<The previous article in this series | The table of contents of this series |
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
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
\(h^`\): \(: f''' (\mathbb{H}^n) \to h (f''' (\mathbb{H}^n))\), \(= \text{ the domain codomain restriction of } h\)
\(h'\): \(: \mathbb{R}^{4 n} \to \mathbb{R}^{8 n}, (a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n) \mapsto (a_1, b_1, - c_1, d_1, c_1, d_1, a_1, - b_1, ..., a_n, b_n, - c_n, d_n, c_n, d_n, a_n, - b_n)\)
\(h'^`\): \(: \mathbb{R}^{4 n} \to h' (\mathbb{R}^{4 n})\), \(= \text{ the codomain restriction of } h'\)
//
Statements:
\({h'^`}^{- 1} \circ h^` \circ f'''^`: \mathbb{H}^n \to \mathbb{R}^{4 n}, (a_1 + b_1 i + c_1 j + d_1 k, ..., a_n + b_n i + c_n j + d_n k) \mapsto (a_1, b_1, c_1, d_1, ..., a_n, b_n, c_n, d_n) \in \{\text{ the homeomorphisms }\}\)
//
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.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that multiplication on finite-product of real matrices topological spaces is continuous
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the multiplication on any multiplication-possible finite-product of real matrices topological spaces is continuous.
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:
\(M_{n_1 \times n_2} (\mathbb{R})\): \(= \text{ the real matrices topological space }\)
\(M_{n_2 \times n_3} (\mathbb{R})\): \(= \text{ the real matrices topological space }\)
\(...\):
\(M_{n_m \times n_{m + 1}} (\mathbb{R})\): \(= \text{ the real matrices topological space }\)
\(m\): \(: M_{n_1 \times n_2} (\mathbb{R}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{R}) \to M_{n_1 \times n_{m + 1}} (\mathbb{R})\), \(= \text{ the multiplication }\)
//
Statements:
\(m \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take the homeomorphisms, \(h_{n_j, n_l}: M_{n_j, n_l} (\mathbb{R}) \to \mathbb{R}^{n_j n_l}\), and see that \(m = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\); Step 2: see that \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) is continuous; Step 3: conclude the proposition.
Step 1:
Let us take the homeomorphisms, for each \(n_j, n_l \in \mathbb{N} \setminus \{0\}\), \(h_{n_j, n_l}: M_{n_j, n_l} (\mathbb{R}) \to \mathbb{R}^{n_j n_l}\), by which the topology of \(M_{n_j, n_l} (\mathbb{R})\) is defined.
\(m = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\), because for each \((M_1, ..., M_m) \in M_{n_1 \times n_2} (\mathbb{R}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{R})\), \({h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}}) (M_1, ..., M_m) = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m (({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) (h_{n_1, n_2} (M_1), ..., h_{n_m, n_{m + 1}} (M_m))) = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m ({h_{n_1, n_2}}^{- 1} (h_{n_1, n_2} (M_1)), ..., {h_{n_m, n_{m + 1}}}^{- 1} (h_{n_m, n_{m + 1}} (M_m))) = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m (M_1, ..., M_m) = m (M_1, ..., M_m)\).
Step 2:
Let us see that \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) is continuous.
\(\mathbb{R}^{n_1 n_2} \times ... \times \mathbb{R}^{n_m n_{m + 1}}\) is a topological subspace of the finite-product \(C^{\infty}\) manifold, \(\mathbb{R}^{n_1 n_2} \times ... \times \mathbb{R}^{n_m n_{m + 1}}\), and \(\mathbb{R}^{n_1 n_{m + 1}}\) is a topological subspace of the \(C^{\infty}\) manifold, \(\mathbb{R}^{n_1 n_{m + 1}}\), and by the standard charts for \(\mathbb{R}^{n_1 n_2} \times ... \times \mathbb{R}^{n_m n_{m + 1}}\) and \(\mathbb{R}^{n_1 n_{m + 1}}\), each component of each \((v_1, ..., v_m) \in \mathbb{R}^{n_1 n_2} \times ... \times \mathbb{R}^{n_m n_{m + 1}}\), which is a component of a \(v_j \in \mathbb{R}^{n_j n_{j + 1}}\), becomes a component of \({h_{n_j, n_{j + 1}}}^{- 1} (v_j)\), each component of \(m ({h_{n_1, n_2}}^{- 1} (v_1), ..., {h_{n_m, n_{m + 1}}}^{- 1} (v_m))\) is a polynomial of the components of \({h_{n_j, n_{j + 1}}}^{- 1} (v_j)\) s, which is a polynomial of the components of \(v_j\) s, and those components of \(m ({h_{n_1, n_2}}^{- 1} (v_1), ..., {h_{n_m, n_{m + 1}}}^{- 1} (v_m))\) become the components of \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) (v_1, ..., v_m)\), so, the components of the components function of \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) are some polynomials, which are continuous.
So, \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) 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.
Step 3:
\((h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
\({h_{n_1, n_{m + 1}}}^{- 1}\) is continuous, because \(h_{n_1, n_{m + 1}}\) is a homeomorphism.
So, \(m = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\) 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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that multiplication on finite-product of quaternion matrices topological spaces is continuous
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the multiplication on any multiplication-possible finite-product of quaternion matrices topological spaces is continuous.
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:
\(M_{n_1 \times n_2} (\mathbb{H})\): \(= \text{ the quaternion matrices topological space }\)
\(M_{n_2 \times n_3} (\mathbb{H})\): \(= \text{ the quaternion matrices topological space }\)
\(...\):
\(M_{n_m \times n_{m + 1}} (\mathbb{H})\): \(= \text{ the quaternion matrices topological space }\)
\(m\): \(: M_{n_1 \times n_2} (\mathbb{H}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{H}) \to M_{n_1 \times n_{m + 1}} (\mathbb{H})\), \(= \text{ the multiplication }\)
//
Statements:
\(m \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take the homeomorphisms, \({f'^`}_{n_j, n_l}: M_{n_j, n_l} (\mathbb{H}) \to f' (M_{n_j, n_l} (\mathbb{H})) \subseteq M_{2 n_j, 2 n_l} (\mathbb{C})\), and see that \(m = {{f'^`}_{n_1, n_{m + 1}}}^{- 1} \circ m' \circ ({f'^`}_{n_1, n_2}, ..., {f'^`}_{n_m, n_{m + 1}})\) where \(m'\) is the restriction of the multiplication on \(M_{2 n_1, 2 n_2} (\mathbb{C}) \times ... \times M_{2 n_m, 2 n_{m + 1}} (\mathbb{C})\); Step 2: see that \(m'\) is continuous; Step 3: conclude the proposition.
Step 1:
Let us take the homeomorphisms, for each \(n_j, n_l \in \mathbb{N} \setminus \{0\}\), \({f'^`}_{n_j, n_l}: M_{n_j, n_l} (\mathbb{H}) \to f'_{n_j, n_l} (M_{n_j, n_l} (\mathbb{H})) \subseteq M_{2 n_j, 2 n_l} (\mathbb{C})\), by which the topology of \(M_{n_j, n_l} (\mathbb{H})\) is defined.
Let us see that for each \((M_1, ..., M_m) \in M_{n_1 \times n_2} (\mathbb{H}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{H})\), \(f'_{n_1, n_{m + 1}} (M_1 ... M_m) = f'_{n_1, n_2} (M_1) ... f'_{n_m, n_{m + 1}} (M_m)\) (which is really parallel to Step 2 of Proof of the proposition that the ring of the \(n \times n\) quaternion matrices is 'rings - homomorphisms' isomorphic to a subring of the \(2 n \times 2 n\) complex matrices, via the between-the-quaternions-division-associative-algebra-and-a-subring-of-\(2 \times 2\)-complex-matrices 'rings - homomorphisms' isomorphism).
Let us see that for the case, \(m = 2\), which is about \(f'_{n_1, n_3} (M_1 M_2) = f'_{n_1, n_2} (M_1) f'_{n_2, n_3} (M_2)\).
As for \(f'_{n_1, n_3} (M_1 M_2)\), for each \((M_1 M_2)^j_l\), \(f ((M_1 M_2)^j_l) = f ({M_1}^j_p {M_2}^p_l) = \sum_{p \in \{1, ..., n_2\}} f ({M_1}^j_p {M_2}^p_l) = \sum_{p \in \{1, ..., n_2\}} f ({M_1}^j_p) f ({M_2}^p_l)\), because \(f\) is a ring homomorphism, which is the \((j, l)\) \(2 \times 2\) block of \(f'_{n_1, n_3} (M_1 M_2)\), but the corresponding \(2 \times 2\) block of \(f'_{n_1, n_2} (M_1) f'_{n_2, n_3} (M_2)\) is \(\sum_{p \in \{1, ..., n_2\}} f ({M_1}^j_p) f ({M_2}^p_l)\), by the the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise, so, \(f'_{n_1, n_3} (M_1 M_2) = f'_{n_1, n_2} (M_1) f'_{n_2, n_3} (M_2)\).
Supposing that \(f'_{n_1, n_q} (M_1 ... M_{q - 1}) = f'_{n_1, n_2} (M_1) ... f'_{n_{q - 1}, n_q} (M_{q - 1})\) holds for any \(q \in \mathbb{N} \setminus \{0\}\), \(f'_{n_1, n_{q + 1}} (M_1 ... M_q) = f'_{n_1, n_{q + 1}} ((M_1 ... M_{q - 1}) M_q) = f'_{n_1, n_q} (M_1 ... M_{q - 1}) f'_{n_q, n_{q + 1}} (M_q)\), by the \(m = 2\) case (\(M_1 ... M_{q - 1} \in M_{n_1 \times n_q} (\mathbb{H})\) and \((n_1, n_q, n_{q + 1})\) is taken as \((n_1, n_2, n_3)\)), \(= f'_{n_1, n_2} (M_1) ... f'_{n_{q - 1}, n_q} (M_{q - 1}) f'_{n_q, n_{q + 1}} (M_q)\).
So, by the induction principle, \(f'_{n_1, n_{m + 1}} (M_1 ... M_m) = f'_{n_1, n_2} (M_1) ... f'_{n_m, n_{m + 1}} (M_m)\) for each \(m \in \mathbb{N} \setminus \{0\}\).
Now, \(m = {{f'^`}_{n_1, n_{m + 1}}}^{- 1} \circ m' \circ ({f'^`}_{n_1, n_2}, ..., {f'^`}_{n_m, n_{m + 1}})\) where \(m': f'_{n_1, n_2} (M_{n_1, n_2} (\mathbb{H})) \times ... \times f'_{n_m, n_{m + 1}} (M_{n_m, n_{m + 1}} (\mathbb{H})) \to f'_{n_1, n_{m + 1}} (M_{n_1, n_{m + 1}} (\mathbb{H}))\) is the restriction of the multiplication on \(M_{2 n_1, 2 n_2} (\mathbb{C}) \times ... \times M_{2 n_m, 2 n_{m + 1}} (\mathbb{C})\), because \({{f'^`}_{n_1, n_{m + 1}}}^{- 1} \circ m' \circ ({f'^`}_{n_1, n_2}, ..., {f'^`}_{n_m, n_{m + 1}}) (M_1, ..., M_m) = {{f'^`}_{n_1, n_{m + 1}}}^{- 1} \circ m' ({f'^`}_{n_1, n_2} (M_1), ..., {f'^`}_{n_m, n_{m + 1}} (M_m)) = {{f'^`}_{n_1, n_{m + 1}}}^{- 1} ({f'^`}_{n_1, n_2} (M_1) ... {f'^`}_{n_m, n_{m + 1}} (M_m)) = {{f'^`}_{n_1, n_{m + 1}}}^{- 1} (f'^`_{n_1, n_{m + 1}} (M_1 ... M_m))\), which has been seen above, \(= M_1 ... M_m = m (M_1, ..., M_m)\).
Step 2:
\(m'\) is continuous, because the multiplication on \(M_{2 n_1, 2 n_2} (\mathbb{C}) \times ... \times M_{2 n_m, 2 n_{m + 1}} (\mathbb{C})\) is continuous, by the proposition that the multiplication on any multiplication-possible finite-product of complex matrices topological spaces is continuous, and the proposition that any restriction of any continuous map on the domain and the codomain is continuous applies.
Step 3:
\(({f'^`}_{n_1, n_2}, ..., {f'^`}_{n_m, n_{m + 1}})\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
\({f'^`_{n_1, n_m}}^{- 1}\) is continuous, because \(f'^`_{n_1, n_m}\) is a homeomorphism.
So, \(m = {f'^`_{n_1, n_m}}^{- 1} \circ m' \circ ({f'^`}_{n_1, n_2}, ..., {f'^`}_{n_m, n_{m + 1}})\) 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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that multiplication on finite-product of complex matrices topological spaces is continuous
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the multiplication on any multiplication-possible finite-product of complex matrices topological spaces is continuous.
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:
\(M_{n_1 \times n_2} (\mathbb{C})\): \(= \text{ the complex matrices topological space }\)
\(M_{n_2 \times n_3} (\mathbb{C})\): \(= \text{ the complex matrices topological space }\)
\(...\):
\(M_{n_m \times n_{m + 1}} (\mathbb{C})\): \(= \text{ the complex matrices topological space }\)
\(m\): \(: M_{n_1 \times n_2} (\mathbb{C}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{C}) \to M_{n_1 \times n_{m + 1}} (\mathbb{C})\), \(= \text{ the multiplication }\)
//
Statements:
\(m \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take the homeomorphisms, \(h_{n_j, n_l}: M_{n_j, n_l} (\mathbb{C}) \to \mathbb{C}^{n_j n_l} \to \mathbb{R}^{2 n_j n_l}\), and see that \(m = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\); Step 2: see that \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) is continuous; Step 3: conclude the proposition.
Step 1:
Let us take the homeomorphisms, for each \(n_j, n_l \in \mathbb{N} \setminus \{0\}\), \(h_{n_j, n_l}: M_{n_j, n_l} (\mathbb{C}) \to \mathbb{C}^{n_j n_l} \to \mathbb{R}^{2 n_j n_l}\), by which the topology of \(M_{n_j, n_l} (\mathbb{C})\) is defined.
\(m = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\), because for each \((M_1, ..., M_m) \in M_{n_1 \times n_2} (\mathbb{C}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{C})\), \({h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}}) (M_1, ..., M_m) = {h_{n_1, n_{m + 1}}}^{- 1} (h_{n_1, n_{m + 1}} (m (({h_{n_1, n_2}}^{- 1} (h_{n_1, n_2} (M_1)), ..., {h_{n_m, n_{m + 1}}}^{- 1} (h_{n_m, n_{m + 1}} (M_m)))))) = {h_{n_1, n_{m + 1}}}^{- 1} (h_{n_1, n_{m + 1}} (m (M_1, ..., M_m))) = m (M_1, ..., M_m)\).
Step 2:
Let us see that \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) is continuous.
\(\mathbb{R}^{2 n_1 n_2} \times ... \times \mathbb{R}^{2 n_m n_{m + 1}}\) is a topological subspace of the finite-product \(C^{\infty}\) manifold, \(\mathbb{R}^{2 n_1 n_2} \times ... \times \mathbb{R}^{2 n_m n_{m + 1}}\), and \(\mathbb{R}^{2 n_1 n_{m + 1}}\) is a topological subspace of the \(C^{\infty}\) manifold, \(\mathbb{R}^{2 n_1 n_{m + 1}}\), and by the standard charts for \(\mathbb{R}^{2 n_1 n_2} \times ... \times \mathbb{R}^{2 n_m n_{m + 1}}\) and \(\mathbb{R}^{2 n_1 n_{m + 1}}\), each component of each \((v_1, ..., v_m) \in \mathbb{R}^{2 n_1 n_2} \times ... \times \mathbb{R}^{2 n_m n_{m + 1}}\), which is a component of a \(v_j \in \mathbb{R}^{2 n_j n_{j + 1}}\), becomes the real or imaginary part of a component of \({h_{n_j, n_{j + 1}}}^{- 1} (v_j)\), each component of \(m ({h_{n_1, n_2}}^{- 1} (v_1), ..., {h_{n_m, n_{m + 1}}}^{- 1} (v_m))\) is a polynomial of the components of \({h_{n_j, n_{j + 1}}}^{- 1} (v_j)\) s, whose real and imaginary parts are some polynomials of the components of \(v_j\) s, and those real and imaginary parts of the components of \(m ({h_{n_1, n_2}}^{- 1} (v_1), ..., {h_{n_m, n_{m + 1}}}^{- 1} (v_m))\) become the components of \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) (v_1, ..., v_m)\), so, the components of the components function of \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) are some polynomials, which are continuous.
So, \(h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1})\) 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.
Step 3:
\((h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
\({h_{n_1, n_{m + 1}}}^{- 1}\) is continuous, because \(h_{n_1, n_{m + 1}}\) is a homeomorphism.
So, \(m = {h_{n_1, n_{m + 1}}}^{- 1} \circ h_{n_1, n_{m + 1}} \circ m \circ ({h_{n_1, n_2}}^{- 1}, ..., {h_{n_m, n_{m + 1}}}^{- 1}) \circ (h_{n_1, n_2}, ..., h_{n_m, n_{m + 1}})\) 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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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 \(B\)
//
Conditions:
\(g \in \{\text{ the homeomorphisms }\}\)
//
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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of \(n \times n\) orthogonal matrices topological group
Topics
About:
group
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of \(n \times n\) orthogonal 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:
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\( M_n (\mathbb{R})^{\times}\): \(= \text{ the } n \times n \text{ invertible real matrices topological group }\)
\(*O (n)\): \(\subseteq M_n (\mathbb{R})^{\times}\), as the subgroup and the topological subspace
//
Conditions:
//
2: Note
\(O (n)\) 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 \(O (n)\) is a closed subspace of \(M_n (\mathbb{R})\).
Let \(t\) be the transposition on \(M_n (\mathbb{R})\), which is continuous, by Note for the definition of \(m \times n\) real matrices topological space.
Let us take \(h: M_n (\mathbb{R}) \to M_n (\mathbb{R}), M \to M^t M\).
\(h = m \circ \phi\), where \(\phi: M_n (\mathbb{R}) \to M_n (\mathbb{R}) \times M_n (\mathbb{R}), M \mapsto (t (M), id (M))\).
\(\phi\) is continuous, because \(t\) 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 \(m \times n\) real 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.
\(O (n) = h^{- 1} (\{I\})\), but \(\{I\} \subseteq M_n (\mathbb{R})\) is a closed subset, by the proposition that for any Hausdorff topological space, any 1 point subset is closed: \(M_n (\mathbb{R})\) is Hausdorff, because it is homeomorphic to \(\mathbb{R}^{n^2}\), which is Hausdorff.
\(O (n)\) is compact, because while there is the canonical homeomorphism, \(h: M_n (\mathbb{R}) \to \mathbb{R}^{n^2}\), \(h (O (n))\) is bounded, by the proposition that each component of any unitary matrix has an absolute value equal to or smaller than \(1\), and \(h (O (n))\) is closed on \(\mathbb{R}^{n^2}\), because \(O (n)\) is closed on \(M_n (\mathbb{R})\), so, \(h (O (n))\) 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), and so, \(h^{- 1} (h (O (n)))\) is compact, because \(h\) is a homeomorphism.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of \(n \times n\) symplectic matrices topological group
Topics
About:
group
About:
topological space
The table of contents of this article
Starting Context
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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
