definition of special linear group of finite-dimensional vectors space
Topics
About: vectors space
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of group.
- The reader knows a definition of linear map.
- The reader knows a definition of bijection.
- The reader admits 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.
Target Context
- The reader will have a definition of special linear group of finite-dimensional 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:
\( F\): \(\in \{\text{ the fields }\}\)
\( d\): \(\in \mathbb{N} \setminus \{0\}\)
\( V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\( GL (V)\): \(= \text{ the general linear group of } V\)
\( g\): \(: GL (V) \to M_d (F)^{\times}\), \(= \text{ the canonical 'groups - homomorphisms' isomorphism }\) 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 any basis for \(V\)
\(*SL (V)\): \(= \{f \in GL (V) \vert det g (f) = 1\}\), \(\in \{\text{ the subgroups of } GL (V)\}\)
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Conditions:
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2: Note
\(SL (V)\) looks as though it depended on the choice of basis for \(V\), but it really does not, because for any another basis for \(V\), \(g (f)\) becomes \(N g (f) N^{- 1}\), 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, and \(det (N g (f) N^{- 1}) = det N det g (f) det N^{- 1}\), by the proposition that over any commutative ring, the determinant of the product of any square matrices is the product of the determinants of the matrices, \(= det N 1 det N^{- 1} = det N det N^{- 1} = det (N N^{- 1}) = det I = 1\), so, if \(f \in SL (V)\) with respect to a basis, \(f \in SL (V)\) with respect to any basis.
It is indeed a subgroup of \(GL (V)\), because for each elements, \(f_1, f_2 \in SL (M)\), \(det g (f_2 \circ f_1) = det (g (f_2) g (f_1))\), because \(g\) is a 'groups - homomorphisms' isomorphism, \(= det g (f_2) det g (f_1)\), by the proposition that over any commutative ring, the determinant of the product of any square matrices is the product of the determinants of the matrices, \(= 1 1 = 1\), and \(det g ({f_1}^{- 1}) = det g (f_1)^{- 1}\), because \(g\) is a 'groups - homomorphisms' isomorphism, \(= 1\), because \(1 = det I = det (g (f_1) g (f_1)^{- 1}) = det g (f_1) det g (f_1)^{- 1} = 1 det g (f_1)^{- 1} = det g (f_1)^{- 1}\), and the proposition that for any group, any nonempty subset is a subgroup if and only if it is closed under the operation and the inversion applies.
This concept is defined only for any vectors space, \(V\), instead for general modules, because the proof of \(SL (V)\) being a subgroup depends on that \(F\) is commutative and the proof of \(SL (V)\) being independent of the choice of basis depends on that the bases of \(V\) have the same cardinality.
\(V\) needs to be finite-dimensional, because we have defined 'determinant' only for finite-dimensional matrices.
