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What is (or where can I find) an explicit formula for the Haar measure of the group of linear symplectic transformations of $\mathbb{R}^{2n}$?

Added 13/05/2014. Some clarifying remarks:

(1) by symplectic group I mean the group of linear transformations that preserve the canonical (symplectic) two-form in $\mathbb{R}^{2n}$.

(2) "explicit" is very subjective, but what I have in mind is the formula for the Haar measure of $SL(2,\mathbb{R})$ in terms of the invariant area on the hyperbolic plane (KAN decomposition seen geometrically).

(3) the aim is to have some intuition for the measure of certain geometrically-defined subsets of the symplectic group such as the set of all linear symplectic transformations that send the unit ball into the ball or cylinder of radius 2. The arbitrary constant inherent in the definition of the Haar measure can be dealt with by looking at quotients of measures of geometrically-defined subsets.

(4) probably once I get a hold of the references Robert Bryant and Jim Humphreys have proposed I'll have no other questions (or the question will be more precise) ...

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    $\begingroup$ Well, the obvious answer is to just wedge together a basis for the left-invariant $1$-forms, which will give you (up to a normalizing constant) the $(2n^2{+}n)$-form whose integral gives you the Haar measure. If you want something more 'explicit' than this, you need to specify which 'explicit' parametrization or factorization (such as KAN, etc.) of the group in question you want to use. $\endgroup$ – Robert Bryant May 12 '14 at 19:03
  • $\begingroup$ @RobertBryant: I was hoping that the parametrization would be part of the answer or that the measure could be simply described in terms of the polar decomposition of symplectic matrices. $\endgroup$ – alvarezpaiva May 12 '14 at 19:37
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    $\begingroup$ Well, the general case of what you appear to want is treated in Chapter VIII ("Integration") of A. Knapp's "Lie Groups Beyond an Introduction". Particularly see Sections 3 and 4 of that Chapter. You should have no trouble specializing these formulae to the decomposition of $\mathrm{Sp}(n,\mathbb{R})$ that you want to use for your purposes. $\endgroup$ – Robert Bryant May 12 '14 at 21:03
  • $\begingroup$ @RobertBryant: thanks for the reference. $\endgroup$ – alvarezpaiva May 13 '14 at 8:32
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    $\begingroup$ For the 'unit ball into larger ball' problem, you will most likely find a KAK decomposition more useful than a KAN decomposition. Using KAK, it should be easy. For the 'unit ball into cylinder', I'm not sure what would be the best strategy, but this will clearly be a more difficult problem. $\endgroup$ – Robert Bryant May 13 '14 at 16:58
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Once you have made the polar decomposition, it is sufficient to find the Haar measure on the compact symplectic group. This can be calculated starting from your favorite parameterization $U(\{\alpha_i\})$ of the unitary symplectic matrix $U$, via the metric tensor,

$$g_{ij}=-{\rm tr}\,U^{\dagger}(\partial U/\partial\alpha_i)U^{\dagger}(\partial U/\partial\alpha_j),$$

and then the Haar measure is $d\mu= \sqrt{{\rm det}\,g}\prod_i d\alpha_i$.

Alternatively, you can use a computer to generate random matrices with the desired measure, as explained by Franco Mezzadri.

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  • $\begingroup$ What is "the compact symplectic group"? $\endgroup$ – Igor Rivin May 12 '14 at 20:44
  • $\begingroup$ Ah, I see the "hyperunitary group". $\endgroup$ – Igor Rivin May 12 '14 at 20:47
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    $\begingroup$ @CarloBeenakker: A notational point. You seem to be using '$n$' as an index (running over the dimension of the group?). The casual reader might confuse this with the OP's dimensional use of '$n$'. It seems that your answer amounts to the fact that Haar measure is just integration of the density of the bi-invariant pseudo-Riemannian metric on the group. That's true, of course, but it doesn't depend on considering a maximal compact or anything else, other than that the bi-invariant metric on $\mathrm{Sp}(n,\mathbb{R})$ is the restriction of the standard one on $\mathrm{GL}(2n,\mathbb{R})$. $\endgroup$ – Robert Bryant May 12 '14 at 23:26
  • $\begingroup$ @RobertBryant -- thanks, $n\mapsto i$. $\endgroup$ – Carlo Beenakker May 13 '14 at 6:31
  • $\begingroup$ @CarloBeenakker: I was referring to the "other" symplectic group (the one preserving the symplectic two-form). $\endgroup$ – alvarezpaiva May 13 '14 at 8:32

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