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
- The reader knows a definition of %ring name% matrices space.
- The reader knows a definition of \(m \times n\) quaternion matrices topological space.
- The reader knows a definition of product topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader admits the proposition that the multiplication on any multiplication-possible finite-product of complex matrices topological spaces is continuous.
- 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 the product map of any finite number of continuous maps is continuous by the product topologies.
- The reader admits 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.
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:
\(M_{n_1 \times n_2} (\mathbb{H})\): \(= \text{ the quaternion matrices topological space }\)
\(M_{n_2 \times n_3} (\mathbb{H})\): \(= \text{ the quaternion matrices topological space }\)
\(...\):
\(M_{n_m \times n_{m + 1}} (\mathbb{H})\): \(= \text{ the quaternion matrices topological space }\)
\(m\): \(: M_{n_1 \times n_2} (\mathbb{H}) \times ... \times M_{n_m \times n_{m + 1}} (\mathbb{H}) \to M_{n_1 \times n_{m + 1}} (\mathbb{H})\), \(= \text{ the multiplication }\)
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Statements:
\(m \in \{\text{ the continuous maps }\}\)
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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.
