definition of \(n \times n\) symplectic matrices topological group
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of \(n \times n\) invertible quaternion matrices topological group.
- The reader knows a definition of \(n \times n\) symplectic group.
Target Context
- The reader will have a definition of \(n \times n\) symplectic matrices topological group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\( M_n (\mathbb{H})^{\times}\): \(= \text{ the } n \times n \text{ invertible quaternion matrices topological group }\)
\(*Sp (n)\): \(\subseteq M_n (\mathbb{H})^{\times}\), as the subgroup and the topological subspace
//
Conditions:
//
2: Note
\(Sp (n)\) is indeed a topological group, by the proposition that for any topological group, any subgroup as the topological subspace is a topological group.
Let us see that \(Sp (n)\) is a closed subspace of \(M_n (\mathbb{H})\).
Let \(c\) be the quaternion-conjugation-transposition on \(M_n (\mathbb{H})\), which is continuous, by Note for the definition of \(m \times n\) quaternion matrices topological space.
Let us take \(h: M_n (\mathbb{H}) \to M_n (\mathbb{H}), M \to M^* M\).
\(h = m \circ \phi\), where \(\phi: M_n (\mathbb{H}) \to M_n (\mathbb{H}) \times M_n (\mathbb{H}), M \mapsto (c (M), id (M))\).
\(\phi\) is continuous, because \(c\) and \(id\) are continuous, and the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is continuous applies.
\(m\) is continuous, by Note for the definition of \(m \times n\) quaternion matrices topological space.
\(h\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(Sp (n) = h^{- 1} (\{I\})\), but \(\{I\} \subseteq M_n (\mathbb{H})\) is a closed subset, by the proposition that for any Hausdorff topological space, any 1 point subset is closed: \(M_n (\mathbb{H})\) is Hausdorff, because \(M_n (\mathbb{H})\) is homeomorphic to a subspace of \(\mathbb{R}^{2 (2 n)^2}\), which is Hausdorff.
So, \(Sp (n)\) is closed on \(M_n (\mathbb{H})\).
\(f' (Sp (n))\) is closed on \(f' (M_n (\mathbb{H}))\), because \(f'^`: M_{m \times n} (\mathbb{H}) \to f' (M_{m \times n} (\mathbb{H}))\) is a homeomorphism.
As \(f' (M_n (\mathbb{H}))\) is closed on \(M_{2 n} (\mathbb{C})\), by Note for the definition of \(m \times n\) quaternion matrices topological space, \(f' (Sp (n))\) is closed on \(M_{2 n} (\mathbb{C})\), by the proposition that any closed set on any closed topological subspace is closed on the base space.
While there is the canonical homeomorphism, \(h': M_{2 n} (\mathbb{C}) \to \mathbb{R}^{2 (2 n)^2}\), \(h' (f' (Sp (n))) \subseteq \mathbb{R}^{2 (2 n)^2}\) is bounded, by the proposition that each component of any symplectic matrix has an absolute value equal to or smaller than \(1\), and it is closed, so, \(h' (f' (Sp (n)))\) is compact, by the proposition that any subset of any Euclidean metric topological space is compact if and only if the subset is closed and bounded (the Heine-Borel theorem).
So, \(f' (Sp (n)) = {h'}^{- 1} h' (f' (Sp (n))) \subseteq M_{2 n} (\mathbb{C})\) is compact, because \(h'\) is a homeomorphism, and is compact on \(f' (M_n (\mathbb{H}))\), by the proposition that for any topological space, any compact subset of the space that is contained in any subspace is compact on the subspace.
So, \(Sp (n) = {f'^`}^{- 1} (f' (Sp (n)))\) is compact on \(M_{n} (\mathbb{H})\), because \(f'^`\) is a homeomorphism.
