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.
Are there any good ways to see the universal cover of $GL^{+}(2,\mathbb{R})$?
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0That embdding will not give you an embedding of groups. Every manifold can be embedded in some way in a space of matrices, that follows from Whitney's embedding theorems; but here we have the group structure to take care of too. – 2013-05-07
1 Answers
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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0Now corrected, thank you. – 2013-05-08