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When we define the volume of a compact matrix lie group (subgroup of $M_n(C)$) by viewing it as a subspace of $R^{n^{2}}$ and applying the usual Lebesgue measure, what's the volume of SO(n), SU(n), Sp(n) and...?

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3 Answers 3

up vote 6 down vote accepted

SO(n) is a compact manifold of dimension $n(n-1)/2$ in $\mathbb R^{n^2}$. It follows that its $n^2$-dimensional Lebesgue measure is $0$. The same argument applies to the other examples.

I think the right question would be what is the volume of these groups with respect to the Lebesgue measure on the manifold itself. Computing the volume of things like SO(2) is straight forward, since we have a good parametrization of this group.

The volume of SO(n) is computed here: http://arxiv.org/pdf/0809.0808.pdf

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Thanks Stefan, I do mean the Lebesgue measure on the manifold itself. Thank you for the paper and it seems that my first two questions are equivalent to the problem of volume of Grassmann and Stiefel manifolds. –  Jun Su Aug 16 '12 at 12:29

One book in which most of these are computed is Robb Muirhead: "Aspects of multivariate Statistical Analysis"

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I think the following paper of Mkrtchyan and Veselovmay be relevant to your question:


It computes the volume of a compact Lie group when normalized using the Cartan-Killing form of its Lie algebra. Now if your group is a compact real form of a classical Lie group then, up to a constant, the Cartan-Killing form is equivalent to the form $(X,Y)=Tr X^{\ast}Y$ where $X$ and $Y$ are elements of your Lie algebra. This latest form, of course, allows you to compute the volume of your group as a subset of the space of matrices. If you compute the constant relating the Cartan-Killing form to the form I defined above, then you can read the answer using the paper in the link.

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