definition of general linear group of finite-dimensional real vectors space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of general linear group of vectors space.
- The reader knows a definition of tensors space with respect to field and \(k\) vectors spaces and vectors space over field.
- The reader knows a definition of canonical topology for finite-dimensional real vectors space.
- The reader knows a definition of canonical \(C^\infty\) atlas for finite-dimensional real vectors space.
- The reader knows a definition of open submanifold with boundary of \(C^\infty\) manifold with boundary.
- The reader admits the proposition that from any module with any basis into any module, a linear map can be defined by mapping the basis and linearly expanding the mapping.
Target Context
- The reader will have a definition of general linear group of finite-dimensional real vectors 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:
\( \mathbb{R}\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the } d \text{ -dimensional } \mathbb{R} \text{ vectors spaces }\}\)
\(*GL (V)\): \(= \text{ the general linear group of } V\), \(\in \{\text{ the } d^2 \text{ -dimensional } C^\infty \text{ manifolds }\}\)
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Conditions:
\(GL (V)\) is the open submanifold of the \(C^\infty\) manifold, \(L (V: V)\), with the canonical topology and the canonical \(C^\infty\) atlas
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2: Note
The reason why this special case of general linear group of vectors space is taken up specifically is that this concept is also a \(C^\infty\) manifold: the field needs to be \(\mathbb{R}\) and \(V\) needs to be finite-dimensional for this concept.
Let us see that \(GL (V)\) is indeed an open submanifold of \(L (V: V)\).
Let us see that \(L (V: V)\) is indeed a \(d^2\)-dimensional \(\mathbb{R}\) vectors space.
Note for the definition of tensors space with respect to field and \(k\) vectors spaces and vectors space over field has already shown that it is an \(\mathbb{R}\) vectors space.
Let any basis of \(V\) be \(\{b_1, ..., b_d\}\).
Let \(b^j_k \in L (V: V)\) be the one that maps \(b_j\) to \(b_k\) and maps any other basis element to \(0\), which is indeed in \(L (V: V)\), by the proposition that from any module with any basis into any module, a linear map can be defined by mapping the basis and linearly expanding the mapping.
Let us see that \(\{b^j_k \vert j \in \{1, ..., d\}, k \in \{1, ..., d\}\}\) is a basis for \(L (V: V)\).
Let us suppose that \(v^k_j b^j_k = 0\) for any \(v^k_j \in \mathbb{R}\). Let it operate on \(b_l\). \(v^k_j b^j_k (b_l) = 0 (b_l) = 0\), but \(v^k_j b^j_k (b_l) = v^k_l b^l_k (b_l) = v^k_l b_k\), which implies that \(v^k_l = 0\). So, \(\{b^j_k\}\) is linearly independent.
For each fixed \(j\), \(v^k_l b^l_k (b_j) = v^k_j b_k\) spans \(V\) by choosing \(v^k_j\) s. So, each element of \(L (V: V)\) is expressed as a \(v^k_l b^l_k\).
So, \(\{b^j_k\}\) is a basis for \(L (V: V)\), and so, \(L (V: V)\) is \(d^2\)-dimensional.
So, \(L (V: V)\) can be made a \(C^\infty\) manifold, by the definition of canonical topology for finite-dimensional real vectors space and the definition of canonical \(C^\infty\) atlas for finite-dimensional real vectors space.
By the definition of canonical \(C^\infty\) atlas for finite-dimensional real vectors space, \((L (V: V) \subseteq L (V: V), \phi)\) where \(\phi: L (V: V) \to \mathbb{R}^{d^2}, v = v^k_j b^j_k \mapsto M (v) = \begin{pmatrix} v^1_1 & ... & v^1_d \\ ... \\ v^d_1 & ... & v^d_d \end{pmatrix}\) is a global chart.
\(\phi (GL (V)) := S = \{M (v) \in \mathbb{R}^{d^2} \vert det M (v) \neq 0\} = \mathbb{R}^{d^2} \setminus \{M (v) \in \mathbb{R}^{d^2} \vert det M (v) = 0\}\).
But \(\{M (v) \in \mathbb{R}^{d^2} \vert det M (v) = 0\}\) is a closed subset of \(\mathbb{R}^{d^2}\), because \(det: \mathbb{R}^{d^2} \to \mathbb{R}\) is continuous and \(\{M (v) \in \mathbb{R}^{d^2} \vert det M (v) = 0\} = det^{-1} (\{0\})\), where \(\{0\} \in \mathbb{R}\) is a closed subset.
So, \(S\) is an open subset of \(\mathbb{R}^{d^2}\). \(GL (V) = \phi^{-1} (S)\) is an open subset of \(L (V: V)\).
So, \(GL (V)\) is an open submanifold of \(L (V: V)\).
