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In the theory of Bridgeland stability conditions one has an action of the universal cover $G'$ of $G = GL^+(2,\mathbb R)$.

What is G'?

I know there is concrete description in terms of pairs (M,f) with M in G and f a function (see Lemma 2.14 of Huybrechts http://arxiv.org/pdf/1111.1745v2.pdf). But I'd like to know if G' is diffeomorphic to some other manifold.

At some point I convinced myself that G was actually diffeomorphic to $\mathbb C^* \times \mathbb C$ and therefore $G'$ was $\mathbb C^2$. But I've never seen this written down, so I'm getting suspicious.

EDIT: For those who have voted to close: I realize this question was very elementary. However, none of the resources on stability conditions I know of state this fact explicitly, which is quite frustrating. I think it would be very useful for people looking for a quick answer to have a place where this fact appears unambiguously.

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    $\begingroup$ $GL_2^{+}(\mathbb{R})$, as a manifold, is $\mathbb{R}^3 \times SO(2)$. (More generally, any connected Lie group, as a manifold, is its maximal compact times some $\mathbb{R}^n$.) So its universal cover, as a manifold, is $\mathbb{R}^4$. $\endgroup$ – Qiaochu Yuan Apr 28 '16 at 22:26
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    $\begingroup$ Every right-handed basis of $\mathbb R^2$ can be uniquely rotated so that its first vector is on the positive $x$-axis. Thus every element of $G$ is uniquely expressible as the product of an element of $SO(2)$ and an upper-triangular matrix with positive diagonal entries. This makes $G$ diffeomorphic to $S^1\times \mathbb R^3$. $\endgroup$ – Tom Goodwillie Apr 28 '16 at 23:29
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    $\begingroup$ Concerning the close votes: The risk that this question keeps popping to the top is very small. An answer is accepted, the question will disappear somewhere in the MO dungeons. Why not keep it there? It does no harm and it is possible that someone might be happy about it some day. $\endgroup$ – Helene Sigloch Apr 29 '16 at 12:31
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    $\begingroup$ I agree with @HeleneSigloch: why on earth does this need to be closed? This seems like a perfectly fine question to me. Sure, the answer is pretty easy for people who think about these sorts of things. I thought that was precisely the point of this site: to allow people to put questions far from their expertise in front of those who have such expertise. $\endgroup$ – Charles Rezk Apr 29 '16 at 14:02
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    $\begingroup$ In other words, why not just answer the question and be done with it, instead of playing closure-politics with it. $\endgroup$ – Charles Rezk Apr 29 '16 at 14:03
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For the sake of completeness here is an explicit proof. Let $m\in GL_{2}(\mathbb{R})^{+}$, then $m\rightarrow \frac{m}{\det(m)}$ maps it to an element in $SL_{2}(\mathbb{R})$. And we know $SL_{2}(\mathbb{R})\cong \mathbb{R}^2\times \mathbb{S}^1$ via appropriate parametrization. Thus $GL_{2}(\mathbb{R})^{+}\cong \mathbb{R}^{+}\times \mathbb{R}^{2}\times \mathbb{S}^1\cong \mathbb{R}^{3}\times \mathbb{S}^1$ as a manifold. And its universal cover is $\mathbb{R}^4$.

It might be overkill, but I think others were alluding to Iwasawa decomposition.

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  • $\begingroup$ The OP also says "and therefore $G'$ was $\mathbb{C}^2$" so I think it is clear he meant $\mathbb{C}^{\ast} = \mathbb{C} \setminus \{ 0 \}$. (I prefer the notation $\mathbb{C}^{\times}$.) $\endgroup$ – Qiaochu Yuan Apr 29 '16 at 5:14
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    $\begingroup$ The KAN decomposition of $SL(2, \bf R)$ is $G=KAN$ with $K=SO(2)$ $A$ the set of diagonal matrices, $N$ the set of upper triangular matrices with $1$ on the diagonal. The universal cover map being explicitely given by ${\bf R} \times {\bf R}^+ \times {\bf R} \to Sl(2, {\bf R})$, $F(\theta, \lambda, u)=$$\left( \begin{array}{cc} cos \theta & sin \theta \\ -sin \theta & cos \theta\end{array} \right)\left( \begin{array}{cc} \lambda & 0 \\ 0 & \lambda ^{-1}\end{array} \right)\left( \begin{array}{cc} 1 & u \\ 0 & 1\end{array} \right)$ $\endgroup$ – Thomas Apr 29 '16 at 8:07
  • $\begingroup$ @Thomas: Yes! I just realized this has been written down at here: math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,R).pdf $\endgroup$ – Bombyx mori Apr 29 '16 at 8:19

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