definition of general linear topological group of quaternion module with finite basis
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of quaternions division associative algebra.
- The reader knows a definition of general linear group of module.
- The reader knows a definition of \(n \times n\) invertible quaternion matrices topological group.
Target Context
- The reader will have a definition of general linear topological group of quaternion module with finite basis.
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:
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\( M\): \(\in \{\text{ the } \mathbb{H} \text{ modules }\}\), with any basis, \(B\), such that \(\vert B \vert = n\)
\(*GL (M)\): \(= \text{ the general linear group of } M\), \(\in \{\text{ the topological groups }\}\)
\( M_n (\mathbb{H})^{\times}\): \(\in \{\text{ the topological groups }\}\)
\( g\): \(: GL (M) \to M_d (\mathbb{H})^{\times}\), \(= \text{ the canonical 'groups - homomorphisms' isomorphism }\) defined in the proposition that for any module with any \(d\)-elements basis and the general linear group of the module, there is the canonical 'groups - homomorphisms' isomorphism with respect to the basis from the general linear group onto the group of the invertible \(d \times d\) ring matrices with respect to \(B\)
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Conditions:
\(g \in \{\text{ the homeomorphisms }\}\)
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The topology of \(GL (M)\) is in fact independent of the choice of \(B\), as will be seen in Note.
2: Note
In other words, \(GL (M)\) is given the topology that makes \(g\) a homeomorphism, which is valid, because \(g\) is a bijection: any \(S \subseteq GL (M)\) is open if and only if \(g (S) \subseteq M_d (\mathbb{H})^{\times}\) is open.
\(GL (M)\) is indeed a topological group, by the proposition that for any group topological space, if there is any 'groups - homomorphisms' isomorphism homeomorphism from the space onto any topological group, the space is a topological group.
Let us see that the topology of \(GL (M)\) is in fact independent of the choice of \(B = \{b^1, ..., b^n\}\).
Note that as \(\mathbb{H}\) is a division ring, any other basis has the cardinality, \(n\), by Note for the proposition that for any module over any division ring with any finite basis, there is no basis that has any larger cardinality.
Let \(B' = \{b'^1, ..., b'^n\}\) be any other basis for \(M\).
Let \(g': GL (M) \to M_d (\mathbb{H})^{\times}\) be the canonical 'groups - homomorphisms' isomorphism with respect to \(B'\).
Let \(GL' (M)\) denote the group of \(GL (M)\) with the topology induced by \(g'\).
\(b'^l = N^l_j b^j\) for an \(n \times n\) \(\mathbb{H}\) matrix.
For each \(f \in GL (M)\), \(g' (f) = N g (f) N^{- 1}\), by the proposition that for any module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this
Let us take \(e: M_d (\mathbb{H})^{\times} \to M_d (\mathbb{H})^{\times}, M \mapsto N M N^{- 1}\), which is continuous, because \(M_d (\mathbb{H})^{\times}\) is a topological group and \(e (M) = m (N, m (M, N^{- 1}))\) where \(m\) is the multiplication on \(M_d (\mathbb{H})^{\times}\), which is continuous: \(e = m (N, \bullet) \circ m (\bullet, N^{-1})\) and \(m (N, \bullet)\) and \(m (\bullet, N^{-1})\) are continuous, by the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous and 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 applies.
There is \(e^{- 1}: M_d (\mathbb{H})^{\times} \to M_d (\mathbb{H})^{\times}, M \mapsto N^{- 1} M N\), because \(e^{- 1} \circ e (M) = N^{- 1} e (M) N = N^{- 1} N M N^{- 1} N = M = id (M)\) and \(e \circ e^{- 1} (M) = N e^{- 1} (M) N^{- 1} = N N^{- 1} M N N^{- 1} = M = id (M)\), which is continuous, as before.
So, \(e\) is a homeomorphism.
\(g'^{- 1} \circ e \circ g: GL (M) \to GL' (M) = id\), because \(g'^{- 1} \circ e \circ g (f) = g'^{- 1} (N g (f) N^{- 1}) = f\), because \(g' (f) = N g (f) N^{- 1}\).
\(g'^{- 1} \circ e \circ g\) is a homeomorphism, because \(g\), \(e\), and \(g'^{- 1}\) are some homeomorphisms.
So, for each \(S \subseteq GL (M)\), \(S = id (S) = g'^{- 1} \circ e \circ g (S) \subseteq GL' (M)\) is open if and only if \(S \subseteq GL (M)\) is open.
So, \(GL (M)\) and \(GL' (M)\) have the same topology.
