Understanding the relationship between $Sp(n)$ and $Sp(2n,\mathbb{C})$

Question

The symplectic group $Sp(2n,\mathbb{C})$ is defined as $A\in\mathbb{C}^{2n\times 2n}$ such that $A^TJA=J$, where: $J=\left(\begin{array}{cc} 0& I_n \\ -I_n & 0 \end{array}\right)$ and $I_n$ is the identity matrix in $\mathbb{C}^n$. In other words, a matrix is a member of the symplectic group if it preserves the bilinear form for $x,y\in\mathbb{C}^{2n}$:

$\langle x,y\rangle=x^TJy$ (1)

That is my understanding from the appendix of Tu's 'An Introduction to Manifolds (Second edition)'.

What is the motivation for this? The symplectic group $Sp(n)$ is defined as elements of $\mathbb{H}^n$ which preserve sesquilinear the form for $p,q\in\mathbb{H}^n$:

$\langle p,q\rangle=\overline{p}^Tq$ (2)

Suppose we have some quarternion $q=a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k}$. Using (2) we find:

$\langle q,q\rangle=(a-b\mathrm{i}-c\mathrm{j}-d\mathrm{k})(a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k})=a^2+b^2+c^2+d^2$$

However if we use the mapping $\mathbb{H}\rightarrow\mathbb{C}^2$: $a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k}=(a+b\mathrm{i}) + \mathrm{j}(c-d\mathrm{i})\rightarrow(a+b\mathrm{i},c-d\mathrm{i})$ then under (1):

$\langle p,p\rangle=\left(\begin{array}{cc} a+b\mathrm{i} & c-d\mathrm{i}\end{array}\right)J\left(\begin{array}{c}a+b\mathrm{i} \\ c-d\mathrm{i}\end{array}\right)=\left(\begin{array}{cc} a+b\mathrm{i} & c-d\mathrm{i}\end{array}\right)\left(\begin{array}{c}c-d\mathrm{i} \\ -a-b\mathrm{i}\end{array}\right)=0$

The two forms (1) and (2) seem different. How are they related to each other, and what is the motivation for (1)?

I thought I had, but I had been looking at the page for 'Symplectic matrix', which was different. Thanks, that cleared things up. — Ruvi Lecamwasam, Jan 02 '14 at 22:25
You ment that $Sp(n)$ is the group of matrices $A \in GL(n,\mathbb{H}) \subset M_{n \times n}(\mathbb{H})$ such that $\langle p , q \rangle = \langle Ap , Aq \rangle$. — LinAlgMan, Jul 17 '14 at 12:15
You have a mistake in the bottom equation. You should have taken the $\mathbb{H}$-conjugate of the row vector. — LinAlgMan, Jul 17 '14 at 12:24

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

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Thanks to Grigory M I found the relevant information on the Wikipedia article Symplectic Group (I had previously been looking at the article 'Symplectic matrix', which was different. From wikipedia:

The relationship between the groups Sp(2n, C), Sp(2n, R) and Sp(n) is most evident at the level of their Lie algebras. It turns out that the first of these Lie algebras is a complexification of the Lie algebras of either of the latter two groups.

Stated slightly differently, the complex Lie algebra sp(2n, C) of the complex Lie group Sp(2n, C) has several different real forms:

The compact form, sp(n), which is the Lie algebra of Sp(n),

The algebras, sp(p, n − p), which are the Lie algebras of Sp(p, n − p), the indefinite signature equivalent to the compact form,

The normal form (or split form), sp(2n, R), which is the Lie algebra of Sp(2n, R).

edited Jun 12 '20 at 10:38

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answered Jan 02 '14 at 22:28

Ruvi Lecamwasam

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I think is looking for the proof or at least some explanation and not merely the statement of these facts. – LinAlgMan Jul 17 '14 at 12:17
@AndrewLedesma - Hi Andrew, do you want to mark the question as answered? – Matta Feb 02 '15 at 18:04
Thanks, I just did that. At the time I was apprehensive of accepting my own answer, but I guess it is more helpful to see that the question is answered. – Ruvi Lecamwasam Feb 12 '15 at 20:06

Understanding the relationship between $Sp(n)$ and $Sp(2n,\mathbb{C})$

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