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Are there any good ways to see the universal cover of $GL^{+}(2,\mathbb{R})$? Here $GL^{+}(2,\mathbb{R})$ stands for the identity component of $GL(2,\mathbb{R})$, i.e. positive determinant matrices. I am looking for an explicit description of $GL^{+}(2,\mathbb{R})$. Thanks.

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As a space or as a group? –  Qiaochu Yuan Sep 11 '12 at 5:53
    
Excuse for my low level question: do we have a good definition of universal covers of Lie groups other than viewing them as subsets of $R^{n^{2}}$ and using the inherit topology? Do you imply we can put a different topology by viewing them abstractly as topological groups? (and thus can have different universal covers). –  Bombyx mori Sep 11 '12 at 5:55
    
Yes, as a group. I think that the universal cover, whatever the construction, is automatically group. –  M. K. Sep 11 '12 at 6:47
    
@user32240, it is not true that the universal covering of every Lie group is a Lie group of matrices: the standard example is $SL(2,\mathbb R)$, whose universal cover is not a linear group. –  Mariano Suárez-Alvarez May 7 '13 at 22:14
    
@MarianoSuárez-Alvarez: I see. For unknown reason someone revived this thread, thank you. –  Bombyx mori May 7 '13 at 22:37

1 Answer 1

This is borrowed from Clifford Taubes's differential geometry:

The group $SL(2,\mathbb{R})$ is diffeomorphic to $\mathbb{S}^{1}\times \mathbb{R}^{2}$. This can be seen by using a linear change of coordinates on $M(2,\mathbb{R})$ that writes the entires in terms of $(x,y,u,v)$ as follows $$M_{1,1}=x-u, M_{22}=x+u,M_{12}=v-y,M_{21}=v+y$$ The condition $\det(M)=1$ now says that $x^{2}+y^{2}=1+u^{2}+v^{2}$. This understood, the diffeomorphism from $\mathbb{S}^{1}\times \mathbb{R}^{2}$ to $SL(2,\mathbb{R})$ sends a triple $(\theta,a,b)$ to the matrix determined by $$x=(1+a^{2}+b^{2})^{1/2}\cos[\theta],y=(1+a^{2}+b^{2})^{1/2}\sin[\theta],u=a,v=b$$ Here $\theta\in [0,2\pi]$ is the angular coordinate for $\mathbb{S}^{1}$.

And it should not be difficult for you to see the universal cover of this space is $\mathbb{R}^{3}$.

The typo in the solution was corrected by Taubes' email.

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Thank you for the reply, but I don't quite understand why you construct the universal cover of $SL(2,\mathbb{R})$. How are $GL^{+}(2,\mathbb{R})$ and $SL(2,\mathbb{R})$ related? –  M. K. Sep 11 '12 at 6:52
    
You can homotopically deform $GL^{+}(2,\mathbb{R})$ into $SL(2,\mathbb{R})$ by scaling the determinant. So they should be homotopically equivalent. Therefore knowing the universal cover of $SL(2,\mathbb{R})$ is enough. –  Bombyx mori Sep 11 '12 at 13:45
    
I do not understand the diffeomorphism: the determinant of the matrix will be zero! –  user76701 May 7 '13 at 21:36
    
I suspect you did not understand we translated $det(M)=1$ to the other condition. –  Bombyx mori May 7 '13 at 22:03
    
But the other condition then reads $1 + a^2 + b^2 = a^2 + b^2$, doesn't it? –  user76701 May 8 '13 at 6:45

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