description/proof of that quaternion Euclidean topological space is canonically homeomorphic to corresponding Euclidean topological space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of quaternion Euclidean topological space.
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of homeomorphism.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some \(C^\infty\) manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any quaternion Euclidean topological space is canonically homeomorphic to the corresponding Euclidean topological space.
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:
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{H}^n\): \(= \text{ the quaternion Euclidean topological space }\)
\(\mathbb{R}^{4 n}\): \(= \text{ the Euclidean topological space }\)
\(f'''^`\): \(: \mathbb{H}^n \to f''' (\mathbb{H}^n) \subseteq M_{2 \times 2 n} (\mathbb{C})\), \(= \text{ the homeomorphism }\) by which the topology of \(\mathbb{H}^n\) is defined
\(h\): \(: M_{2 \times 2 n} (\mathbb{C}) \to \mathbb{C}^{4 n} \to \mathbb{R}^{8 n}\), \(= \text{ the homeomorphism }\) by which the topology of \(M_{2 \times 2 n} (\mathbb{C})\) is defined
\(h^`\): \(: f''' (\mathbb{H}^n) \to h (f''' (\mathbb{H}^n))\), \(= \text{ the domain codomain restriction of } h\)
\(h'\): \(: \mathbb{R}^{4 n} \to \mathbb{R}^{8 n}, (a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n) \mapsto (a_1, b_1, - c_1, d_1, c_1, d_1, a_1, - b_1, ..., a_n, b_n, - c_n, d_n, c_n, d_n, a_n, - b_n)\)
\(h'^`\): \(: \mathbb{R}^{4 n} \to h' (\mathbb{R}^{4 n})\), \(= \text{ the codomain restriction of } h'\)
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Statements:
\({h'^`}^{- 1} \circ h^` \circ f'''^`: \mathbb{H}^n \to \mathbb{R}^{4 n}, (a_1 + b_1 i + c_1 j + d_1 k, ..., a_n + b_n i + c_n j + d_n k) \mapsto (a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n) \in \{\text{ the homeomorphisms }\}\)
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2: Note
So, while we have defined the topology of \(\mathbb{H}^n\) by \(f'''^`\) instead of the direct mapping from \(\mathbb{H}^n\) onto \(\mathbb{R}^4\), in fact, \(\mathbb{H}^n\) is homeomorphic to \(\mathbb{R}^4\) by the direct mapping.
3: Proof
Whole Strategy: Step 1: see that \({h'^`}^{- 1} \circ h^` \circ f'''^`\) is valid: Step 2: see that \(h^`\) and \(h'^`\) are some homeomorphisms; Step 3: conclude the proposition.
Step 1:
Let us see that \({h'^`}^{- 1} \circ h^` \circ f'''^`\) is valid.
The codomain of \(f'''^`\) equals the domain of \(h^`\), but the issue is whether the codomain of \(h^`\), \(h (f''' (\mathbb{H}^n))\), equals the domain of \({h'^`}^{- 1}\), \(h' (\mathbb{R}^{4 n})\).
\(\mathbb{H}^n = \{(a_1 + b_1 i + c_1 j + d_1 k, ..., a_n + b_n i + c_n j + d_n k) \vert a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n \in \mathbb{R}\}\).
\(f''' (\mathbb{H}^n) = \{\begin{pmatrix} a_1 + b_1 i & c_1 + d_1 i & ... & a_n + b_n i & c_n + d_n i \\ - c_1 + d_1 i & a_1 - b_1 i & ... & - c_n + d_n i & a_n - b_n i \end{pmatrix} \vert a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n \in \mathbb{R}\}\).
\(h (f''' (\mathbb{H}^n)) = \{(a_1, b_1, - c_1, d_1, c_1, d_1, a_1, - b_1, ..., a_n, b_n, - c_n, d_n, c_n, d_n, a_n, - b_n) \vert a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n \in \mathbb{R}\}\).
\(h' (\mathbb{R}^{4 n}) = \{(a_1, b_1, - c_1, d_1, c_1, d_1, a_1, - b_1, ..., a_n, b_n, - c_n, d_n, c_n, d_n, a_n, - b_n) \vert a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n \in \mathbb{R}\}\).
So, \(h (f''' (\mathbb{H}^n)) = h' (\mathbb{R}^{4 n})\).
Each \((a_1 + b_1 i + c_1 j + d_1 k, ..., a_n + b_n i + c_n j + d_n k) \in \mathbb{H}^n\) is mapped to \(\begin{pmatrix} a_1 + b_1 i & c_1 + d_1 i & ... & a_n + b_n i & c_n + d_n i \\ - c_1 + d_1 i & a_1 - b_1 i & ... & - c_n + d_n i & a_n - b_n i \end{pmatrix}\) by \(f'''^`\), which is mapped to \((a_1, b_1, - c_1, d_1, c_1, d_1, a_1, - b_1, ..., a_n, b_n, - c_n, d_n, c_n, d_n, a_n, - b_n)\) by \(h^`\), which is mapped to \((a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n)\) by \({h'^`}^{- 1}\).
So, \({h'^`}^{- 1} \circ h^` \circ f'''^`\) is valid.
Step 2:
\(f'''^`\) is defined to be a homeomorphism: the topology of \(\mathbb{H}^n\) has been defined to make \(f'''^`\) a homeomorphism.
\(h\) is defined to be a homeomorphism: the topology of \(M_{2 \times 2 n} (\mathbb{C})\) has been defined to make \(h\) a homeomorphism
\(h^`\) is a homeomorphism, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
\(h'\) is obviously an injection.
So, \(h'^`\) is a bijection.
So, \({h'^`}^{- 1}\) exists.
\(\mathbb{R}^{4 n}\) is a topological subspace of the \(C^\infty\) manifold, \(\mathbb{R}^{4 n}\), and \(\mathbb{R}^{8 n}\) is a topological subspace of the \(C^\infty\) manifold, \(\mathbb{R}^{8 n}\), and by the standard charts, the components function of \(h'^`\) is nothing but \(: (a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n) \mapsto (a_1, b_1, - c_1, d_1, c_1, d_1, a_1, - b_1, ..., a_n, b_n, - c_n, d_n, c_n, d_n, a_n, - b_n)\), which is continuous.
So, \(h'\) is continuous, by the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some \(C^\infty\) manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous.
\(h'^`\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Let us think of \(h'': \mathbb{R}^{8 n} \to \mathbb{R}^{4 n}, (a_1, b_1, - c_1, d_1, c'_1, d'_1, a'_1, - b'_1, ..., a_n, b_n, - c_n, d_n, c'_n, d'_n, a'_n, - b'_n) \mapsto (a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n)\).
\(h'' \vert_{h' (\mathbb{R}^{4 n})} = {h'^`}^{- 1}\).
\(h''\) is obviously continuous, so, \(h'' \vert_{h' (\mathbb{R}^{4 n})}\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
So, \({h'^`}^{- 1}\) is continuous, by the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some \(C^\infty\) manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous.
So, \(h'^`\) is a homeomorphism, and \({h'^`}^{- 1}\) is a homeomorphism.
Step 3:
So, \({h'^`}^{- 1} \circ h^` \circ f'''^`\) is a homeomorphism.
