definition of \(n \times n\) invertible quaternion matrices topological group right action
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 quaternion Euclidean topological space.
- The reader knows a definition of topological group right action.
Target Context
- The reader will have a definition of \(n \times n\) invertible quaternion matrices topological group right action.
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:
\( M_n (\mathbb{H})^{\times}\): \(= \text{ the \(n \times n\) invertible quaternion matrices topological group }\)
\( \mathbb{Q}^n\): \(= \text{ the quaternion Euclidean topological space }\)
\(*f\): \(: \mathbb{H}^n \times M_n (\mathbb{H})^{\times} \to \mathbb{H}^n, (t, M) \mapsto t M\), \(\in \{\text{ the topological group right actions }\}\)
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Conditions:
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2: Note
Let us see that \(f\) is indeed a topological group right action.
Note that \(\mathbb{H}^n\) is regarded to be \(M_{1 \times n} (\mathbb{H})\).
\(\forall M_1, M_2 \in M_n (\mathbb{H})^{\times}, \forall t \in \mathbb{H}^n (f (f (t, M_1), M_2) = f (t, M_1 M_2))\): \(f (f (t, M_1), M_2) = (t M_1) M_2 = t (M_1 M_2)\), by the proposition that for any ring, the multiplications of any matrices over the ring are associative, \(= f (t, M_1 M_2)\).
\(\forall t \in \mathbb{H}^n (f (t, 1) = t)\): \(1 = I\), and \(f (t, 1) = t I = t\).
So, \(f\) is a group right action.
\(f\) is continuous, because \(f\) is the restriction of \(f': \mathbb{H}^n \times M_n (\mathbb{H}) \to \mathbb{H}^n, (t, M) \mapsto t M\), which is continuous, by the proposition that the multiplication on any multiplication-possible finite-product of quaternion matrices topological spaces is continuous, and the proposition that any restriction of any continuous map on the domain and the codomain is continuous applies: \(\mathbb{H}^n \times M_n (\mathbb{H})^{\times}\) is the topological subspace of \(\mathbb{H}^n \times M_n (\mathbb{H})\), by the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
So, \(f\) is a topological group right action.
\(f\) is not transitive, because \(M_n (\mathbb{H})^{\times} (0) = {0} \subset \mathbb{H}^n\), for example.
\(f\) is not free, because \(M_n (\mathbb{H})^{\times}_0 = M_n (\mathbb{H})^{\times} \subset \{1\}\), for example.
With \(\mathbb{H}^n\) regarded as the quaternion Euclidean module, the map induced from \(f\) with the group element fixed, \(f^`_M: \mathbb{H}^n \to \mathbb{H}^n, t \mapsto t M\), is \(\mathbb{H}\) linear ('group action' does not require such linearity, but this case happens to have this linearity), because for each \(r_1, r_2 \in \mathbb{H}\) and each \(t_1, t_2 \in \mathbb{H}^n\), \(f^`_M (r_1 t_1 + r_2 t_2) = (r_1 t_1 + r_2 t_2) M = r_1 t_1 M + r_2 t_2 M = r_1 f^` (t_1) + r_2 f^` (t_2)\).
We do not explicitly define "\(n \times n\) invertible quaternion matrices topological group left action" (at least, right now), because that could be the group left action that corresponds to \(n \times n\) invertible quaternion matrices topological group right action, \(: M_n (\mathbb{H})^{\times} \times \mathbb{H}^n \to \mathbb{H}^n, (M, t) \mapsto t M^{- 1}\), or \(: M_n (\mathbb{H})^{\times} \times \mathbb{H}^n \to \mathbb{H}^n, (M, t) \mapsto M t\), which are different, but the former seems structurely better, because \((r_1 t_1 + r_2 t_2) M^{- 1} = r_1 (t_1 M^{- 1}) + r_2 (t_2 M^{- 1})\) holds while "\(M (r_1 t_1 + r_2 t_2) = r_1 (M t_1) + r_2 (M t_2)\)" does not hold in general.
