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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0As a space or as a group? – 2012-09-11
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0Excuse 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). – 2012-09-11
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0Yes, as a group. I think that the universal cover, whatever the construction, is automatically group. – 2012-09-11
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0@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. – 2013-05-07
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0@MarianoSuárez-Alvarez: I see. For unknown reason someone revived this thread, thank you. – 2013-05-07
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0@MarianoSuárez-Alvarez: I am reading Jim Belk's proof on the other thread, but since $SL_{2}(\mathbb{R})\cong \mathbb{S}^{1}\times \mathbb{R}^{2}$, its universal cover should be $\mathbb{R}^{3}$. Can't I embedd it into $M_{3,3}$ via the diagonal mapping? – 2013-05-07
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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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0Thank 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? – 2012-09-11
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0You 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. – 2012-09-11
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0I do not understand the diffeomorphism: the determinant of the matrix will be zero! – 2013-05-07
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0I suspect you did not understand we translated $det(M)=1$ to the other condition. – 2013-05-07
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0But the other condition then reads $1 + a^2 + b^2 = a^2 + b^2$, doesn't it? – 2013-05-08
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0You are right. I suspect Taubes made a mistake here. But I do not know how to fix it. – 2013-05-08
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0I emailed Taubes, just in case we am wrong. – 2013-05-08
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0Now corrected, thank you. – 2013-05-08