Nondegeneracy of simple Lie algebras of real rank 1

By NTZung, on January 25th, 2012

This is work in progress with Ph. Monnier.

The nondegeneracy here is in the sense of Weinstein: a real Lie algebra g is called smoothly nondegenerate if any smooth Poisson structure whose linear part is g* is locally smoothly linearizable (i.e. isomorphic to g*)

The theorem that we want to prove gives a complete list of smoothly nondegenerate simple Lie algebras:

Theorem. Let g be a simple Lie algebra over R. If g is compact, of if reak rank of g is 1 and g does not belong to the series su(n,1), then g is smoothly nondegenerate. In the opposite case (i.e. g belongs to the series su(n,1), or the real rank of g is at least 2), g is smoothly degenerate.

The compact case is due to Conn (1984?), the real-rank case >= 2 is due to Weinstein (around 1997 ?), the case su(n,1) was shown to be degenerate by Philippe and myself. It remains to treat the other simple Lie algebras of real rank 1.

There are only 3 series of real simple Lie algebras of real rank 1: so(n,1), su(n,1), sp(n,1). So actually only so(n,1) and sp(n,1) have not been treated before. We will treat these cases here. First we look at so(n,1), whih are less complicated than sp(n,1)

The Lie algebra so(n,1), with n>= 3.

Here we will assume that n>=3, because so(2,1) = su(1,1) = sl(2,R) is degenerate.

A canonical linear basis:

$x_{ij} = x_i \partial_{x_j} - x_j \partial_{x_i}$ , ( $i,j = 1,\hdots, n$ and $i \neq j$ ),

$y_i = y\partial_{x_i} + x_i \partial_{y}$

This basis is written in the form of linear vector fields which preserve the quadratic from

$Q = y^2 - \sum x_i^2$ on $R^{n+1}$

Assume that our nonlinear Poisson structure has this linear part. Then, applying our smooth Levi decomposition theorem for $so(n) \subset so(n,1)$ , we may assume that our smooth coordinates already satisfy the following conditions wrt to the Poisson bracket:

$\{x_{ij},x_{jk}\} = x_{ik}$ and the other brackets involving the $x_{ij}$ are 0.

$\{y_i,x_{ij}\} = y_j$ and $\{y_i, x_{jk}\} = 0$ if both j and k are different from i.

$\{y_i,y_j\} = x_{ij} + \epsilon_{ij}$

Here $\epsilon_{ij}$ are the error terms, which may be assumed to be flat, according to the formal Levi decomposition theorem for $so(n,1)$ .

The problem here is to eliminate the $\epsilon_{ij}$ by a change of coordinates. Need a cohomological equation which can be solved smoothly. What is the cohomological equation here ?

Some properties of $\epsilon_{ij}$ :

$\{x_{ij},\epsilon{ik}\} = \epsilon_{jk}$ , and $\{x_{ij},\epsilon{kl}\} = 0$ if $i,j \neq k, l$ .

We want to change the $y_i$ without changing the $x_{ij}$ . Note that $y_i$ is determined by $y_1$ via the formula $y_i = \{y_1,x_{1i}\}$ .

Properties of “admissible” $y_1$ :

$\{x_{1i}, \{x_{1i},y_1\}\} =-y_1$ for all $i=2,\hdots,n$ ,

$\{x_{1i}, \{x_{1j},y_1\}\} = 0$ for all $i, j=2,\hdots,n$ and $i\neq j$ ,

$\{x_{ij},y_1\} = 0$ for all $i,j \geq 2$ ,

In other words:

$y_1$ is invariant wrt to the action of so(n-1), and is an eigenfunction of the operators $\{x_{1i}, \{x_{1i},*\}\}$ and $\{x_{1i}, \{x_{1j},y_1\}\}$ with eigenvalues equal to -1 and 0 respectively.

Steps for linearizing the Poisson structure $atex \Pi$:

1) Smooth Levi as above

2) Linearization of the symplectic leaves in a so(n)-equivariant way, using hyperbolicity (Chaperon, Belitskii-Kopanskii, …). The fact that the so(n) action remains intact means that we sill have Levi after the linearization of the foliation.

A way to linearize the foliation:

2a. Construct the Casmimir functions by extending from $\{y_i=0 \forall i\}$ (this set can be identified with $son(n)$ ) to the whole space by following the Hamiltonian vector fields. This way obtain the Casimirs.

2b. One we have the Casimir functions, normalize them –> linearization of the foliation.

3) Show that, after Step 2, the Poisson structure has the form:

$\Pi = \Pi^{(1)} + \sum_{ij} c_{ij} X_{H_i} \wedge X_{H_j}$

where $H_i$ are pull-back of Casimir on $so(n)*$ , and $c_{ij}$ are also pull-back of Casimirs on $son(n)*$ . This formula is similar to the one in our paper with Dufour on the nondegeneracy of aff(n).

For Step 3, look at $X_{y_1}$ for example:

$X_{y_1} = \sum_j (x_{1j} \partial_{y_j} + y_j \partial x_{1j}) + \sum_j \epsilon_{1j} \partial y_j$

The foliation is linearized –> the Casimirs for the linear strucure are also Casimirs here –> $X_{y_1}$ preserves this Casimir –> equations on $\epsilon_{1j}$ –> the number of free parameters is restricted. On the other hand, $\Pi^{(1)} + \sum_{ij} c_{ij} X_{H_i} \wedge X_{H_j}$ provides sufficiently parameters –> the Poisson structure is of this form.

$c_{ij}$ can be written as functions of n variables: the Casimirs of so(n) and the Casimir of so(n,1)

4) Now use Moser’s path method, similar to Dufour-Z aff(n) 2002.

Seems to work very well ! Everything fits!

Important idea: reuse the ideas of Dufour-Z, for so(n,1), and also for sp(n,1) !

The Lie algebra sp(n,1)

It maximal compact subalgebra is $sp(n) \times sp(1)$ , whih has the same rank as it (?)

The leaves contain 2 2-cycles –> prove existence of 2 smooth Casimirs –> linearization of the foliation (would not be very hard)

Assume now that the structure is in Levi form and the symplectic folation is linearized.

Question: it is true that the Poisson structure is compatible with its linear part then ?

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Nondegeneracy of simple Lie algebras of real rank 1

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