International Journal of Mathematics and Mathematical Sciences

Volume 2011 (2011), Article ID 135167, 10 pages

http://dx.doi.org/10.1155/2011/135167

## A Note on Iwasawa-Type Decomposition

^{1}CEGEP, Champlain St. Lawrence, QC, G1V 4K2, Canada^{2}Department of Mathematics, The University of Arizona, Tucson, AZ 85721-0089, USA

Received 2 December 2010; Accepted 25 February 2011

Academic Editor: Hans Engler

Copyright © 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.

#### Abstract

We study the Iwasawa-type decomposition of an open subset of SL as SU. We show that the dressing action of SU is globally defined on the space of admissible elements in . We also show that the space of admissible elements is a multiplicative subset of . We establish a geometric criterion: the symmetrization of an admissible element maps the positive cone in into itself.

#### 1. Introduction

In Poisson geometry, the groups and (the upper-triangular subgroup of 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 is globally defined on the open subset of the so-called admissible elements of (see Section 2).

We also show that the admissible elements can be characterized as follows: these are exactly those elements of whose symmetrization maps the closure of the positive cone in into the positive cone. In addition, we establish a useful fact that the set of admissible elements is a multiplicative subset of .

#### 2. Admissible Elements

Let and be positive integers and let . Consider the space and the group , the group is the subgroup of , which preserves the following sesquilinear pairing on :
Denote the corresponding norm by . A vector is said to be *timelike* if the square of its norm is positive, and *spacelike*, if it is negative.

Let be the subgroup of positive real diagonal matrices and be the unipotent upper triangular subgroup. Denote by , , , and the Lie algebras of , , , and . The Iwasawa-type decomposition for states that for an open dense subset of one has [2, page 167]: where is the Weyl group, is the subgroup of with representatives in , and is a representative of in .

An important question is to determine which elements of allow such a decomposition for a given choice of . The case of particular interest is when , since it is related to the dressing action in Poisson geometry.

Let be the diagonal matrix . Introduce the following involution on the space of matrices: where is the usual conjugate transpose. The Lie algebra splits, as a vector space, into the direct sum of -eigenspaces of : . Let also be the submanifold of elements satisfying . Clearly, .

The next definition is quintessential for this paper. We say that
is *admissible* if for all . Clearly, using the action of , one can arrange 's and 's in the nonincreasing order, and the condition of admissibility will become simply . Same definition applies for . Next, we say that an element is admissible, if it is -conjugate to an admissible element in . The set of admissible elements in form an open cone. Define the admissible elements in as the exponents of those in . Finally, an element is called admissible, if it obtained from an admissible element of 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 on . Let and . Assume there exist and such that . In this case we write .

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. *Let be admissible and let be timelike. Then, is timelike as well. *

* Proof. *Decompose , where and is admissible, so that we have for all and .

Let . Since the group preserves the cone of timelike elements, is timelike as well and, denoting its coordinates
we see that

Now, let us show that is timelike, by using the above equation and expressing in terms of the other coordinates:
The last step in the proof is to notice that preserves the timelike cone, and thus is timelike.

Later on, in Proposition 4.4, 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 if is admissible, then the whole orbit admits a global decomposition, that is, . Therefore, the right dressing action is globally defined on the set of admissible elements in . Note that this is not true for the left dressing action, as a simple example can show.

Next, we will use the Gram-Schmidt orthogonalization process to show that for any and admissible , we have . A short proof of this will be given in Section 4. Let us denote the columns of by . The columns of are (pseudo-) orthonormal with respect to , and the first columns are timelike, and the last are spacelike.

Denote the columns of by . This element of is obtained from by multiplying the first row by , the th row by , the st row by , and the last row by .

An important observation is that due to the admissibility of , the first columns of will remain timelike, however nothing definite can be said about the last .

The decomposition with and is an analogue of the Gram-Schmidt orthogonalization process for the pseudometric .

Denote the columns of by . Proving the existence of such decomposition amounts to showing that if we follow the Gram-Schmidt process, the first columns of will be timelike, and the last will be spacelike, and that the diagonal entries of will be positive real numbers. Let us denote the diagonal entries of by and the off-diagonal by , where for .

Consider the first step of the Gram-Schmidt process, namely that and . Since is timelike, we see that is real positive and that is timelike.

Now we move to the second step, (obviously, under the assumption that ), which we consider in detail, because it lays the foundation for the other columns: Here and . In order to complete this step we need to show that the vector is timelike.

Note that so we just need to show that this number is positive. Since , this is equivalent to showing that or, after multiplying both sides by , that

Denote the coordinates of the vector by , and the coordinates of by . We have

The coordinates of the vector are given by and by

The RHS of (3.4) now becomes as follows, using (3.5): which is strictly greater than which in turn, by Cauchy-Schwarz, is greater or equal than which is exactly , if we take into account (3.6).

Now, we will similarly show that is timelike for any . We have where .

This amounts to showing that

Consider the subspace spanned by , which is the same subspace that is spanned by . It is isomorphic to the subspace spanned by , and the explicit isomorphism is given by multiplying the first coordinate by , the th by , the st by , and the last by .

Denote by If , then by definition of and this would imply that is orthogonal to all the for , and thus all and, similar to the case , the vector is timelike as desired.

Thus, we can assume that

Define This is an essentially unique element of (up to a circle factor) with the property that: Let also be the unit vector in , which is the rescaled image of under the above homomorphism. The previous proof for the case now applies verbatim, with the role of played by and the role of played by .

Thus we have established that the first columns of are timelike vectors, and that 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 spacelike vectors—therefore if we continue the Gram-Schmidt orthogonalization process, then the last columns of will be spacelike as required.

#### 4. Iwasawa-Type Decomposition

Let us introduce more notation. Denote by and the sets of admissible elements in and , respectively. Recall that . The following result is straightforward.

Lemma 4.1. *On these sets of admissible elements the map is a diffeomorphism.*

Now, consider the symmetrization map This map sends the orbits of the dressing action to the conjugation orbits by the action of 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, be admissible, let , and let us find an explicit decomposition with , , and . Notice that the symmetrization map yields 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 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 are the ratios of the leading principal minors:

Thus we also see that the entries of are positive, as required. Thus, we have established the following.

Proposition 4.2. *If is admissible, then the whole orbit admits a global decomposition, that is, .*

Another interesting property of is that it is a multiplicative set.

Proposition 4.3. *If two elements and from are admissible, , then their product is admissible as well.*

* Proof. *Applying the dressing action, we can actually assume that . This follows from the fact that for and :
Applying the symmetrization map to , we obtain a new matrix , which we need to prove admissible.

Since the space is clearly connected and is admissible, consider a path such that , and for . Also denote .

Assume, on the contrary, that is not admissible and let be the (nonempty) subset defined by the property that for is not admissible. Then consider .

We recall [3] that the space forms a convex cone in , therefore a simple infinitesimal computation can show that is a connected open subset of and any element from the boundary of , such as , has the property that its eigenvalues remain real and, moreover, the lowest eigenvalue corresponding to the timelike cone and the highest eigenvalue corresponding to the spacelike cone collide: . Thus, there exists a whole plane of eigenvectors containing vectors from both and . Thus it must contain an eigenvector, which we denote by , from the null cone.

For this eigenvector we have the following equation: . Or, equivalently,
Note that both and , being admissible, map the timelike cone into itself by Lemma 2.1. Similar proof shows that they map the null cone minus the origin inside the timelike cone. Therefore, the element also maps inside of and thus cannot be its eigenvector.

In geometric terms, we have established the following characterization of the space of admissible elements .

Proposition 4.4. *An element is admissible if and only if it has real positive eigenvalues and maps the closure of the timelike cone into the timelike cone (plus the origin).*

Notice that since from the above proof is admissible diagonal, the pseudohermitian analogue of the Rayleigh-Ritz ratio for the matrices and will have the following properties. Let be timelike, then, as we saw earlier, is timelike as well with a bigger norm by Lemma 2.1.

Thus, for , we have It follows since maps to itself, that From [5, Theorem 2.1], it follows that . Similar inequalities can be established for other eigenvalues.

The notion of admissibility can be extended to the whole group . We say that is admissible, if it decomposes as , with and an admissible . Such a decomposition, if exists, is clearly unique. Note, moreover, that . The singular admissible spectrum of are the square roots of the spectrum of . (Note of warning: one should not define by inducing this definition from , but rather as we did previously.)

Suppose that , are two admissible elements from such that and . Then we have , where the powers mean the dressing actions. This is possible, because is admissible. Thus we have established that the product is admissible if and only if the product is such.

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

The Poisson geometry related to the group was considered in detail in [6]. Here we just recall several facts to illustrate the results of this paper. First of all, the space is the following convex cone of matrices Next, we define . Consider an element where , and satisfy the determinant condition . 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: Next, an element is admissible if and only if

Now, let us write down the Iwasawa-type decomposition explicitly. Let us consider a general element of : The condition that this general element admits such a decomposition is simply : Thus, it is quite easy to see that for any element and , the element . Explicitly, if then is an admissible element of .

The statemement that the admissibility of two elements and of implies the admissibility of their product is also a short computational affair.

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