description/proof of that finite-dimensional complex vectors space topology defined based on coordinates space does not depend on choice of basis
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of canonical topology for finite-dimensional complex vectors space.
- The reader admits the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
- The reader admits the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional complex vectors space, the topology defined by the complex Euclidean topology of the coordinates space based on any basis does not depend on the choice of 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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional complex vectors spaces }\}\)
\(B\): \(= \{b_1, ..., b_d\} \subseteq V\), \(\in \{\text{ the bases for } V\}\)
\(B'\): \(= \{b'_1, ..., b'_d\} \subseteq V\), \(\in \{\text{ the bases for } V\}\)
\(\mathbb{C}^d\): \(= \text{ the complex Euclidean topological space }\)
\(f\): \(: V \to \mathbb{C}^d\), \(v = v^j b_j \mapsto (v^1, ..., v^d)\)
\(f'\): \(: V \to \mathbb{C}^d\), \(v = v'^j b'_j \mapsto (v'^1, ..., v'^d)\)
\(O\): \(= \{U \subseteq V \vert f (U) \in \text{ the topology of } \mathbb{C}^d\}\)
\(O'\): \(= \{U \subseteq V \vert f' (U) \in \text{ the topology of } \mathbb{C}^d\}\)
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Statements:
\(O = O'\)
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2: Proof
Whole Strategy: Step 1: see that \(f' = \phi \circ f\), where \(\phi: \mathbb{C}^d \to \mathbb{C}^d\) is the coordinates transition map, and see that \(\phi\) is a homeomorphism; Step 2: conclude the proposition.
Step 1:
\(f\) is a bijection.
\(f'\) is a bijection.
\(f' = \phi \circ f\), where \(\phi: \mathbb{C}^d \to \mathbb{C}^d\) is the bijective coordinates transition map, where there is a constant invertible matrix, \(M\), such that \(\phi: (c^1, ..., c^d)^t \mapsto (M^1_j c^j, ..., M^d_j c^j)^t\), by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.
Let \(g: \mathbb{R}^{2 d} \to \mathbb{C}^d, (x^1, ..., x^{2 d}) \mapsto (x^1 + x^2 i, ..., x^{2 (d - 1) + 1} + x^{2 d} i)\) be the homeomorphism cited in the definition of complex Euclidean topological space.
Let us see that \(g^{-1} \circ \phi \circ g: \mathbb{R}^{2 d} \to \mathbb{R}^{2 d}\) is a homeomorphism.
\(g^{-1} \circ \phi \circ g: (x^1, ..., x^{2 d}) \mapsto (x^1 + x^2 i, ..., x^{2 (d - 1) + 1} + x^{2 d} i) \mapsto (M^1_j (x^{2 (j - 1) + 1} + x^{2 (j - 1) + 2} i), ..., M^d_j (x^{2 (j - 1) + 1} + x^{2 (j - 1) + 2} i)) \mapsto (M^1_j x^{2 (j - 1) + 1}, M^1_j x^{2 (j - 1) + 2}, ..., M^d_j x^{2 (j - 1) + 1}, M^d_j x^{2 (j - 1) + 2})\).
Let \(\mathbb{R}^{2 d}\) have the Euclidean atlas being the Euclidean \(C^\infty\) manifold.
The formula for \(g^{-1} \circ \phi \circ g\) means that \(g^{-1} \circ \phi \circ g\) is continuous in the norm sense for the coordinates function.
So, by the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, \(g^{-1} \circ \phi \circ g\) is continuous in the topological sense.
The inverse of \(g^{-1} \circ \phi \circ g\) is \(g^{-1} \circ \phi^{-1} \circ g: \mathbb{R}^{2 d} \to \mathbb{R}^{2 d}, (x^1, ..., x^{2 d}) \mapsto ({M^{-1}}^1_j x^{2 (j - 1) + 1}, {M^{-1}}^1_j x^{2 (j - 1) + 2}, ..., {M^{-1}}^d_j x^{2 (j - 1) + 1}, {M^{-1}}^d_j x^{2 (j - 1) + 2})\), likewise.
So, \(g^{-1} \circ \phi^{-1} \circ g\) is continuous, likewise.
So, \(\phi = g \circ g^{-1} \circ \phi \circ g \circ g^{-1}\) is a homeomorphism.
Step 2:
So, for each \(U \in O\), \(f' (U) = \phi \circ f (U) \subseteq \mathbb{C}^d\) is open, because \(f (U) \subseteq \mathbb{C}^d\) is open and \(\phi\) is a homeomorphism.
That means \(U \in O'\).
By symmetry, for each \(U' \in O'\), \(U' \in O\).
So, \(O = O'\).
