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I have read John Baez on symplectic but not understood the relation between the real symplectic group $Sp(2n,\mathbb R)$ and the quaternion group $Sp(n)$. They seem to both have real dimension $n(2n+1)$ which I understand in the real case but not in the quaternion case. I know that there is a matrix representation of a quaternion as a $4 \times 4$ real orthogonal matrix or a $2 \times 2$ complex matrix so that the quaternion case would be isomorph to $Sp(4n,\mathbb R)$ and $Sp(2n,\mathbb C)$ respectively.

Another confusion is how a quaternion inner product decomposes into an orthogonal and symplectic case. This is clear to me in the complex case where the orthogonal structure is the real part and the imaginary part is the symplectic one. The unitary group keeps these two parts invariant seperatly. But in the quaternion case there are three symplectic forms for each of $i$, $j$ and $k$ imaginary parts. What is the equivalent to the unitary group?

glurks
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  • There is a group called the complex symplectic group, of complex dimension $n(2n+1)$, and these are two of its "real forms": the real symplectic group is its so-called split form, which is in some sense as noncompact as possible, and the quaternion group is its compact form, which is as compact as possible. – Qiaochu Yuan Oct 23 '17 at 18:05
  • Meanwhile I have found this where in section 4 titled "Natural Inclusions" the doubling of dimension is described in (4.2). But I'm still confused what exactly Sp(n) is. – glurks Oct 26 '17 at 14:59
  • I have found this paper which describes the three symplectic forms that confused me on page 33. In section "Quaternionic vector spaces, some group theory 2" it gives an explicit construction of $Sp(n)$. – glurks Nov 02 '17 at 15:02

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As Qiaochu says, there is a complex symplectic group $G = Sp(2n,\mathbb{C})$ and these are both real forms of it. To see this we could consider $\mathbb{C}^{2n}$ equipped with a symplectic form $\omega$. Then a real structure on $\mathbb{C}^{2n}$ is some antilinear endomorphism $\sigma$ with $\sigma^2 =1$, and a quaternionic structure is also an antilinear endomorphism $j$ but with $j^2=-1$. Then we either look at $\mathrm{Fix}(\sigma)\subset\mathbb{C}^{2n}$ which is a real subspace or we view $\mathbb{C}^{2n}$ as $\mathbb{H}^{n}$ using the original complex $i$, our new $j$ and $k:=ij$.

Now for these to give us real forms of $G$ we need these structures to play nicely with $\omega$. The two cases you mention are the real case where $\omega$ descends to a symplectic form on $\mathrm{Fix}(\sigma) \cong \mathbb{R}^{2n}$, and the case that $\omega$ forms a Hermitian inner product on $\mathbb{H}^n \cong \mathbb{C}^{2n}$ (note that there is no non-degenerate symplectic form on $\mathbb{H}^n$). There are a few other real forms ($Sp(p,q)$) where $\omega$ forms "inner products" of various signatures on $\mathbb{H}^n$. The dimension of all these real forms must be the same.

In general we can understand the majority of the real forms of Lie groups by applying some real quaternionic or Hermitian structure in the defining representation of the complex group.

Callum
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  • Can you expand or offer references as to your note "there is no non-degenerate symplectic form on $\mathbb{H}^n$"? Is this because of the anti-commuting nature of the basis? – Craig May 13 '23 at 23:02
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    @Craig Note the basis doesn't fully anti commute but yes that is a problem. $\omega(ia,jb) = ij \omega(a,b)$ but skew symmetry says this is also $-\omega(jb,ia)= - ji\omega(b,a) = ji\omega(a,b)$ – Callum May 14 '23 at 11:07
  • Wow! Fun, this works for orthogonal structures too! Not only are they degenerate they are identically zero. Thanks :) – Craig May 14 '23 at 17:56