GL(2;C) is the group of linear transformations on C^2

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Incase any one doesn't understand the notation, GL(2;C) is the group of linear transformations on C^2 which are invertible. Another way of looking at it is all complex 2x2 matrices with non-zero determinant.

It is fairly easy to show that GL(2;C) is not simply connected (just define a homomorphism that maps an element to its determinant). So here's my question. If it is not simply connected, than what is its universal covering group? All the textbooks I've looked at give many examples of universal covering groups but they never give you any sort of strategy of how to find one, so I have no idea where to start. Anyone know the answer, or have any hints so I can try to figure it out? Thanks
 
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Hurkyl
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There might not be an easy answer. e.g. Wikipedia's page on covering groups indicates that the universal cover of SL(2, R) isn't a matrix group!

If I had to guess, I think the homomorphism you mention gives a big hint. The determinant is a map
GL(2, C) --> C*
so the universal cover of GL(2, C) has to have a morphism to the universal cover of C*. (I think)

You've probably seen the universal cover of C*; I'm pretty sure it's a corkscrew shape: the graph of the (multivalued) complex logarithm function. Happily, we can unwind it and get that the universal cover is the exponential map
exp:C-->C*.​
(C viewed as an additive group)

So, if I had to guess, I would claim that the (matrix) exponential
exp:M(2,C)-->GL(2,C)​
is related to the universal cover. (M(2,C) viewed as an additive group) Alas, this isn't a group homomorphism. :frown:
 
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