What is a "nice" way of choosing coset representatives for the symplectic group $Sp_{2k}(\mathbb{C})$ in the general linear group $GL_{2k}(\mathbb{C})$?
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asked Dec 12, 2009 at 5:55
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6$\begingroup$ They are at least parametrized by the symplectic forms on $\mathbb C^{2n}$, for the big group acts transitively on them, and the small group is the stabilizer of one of them... $\endgroup$– Mariano Suárez-ÁlvarezCommented Dec 12, 2009 at 6:04
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$\begingroup$ I think that in English people say the "little" group :) $\endgroup$– José Figueroa-O'FarrillCommented Dec 12, 2009 at 17:57
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$\begingroup$ Actually, in English, people would most likely say "the smaller group" (to me "little" would sound quite strange, though I have no explanation for this fact). $\endgroup$– Ben Webster ♦Commented Dec 12, 2009 at 18:55
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$\begingroup$ Actually, the term "little group" is due to Wigner. In its origin it was the maximal compact subgroup of the stabilizer of a character of the Poincaré group. These days it's synomymous with stabilizer. $\endgroup$– José Figueroa-O'FarrillCommented Dec 12, 2009 at 23:01
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3 Answers
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I suppose that "nice" is very much application-dependent, but let me give it a try.
The first thing to notice is that, of course, you will not be able to find a global coset representative, since the principal bundle $$\mathrm{Sp}(2k,\mathbb{C}) \to \mathrm{GL}(2k,\mathbb{C}) \to M = \mathrm{GL}(2k,\mathbb{C})/\mathrm{Sp}(2k,\mathbb{C})$$ is not trivial and hence it has no global (continous) section. So the best you can do is find a section over some $U \subset M$.
The subgroup $\mathrm{Sp}(2k,\mathbb{C})$ is the stabilizer of a symplectic structure $\Omega$ on $\mathbb{C}^{2k}$. A convenient choice is $$\Omega = \pmatrix{ 0 & -\mathbf{1} \cr \mathbf{1} & 0}$$ where $\mathbf{1}$ is the $k\times k$ identity matrix.
The Lie algebra $\mathfrak{sp}(2k,\mathbb{C})$ consists of those $2k \times 2k$ complex matrices $X$ such that $\Omega X$ is symmetric. A complementary vector subspace of $\mathfrak{sp}(2k,\mathbb{C})$ in $\mathfrak{gl}(2k,\mathbb{C})$ is given by $$\mathfrak{sp}(2k,\mathbb{C})^\perp := \big\lbrace \Omega X \mid X \in \mathfrak{so}(2k,\mathbb{C})\big\rbrace$$
Explicitly and for our choice of $\Omega$ above, this subspace consists of the matrices of the form $$\pmatrix{B^t & C \cr A & B}$$ for $k\times k$ matrices $A,B,C$ with $A$ and $C$ skewsymmetric.
You can now exponentiate these matrices to find a coset representative. Depending on the calculation, though, you might it easier to write the coset representative as a product of exponentials,...
answered Dec 12, 2009 at 17:56
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There's a slight issue, even when $k=2$. In this case, $Sp_2=SL_2$, so you want to find coset reps. of $SL_2$ in $GL_2$. The natural choice is the diagonal matrices, which is $\mathbb{C}^{\ast} I$, but there is an ambiguity, since $-I \in SL_2$. Essentially, you need to take a "branch cut" of $\sqrt{}$ to get a unique representative.
In general, I think one can get an analogue of the QR matrix decomposition by performing a version of Gram-Schmidt orthogonalization to make a symplectic form standard, but I'm not sure how canonical this decomposition may be chosen. It would be interesting if there is an analogue of the Polar decomposition.
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1$\begingroup$ Depending what you mean by "analogue", I would say that there is not. The reason the polar decomposition exists is that you are looking for cosets with respect to the maximal compact group and the homotopy type of a connected Lie group is that of its maximal compact subgroup. In this case, $\mathrm{GL}(2k,\mathbb{C})/\mathrm{Sp}(2k,\mathbb{C})$ is not homotopy equivalent to a point, whence you can only hope for local coset representatives. Granted that, there is a sort of local polar decomposition, as in my answer below. $\endgroup$ Commented Dec 12, 2009 at 18:31
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There is a one to one correspondence between 2n*2n complex matrices and n*n quaternionic matrices. In the n*n quaternionic representation the coset representatives of GL(2n,C)/Sp(2n,C) are self dual n*n quaternionic matrices. This can be seen from the 2n*2n complex matrix representation A of a self dual quaternionic matrix which satisfies the constraint:
A = - J A^t J
Where J is the block skew symmetric symplectic form.
From this expression, one can see that the isotropy group of a fixed A in GL(2n, C) is Sp(2n,C).
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$\begingroup$ This is not correct, I'm afraid. The $n\times n$ quaternionic matrices are in one-to-one correspondence with $\mathbb{H}$-linear endomorphisms of $\mathbb{H}^n$, whereas $2n \times 2n$ complex matrices are in one-to-one correspondence with $\mathbb{C}$-linear endomorphisms of $\mathbb{C}^{2n}$. Although as complex vector spaces $\mathbb{H}^n \cong \mathbb{C}^{2n}$, $\mathbb{H}$-linearity is a strong condition. Consider $n=1$. A $1\times 1$ quaternionic matrix is a quaternion, which is 2-dimensional over $\mathbb{C}$, whereas a $2\times 2$ complex matrix is 4-dimsensional over $\mathbb{C}$. $\endgroup$ Commented Dec 12, 2009 at 16:31
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$\begingroup$ Sorry that I haven't given more detail. I meant quaternionic matrices where each element is a complex quaternion: a 1 + b i + c j + d k , where a, b, c, d are complex numbers. Now, this parameterization is not new, a similar parameterization is used in the circular classical ensemble, with an additional constraint of unitarity. There, the coset representaties are self dual quaternionic matrices with an additional unitarity constraint. $\endgroup$ Commented Dec 12, 2009 at 19:27
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$\begingroup$ I meant the circular symplectic ensemble (not the circular clasical ensemble). Anyway, I hope that the above remark clarified the matter. $\endgroup$ Commented Dec 12, 2009 at 19:36