First let me fix some notation:

Let $O(n)$ be the group of $n \times n$ real matrices $T$ which are "orthogonal", $U(n)$ be the group of $n \times n$ complex matrices $T$ which are "unitary" and $Sp(n)$ be the group of $n \times n$ quaternionic matrices $T$ which are "symplectic" (in all three cases $T^hT=TT^h=I$).

Let $Sp(2n,F)$ be the group of $2n \times 2n$ matrices that preserve a non-degenerate skew-symmetric bilinear form on $F^{2n}$, where $F$ is the field of real $\mathbb{R}$, complex $\mathbb{C}$ or quaternion $\mathbb{H}$ numbers (skew-field in the case of quaternions).

The following are true:

$O(2n) \cap Sp(2n,\mathbb{R}) = U(n)$

$U(2n) \cap Sp(2n,\mathbb{C}) = Sp(n)$

So my question is about the next logical step. Clearly both $Sp(2n)$ and $Sp(2n,\mathbb{H})$ are groups acting on $\mathbb{H}^{2n}$ but do they intersect to a non-empty group? In other words what is $X(n)$ below (if anything)?

$Sp(2n) \cap Sp(2n,\mathbb{H}) = X(n)$?

PS This is a question I naturally asked myself after reading Baez's "Symplectic, Quaternionic, Fermionic" blog posting:  http://math.ucr.edu/home/baez/symplectic.html

First let me fix some notation:

Let $O(n)$ be the group of $n \times n$ real matrices $T$ which are "orthogonal", $U(n)$ be the group of $n \times n$ complex matrices $T$ which are "unitary" and $Sp(n)$ be the group of $n \times n$ quaternionic matrices $T$ which are "symplectic" (in all three cases $T^hT=TT^h=I$).

Let $Sp(2n,F)$ be the group of $2n \times 2n$ matrices that preserve a non-degenerate skew-symmetric bilinear form on $F^{2n}$, where $F$ is the field of real $\mathbb{R}$, complex $\mathbb{C}$ or quaternion $\mathbb{H}$ numbers (skew-field in the case of quaternions).

The following are true:

$O(2n) \cap Sp(2n,\mathbb{R}) = U(n)$

$U(2n) \cap Sp(2n,\mathbb{C}) = Sp(n)$

So my question is about the next logical step. Clearly both $Sp(2n)$ and $Sp(2n,\mathbb{H})$ are groups acting on $\mathbb{H}^{2n}$ but do they intersect to a non-empty group? In other words what is $X(n)$ below (if anything)?

$Sp(2n) \cap Sp(2n,\mathbb{H}) = X(n)$?

PS 1 This is a question I naturally asked myself after reading Baez's "Symplectic, Quaternionic, Fermionic" blog posting:  http://math.ucr.edu/home/baez/symplectic.html

PS 2 By writing $X(n)$ instead of $X(2n)$ above I am hinting something related to Octonions but I don't want to scare off anyone.

First lets fix some notation:

Let $O(n)$ be the group of $n \times n$ real matrices $T$ which are "orthogonal", $U(n)$ be the group of $n \times n$ complex matrices $T$ which are "unitary" and $Sp(n)$ be the group of $n \times n$ quaternionic matrices $T$ which are "symplectic" (in all three cases $T^hT=TT^h=I$).

Let $Sp(2n,F)$ be the group of $2n \times 2n$ matrices that preserve a non-degenerate skew-symmetric bilinear form on $F^{2n}$, where $F$ is the field of real ($R$), complex ($C$) or quaternion ($H$) numbers (skew-field in the case of quaternions).

The following are accepted to be true (see Baez below)

$O(2n) \cap Sp(2n,R) = U(n)$

$U(2n) \cap Sp(2n,C) = Sp(n)$

So my question is about the next logical step. Clearly both $Sp(2n)$ and $Sp(2n,H)$ are groups acting on $H^{2n}$ but do they intersect to a non-empty group? In other words what is $X(n)$ below (if anything)?

$Sp(2n) \cap Sp(2n,H) = X(n)$?

PS This is a question I naturally asked myself after reading Baez's "Symplectic, Quaternionic, Fermionic" blog posting: http://math.ucr.edu/home/baez/symplectic.html

First let me fix some notation:

Let $O(n)$ be the group of $n \times n$ real matrices $T$ which are "orthogonal", $U(n)$ be the group of $n \times n$ complex matrices $T$ which are "unitary" and $Sp(n)$ be the group of $n \times n$ quaternionic matrices $T$ which are "symplectic" (in all three cases $T^hT=TT^h=I$).

Let $Sp(2n,F)$ be the group of $2n \times 2n$ matrices that preserve a non-degenerate skew-symmetric bilinear form on $F^{2n}$, where $F$ is the field of real $\mathbb{R}$, complex $\mathbb{C}$ or quaternion $\mathbb{H}$ numbers (skew-field in the case of quaternions).

The following are true:

$O(2n) \cap Sp(2n,\mathbb{R}) = U(n)$

$U(2n) \cap Sp(2n,\mathbb{C}) = Sp(n)$

So my question is about the next logical step. Clearly both $Sp(2n)$ and $Sp(2n,\mathbb{H})$ are groups acting on $\mathbb{H}^{2n}$ but do they intersect to a non-empty group? In other words what is $X(n)$ below (if anything)?

$Sp(2n) \cap Sp(2n,\mathbb{H}) = X(n)$?

PS This is a question I naturally asked myself after reading Baez's "Symplectic, Quaternionic, Fermionic" blog posting: http://math.ucr.edu/home/baez/symplectic.html