GL(n,R)
in [Math]
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From: GeometricGroup on 19 Jun 2010 19:44 This a question from "Ask an algebraist". http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1513 "Is the general linear group GL(n,R) has a discontinuous automorphisms? where GL(n,R) isomorphic to the semidirect product of the special linear group and the R\{0}." =============================================== I am happened to see this question and curious about the answer. 1. If GL(n, R) = SL(n, R) \semidirect R\{0}, then how R\{0} acts on SL(n, R), where R is the set of real numbers. 2.Why GL(n,R) is a open affine subvariety of Mn(R) (a non-empty open subset of Mn(R) in the Zariski topology), and therefore a smooth manifold of the same dimension? http://en.wikipedia.org/wiki/General_linear_group 3. Is the general linear group GL(n,R) has a discontinuous automorphisms? Thanks
From: José Carlos Santos on 20 Jun 2010 03:56 On 20-06-2010 4:44, GeometricGroup wrote: > This a question from "Ask an algebraist". > > http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1513 > > > "Is the general linear group GL(n,R) has a discontinuous automorphisms? > > where GL(n,R) isomorphic to the semidirect product of the special linear group and the R\{0}." > > =============================================== > > I am happened to see this question and curious about the answer. > > 1. If GL(n, R) = SL(n, R) \semidirect R\{0}, then how R\{0} acts on SL(n, R), where R is the set of real numbers. Consider, for instance, A:R\{0} ---> SL(n, R) defined by x 0 0 ... 0 0 1 0 ... 0 A(x) = 0 0 1 ... 0 0 0 0 ... 1 and the action f(x)(M) = A(x).M.A(x)^{-1}. > 2.Why GL(n,R) is a open affine subvariety of Mn(R) (a non-empty open subset of Mn(R) in the Zariski topology), and therefore a smooth manifold of the same dimension? Because it is the set of non-zeros of the determinant function. > http://en.wikipedia.org/wiki/General_linear_group > > 3. Is the general linear group GL(n,R) has a discontinuous automorphisms? Yes, because (R\{0},.) has discontinuous automorphisms. This uses the axiom of choice. Best regards, Jose Carlos Santos
From: GeometricGroup on 20 Jun 2010 02:54 > On 20-06-2010 4:44, GeometricGroup wrote: > > > This a question from "Ask an algebraist". > > > > > http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebrais > t;task=show_msg;msg=1513 > > > > > > "Is the general linear group GL(n,R) has a > discontinuous automorphisms? > > > > where GL(n,R) isomorphic to the semidirect product > of the special linear group and the R\{0}." > > > > =============================================== > > > > I am happened to see this question and curious > about the answer. > > > > 1. If GL(n, R) = SL(n, R) \semidirect R\{0}, then > how R\{0} acts on SL(n, R), where R is the set of > real numbers. > > Consider, for instance, A:R\{0} ---> SL(n, R) defined > by > > x 0 0 ... 0 > > 0 1 0 ... 0 > A(x) = > 0 0 1 ... 0 > > 0 0 0 ... 1 > > and the action f(x)(M) = A(x).M.A(x)^{-1}. > > > 2.Why GL(n,R) is a open affine subvariety of Mn(R) > (a non-empty open subset of Mn(R) in the Zariski > topology), and therefore a smooth manifold of the > same dimension? > > Because it is the set of non-zeros of the determinant > function. > > > http://en.wikipedia.org/wiki/General_linear_group > > > > 3. Is the general linear group GL(n,R) has a > discontinuous automorphisms? > > Yes, because (R\{0},.) has discontinuous > automorphisms. This uses the > axiom of choice. > > Best regards, > > Jose Carlos Santos Thanks for your answer. I am confused about 2. >Because it is the set of non-zeros of the determinant >function. Especially, the concept of "open affine subvariety". If a variety is a closed set in the Zariski topology, what is the meaning of the "open affine subvariety" here? Is it simply an open subset of M_n(R) in the Zariski topology? In what case of a field F, a GL(n, F) has a smooth manifold property? Thanks.
From: José Carlos Santos on 20 Jun 2010 09:35 On 20-06-2010 11:54, GeometricGroup wrote: >>> 2.Why GL(n,R) is a open affine subvariety of Mn(R) >> (a non-empty open subset of Mn(R) in the Zariski >> topology), and therefore a smooth manifold of the >> same dimension? >> >> Because it is the set of non-zeros of the determinant >> function. > > Thanks for your answer. > I am confused about 2. > > Especially, the concept of "open affine subvariety". If a variety is a closed set in the Zariski topology, what is the meaning of the "open affine subvariety" here? Is it simply an open subset of M_n(R) in the Zariski topology? I must admit that I don't know what an "open affine subvariety" is; what I proved is that (as you noticed) that it is an open subset of M_n(R). However, please note that you can see GL(n,R) as a closed subset of M_{n + 1}(R). Just consider the map F from GL(n,R) into M_{n + 1}(R) such that, for each A in GL(n,,R), 1) the left upper corner of F(A) is 1/det(A); 2) the rest of the first row and of the first column are just 0's; 3) the rest of the matrix is just A. The map F is clearly injective. Besides, its image is closed with respect to the Zariski topology, since it is the intersection of these closed sets: for each k in {2,...,n+1}, the set of those matrices (a_{kl}) such that a_{k1} = 0, the set of those matrices (a_{kl}) such that a_{k1} = 0 and, finally, the set of those matrices such that a_{11}*det(a_{kl}_{2 <= k,l <= n+1} = 1. Best regards, Jose Carlos Santos
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