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).

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    $\begingroup$ 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
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    $\begingroup$ @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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