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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lie-groups
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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