1 The Iwasawa Decomposition for Symmetric Spaces of Non-Compact Type Let M be a symmetric space of non-compact type and denote G = Iso(M)◦. Fix p 2 M and denote K = StabG(p). Consider the map −1 σ : G ! G; g 7! spgsp ; where sp is the geodesic symmetry around p. This makes (G; K; σ) into a Riemannian-symmetric pair. We have the Cartan involution θ := Deσ : g ! g and the Cartan decomposition g = E1(θ) ⊕ E−1(θ) = k ⊕ p and consider the positive definite symmetric bilinear form hX; Y i = −Bg(X; θ(Y )): This bilinear form was the basis for our discussion of roots and root systems. Fix a maximal abelian subspace a. Recall the following definition and theorem: Definition. An element X 2 p is called regular if Z(X) \ p is abelian. Theorem. For every maximal abelian subspace a ⊂ p there exists X 2 p such that a = Z(X) \ p: In particular, X is regular. Furthermore for any two maximal abelian subspaces a; a0 ⊂ p there is k 2 K such that 0 AdG(k)(a) = a : This Theorem is proved for symmetric spaces of compact type by using that the Lie subgroup associated to a is in fact a torus and so we can find an element in p such that the geodesic in the direction of this element is dense. Then one easily sees that this element commutes with a. One then traverses to symmetric spaces of non-compact type by using the duality between symmetric spaces of compact and non-compact type. Note that fad(H): H 2 ag is simultaneously diagonalizable so for λ 2 a∗ we write 0 6= gλ = fX 2 g : (ad(H))X = λ(H)X for all H 2 ag: By choosing a basis we can define Σ+ the set of positive roots and denote n = ⊕λ2Σ+ gλ: Note n is a nilpotent Lie algebra as [gα; gβ] ⊂ gα+β: and there are only finitely may λ such that gλ 6= 0. 2 Theorem. (Iwasawa decomposition) With the above notations we can write g = k ⊕ a ⊕ n: Denote by A the Lie group associated to a and by N the Lie group associates to n. Then the groups A and N are simply connected and furthermore the multiplication map K × A × N −! G; (k; a; n) 7−! kan is a diffeomorphism. Remark. The Iwasawa decomposition holds for all semisimple Lie groups with the same proof. The proof of the Lie algebra decomposition just uses symmetric positive definite bilinear form hX; Y i = −B(X; θY ): As one can find such a positive definite bilinear form on any semisimple Lie algebra, the proof then carries over. Fix x 2 M(1) and let X 2 p be the unit vector such that exp(tX):p = γxp, where γxp is the unique unit speed geodesic with γpx(0) = p that is asymptotic to x. Recall that we have the following commutative diagram: Deπ p TpM exp expp G π M This commutative diagram holds as we can associate to each geodesic γ the following one-parameter subgroup of transvections Tt = sγ(t=2)γ(0) and then using the fact that one parameter subgroups are precisely given by the exponential map. So there exists a unique X 2 p such that γpx(t) = exp(tX):p. Then we consider the parabolic subgroup at x Gx = fg 2 G : g ◦ γ 2 [x] for all γ 2 [x]g = fg 2 G : lim exp(−tX)g exp(tX) existsg t!1 We hence have a well defined map Tx : Gx ! G; g 7! exp(−tX)g exp(tX): Proposition. The map Tx is a continuous idempotent homomorphism. Proof. Let g; h 2 Gx, then we have that Tx(gh) = lim exp(−tX)gh exp(tX ) t!1 = lim exp(−tX)g exp(tX) exp(−tX)h exp(tX ) = Tx(g)Tx(h); t!1 so Tx is a group homomorphism. To show that Tx is idempotent we note that Tx(Tx(g)) = lim exp(−tX)Tx(g) exp(tX ) t!1 = lim lim exp(−tX) exp(−sX)g exp(sX) exp(tX) t!1 s!1 = lim lim exp(−(s + t)X)g exp((s + t)X) = Tx(g): t!1 s!1 We will prove later that Tx is continuous. 3 We now introduce the subgroups needed in the more general Iwasawa decom- position: Nx := ker(Tx); Zx := im(Tx); Kx := K \ Zx; Ax := exp(Z(X) \ p): We are now ready to state the main theorem of this talk. Theorem. (Generalized Iwasawa decomposition) We have diffeomorphisms given by multiplication Kx × Ax × Nx ! Gx and K × Ax × Nx ! G: Remark. Note that Ax is not necessarily a subgroup by the above. However, if X 2 p is regular then Z(X) \ p is maximal abelian and the above is precisely the normal Iwasawa decomposition. Example. Let H be the upper half plane and let p = i 2 H. Then we have that G = PSL2(R) and K = PSO2(R). We consider x 2 H(1) to we boundary point at infinity which is above p and then we have that X is the unit vector at i that t points upwards. So γi1(t) = e · i and so 1 2 0 γi1(t) = exp t 1 :i: 0 − 2 So 1 2 0 X = 1 0 − 2 a b and for g = 2 PSL ( ) we have c d 2 R a e−tb exp(−tX)g exp(tX) = etc d so a b G = for a 2 nf0g and b 2 ={±I g x 0 a−1 R R 2 and 1 b N = for b 2 ={±I g; x 0 1 R 2 a 0 Z = for a 2 nf0g ={±I g: x 0 a−1 R 2 Then 1 0 K = f g={±I g and A = Z : x 0 1 2 x x 4 Example. Now consider M = SLn(R)=SOn(R) and G = PSLn(R) and K = PSOn(R) then we have that θ : g ! g;X 7! −XT and the Cartan decomposition g = E1(Deσ) ⊕ E−1(Deσ) = k ⊕ p with T k = fX 2 sln(R): X + X = 0g T p = fX 2 sln(R): X = X g: So we take some element X 2 p. By some base change we can assume that X is a diagonal matrix. So write X = diag(λ1; : : : ; λn) with λ1 ≥ ::: ≥ λn. Note that X is regular if and only if all the λ are distinct. However we don't assume that for now. Without loss of generality we say that λ1 ≥ ::: ≥ λn. Note then that t(λj −λi) (exp(−tX)g exp(tX))ij = e gij: So we have that Gx = fg 2 PSLn(R): gij = 0 if λj > λig and we have Zx = fg 2 PSLn(R): gij = 0 if λj 6= λig: For n = (nij) 2 Nx we have that t(λj −λi) e ! δij: So we get the following characterization of elements of Nx: 1. If i = j, then nii = 1. 2. If i > j, then λi ≥ λj and so nij = 0 3. if i < j, then λi ≤ λj. If λi = λj, then nij = 0. The centralizer of X 2 p the turns out to be Z(X) \ p = fA 2 p : aij = 0 if λi 6= λjg To be even more explicit let n = 3 and consider X = (1; 1; −2). Then we have that 80 1 9 < ∗ ∗ ∗ = Gx = @∗ ∗ ∗A 2 SL3(R) ; : 0 0 ∗ ; 80 1 9 < ∗ ∗ 0 = Zx = @∗ ∗ 0A 2 SL3(R) ; : 0 0 ∗ ; 80 1 9 < 1 0 ∗ = Nx = @0 1 ∗A 2 SL3(R) ; : 0 0 1 ; 80 1 9 < ∗ ∗ 0 = Z(X) \ p = @∗ ∗ 0A 2 sl3(R) : : 0 0 ∗ ; 5 We now prove the first part of the Iwasawa decomposition, namely that Gx = KxAxNx. Proof. (First Part of Iwasawa decomposition) The proof comprises several steps: 1. Gx = ZxNx, where the decomposition is unique. To see this note that any g 2 Gx can be written as −1 g = Tx(g)Tx(g) g; −1 where Tx(g) 2 Zx and Tx(g) g 2 Nx as Tx is idempotent. To see uniqueness assume g = zn then Tx(g) = z and so z is determined by the choice of g. So also n is unique. This in fact also shows that Tx is continuous: It suffices to check continuity at e 2 Gx. So let gi ! e as n ! 1. Then we have by the last corollary gi = zini with i ! 1. By uniqueness of the decomposition it follows that zi ! e and ni ! e. So we have Tx(gi) = zi ! e = Tx(e): Hence Tx is continuous. 2. Zx = fg 2 Gx : exp(−tX)g exp(tX) = g for all t 2 Rg. To see this let g 2 Zx. Then g = Tx(h) for h 2 Zx. Thus we have that Zx ⊂ im(Tx). Conversely if y 2 im(Tx), then we have that some g 2 Gx such that Tx(g) = y. Then note that for any s 2 R exp(−sX)y exp(sX) = exp(−sX)Tx(g) exp(sX) = Tx(exp(−sX))Tx(g)Tx(exp(sX)) = Tx(exp(−sX)g exp(sX)) = lim exp(−(s + t)X) t!1 so indeed y 2 Zx. 3. Ax ⊂ Zx: To see this note that for Y 2 Z(X) \ p we have that [tX; Y ] = 0 for all t 2 R and hence exp(Y ) exp(tX) = exp(Y + tX) = exp(tX + Y ) = exp(tX) exp(Y ): So indeed exp(Y ) 2 Zx.