations; i.e. the correspondence Map^{+}(ℤ^{2}, ℂ)/C^{∗} → ℍ^{2} is an an SL_{2}(ℤ)-equivariant bijection.

Remark. _{GL_{2}(ℤ)}\^{Map^{+}(ℤ^{2},C)} ≅_{SL_{2}(ℤ)} \^{Map^{+}(ℤ^{2},ℂ)} can be identified with the space of lattices in ℂ, or with the space of “torus group quotients” of ℂ. In fact the quotient

MV =_{P SL_{2}(ℤ)} \^{Map^{+}(ℤ^{2},ℂ)}/_{ℂ^{∗}}

is the space of “complex structures on the torus”; also called the modular variety. Accordingly, P SL_{2}(ℤ) is also called the modular group.

Question. What does the space MV look like?

To answer this, we will study/find a fundamental domain for the action of P SL_{2}(ℤ) on ℍ^{2} (a fundamental domain is a subset of ℍ^{2} that meets each (generic) orbit in exactly one point). (Note: Although P SL_{2}(ℝ) acts transitively on ℍ^{2}, the subgroup P SL_{2}(ℤ) ⩽ P SL_{2}(ℝ) is discrete and therefore cannot act transitively.)

Claim 6.2. Let F = \{z ∈ ℍ^{2} : |z| ≥ 1\} ∩ \{z ∈ ℍ^{2} : −\frac{1}{2} ≤ Im(z) ≤ \frac{1}{2}\}. Then F is a fundamental domain for the action of P SL_{2}(ℤ) on ℍ^{2}; more precisely, every orbit of P SL_{2}(ℤ) in ℍ^{2} meets F in exactly one interior point, exactly two boundary points, or the point i.

10 COURSE BY URI BADER JULY 16-27, 2007 NOTES TYPED BY SPENCER DOWDALL

Proof. (sketch) Consider a lattice Λ = φ(ℤ^{2}) for φ ∈ Map^{+}(ℤ^{2}, ℂ). Up to the action of ℂ^{∗}, we may assume that φ ( \begin{matrix}1 \\ 0 \end{matrix} ) = 1. Up to the SL_{2}(ℤ) action, we may assume that (i) φ ( \begin{matrix}1 \\ 0 \end{matrix} ) is of shortest length in φ(ℤ^{2}) and that (ii) |Re(φ ( \begin{matrix}0 \\ 1 \end{matrix} ))| ≤ \frac{1}{2}; whence x = φ ( \begin{matrix}0 \\ 1 \end{matrix} ) ∈ F. Since \{φ ( \begin{matrix}1 \\ 0 \end{matrix} ) , φ ( \begin{matrix}0 \\ 1 \end{matrix} )\} forms a basis of φ(ℤ^{2}), it follows that (iii) Im(x) is minimal in φ(ℤ^{2}). (Note that facts (i)-(iii) uniquely determine a vector v such that {1, v} is a basis.) Therefore, to the lattice Λ we associate the unique point x = φ ( \begin{matrix}0 \\ 1 \end{matrix} ) ∈ Λ such that x ∈ F and Im(x) is minimal in Λ. Conversely, given an interior point y ∈ F we associate the lattice with basis {1, y}. Thus we have a bijection:

\xymatrix{lattices \ar[r]^{the above argument} & (interior) points x ∈ F \ar[l]^{lattice generated by \{1,x\}}}

However, this is not a bijection on ∂F where we have the identifications z ↔ z+1 for z ∈ F ∩ \{Re(z) = −\frac{1}{2}\} (corresonding to the action by the group element ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) ∈ P SL_{2}(ℤ)), and z ↔ −\frac{1}{z} for z ∈ F ∩ \{|z| = 1\} (corresponding to the action by the group element ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ∈ P SL_{2}(ℤ)). Making these identifications, we see that the modular variety (recall that MV = the quotient space _{P SL_{2}(ℤ)} \^{Map^{+}(ℤ^{2},ℂ)} /_{ℂ^{∗}})

is shaped like a “triangular pillowcase” with cusps at i, (\frac{1}{2}, \frac{\sqrt{3}}{2}), and ∞. □

This picture of the modular variety should help us understand the algebraic structure of P SL_{2}(ℤ).

7. The Structure of P SL_{2}(ℤ)

Consider the elements S = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , T = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) ∈ P SL_{2}(ℤ) corresponding to the Möbius transformations S(z) = −\frac{1}{z}, and T (z) = z + 1.

Claim 7.1. The elements S and T generate P SL_{2}(ℤ).

Proof. (sketch) Observe that the stabilizer of 2i ∈ F is trivial in P SL_{2}(ℤ). (Exercise: show this. Idea: if \frac{a(2i)+b}{c(2i)+d} = 2i with a, b, c, c ∈ ℤ, then conclude ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}±1 & 0 \\ 0 & ±1 \end{matrix} ).) Now, consider g ∈ P SL_{2}(ℤ) and let y = g(2i). The goal is to use S and T to send y into F, i.e. to find a word w in S and T such that wy ∈ F. Since Orbit_{2i} ∩ F = {2i}, we have that 2i = wy = (wg)2i which implies that g = w^{−1} is a word in S and T since the stabilizer of 2i is trivial.

We now describe the algorithm to construct the word w:

(i) If Im(y) > 1, then T^{n}y ∈ F for some n ∈ ℤ (ii) If Im(y) < 1, then we apply T^{n} to get either into F or into the open hemisphere {|z| < 1}. In the first case we are done, in the second, apply S to increase Im(y).

(iii) Repeat. We need only check that Im(y) increases by a sufficient amount each time so that we always eventually obtain Im(y) > 1 (i.e., so that the algorithm terminates). This is made more precise by the following compactness arbument: If I start at y = a + bi, then when I apply S and T as in the above algorithm, I always end up in \{z : |z| ≤ \frac{1}{b}\} ∩ \{z : Im(z) ≥ b\}, which is a compact set. This compact set can only intersect finitely many copies of the fundamental domain (in the tiling of ℍ^{2} by the images of F under S and T), so the algorithm must terminate. □

Corollary 7.2. Set U = ST. Then P SL_{2}(ℤ) is generated by S and U.

Observe that S(z) = −\frac{1}{z} ↔ ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) and U(z) = \frac{−1}{z+1} ↔ ( \begin{matrix}0 & −1 \\ 1 & 1 \end{matrix} ). One easily checks that S^{2} = 1 = U^{3}.

Consider the group G given by the presentation G = ⟨S̃, Ũ : S̃^{2} = Ũ^{3} = 1⟩. Thus G is the (universal) group generated by the two elements S̃, Ũ under the relations S̃ = 1 = Ũ^{3}. (Note that this is the same as the free product: G = C_{2} ∗ C_{3}, where C_{n} is the cyclic group of order n.)

Such a group can be constructed as follows: Take the free group F (S̃, Ũ) on the two symbols S̃ and Ũ. That is

F (S̃, Ũ) = {all reduced words in S̃, Ũ, S̃^{−1}, Ũ^{−1}}

(reduced means that we cancel all terms S̃S̃^{−1}, S̃^{−1}S̃, ŨŨ^{−1}, Ũ^{−1}Ũ). Note that this F is a functor from the category of sets to the category of groups. It is left adjoint to the forgetful functor U from the category of groups to the category of sets.

Let N be the minimal normal subgroup of F (S̃, Ũ) containing the elements S̃^{2} and Ũ^{3} (e.g, take N to be the intersection of all such subgroups). The presentation ⟨S̃, Ũ|S̃^{2} = Ũ^{3} = 1⟩ is defined to be the quotient by N:

G = ⟨S̃, Ũ|S̃^{2} = Ũ^{3} = 1⟩ := F (S̃, Ũ)/N

The groups F (S̃, Ũ) and G satisfy the following “universal” properties:

(i) For every group Γ, and any two elements s, u ∈ Γ, there exists a unique homomorphism Θ : F (S̃, Ũ) → Γ such that S̃ ↦ s and Ũ ↦ u. Moreover, this map is onto iff Γ is generated by s and u.

(ii) If in Γ we have s^{2} = u^{3} = 1, then Θ factors through the quotient group G:

\xymatrix{F (S̃, Ũ) \ar[d] \ar[dr]^{Θ} \\ G = F (S̃, Ũ)/N \ar@{-->}[r]^{∃!θ} & Γ}

and gives a unique map θ : G → Γ such that S̃ ↦ s and Ũ ↦ u.

Corollary 7.3. There is a unique (surjective) map Ψ : G = ⟨S̃, Ũ|S̃^{2} = Ũ^{3} = 1⟩ → P SL_{2}(ℤ) such that Ψ(S̃) = S and Ψ(Ũ) = U.

Theorem 7.4. The natrual map Ψ : G → P SL_{2}(ℤ) is an isomorphism.

Proof. It remains to prove that Ψ is injective.

First, let’s consider the action of P LS_{2}(ℤ) on ℙ^{1}(ℝ). Consder 𝕊^{1} ≅ ℙ^{1}(ℝ) ⊂ ℙ^{1}(ℂ) ≅ 𝕊^{2}. If we let 0, ∞ ∈ ℙ^{1}(ℂ) be the south and north poles of 𝕊^{2} respectively, then ℙ^{1}(ℝ) is a great meridian of the sphere. Then ℙ^{1}(ℝ) − {0, ∞} = L′ ∪ R′ where L′ and R′ are the two (open) circular arcs from 0 to ∞, labelled such that −1 ∈ L′ and 1 ∈ R′ (i.e. the left and right halfs of the circle ℙ^{1}(ℝ)). Let L = \overline{L′} = L∪\{0, ∞\} and, similarly, R = \overline{R′} = R′ ∪ \{0, ∞\}. We now consider the images of L and R under the maps S, U, and U^{2}:

S(z) = −\frac{1}{z}. Thus S maps 0 ↦ ∞, ∞ ↦ 0, 1 ↦ −1, and −1 ↦ 1; whence S(L) ⊂ R. U(z) = \frac{−1}{z+1}. Thus U maps 0 ↦ −1, ∞ ↦ 0, 1 ↦ −\frac{1}{2}, and −1 ↦ ∞; whence U(R) ⊂ L. Finally, U^{2}(z) = \frac{z+1}{−z}. Thus U^{2} maps 0 ↦ ∞, ∞ ↦ −1, 1 ↦ −2, and −1 ↦ 0; whence U^{2}(R) ⊂ L as well.

The point is that S sends the left half to the right half, whereas U and U^{2} send the right half to the left half. Therefore, in their action on ℙ^{1}(ℝ), we may imgine that S is playing ping-pong against the U’s.

We now prove the claim: If Ψ is not injective, then there is an element w ∈ G with Ψ(w) = 1 ∈ P SL_{2}(ℤ). As w ∈ ⟨S̃, Ũ|S̃^{2} = Ũ^{3} = 1⟩, w can be written as a reduced word in S̃, Ũ, and Ũ^{2} = Ũ^{−1} in which no two Ũ terms next appear to each other. As Ψ(gwg^{−1}) = 1 for all g ∈ G, by conjugating by an appropriate g ∈ G, we may assume that w starts and ends with Ũ and Ũ^{−1} (i.e. that S̃ is neither the first nor last letter in w), hence w = Ũ^{ε_{1}} S̃Ũ^{ε_{2}} S̃ ⋯ S̃Ũ^{ε_{n}} for some n with ε_{i} = ±1 for each i. Then U^{ε_{1}}SU^{ε_{2}}S ⋯ SU^{ε_{n}} = Ψ(w) = 1 acts as the identity on ℙ^{1}(ℝ). However, this contradicts the ping-pong game that S and the U’s are playing against each other: since there are n U letters and n − 1 S letters, we see that Ψ(w) maps R into L. In particular have 1 = Id_{ℙ^{1}(ℝ)}(1) = (Ψ(w))(1) ∈ L, which is a contradiction. Therefore, there is no such w ∈ G and Ψ must be injective. □

Remark. People take this ping-pong idea to great extents; there is even a whole abstract formulation and a “ping-pong lemma” and such. Indeed, by utilizing the ping-pong lemma, Jacques Tits proved the following:

Theorem 7.5. (Tit’s Alternative) Over any filed K, for any dimension n, and any group Γ < GL_{n}(K), either Γ contains a copy of 𝔽_{2} (the free group on 2 generators), or Γ is virtualy solvable (i.e. contains a solvable subgroup of finite index).

Exercise. Prove, using the same techniques, that SL_{2}(ℤ) ≅ ⟨a, b|a^{4} = b^{6} = 1⟩.