DECOMPOSING SL 2 (R)

DECOMPOSING SL 2 R KEITH CONRAD Introduction The group SL 2 R is not easy to visualize: it naturally lies in M 2 R, which is 4- dimensional the entries of a variable 2 2 real matrix are 4 free parameters We will derive a product decomposition for SL 2 R use it to get a concrete image of SL 2 R Inside SL 2 R are the following three subgroups: { K sin θ cos θ }, A { /r } { x : r > 0, N } Theorem We have a decomposition SL 2 R KAN: every g SL 2 R has a unique representation as g kan where k K, a A, n N This formula SL 2 R KAN is called the Iwasawa decomposition of the group Don t confuse the use of a in Theorem as the label for a matrix in A with a as a real number in the matrix a c d b The distinction should always be clear from the context Since SL 2R is defined by the single equation ad bc inside of M 2 R, it is a manifold of dimension 4 3 The subgroups K, A, N are each -dimensional K S, A R >0, N R, Theorem shows they fully account for the 3 dimensions of SL 2 R The subgroups in the Iwasawa decomposition are related to conjugacy classes We will see that a matrix in SL 2 R is, up to sign, conjugate to a matrix in K, A, or N 2 Iwasawa decomposition To derive the Iwasawa decomposition of SL 2 R we will use an action of this group on bases in R 2 For g a c d b in SL 2R, apply it to the stard basis e, e 2 The vectors ge a c, ge 2 are also a basis of R 2 We will pass from this new basis of R 2 back to the stard basis e, e 2 of R 2 by a sequence of transformations in SL 2 R that amounts to something like the Gram-Schmidt process which turns any basis of R n into an orthonormal basis of R n Let θ be the angle from the positive x-axis to ge Let ρ θ be the counterclockwise rotation of the plane around the origin by θ, so ρ θ ge is on the positive x-axis Because det g is positive, the ordered pair of vectors ge, ge 2 has the same orientation as the ordered pair e, e 2, so ρ θ ge 2 is in the upper rather than lower half-plane Since ρ θ ge is a positive scalar multiple of e, we want to divide ρ θ ge by its length so it becomes e Its length is r ρ θ ge ge a 2 + c 2 Applying / /r will have the desired effect ρ θ ge e, but this matrix doesn t have determinant On the other h, / 0 r also has the desired effect on ρ θge has determinant So b d

2 KEITH CONRAD apply the matrix / 0 r to R2 It sends ρ θ ge to / 0 r ρ θge e What does it do to ρ θ ge 2? The vector / 0 r ρ θge 2 is in the upper half-plane because / 0 r has positive determinant along with e it forms two edges of a parallelogram with area because / 0 r has determinant ± A parallelogram with area having base e must have height, so / 0 r ρ θge 2 x for some x Any horizontal shear transformation t, which has determinant, fixes the x-axis acts as a stretching along each horizontal line Applying the horizontal shear transformation 0 x to R2 takes x to 0 e2 fixes e We have finally returned to the stard basis e, e 2 from the basis ge, ge 2 by a sequence of transformations in SL 2 R Our overall composite transformation is x so x / 0 r / 0 r ρ θ, ρ θ g sends e to e e 2 to e 2 A linear transformation on R 2 is determined by what it does to a basis, so x / 0 ρ 0 r θ g Solving for g, /r x g ρ θ 0 r sin θ cos θ /r KAN x Such an expression for g as a product kan with k K, a A, n N is called the Iwasawa decomposition of g To check this decomposition is unique, for any angle θ, r > 0, x R, set r x r cos θ xr cos θ /r sin θ g sin θ cos θ /r r sin θ xr sin θ + /r cos θ If this is a b c d SL 2R then 2 r a 2 + c 2 > 0, cos θ a r, sin θ c r, { b+/r sin θ x r cos θ, if cos θ 0, d /r cos θ r sin θ, if sin θ 0 Substituting the formulas for cos θ sin θ into the formula for x, using ad bc, we obtain the uniform formula ab + cd 22 x a 2 + c 2 All the parameters in the matrices making up the Iwasawa decomposition of a c d b are determined in 2 22, so the Iwasawa decomposition is unique This completes

DECOMPOSING SL 2 R 3 the proof of Theorem In an appendix we derive the Iwasawa decomposition using a different action of SL 2 R, on the upper half-plane The Iwasawa decomposition for SL 2 R extends to higher dimensions: SL n R KAN where K SO n R {T GL n R : T T I n, det T }, A is the group of diagonal matrices with positive diagonal entries determinant N is the group of uppertriangular matrices with s along the main diagonal While A N are both isomorphic to R when n 2, N becomes nonabelian for n > 2 The group K is compact, the group A R >0 n R n is abelian, N is a nilpotent group This explains the notation A N, for abelian nilpotent Since r x r 23 2 x, /r /r we can move any element of A past an element of N on either side at the cost of changing the element of N Therefore AN NA is a subgroup of SL 2 R Explicitly, { } y x 24 AN : y > 0, x R /y The Iwasawa decomposition KAN KAN for SL 2 R is the analogue of the polar decomposition S R >0 for C In the Iwasawa decomposition, neither K nor AN nor A or N is normal in SL 2 R For example, the conjugate of an element of K by is usually not in K the conjugate of an element of AN by 0 is usually not in AN Because of the non-normality, it is not easy to describe the group operation in SL 2 R in terms of its Iwasawa decomposition This decomposition is important for other purposes, such as the following Corollary 2 As a topological space, SL 2 R is homeomorphic to the inside of a solid torus Proof Let f : K A N SL 2 R by fk, a, n kan This is continuous, by Theorem it is surjective We can write down an inverse function using the computations at the end of the proof of Theorem For g a c d b in SL 2R, define rg a 2 + c 2 a/rg c/rg rg 0 kg, ag, c/rg a/rg /rg ab + cd/a ng 2 + c 2 The function g kg, ag, ng from SL 2 R to K A N is continuous is an inverse to f Topologically, K S, A R >0 R, N R Therefore topologically, SL2 R S R 2 The plane R 2 is homeomorphic to the open unit disc D by v v/ + v 2 with inverse w w/ w 2, where is the usual length function on R 2, so as a topological space SL 2 R is homeomorphic to S D, which is the inside of a solid torus As an alternate ending, on the decomposition K A N S R >0 R treat the product R >0 R as the right half plane {x + iy : x > 0} identify it with the open unit disc D by the Cayley transformation z z /z + Vertical lines in the half-plane are sent to circles inside D that are tangent to the unit circle at

4 KEITH CONRAD Although the proof of Corollary 2 shows SL 2 R S R 2 are homeomorphic as topological spaces, they are not isomorphic as groups Equivalently, the homeomorphism K A N SL 2 R in Corollary 2 is not a group homomorphism The Iwasawa decomposition of a matrix in K, A, or N is the obvious one For a lower triangular matrix y 0, which is in none of these subgroups, the inverse map in the proof of Corollary 2 gives us the decomposition / + y 2 y/ + y 2 y/ + y 2 / + y 2 + y 2 0 / + y 2 y/ + y 2 Remark 22 The inside of a solid torus has a circle as a strong deformation retract, so the fundamental group of SL 2 R is the same as that of a circle, namely Z In terms of SL 2 R itself, one circle inside it is the subgroup K a strong deformation retract to K can be written down using the Iwasawa decomposition make a suitable path in the group from kan to k From the connection between covering spaces subgroups of the fundamental group, SL 2 R admits a unique covering space of degree d for each positive integer d the universal covering space of SL 2 R is the inside of a solid cylinder R D homeomorphic to R 3 The degree-2 cover of SL2, R is an important group called the metaplectic group That the Iwasawa decomposition gives us a picture of SL 2 R is a striking geometric application Here is an algebraic application whose punchline is the corollary Theorem 23 The only continuous homomorphism SL 2 R R is the trivial homomorphism Proof Let f : SL 2 R R be a continuous homomorphism Then fkan fk + fa + fn We will show f is trivial on K, A, N, thus f is trivial on KAN SL 2 R Since K S, the elements of finite order in K are dense Since R has no elements of finite order except 0, f is trivial on a dense subset of K thus is trivial on K by continuity As an alternate argument, since K is a compact group so is fk, the only compact subgroup of R is {0} Now we look at f on A N Since A R >0 R by /r log r N R by 0 x x, both algebraically topologically, describing the continuous homomorphisms from A N to R is the same as describing the continuous homomorphisms from R to R Any continuous homomorphism R R has the form x tx for some real number t see where goes, call that t, then appeal to the denseness of Q in R Therefore x f t log r, f t /r x for some t t Applying f to both sides of 23, t log r + t x t r 2 x + t log r, so t x t r 2 x for all r > 0 x R Thus t 0 eg, take x r 2 to see this This shows f is trivial on N It remains to show f is trivial on A For this we appeal to the conjugation relation 0 0 / 0 /r 0 0 r /r Applying f, we get f /r f /r, so f /r 0

DECOMPOSING SL 2 R 5 Corollary 24 Any continuous homomorphism SL 2 R GL n R has image in SL n R Proof Let f : SL 2 R GL n R be a continuous homomorphism Composing f with the determinant map GL n R R gives a continuous homomorphism det f : SL 2 R R Since SL 2 R is connected Corollary 2, its image under det f is a connected subgroup of R, hence lies in R >0 As R >0 R both topologically algebraically, det f is trivial by Theorem 23 Thus detfg for all g SL 2 R, so fg SL n R for all g SL 2 R Example 25 We will construct a continuous homomorphism GL 2 R GL 3 R see its restriction to SL 2 R has values in SL 3 R Let V Rx 2 + Rxy + Ry 2 be the vector space of homogeneous polynomials in x y of degree 2 It is 3-dimensional, with basis x 2, xy, y 2 For a matrix g a c d b GL 2R Qx, y V, let gqx, y Qax+cy, bx+dy A calculation shows g g 2 Q g g 2 Q, so GL 2 R acts on V This action is a linear change of variables on V given by the entries of 2 2 matrices Since gq + Q gq + gq gsq sgq for s R, the action of g on V is a linear transformation necessarily invertible, since the action of g on V is its inverse Using the basis x 2, xy, y 2 of V, we can compute a matrix representation of g on V : we have gx 2 ax + cy 2 a 2 x 2 + 2acxy + c 2 y 2, gxy ax + cybx + dy abx 2 + ad + bcxy + cdy 2, gy 2 bx + dy 2 b 2 x 2 + 2bdxy + d 2 y 2, so the matrix of g with respect to the basis x 2, xy, y 2 is a2 ab b 2 2ac ad + bc 2bd c 2 cd d 2 Call this matrix fg, so f : GL 2 R GL 3 R is a homomorphism the formula for fg shows f is continuous By a calculation, detfg ad bc 3 deg g 3, so when g has determinant so does fg Example 25 can be generalized For any integer n, the space V n n i0 Rxn i y i of homogeneous 2-variable polynomials of degree n has dimension n + GL 2 R acts on this space by linear changes of variables The restriction of this action to SL 2 R on V n accounts for essentially all interesting actions of SL 2 R on finite-dimensional vector spaces Theorem 26 The homomorphism SL 2 Z[x, y] SL 2 Z[x, y]/x 2 + y 2 is not surjective Proof Let x y be the cosets of x y in Z[x, y]/x 2 + y 2, so x 2 + y 2 An obvious matrix in SL 2 Z[x, y]/x 2 + y 2 is x y 25 y x Suppose there is a matrix Ax, y ax, y bx, y cx, y dx, y

6 KEITH CONRAD in SL 2 Z[x, y] which reduces to the matrix 25 in SL 2 Z[x, y]/x 2 + y 2 A continuous map S SL 2 R is given by u, v v u u v Au, v The image is a loop in SL 2 R which generates π SL 2 R Let h: [0, ] S SL 2 R by ht, u, v Atu, tv Then h0, u, v is a constant map h, u, v is the stard loop in SL 2 R, so we can shrink the loop to a point continuously This is false, so there is no such map 3 Conjugacy Classes The conjugacy class of a matrix in SL 2 R is nearly determined by its eigenvalues, but we have to be a little bit careful so we don t confuse conjugacy in SL 2 R with conjugacy in the larger group GL 2 R For example, its inverse are conjugate in GL 2R by 0 0, whose determinant is These two matrices are not conjugate in SL 2R, since any SL 2 R-conjugate of has a positive upper right entry, by an explicit calculation Theorem 3 Let T SL 2 R If Tr T 2 > 4 then T is conjugate to a unique matrix of the form λ /λ 0 with λ > If Tr T 2 4 then T is conjugate to exactly one of ±I 2, ±, or ± 0 If Tr T 2 < 4 then T is conjugate to a unique matrix of the form sin θ cos θ other than ±I 2 Proof For T SL 2 R, its eigenvalues are roots of its characteristic polynomial, which is X 2 tx +, where t TrT The nature of the eigenvalues of T are determined by the discriminant of this polynomial, t 2 4: two distinct real eigenvalues if t 2 > 4, a repeated eigenvalue if t 2 4, two complex conjugate eigenvalues if t 2 < 4 We will find a representative for the conjugacy class of T based on the sign of t 2 4 Of course matrices with different t s are not conjugate In what follows, if v w are vectors in R 2 whose specific coordinates are not important to make explicit, we will write [v] [w] for the matrix a c d b This matrix is invertible when v w are linearly independent Suppose t 2 > 4 Then T has distinct real eigenvalues λ /λ Let v v be eigenvectors in R 2 for these eigenvalues: T v λv T v /λv In coordinates from the basis v v, T is represented by λ /λ 0 λ 0, so T is conjugate to /λ by the 2 2 matrix [v] [v ] Scaling v keeps it as an eigenvector of T, by a suitable nonzero scaling the matrix [v] [v ] has determinant Therefore T is conjugate to λ /λ 0 in SL 2R We did not specify an ordering of the eigenvalues, so λ /λ 0 /λ 0 have to be conjugate 0 λ to each other in SL 2 R Explicitly, 0 λ /λ 0 0 0 /λ 0 Conjugate matrices 0 λ have the same eigenvalues, so λ /λ 0 µ 0 is conjugate to /µ only when µ equals λ or /λ We can therefore pin down a representative for the conjugacy class of T in SL 2 R as λ /λ 0 with λ > Now suppose t 2 4 The roots of X 2 tx + are both if t 2 or both if t 2 Let λ ± be the eigenvalue for T Extend v to a basis {v, v } of R 2 Scaling v, we may assume the matrix [v] [v ] has determinant Conjugating T by this matrix expresses it in the basis v v as λ 0 x y Since the determinant is, y /λ λ ± Therefore T is conjugate in SL 2 R to a matrix of the form 0 x x or 0 If x 0 these matrices are ±I 2, which are in their own conjugacy class The formulas r r 2 /r /r

/r DECOMPOSING SL 2 R 7 /r r 2 show 0 x is conjugate to either or, depending on the sign of x Similarly 0 x is conjugate to either 0 or 0 The four matrices 0, 0, 0 are nonconjugate in SL 2R, eg, an SL 2 R-conjugate of looks like x with x a perfect square Other cases are left to the reader Finally, suppose t 2 < 4 Now T has complex conjugate eigenvalues which, by the quadratic formula for X 2 tx +, are of absolute value are not ± since t ±2 We can write the eigenvalues as e ±iθ, with sin θ 0 Pick an eigenvector v in C 2 such that T v e iθ v Since e iθ is not real, v R 2 Let v be the vector with coordinates that are complex conjugate to those of v, so T v e iθ v Then v + v iv v are in R 2, with T v + v cos θv + v + sin θiv v T v v sin θv + v + cos θiv v Therefore conjugating T by the invertible real matrix [v + v] [iv v] turns T into We don t know the determinant of [v + v] [iv v], but scaling v by a real sin θ cos θ number v by the same amount, to keep it conjugate can give this conjugating matrix determinant ± If the determinant is then T is conjugate to sin θ cos θ in SL 2R If the determinant is, then reverse the order of the columns in the conjugating marix to give it determinant then T is conjugate to cos θ sin θ cos θ sin θ sin θ cos θ sin θ cos θ in SL 2 R Two matrices cos θ sin θ sin θ cos θ cos ϕ sin ϕ sin ϕ cos ϕ in K can be conjugate only when ϕ ±θ mod 2πZ, by looking at eigenvalues, a direct calculation shows the SL 2 R- conjugate of cos θ sin θ cos θ sin θ sin θ cos θ never equals unless sin θ 0, but we are in a sin θ cos θ case when sin θ 0 When T SL 2 R satisfies TrT 2 > 4 we say T is hyperbolic, when Tr T 2 4 we say T is parabolic, when Tr T 2 < 4 we say T is elliptic This terminology is borrowed from the shape of a plane conic ax 2 + bxy + cy 2 in terms of its discriminant d b 2 4ac: it is a hyperbola when d > 0, a parabola when d 0, an ellipse when d < 0 Up to sign, the hyperbolic conjugacy classes in SL 2 R are represented by matrices in A besides I 2, the elliptic conjugacy classes are represented by matrices in K besides ±I 2, the parabolic conjugacy classes are represented by matrices in N Appendix A Acting on the Upper Half-Plane We will use an action of SL 2 R on the upper half-plane h to obtain the Iwasawa decomposition of SL 2 R in a more efficient manner than the first proof that used an action on bases of R 2 For a c d b GL 2R a non-real complex number z, set a b A z az + b c d cz + d C R

8 KEITH CONRAD By a calculation left to the reader, g g 2 z g g 2 z for g g 2 in GL 2 R, az + b ad bc Imz Im cz + d cz + d 2 Therefore when ad bc > 0, z az + b/cz + d have the same sign for their imaginary parts In particular, if g SL 2 R z is in the upper half-plane then so is gz, so A is an action of the group SL 2 R on the set h This action has one orbit since we can get anywhere in h from i using SL 2 R: A2 y x/ y / y i x + iy Notice that the matrix used here to send i to x + iy is in the subgroup AN see 24 Let s determine the stabilizer of i Saying a c d b i i is equivalent to ai+b/ci+d i, so ai + b di c Therefore d a b c, so a c d b a c c a with a2 + c 2 We can therefore write a cos θ c sin θ, which shows the stabilizer of i is the set of matrices sin θ cos θ This is the subgroup K so h can be viewed as the coset space SL 2R/K on which SL 2 R acts by left multiplication Now we are ready to derive the Iwasawa decomposition For g SL 2 R, write gi x + iy h Using A2, y x/ y x y 0 A3 gi x + iy / i y / i NAi y Since g acts on i in the same way as an element of NA, the stabilizer of i is K, g NAK Thus SL 2 R NAK Applying inversion to this decomposition, SL 2 R KAN That settles the existence of the Iwasawa decomposition To prove uniqueness, assume nak n a k Applying both sides to i, k k fix i so nai n a i For n x a /r, na r x/r /r, so nai x + r2 i In particular, knowing nai tells us the parameters determining n a Hence n n a a, so k k The upper half-plane action of SL 2 R leads in a second way to the formulas 2 22 for the matrix entries in the factors of the Iwasawa decomposition for g a c d b SL 2 R We have just proved anew the existence uniqueness of this decomposition Write, as in Section 2, A4 g kan sin θ cos θ /r x We want to determine the entries of these matrices in terms of the entries of g We will work with g n a k since the SL 2 R-action on h leads to the decomposition NAK rather than KAN: g d b i c a i di b + cd ab ci + a a 2 + c 2 + a 2 + c 2 i Writing this as u + iv, from A3 with g in place of g u + iv in place of x + iy we get n u ab + cd/a 2 + c 2

DECOMPOSING SL 2 R 9 a v 0 / / a 2 + c 2 0 v 0 a 2 + c 2, so ab + cd/a n 2 + c 2 a, a 2 + c 2 0 / a 2 + c 2 Since g kan, k gn a a/ a 2 + c 2 c/ a 2 + c 2 c/ a 2 + c 2 a/ a 2 + c 2 The formulas for the entries of k, a, n match those in 2 22 It is interesting to compare the role of the group K in the geometry of R 2 h As a transformation of R 2, an element of K is a rotation around the origin This is an isometry of R 2 using the Euclidean metric, the K-orbit of a nonzero vector in R 2 is the circle which passes through that vector is centered at the origin As a transformation of h, an element of K is a rotation around i relative to the hyperbolic metric on h This is a hyperbolic isometry of h, the K-orbit of a point in h is the circle through that point which is centered at i relative to the hyperbolic metric The conjugacy class of a matrix T SL 2 R was determined in Theorem 3 in terms of Tr T 2 4, which is the discriminant of the characteristic polynomial of T The sign of this quantity tells us whether T has real or non-real eigenvectors The difference Tr T 2 4 is also relevant to the action of T on the upper half-plane, with fixed points replacing eigenvectors When T a c d b, the fixed-point condition T z z is equivalent to az + b cz + dz, which says cz 2 + d az b 0 The discriminant of this equation, which tells us the number of real roots, is d a 2 + 4bc d 2 2da + a 2 + 4ad a + d 2 4 Tr T 2 4