# A Note on Iwasawa-Type Decomposition

Authors: Philip Foth
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We study the Iwasawa-type decomposition of an open subset of SL(n,ℂ) as SU(p,q)AN. We show 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 ℂn into itself.


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## Body

## 1. Introduction

In Poisson geometry, the groupsSU(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 ofAN whose symmetrization maps the closure of the positive cone in ℂn 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

Letp and q be positive integers and let n=p+q. Consider the space ℂn and the group G0=SU(p,q), the group G0 is the subgroup of G=SL(n,ℂ), which preserves the following sesquilinear pairing on ℂn: (2.1)〈x,y〉=∑i=1pxiy¯i-∑j=p+1nxjy¯j.
Denote the corresponding norm by ∥x∥. A vector x∈ℂn is said to be timelike if the square of its norm is positive, and spacelike, if it is negative.LetA⊂G be the subgroup of positive real diagonal matrices and N be the unipotent upper triangular subgroup. Denote by 𝔤, 𝔞, 𝔫, and 𝔤0 the Lie algebras of G, A, N, and G0. The Iwasawa-type decomposition for G states that for an open dense subset G∨ of G one has [2, page 167]: (2.2)G∨=∐w∈W/W0G0ẇAN,
where W=W(𝔤,𝔞) is the Weyl group, W0 is the subgroup of W with representatives in K=SU(n)∩G0, and ẇ is a representative of w∈W/W0 in SU(n).An important question is to determine which elements ofG 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.LetJp,q be the diagonal matrix Jp,q=diag(1,1,…,1︸p,-1,-1,…,-1︸q). Introduce the following involution on the space of n×n matrices: (2.3)A†=Jp,qA*Jp,q,
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. Let also Q⊂G be the submanifold of elements satisfying A†=A. Clearly, exp(𝔮)⊂Q.The next definition is quintessential for this paper. We say that(2.4)λ⃗=diag⁡(λ1,…,λp,μ1,…,μq)⊂a
is admissible if λi>μj for all i,j. Clearly, using the action of W0, one can arrange λi's and μj's in the nonincreasing order, and the condition of admissibility will become simply λp>μ1. Same definition applies for A. Next, we say that an element X∈𝔮 is admissible, if it is G0-conjugate to an admissible element in 𝔞. The set of admissible elements in 𝔮 form an open cone. Define the admissible elements in Q 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 G0 on AN. Let b∈AN and g∈G0. Assume there exist b′∈AN and g′∈G0 such that bg=g′b′. In this case we write b′=bg.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 2.1.
Lets∈Q be admissible and let x∈ℂn be timelike. Then, sx is timelike as well.Proof.
Decomposes=g-1eλ⃗g, where g∈G0 and λ⃗=diag(λ1,…,λp,μ1,…,μq)⊂𝔞 is admissible, so that we have λi>μ1≥μj for all 1≤i≤p and 2≤j≤q.
Lety=gx. Since the group G0 preserves the cone of timelike elements, y is timelike as well and, denoting its coordinates (2.5)y=(z1,…,zp,w1,…,wq),
we see that
(2.6)‖y‖2=∑i=1p|zi|2-∑j=1q|wj|2=r,r∈R,r>0.
Now, let us show thateλ⃗y is timelike, by using the above equation and expressing |w1|2 in terms of the other coordinates: (2.7)∑i=1pe2λi|zi|2-∑j=1qe2μj|wj|2=∑i=1p(e2λi-e2μ1)|zi|2+∑j=2q(e2μ1-e2μj)|wj|2+re2μ1>0.
The last step in the proof is to notice that g-1 preserves the timelike cone, and thus sx=g-1eλ⃗y is timelike.Later on, in Proposition4.4, 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 ifλ⃗∈𝔞 is admissible, then the whole orbit exp(λ⃗)G0 admits a global decomposition, that is, exp(λ⃗)G0⊂G0AN. 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 2×2 example can show.Next, we will use the Gram-Schmidt orthogonalization process to show that for anyg∈G0 and admissible λ⃗, we have exp(λ⃗)g⊂G0AN. A short proof of this will be given in Section 4. Let us denote the columns of g by w1,…,wn. The columns of g are (pseudo-) orthonormal with respect to 〈·,·〉, and the first p columns are timelike, and the last q are spacelike.Denote the columns ofexp(λ⃗)g by v1,…,vn. This element of G is obtained from g by multiplying the first row by eλ1,…, the pth row by eλp, the (p+1)st row by eμ1,…, and the last row by eμq.An important observation is that due to the admissibility ofλ⃗, the first p columns of exp(λ⃗)g will remain timelike, however nothing definite can be said about the last q.The decompositionexp(λ⃗)g=sb with s∈G0 and b∈AN is an analogue of the Gram-Schmidt orthogonalization process for the pseudometric 〈·,·〉.Denote the columns ofs by u1,…,un. Proving the existence of such decomposition amounts to showing that if we follow the Gram-Schmidt process, the first p columns of s will 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 (r1,…,rn) and the off-diagonal by mij, where mij=0 for i>j.Consider the first step of the Gram-Schmidt process, namely thatr1=∥v1∥ and u1=v1/r1. Since v1 is timelike, we see that r1 is real positive and that u1 is timelike.Now we move to the second step, (obviously, under the assumption thatp>1), which we consider in detail, because it lays the foundation for the other columns: (3.1)v2=m12u1+r2u2.
Here m12=〈v2,u1〉 and r2=∥v2-m12u1∥. In order to complete this step we need to show that the vector v2-m12u1 is timelike.Note that(3.2)‖v2-m12u1‖2=‖v2‖2-|〈v2,u1〉|2,
so we just need to show that this number is positive. Since ∥u1∥2=1, this is equivalent to showing that (3.3)‖v2‖2⋅‖u1‖2>|〈v2,u1〉|2,
or, after multiplying both sides by r12, that(3.4)‖v2‖2⋅‖v1‖2>|〈v2,v1〉|2.Denote the coordinates of the vectorw1 by (a1,…,ap,b1,…,bq), and the coordinates of w2 by (c1,…,cp,d1,…,dq). We have(3.5)∑i=1p|ai|2-∑j=1q|bj|2=∑i=1p|ci|2-∑j=1q|dj|2=1,(3.6)∑i=1paic¯i-∑j=1qbjd¯j=0.The coordinates of the vectorv1 are given by (3.7)(eλ1a1,…,eλpap,eμ1b1,…,eμqbq)
and v2 by (3.8)(eλ1c1,…,eλpcp,eμ1d1,…,eμqdq).The RHS of (3.4) now becomes as follows, using (3.5): (3.9)(∑i=1pe2λi|ai|2-∑j=1qe2μj|bj|2)⋅(∑i=1pe2λi|ci|2-∑j=1qe2μj|dj|2)=(∑i=1p(e2λi-e2μ1)|ai|2+∑j=2q(e2μ1-e2μj)|bj|2+e2μ1)⋅(∑i=1p(e2λi-e2μ1)|ci|2+∑j=2q(e2μ1-e2μj)|dj|2+e2μ1),
which is strictly greater than (3.10)(∑i=1p(e2λi-e2μ1)|ai|2+∑j=2q(e2μ1-e2μj)|bj|2)⋅(∑i=1p(e2λi-e2μ1)|ci|2+∑j=2q(e2μ1-e2μj)|dj|2),
which in turn, by Cauchy-Schwarz, is greater or equal than (3.11)|∑i=1p(e2λi-e2μ1)aic¯i+∑j=2q(e2μ1-e2μj)bjd¯j|2,
which is exactly |〈v1,v2〉|2, if we take into account (3.6).Now, we will similarly show thatuk is timelike for any k≤p. We have (3.12)vk=∑l=1k-1mlkul+rkuk,
where mℓk=〈vk,uℓ〉.This amounts to showing that(3.13)‖vk‖2>∑l=1k-1|〈vk,ul〉|2.Consider the subspaceV⊂ℂn spanned by u1,…,uk-1, which is the same subspace that is spanned by v1,…,vk-1. It is isomorphic to the subspace W spanned by w1,…,wk-1, and the explicit isomorphism is given by multiplying the first coordinate by e-λ1,…, the pth by e-λp, the (p+1)st by e-μ1,…, and the last by e-μq.Denote byyk=∑ℓ=1k-1mℓkuℓ If 〈vk,yk〉=0, then by definition of yk and mℓk this would imply that vk is orthogonal to all the ui for 1≤i≤k-1, and thus all mℓk=0 and, similar to the case k=1, the vector uk=vk/|vk| is timelike as desired.Thus, we can assume that(3.14)αk=〈vk,yk〉=∑l=1k-1|mlk|2>0.Define(3.15)u=ykαk.
This is an essentially unique element of V (up to a circle factor) with the property that: (3.16)〈vk,u〉⋅u=∑l=1k-1mlkul.
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 w1 played by w and the role of v2 played by vk.Thus we have established that the firstp columns of s are timelike vectors, and that r1,…,rp 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 byQadm and 𝔮adm the sets of admissible elements in Q and 𝔮, respectively. Recall that Qadm=exp(𝔮adm). The following result is straightforward.Lemma 4.1.
On these sets of admissible elements the mapexp is a diffeomorphism.Now, consider the symmetrization map(4.1)Sym:AN⟶Q,Y⟼Y†Y.
This map sends the orbits of the dressing action to the conjugation orbits by the action of G0 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,λ⃗∈𝔞 be admissible, let g0∈G0, and let us find an explicit decomposition (4.2)exp⁡(λ⃗)g0=g0′an,
with g0′∈G0, a∈A, and n∈N. Notice that the symmetrization map yields (4.3)g0-1exp⁡(2λ⃗)g0=n†a2n.
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, Proposition 4.1], 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∈Qadm are positive, then the eigenvalues of any leading principal minor would be positive as well, and therefore the Gauss decomposition exists.The diagonal entries ofa2 are the ratios of the leading principal minors:(4.4)(a2)ii=ΔiΔi-1.Thus we also see that the entries ofa2 are positive, as required. Thus, we have established the following.Proposition 4.2.
Ifλ⃗∈𝔞 is admissible, then the whole orbit exp(λ⃗)G0 admits a global decomposition, that is, exp(λ⃗)G0⊂G0AN.Another interesting property of(AN)adm is that it is a multiplicative set.Proposition 4.3.
If two elementsb1 and b2 from AN are admissible, b1,b2∈(AN)adm, then their product b1b2 is admissible as well.Proof.
Applying the dressing action, we can actually assume thatb2=a∈Aadm. This follows from the fact that for g∈G0 and b1,b2∈AN:
(4.5)gb1b2=(gb2)b1,(b1b2)g=b1(gb2)b2g.
Applying the symmetrization map to b1a, we obtain a new matrix f=ab1†b1a∈Q, which we need to prove admissible.
Since the spaceAadm is clearly connected and a is admissible, consider a path a(t) such that a(0)=Id, a(1)=a and a(t)∈Aadm for 0<t≤1. Also denote f(t)=a(t)b1†b1a(t).
Assume, on the contrary, thatf(1)=f is not admissible and let ℰ⊂[0,1] be the (nonempty) subset defined by the property that f(t) for t∈ℰ is not admissible. Then consider τ=infℰ.
We recall [3] that the space 𝔮adm forms a convex cone in q, therefore a simple infinitesimal computation can show that Qadm is a connected open subset of Q and any element from the boundary of Qadm, such as f(τ), has the property that its eigenvalues remain real and, moreover, the lowest eigenvalue λp corresponding to the timelike cone and the highest eigenvalue μ1 corresponding to the spacelike cone collide: λp=μ1=β. Thus, there exists a whole plane of eigenvectors containing vectors from both ℂ+n and ℂ-n. Thus it must 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, (4.6)a(τ)(b1†b1)a(τ)z=βz.
Note that both a(τ) and b1†b1, being admissible, map the timelike cone ℂ+n into itself by Lemma 2.1. Similar proof shows that they map the null cone minus the origin ℂ0n∖{0} inside the timelike cone. Therefore, the element a(τ)(b1†b1)a(τ) also maps ℂ0n∖{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 elementsQadm.Proposition 4.4.
An elements∈Q is admissible if and only if it has real positive eigenvalues and maps the closure of the timelike cone ℂ+n¯=ℂ+n∪ℂ0n into the timelike cone ℂ+n (plus the origin).Notice that sincea from the above proof is admissible diagonal, the pseudohermitian analogue of the Rayleigh-Ritz ratio (4.7)RA=x†Axx†x
for the matrices f and s=b1†b1 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 2.1.Thus, forx∈ℂ+n, we have (4.8)Rasa(x)=x†asaxx†x>(ax)†s(ax)†(ax)†ax=Rs(ax).
It follows since a maps ℂ+n to itself, that (4.9)min⁡x∈C+nRs(x)≤min⁡x∈C+nRs(ax)<min⁡x∈C+nRasa(x).
From [5, Theorem 2.1], it follows that λ1(asa)>λ1(s). Similar inequalities can be established for other eigenvalues.The notion of admissibility can be extended to the whole groupG=SL(n,ℂ). We say that g∈G is admissible, if it decomposes as g=hb, with h∈G0 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 Qadm by inducing this definition from G, but rather as we did previously.)Suppose thatg1, g2 are two admissible elements from G such that g1=h1b1 and g2=h2b2. Then we have g1g2=h1b1h2b2=h1h2b1b1h2b2, where the powers mean the dressing actions. This is possible, because b1 is admissible. Thus we have established that the product g1g2 is admissible if and only if the product b1h2b2 is such.


## 5. Example of the Group SU(1,1)

The Poisson geometry related to the groupSU(1,1) was considered in detail in [6]. Here we just recall several facts to illustrate the results of this paper. First of all, the space 𝔮adm is the following convex cone of matrices (5.1)(zx+iy-x+iy-z):x,y,z∈R,z2-x2-y2>0,z>0.
Next, we define Qadm=exp(𝔮adm). Consider an element (5.2)(t1m-m¯t2)∈Q,
where t1,t2∈ℝ, and m∈ℂ satisfy the determinant condition t1t2+|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: (5.3)t1+t2>2,t1>1.
Next, an element (5.4)(rn0r-1)∈AN
is admissible if and only if (5.5)r>1,r2+r-2-|n|2>2.Now, let us write down the Iwasawa-type decompositionG⊃G0AN explicitly. Let us consider a general element of SL(2,ℂ): (5.6)(abcd):a,b,c,d∈C,ad-bc=1.
The condition that this general element admits such a decomposition is simply |a|>|c|: (5.7)(abcd)=(a|a|2-|c|2c¯|a|2-|c|2c|a|2-|c|2a¯|a|2-|c|2)⋅(|a|2-|c|2(b-c¯|a|2-|c|2)|a|2-|c|2a01|a|2-|c|2).
Thus, it is quite easy to see that for any element g∈G and exp(λ⃗)∈Aadm, the element exp(λ⃗)g∈G0AN. Explicitly, if (5.8)g=(uvv¯u¯),exp⁡(λ⃗)=(eλ00e-λ),λ>0,
then (5.9)exp⁡(λ⃗)g=(eλueλve-λv¯e-λu¯)
is an admissible element of G.The statemement that the admissibility of two elementsb1 and b2 of AN⊂SL(2,ℂ) implies the admissibility of their product is also a short computational affair.



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*Source: 135167-2011-04-19.xml*