##SL(2,\mathbb R)## Lie group as manifold

[cianfa72]

TL;DR

About the $SL(2,\mathbb R)$ Lie group parametrization as manifold

Hi,
consider the set of the following parametrized matrices
$$
\begin{bmatrix}
1+a & b \\
c & \frac {1 + bc} {1 + a} \\
\end{bmatrix}
$$
They are member of the group $SL(2,\mathbb R)$ (indeed their determinant is 1). The group itself is homemorphic to a quadric in $\mathbb R^4$.

I believe the above parametrization is just a chart for the group as manifold. It makes sense only for $a \neq -1$ and this condition yields an open subset in $\mathbb R^3$. On this open set the map is bijective.

That means the are matrices of $SL(2,\mathbb R)$ that cannot be parametrized by the above map (i.e. matrices with the element in the first row/column equal 0).

My question is: the fact that the above parametrization does not cover entirely the group manifold does not rule out in principle that $SL(2,\mathbb R)$ as manifold might be homeomorphic with $\mathbb R^3$. Thanks.

 

[martinbn]

The question is a bit unclear. In any case $SL(2,\mathbb R)$ is not homeomorphic to $\mathbb R^3$.

 

[cianfa72]

In any case $SL(2,\mathbb R)$ is not homeomorphic to $\mathbb R^3$.

Ok, the point is: even if I found a parametrization (a chart) that doesn't cover entirely a manifold, there might be one that instead covers it entirely, right ?

 

[martinbn]

Ok, the point is: even if I found a parametrization (a chart) that doesn't cover entirely a manifold, there might be one that instead covers it entirely, right ?

In general yes. Take an open proper subset of $\mathbb R^3$, it is a chart that doesn't cover the whole space, but there are charts that do.

 

[cianfa72]

Ok, as in post #1 the Lie group $SL(2,\mathbb R)$ is homeorphic with a quadric in $\mathbb R^4$ with the subspace topology from $\mathbb R^4$ -- in some sense this is "tautologically" true since the topology of $SL(2,\mathbb R)$ is defined that way.

I believe we can cover $SL(2,\mathbb R)$ (i.e. the quadric in $\mathbb R^4$) with just 2 charts. One is the chart in post#1, the other one could be
$$
\begin{bmatrix}
a & b \\
\frac {1 + ac} {b} & c \\
\end{bmatrix}
$$
Btw, $SL(2,\mathbb R)$ as topological space should have the same topology as the subspace topology from $GL(2,\mathbb R)$. Indeed the latter is open in $\mathbb R^4$ and of course $SL(2,\mathbb R)$ is a subset of it. Then the open sets in $SL(2,\mathbb R)$ are all and only the intersections of open sets in $GL(2,\mathbb R)$ with the set $SL(2,\mathbb R)$.

 

[fresh_42]

$\operatorname{SL}(2,\mathbb{R})\cong_\mathbb{R} \operatorname{SU}(2,\mathbb{C})\cong_\mathbb{R} \mathbb{S}^3$ - two charts.

 

[martinbn]

$\operatorname{SL}(2,\mathbb{R})\cong_\mathbb{R} \operatorname{SU}(2,\mathbb{C})\cong_\mathbb{R} \mathbb{S}^3$ - two charts.

This is confusing. The special linear group is not compact. It cannot be homeomorphic to a sphere. Also your notation for the unitary group is non standard. Do you mean $SU(2)$ or the complex points of the algebraic group?

 

[fresh_42]

I just thought they should be "equal" for sharing the same Lie algebra, but you are right. And it's a mistake. The corresponding Lie algebras are only complex isomorphic.

It was so wrong, that I even missed the fact that the Lie algebra as the tangent space at $1$ is a local property.

 

[cianfa72]

Does make sense my post#5 ? Thank you.

 

[martinbn]

Does make sense my post#5 ? Thank you.

Yes.

 

[WWGD]

Ok, the point is: even if I found a parametrization (a chart) that doesn't cover entirely a manifold, there might be one that instead covers it entirely, right ?

That can happen if the manifold is globally, not just locally homeomorphic to $\mathbb R^3$.

 

[martinbn]

Ok, the point is: even if I found a parametrization (a chart) that doesn't cover entirely a manifold, there might be one that instead covers it entirely, right ?

That can happen if the manifold is globally, not just locally homeomorphic to $\mathbb R^3$.

It could happen even if it isn't homeomorphic to $\mathbb R^3$. For example take your manifold to be two disjoint open sets in $\mathbb R^3$ then you have a global chart but it is not connected so it is not homeomorphic to the whole $\mathbb R^3$.

 

[cianfa72]

For example take your manifold to be two disjoint open sets in $\mathbb R^3$ then you have a global chart but it is not connected so it is not homeomorphic to the whole $\mathbb R^3$.

Yes, the global chart map in this case is just the Identity map.