4

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.

M. K.
  • 5,021
  • As a space or as a group? – Qiaochu Yuan Sep 11 '12 at 05:53
  • 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 05:55
  • 1
    Yes, as a group. I think that the universal cover, whatever the construction, is automatically group. – M. K. Sep 11 '12 at 06:47
  • @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-Álvarez May 07 '13 at 22:14
  • @MarianoSuárez-Alvarez: I see. For unknown reason someone revived this thread, thank you. – Bombyx mori May 07 '13 at 22:37
  • @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? – Bombyx mori May 07 '13 at 22:44
  • That 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. – Mariano Suárez-Álvarez May 07 '13 at 22:51

1 Answers1

5

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.

Bombyx mori
  • 19,638
  • 6
  • 52
  • 112
  • 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 06:52
  • 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! –  May 07 '13 at 21:36
  • I suspect you did not understand we translated $det(M)=1$ to the other condition. – Bombyx mori May 07 '13 at 22:03
  • But the other condition then reads $1 + a^2 + b^2 = a^2 + b^2$, doesn't it? –  May 08 '13 at 06:45
  • You are right. I suspect Taubes made a mistake here. But I do not know how to fix it. – Bombyx mori May 08 '13 at 09:56
  • I emailed Taubes, just in case we am wrong. – Bombyx mori May 08 '13 at 10:02
  • Now corrected, thank you. – Bombyx mori May 08 '13 at 17:56
  • 1
    Unfortunately the diffeomorphism $\operatorname{Sl}(2, \mathbb{R}) \approx S^1 \times \mathbb{R}^2$ is not a homomorphism of groups, $\operatorname{Sl}(2, \mathbb{R})$ is not abelian. – red_trumpet Jan 02 '20 at 10:19