$GL(n,G)$ General Linear group of a group $G$?












2












$begingroup$


I know you can take the General linear group of some vector space $V$: $GL(V)$.



For example, suppose I have a three-sphere, elements of which I might represent as $SU(2)$ since they're diffeomorphic. I know that the three sphere admits a set of three linearly independent vector fields, the $it{basis}$ of which can be represented by elements $SU(2)$ (or equivalently the $i,j,j$ of the quaternions).



So given such a space, how would I go about finding the General Linear group of that group (maybe that's a bad way to word it)?



Or maybe I should say how do I find the General linear group with that structure imposed upon it. How do I find out how that group differs from one on Euclidean 3-space, in a group theoretic manner. (apologies for poor terminology, my background is in physics).



I was hoping to do this for more general cases and groups like the special affine group of some particular space (for example).










share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    You need to be able to add elements in $G$ to define the product of matrices. You at least need a semiring.
    $endgroup$
    – Pedro Tamaroff
    Dec 23 '18 at 12:56










  • $begingroup$
    As an aside, what makes you think the 'special affine group' is isomorphic to the double cover of the Poincaré group? The dimensions don't seem to match.
    $endgroup$
    – Thomas Bakx
    Dec 23 '18 at 19:45










  • $begingroup$
    @ThomasBakx The special affine group: $$SA(4,R)=SL(4,R)ltimes R^{1,3} $$ (for Minkowskian space) which is isomorphic to: $$SL(2,C)ltimes R^{1,3}$$, which is the connected double cover of the Poincare group. I'm pretty sure that's right. I decided to ask about it here: physics.stackexchange.com/questions/450963/…
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:23












  • $begingroup$
    @ThomasBakx I could be wrong (though i'd like to know either way) math.stackexchange.com/questions/3048029/… It would seem at the very least, The connected double cover is a subgroup of the SAG?
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:30






  • 1




    $begingroup$
    @R.Rankin The identity matrix is special unitary, but twice the special matrix is not. Similarly, the sum of two unit quaternions is not, in general, a unit quaternion.
    $endgroup$
    – Pedro Tamaroff
    Dec 30 '18 at 2:01
















2












$begingroup$


I know you can take the General linear group of some vector space $V$: $GL(V)$.



For example, suppose I have a three-sphere, elements of which I might represent as $SU(2)$ since they're diffeomorphic. I know that the three sphere admits a set of three linearly independent vector fields, the $it{basis}$ of which can be represented by elements $SU(2)$ (or equivalently the $i,j,j$ of the quaternions).



So given such a space, how would I go about finding the General Linear group of that group (maybe that's a bad way to word it)?



Or maybe I should say how do I find the General linear group with that structure imposed upon it. How do I find out how that group differs from one on Euclidean 3-space, in a group theoretic manner. (apologies for poor terminology, my background is in physics).



I was hoping to do this for more general cases and groups like the special affine group of some particular space (for example).










share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    You need to be able to add elements in $G$ to define the product of matrices. You at least need a semiring.
    $endgroup$
    – Pedro Tamaroff
    Dec 23 '18 at 12:56










  • $begingroup$
    As an aside, what makes you think the 'special affine group' is isomorphic to the double cover of the Poincaré group? The dimensions don't seem to match.
    $endgroup$
    – Thomas Bakx
    Dec 23 '18 at 19:45










  • $begingroup$
    @ThomasBakx The special affine group: $$SA(4,R)=SL(4,R)ltimes R^{1,3} $$ (for Minkowskian space) which is isomorphic to: $$SL(2,C)ltimes R^{1,3}$$, which is the connected double cover of the Poincare group. I'm pretty sure that's right. I decided to ask about it here: physics.stackexchange.com/questions/450963/…
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:23












  • $begingroup$
    @ThomasBakx I could be wrong (though i'd like to know either way) math.stackexchange.com/questions/3048029/… It would seem at the very least, The connected double cover is a subgroup of the SAG?
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:30






  • 1




    $begingroup$
    @R.Rankin The identity matrix is special unitary, but twice the special matrix is not. Similarly, the sum of two unit quaternions is not, in general, a unit quaternion.
    $endgroup$
    – Pedro Tamaroff
    Dec 30 '18 at 2:01














2












2








2





$begingroup$


I know you can take the General linear group of some vector space $V$: $GL(V)$.



For example, suppose I have a three-sphere, elements of which I might represent as $SU(2)$ since they're diffeomorphic. I know that the three sphere admits a set of three linearly independent vector fields, the $it{basis}$ of which can be represented by elements $SU(2)$ (or equivalently the $i,j,j$ of the quaternions).



So given such a space, how would I go about finding the General Linear group of that group (maybe that's a bad way to word it)?



Or maybe I should say how do I find the General linear group with that structure imposed upon it. How do I find out how that group differs from one on Euclidean 3-space, in a group theoretic manner. (apologies for poor terminology, my background is in physics).



I was hoping to do this for more general cases and groups like the special affine group of some particular space (for example).










share|cite|improve this question











$endgroup$




I know you can take the General linear group of some vector space $V$: $GL(V)$.



For example, suppose I have a three-sphere, elements of which I might represent as $SU(2)$ since they're diffeomorphic. I know that the three sphere admits a set of three linearly independent vector fields, the $it{basis}$ of which can be represented by elements $SU(2)$ (or equivalently the $i,j,j$ of the quaternions).



So given such a space, how would I go about finding the General Linear group of that group (maybe that's a bad way to word it)?



Or maybe I should say how do I find the General linear group with that structure imposed upon it. How do I find out how that group differs from one on Euclidean 3-space, in a group theoretic manner. (apologies for poor terminology, my background is in physics).



I was hoping to do this for more general cases and groups like the special affine group of some particular space (for example).







group-theory affine-geometry semi-riemannian-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 2:40







R. Rankin

















asked Dec 23 '18 at 12:45









R. RankinR. Rankin

340213




340213








  • 8




    $begingroup$
    You need to be able to add elements in $G$ to define the product of matrices. You at least need a semiring.
    $endgroup$
    – Pedro Tamaroff
    Dec 23 '18 at 12:56










  • $begingroup$
    As an aside, what makes you think the 'special affine group' is isomorphic to the double cover of the Poincaré group? The dimensions don't seem to match.
    $endgroup$
    – Thomas Bakx
    Dec 23 '18 at 19:45










  • $begingroup$
    @ThomasBakx The special affine group: $$SA(4,R)=SL(4,R)ltimes R^{1,3} $$ (for Minkowskian space) which is isomorphic to: $$SL(2,C)ltimes R^{1,3}$$, which is the connected double cover of the Poincare group. I'm pretty sure that's right. I decided to ask about it here: physics.stackexchange.com/questions/450963/…
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:23












  • $begingroup$
    @ThomasBakx I could be wrong (though i'd like to know either way) math.stackexchange.com/questions/3048029/… It would seem at the very least, The connected double cover is a subgroup of the SAG?
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:30






  • 1




    $begingroup$
    @R.Rankin The identity matrix is special unitary, but twice the special matrix is not. Similarly, the sum of two unit quaternions is not, in general, a unit quaternion.
    $endgroup$
    – Pedro Tamaroff
    Dec 30 '18 at 2:01














  • 8




    $begingroup$
    You need to be able to add elements in $G$ to define the product of matrices. You at least need a semiring.
    $endgroup$
    – Pedro Tamaroff
    Dec 23 '18 at 12:56










  • $begingroup$
    As an aside, what makes you think the 'special affine group' is isomorphic to the double cover of the Poincaré group? The dimensions don't seem to match.
    $endgroup$
    – Thomas Bakx
    Dec 23 '18 at 19:45










  • $begingroup$
    @ThomasBakx The special affine group: $$SA(4,R)=SL(4,R)ltimes R^{1,3} $$ (for Minkowskian space) which is isomorphic to: $$SL(2,C)ltimes R^{1,3}$$, which is the connected double cover of the Poincare group. I'm pretty sure that's right. I decided to ask about it here: physics.stackexchange.com/questions/450963/…
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:23












  • $begingroup$
    @ThomasBakx I could be wrong (though i'd like to know either way) math.stackexchange.com/questions/3048029/… It would seem at the very least, The connected double cover is a subgroup of the SAG?
    $endgroup$
    – R. Rankin
    Dec 30 '18 at 1:30






  • 1




    $begingroup$
    @R.Rankin The identity matrix is special unitary, but twice the special matrix is not. Similarly, the sum of two unit quaternions is not, in general, a unit quaternion.
    $endgroup$
    – Pedro Tamaroff
    Dec 30 '18 at 2:01








8




8




$begingroup$
You need to be able to add elements in $G$ to define the product of matrices. You at least need a semiring.
$endgroup$
– Pedro Tamaroff
Dec 23 '18 at 12:56




$begingroup$
You need to be able to add elements in $G$ to define the product of matrices. You at least need a semiring.
$endgroup$
– Pedro Tamaroff
Dec 23 '18 at 12:56












$begingroup$
As an aside, what makes you think the 'special affine group' is isomorphic to the double cover of the Poincaré group? The dimensions don't seem to match.
$endgroup$
– Thomas Bakx
Dec 23 '18 at 19:45




$begingroup$
As an aside, what makes you think the 'special affine group' is isomorphic to the double cover of the Poincaré group? The dimensions don't seem to match.
$endgroup$
– Thomas Bakx
Dec 23 '18 at 19:45












$begingroup$
@ThomasBakx The special affine group: $$SA(4,R)=SL(4,R)ltimes R^{1,3} $$ (for Minkowskian space) which is isomorphic to: $$SL(2,C)ltimes R^{1,3}$$, which is the connected double cover of the Poincare group. I'm pretty sure that's right. I decided to ask about it here: physics.stackexchange.com/questions/450963/…
$endgroup$
– R. Rankin
Dec 30 '18 at 1:23






$begingroup$
@ThomasBakx The special affine group: $$SA(4,R)=SL(4,R)ltimes R^{1,3} $$ (for Minkowskian space) which is isomorphic to: $$SL(2,C)ltimes R^{1,3}$$, which is the connected double cover of the Poincare group. I'm pretty sure that's right. I decided to ask about it here: physics.stackexchange.com/questions/450963/…
$endgroup$
– R. Rankin
Dec 30 '18 at 1:23














$begingroup$
@ThomasBakx I could be wrong (though i'd like to know either way) math.stackexchange.com/questions/3048029/… It would seem at the very least, The connected double cover is a subgroup of the SAG?
$endgroup$
– R. Rankin
Dec 30 '18 at 1:30




$begingroup$
@ThomasBakx I could be wrong (though i'd like to know either way) math.stackexchange.com/questions/3048029/… It would seem at the very least, The connected double cover is a subgroup of the SAG?
$endgroup$
– R. Rankin
Dec 30 '18 at 1:30




1




1




$begingroup$
@R.Rankin The identity matrix is special unitary, but twice the special matrix is not. Similarly, the sum of two unit quaternions is not, in general, a unit quaternion.
$endgroup$
– Pedro Tamaroff
Dec 30 '18 at 2:01




$begingroup$
@R.Rankin The identity matrix is special unitary, but twice the special matrix is not. Similarly, the sum of two unit quaternions is not, in general, a unit quaternion.
$endgroup$
– Pedro Tamaroff
Dec 30 '18 at 2:01










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