Research Article A Note on Iwasawa-Type Decomposition

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1 International Mathematics and Mathematical Sciences Volume 2011, Article ID , 10 pages doi:101155/2011/ Research Article A Note on Iwasawa-Type Decomposition Philip Foth 1, 2 1 CEGEP, Champlain St Lawrence, QC, Canada G1V 4K2 2 Department of Mathematics, The University of Arizona, Tucson, AZ , USA Correspondence should be addressed to Philip Foth, phfoth@gmailcom Received 2 December 2010; Accepted 25 February 2011 Academic Editor: Hans Engler Copyright q 2011 Philip Foth This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited We study the Iwasawa-type decomposition of an open subset of SL n, C as SU p, q ANWeshow that the dressing action of SU p, q is globally defined on the space of admissible elements in AN We also show that the space of admissible elements is a multiplicative subset of AN We establish a geometric criterion: the symmetrization of an admissible element maps the positive cone in C n into itself 1 Introduction In Poisson geometry, the groups SU p, q and AN the upper-triangular subgroup of SL n, C with real positive diagonal entries are naturally dual to each other 1 Therefore, it is important to know the geometry of the orbits of the dressing action We show that the right dressing action of SU p, q is globally defined on the open subset of the so-called admissible elements of AN see Section 2 We also show that the admissible elements can be characterized as follows: these are exactly those elements of AN whose symmetrization maps the closure of the positive cone in C n into the positive cone In addition, we establish a useful fact that the set of admissible elements is a multiplicative subset of AN 2 Admissible Elements Let p and q be positive integers and let n p q Consider the space C n and the group G 0 SU p, q, thegroupg 0 is the subgroup of G SL n, C, which preserves the following sesquilinear pairing on C n :

2 2 International Mathematics and Mathematical Sciences x, y p n x i y i x j y j j p 1 21 Denote the corresponding norm by x Avectorx C n is said to be timelike if the square of its norm is positive, and spacelike, if it is negative Let A G be the subgroup of positive real diagonal matrices and N be the unipotent upper triangular subgroup Denote by g, a, n,andg0 the Lie algebras of G, A, N,andG 0 The Iwasawa-type decomposition for G states that for an open dense subset G of G one has 2, page 167 : G G 0 ẇan, w W/W 0 22 where W W g, a is the Weyl group, W 0 is the subgroup of W with representatives in K SU n G 0,andẇ is a representative of w W/W 0 in SU n An important question is to determine which elements of G allow such a decomposition for a given choice of w The case of particular interest is when w 1, since it is related to the dressing action in Poisson geometry Let J p,q be the diagonal matrix J p,q diag 1, 1,,1, 1, 1,, 1 Introducethe }{{}}{{} p q following involution on the space of n n matrices: A J p,q A J p,q, 23 where A is the usual conjugate transpose The Lie algebra g splits, as a vector space, into the direct sum of ±1-eigenspaces of : g q g0 LetalsoQ G be the submanifold of elements satisfying A A Clearly, exp q Q The next definition is quintessential for this paper We say that λ diag ( λ 1,,λ p,μ 1,,μ q ) a 24 is admissible if λ i >μ j for all i, j Clearly, using the action of W 0, one can arrange λ i s and μ j s in the nonincreasing order, and the condition of admissibility will become simply λ p >μ 1 Same definition applies for A Next, we say that an element X q is admissible, if it is G 0 -conjugate to an admissible element in a The set of admissible elements in q form an open cone Define theadmissibleelementsinq as the exponents of those in q Finally, an element b AN is called admissible, if it obtained from an admissible element of A by the right dressing action The above definition stems from the work of Hilgert, Neeb and others, see, for example, 3 Recall the definition of the right dressing action of G 0 on AN Letb AN and g G 0 Assume there exist b AN and g G 0 such that bg g b Inthiscasewewrite b b g One of the important properties of the set of admissible elements can be observed in the following result, which asserts that admissible elements map the timelike cone into itself Lemma 21 Let s Q be admissible and let x C n be timelike Then, sx is timelike as well

3 International Mathematics and Mathematical Sciences 3 Proof Decompose s g 1 e λ g,whereg G 0 and λ diag λ 1,,λ p,μ 1,,μ q a is admissible, so that we have λ i >μ 1 μ j for all 1 i p and 2 j q Let y gxsincethegroupg 0 preserves the cone of timelike elements, y is timelike as well and, denoting its coordinates y ( z 1,,z p,w 1,,w q ), 25 we see that y 2 p q z i 2 wj 2 r, r R, r > 0 26 j 1 Now, let us show that e λ y is timelike, by using the above equation and expressing w 1 2 in terms of the other coordinates: p z i 2 q e 2μ j j 1 wj 2 p z i 2 q e 2μ j wj 2 r > 0 27 The last step in the proof is to notice that g 1 preserves the timelike cone, and thus sx g 1 e λ y is timelike Later on, in Proposition 44, we will show that conversely, admissible elements can be characterized by this property 3 Gram-Schmidt Orthogonalization and Admissible Elements Here we will give an explicit computational indication that if λ a is admissible, then the whole orbit exp λ G 0 admits a global decomposition, that is, exp λ G 0 G 0 AN Therefore, the right dressing action is globally defined on the set of admissible elements in AN Note that this is not true for the left dressing action, as a simple 2 2examplecanshow Next, we will use the Gram-Schmidt orthogonalization process to show that for any g G 0 and admissible λ,wehaveexp λ g G 0 AN A short proof of this will be given in Section 4 Let us denote the columns of g by w 1,,w n Thecolumnsofg are pseudo- orthonormal with respect to,, andthefirstp columns are timelike, and the last q are spacelike Denote the columns of exp λ g by v 1,,v n This element of G is obtained from g by multiplying the first row by e λ 1,,thepth row by e λ p,the p 1 st row by e μ 1,,andthe last row by e μ q An important observation is that due to the admissibility of λ,thefirstp columns of exp λ g will remain timelike, however nothing definite can be said about the last q The decomposition exp λ g sb with s G 0 and b AN is an analogue of the Gram-Schmidt orthogonalization process for the pseudometric, Denote the columns of s by u 1,,u n Proving the existence of such decomposition amounts to showing that if we follow the Gram-Schmidt process, the first p columns of s will

4 4 International Mathematics and Mathematical Sciences be timelike, and the last q will be spacelike, and that the diagonal entries of b will be positive real numbers Let us denote the diagonal entries of b by r 1,,r n and the off-diagonal by m ij,wherem ij 0fori>j Consider the first step of the Gram-Schmidt process, namely that r 1 v 1 and u 1 v 1 /r 1 Sincev 1 is timelike, we see that r 1 is real positive and that u 1 is timelike Now we move to the second step, obviously, under the assumption that p>1,which we consider in detail, because it lays the foundation for the other columns: v 2 m 12 u 1 r 2 u 2 31 Here m 12 v 2, u 1 and r 2 v 2 m 12 u 1 In order to complete this step we need to show that the vector v 2 m 12 u 1 is timelike Note that v 2 m 12 u 1 2 v 2 2 v 2, u 1 2, 32 so we just need to show that this number is positive Since u 1 2 1, this is equivalent to showing that v 2 2 u 1 2 > v 2, u 1 2, 33 or, after multiplying both sides by r 2 1,that v 2 2 v 1 2 > v 2, v Denote the coordinates of the vector w 1 by a 1,,a p,b 1,,b q, and the coordinates of w 2 by c 1,,c p,d 1,,d q Wehave p q a i 2 p q bj 2 c i 2 dj 2 1, j 1 j 1 p q a i c i b j d j 0 j The coordinates of the vector v 1 are given by (e λ 1 a 1,,e λ p a p,e μ 1 b 1,,e μ q b q ) 37 and v 2 by (e λ 1 c 1,,e λ p c p,e μ 1 d 1,,e μ q d q ) 38

5 International Mathematics and Mathematical Sciences 5 The RHS of 34 now becomes as follows, using 35 : p a i 2 q e 2μ j j 1 p a i 2 p c i 2 bj 2 p c i 2 q e 2μ j j 1 dj 2 q e 2μ j b j 2 q e 2μ j dj 2 e 2μ 1, 39 which is strictly greater than p a i 2 q e 2μ j bj 2 p c i 2 q e 2μ j dj 2, 310 which in turn, by Cauchy-Schwarz, is greater or equal than p a i c i q ( e 2μ j ) b j d j 2, 311 which is exactly v 1, v 2 2, if we take into account 36 Now, we will similarly show that u k is timelike for any k p Wehave k 1 v k m lk u l r k u k, l where m lk v k, u l This amounts to showing that v k 2 k 1 > v k, u l 2 l Consider the subspace V C n spanned by u 1,,u k 1, which is the same subspace that is spanned by v 1,,v k 1 It is isomorphic to the subspace W spanned by w 1,,w k 1, and the explicit isomorphism is given by multiplying the first coordinate by e λ 1,,thepth by e λ p,the p 1 st by e μ 1,,andthelastbye μ q

6 6 International Mathematics and Mathematical Sciences Denote by y k k 1 l 1 m lku l If v k, y k 0, then by definition of y k and m lk this would imply that v k is orthogonal to all the u i for 1 i k 1, and thus all m lk 0 and, similar to the case k 1, the vector u k v k / v k is timelike as desired Thus, we can assume that Define k 1 α k v k, y k m lk 2 > l 1 u y k αk 315 This is an essentially unique element of V up to a circle factor with the property that: k 1 v k, u u m lk u l l Let w also be the unit vector in W, which is the rescaled image of u under the above homomorphism The previous proof for the case k 2 now applies verbatim, with the role of w 1 played by w and the role of v 2 played by v k Thus we have established that the first p columns of s are timelike vectors, and that r 1,,r p are positive real numbers Due to the Sylvester s law of inertia for quadratic forms, the only way to complete this to a pseudo- orthonormal basis with respect to, is to add q spacelike vectors therefore if we continue the Gram-Schmidt orthogonalization process, then the last q columns of s will be spacelike as required 4 Iwasawa-Type Decomposition Let us introduce more notation Denote by Q adm and qadm thesetsofadmissibleelementsin Q and q, respectively Recall that Q adm exp qadm The following result is straightforward Lemma 41 On these sets of admissible elements the map exp is a diffeomorphism Now, consider the symmetrization map Sym : AN Q, Y Y Y 41 This map sends the orbits of the dressing action to the conjugation orbits by the action of G 0 and therefore maps admissible elements to admissible In fact, on the set of admissible elements, this map is, again, bijective, and, moreover, a simple computation of the differential can show that this is a diffeomorphism Let, as before, λ a be admissible, let g 0 G 0, and let us find an explicit decomposition exp λ g 0 g 0 an, 42

7 International Mathematics and Mathematical Sciences 7 with g 0 G 0, a A, andn N Notice that the symmetrization map yields g 1 0 (2 exp λ ) g 0 n a 2 n 43 The right hand side provides the Gauss decomposition also known as the triangular or the LDV-factorization of the left hand side It exists if and only if the leading principal minors of the matrix on the left are nonzero However, it was shown in 4, Proposition41, that the eigenvalues of the principal minors of an admissible element satisfy certain interlacing conditions, similar to the Gelfand-Tsetlin conditions in the Hermitian symmetric case In particular, since all the eigenvalues of s Q adm are positive, then the eigenvalues of any leading principal minor would be positive as well, and therefore the Gauss decomposition exists The diagonal entries of a 2 are the ratios of the leading principal minors: ( a 2) ii Δ i Δ i 1 44 Thus we also see that the entries of a 2 established the following are positive, as required Thus, we have Proposition 42 If λ a is admissible, then the whole orbit exp λ G 0 admits a global decomposition, that is, exp λ G 0 G 0 AN Another interesting property of AN adm is that it is a multiplicative set Proposition 43 If two elements b 1 and b 2 from AN are admissible, b 1,b 2 AN adm, then their product b 1 b 2 is admissible as well Proof Applying the dressing action, we can actually assume that b 2 a A adm This follows fromthefactthatforg G 0 and b 1,b 2 AN: g b 1b 2 g b b1 2, b 1 b 2 g b gb 2 1 b g 45 2 Applying the symmetrization map to b 1 a, we obtain a new matrix f ab 1 b 1a Q, whichwe need to prove admissible Since the space A adm is clearly connected and a is admissible, consider a path a t such that a 0 Id, a 1 a and a t A adm for 0 <t 1 Also denote f t a t b 1 b 1a t Assume, on the contrary, that f 1 f is not admissible and let E 0, 1 be the nonempty subset defined by the property that f t for t Eis not admissible Then consider τ inf E We recall 3 that the space qadm forms a convex cone in q, therefore a simple infinitesimal computation can show that Q adm is a connected open subset of Q and any element from the boundary of Q adm,suchasf τ, has the property that its eigenvalues remain real and, moreover, the lowest eigenvalue λ p corresponding to the timelike cone and the highest eigenvalue μ 1 corresponding to the spacelike cone collide: λ p μ 1 βthus,there

8 8 International Mathematics and Mathematical Sciences exists a whole plane of eigenvectors containing vectors from both C n and C n Thusitmust contain an eigenvector, which we denote by z, from the null cone For this eigenvector we have the following equation: f τ z βz Or, equivalently, a τ b 1 b 1 a τ z βz 46 Note that both a τ and b 1 b 1, being admissible, map the timelike cone C n into itself by Lemma 21 Similar proof shows that they map the null cone minus the origin C n 0 \{0} inside the timelike cone Therefore, the element a τ b 1 b 1 a τ also maps C n 0 \{0} inside of C n and thus z cannot be its eigenvector In geometric terms, we have established the following characterization of the space of admissible elements Q adm Proposition 44 An element s Q is admissible if and only if it has real positive eigenvalues and maps the closure of the timelike cone C n C n C n 0 into the timelike cone C n (plus the origin) Notice that since a from the above proof is admissible diagonal, the pseudohermitian analogue of the Rayleigh-Ritz ratio R A x Ax x x 47 for the matrices f and s b 1 b 1 will have the following properties Let x C n be timelike, then, as we saw earlier, ax is timelike as well with a bigger norm by Lemma 21 Thus, for x C n,wehave R asa x x asax x x > ax s ax ax ax R s ax 48 It follows since a maps C n to itself, that min x C n R s x min R s ax < min R asa x 49 x C n x C n From 5,Theorem21, it follows that λ 1 asa >λ 1 s Similar inequalities can be established for other eigenvalues The notion of admissibility can be extended to the whole group G SL n, C Wesay that g G is admissible, if it decomposes as g hb, withh G 0 and an admissible b AN adm Such a decomposition, if exists, is clearly unique Note, moreover, that g g b b The singular admissible spectrum of g are the square roots of the spectrum of g g Note of warning: one should not define Q adm by inducing this definition from G,butratheraswedid previously

9 International Mathematics and Mathematical Sciences 9 Suppose that g 1, g 2 are two admissible elements from G such that g 1 h 1 b 1 and g 2 h 2 b 2 Thenwehaveg 1 g 2 h 1 b 1 h 2 b 2 h 1 h b 1 2 bh 2 1 b 2, where the powers mean the dressing actions This is possible, because b 1 is admissible Thus we have established that the product g 1 g 2 is admissible if and only if the product b h 2 1 b 2 is such 5 Example of the Group SU(1,1) The Poisson geometry related to the group SU 1, 1 was considered in detail in 6 Herewe just recall several facts to illustrate the results of this paper First of all, the space qadm is the following convex cone of matrices z x iy : x, y, z R, z 2 x 2 y 2 > 0, z>0 51 x iy z Next, we define Q adm exp qadm Consideranelement t1 m Q, m t 2 52 where t 1,t 2 R, andm C satisfy the determinant condition t 1 t 2 m 2 1 It is admissible if and only if its eigenvalues are real and positive, and, moreover, the eigenvalue for the timelike cone is greater than the other one This translates into the following two conditions on the coefficients of this matrix: t 1 t 2 > 2, t 1 > 1 53 Next, an element r n AN 0 r 1 54 is admissible if and only if r>1, r 2 r 2 n 2 > 2 55 Now, let us write down the Iwasawa-type decomposition G G 0 AN explicitly Let us consider a general element of SL 2, C : a b c d : a, b, c, d C, ad bc 1 56

10 10 International Mathematics and Mathematical Sciences The condition that this general element admits such a decomposition is simply a > c : a b c d a a 2 c 2 c a 2 c 2 ( ) c a 2 c 2 a 2 c 2 c a 2 c 2 b a 2 c 2 a a 1 0 a 2 c 2 a 2 c 2 57 Thus, it is quite easy to see that for any element g G and exp λ A adm, the element exp λ g G 0 AN Explicitly, if u v e λ 0 g, exp λ, λ > 0, 58 v u 0 e λ then exp λ g ( e λ u e λ ) v e λ v e λ u 59 is an admissible element of G The statemement that the admissibility of two elements b 1 and b 2 of AN SL 2, C implies the admissibility of their product is also a short computational affair References 1 P Foth and J-H Lu, Poisson structures on complex flag manifolds associated with real forms, Transactions of the American Mathematical Society, vol 358, no 4, pp , N J Vilenkin and A U Klimyk, Representation of Lie Groups and Special Functions Vol 3, vol 75, Kluwer Academic Publishers Group, Dordrecht, The Netherlands, K-H Neeb, Holomorphy and Convexity in Lie Theory, vol28ofde Gruyter Expositions in Mathematics, Walter de Gruyter & Co, Berlin, Germany, P Foth, Polygons in Minkowski space and the Gelfand-Tsetlin method for pseudo-unitary groups, Geometry and Physics, vol 58, no 7, pp , P Foth, Eigenvalues of sums of pseudo-hermitian matrices, Electronic Linear Algebra, vol 20, pp , P Foth and M Lamb, The Poisson geometry of SU 1, 1, Mathematical Physics, vol 51, no 9, p 10, 2010

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