I SL(n,R) Lie group as submanifold of GL(n,R)

[martinbn]

I'm not an expert in this area however, if the Lie group was not connected, then how could one "get/recover" any group element via exponentiation of generators members of the associated Lie algebra (i.e. the tangent space at the group Identity) ?

One cannot. Why should one?

 

[jbergman]

I'm not an expert in this area however, if the Lie group was not connected, then how could one "get/recover" any group element via exponentiation of generators members of the associated Lie algebra (i.e. the tangent space at the group Identity) ?

Who said you could? But, also, all components of a Lie Group are diffeomorphic so one could always get to an element by exponentiation followed by group multiplication.

Lie Groups are a deep subject and this is one of the reasons people are interested in Universal covering spaces of Lie Groups among other things. Lie Algebras are a simplification/linearization of a Lie Group but multiple Lie Groups can share the same Lie Algebra. Just look at su(2) and so(3).

 

[cianfa72]

Who said you could? But, also, all components of a Lie Group are diffeomorphic so one could always get to an element by exponentiation followed by group multiplication.

Ah ok. Basically via exponentiation of Lie algebra's generators, one gets any group element in the connected part of the group containing the Identity. Then one can recover any other Lie group element (in the other parts) via group multiplication.

Lie Groups are a deep subject and this is one of the reasons people are interested in Universal covering spaces of Lie Groups among other things. Lie Algebras are a simplification/linearization of a Lie Group but multiple Lie Groups can share the same Lie Algebra. Just look at su(2) and so(3).

Yes, this point is well explained in those lectures. Indeed $SU(2)$ is the Universal Covering Group -- btw it is isomorphic/homeomorphic to $SL(1,\mathbb Q)$.

 

[fresh_42]

My guess is that uncluding conectedness in the definition excludes finite groups, for which very different tools are needed.

Unfortunately, it also excludes covers.

I think the guy is just a physics guy who doesn't know the precise definition of a Lie Group.

I don't think so, orthogonal groups are essential in physics.

Lie Groups are a deep subject and this is one of the reasons people are interested in Universal covering spaces of Lie Groups among other things.

This.

 

[fresh_42]

Ah ok. Basically via exponentiation of Lie algebra's generators, one gets any group element in the connected part of the group containing the Identity. Then one can recover any other Lie group element (in the other parts) via group multiplication.

More or less, yes.

Yes, this point is well explained in those lectures. Indeed $SU(2)$ is the Universal Covering Group -- btw it is isomorphic/homeomorphic to $SL(1,\mathbb Q)$.

How? $\operatorname{SL}(1,\mathbb{Q})=\{A\in \mathbb{M}(1,\mathbb{Q})=\mathbb{Q}\,|\,\det A=1\}=\{1\}$ and $\operatorname{SU}(2,\mathbb{C})$ is real three-dimensional.

I would forget the discrete topology in the context of Lie groups. It only creates confusion. The essential part of a Lie group is its analytical (real or complex smooth) structure. If you are interested in other fields, then it belongs to a book about linear algebraic groups. If you are interested in other topologies, then it belongs to a topology book. The topological terms that are relevant for Lie groups - besides those directly related to their analytical structures like open neighborhoods or radii of convergence - are first of all compactness, then (path-)connectedness, and coverings:

as in $\operatorname{O}(n,\mathbb{R})=\{A\in \mathbb{M}(n,\mathbb{R})\,|\,A^\tau A =1\wedge \det A=-1\} \cup \operatorname{SO}(n,\mathbb{R})$ where the group structure of $\operatorname{O}(n,\mathbb{R})\big{/} \operatorname{SO}(n,\mathbb{R}) \cong \mathbb{Z}_2$ is more important than possible topologies on $\mathbb{Z}_2.$


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Image source: https://en.wikipedia.org/wiki/Covering_space

 

[cianfa72]

How? $\operatorname{SL}(1,\mathbb{Q})=\{A\in \mathbb{M}(1,\mathbb{Q})=\mathbb{Q}\,|\,\det A=1\}=\{1\}$ and $\operatorname{SU}(2,\mathbb{C})$ is real three-dimensional.

In his lectures, he claims $SL(1,\mathbb Q)$ is simply connected like $SU(2)$ and it is (group) isomorphic and (topologically) homeomorphic to $SU(2)$. Note that $\mathbb Q$ is the field of quaternions.

 

[fresh_42]

In his lectures, he claims $SL(1,\mathbb Q)$ is simply connected like $SU(2)$ and it is (group) isomorphic and (topologically) homeomorphic to $SU(2)$. Note that $\mathbb Q$ is the field of quaternions.

That was confusing. $\mathbb{Q}$ is usually the rational numbers. The quaternions are usually represented by $\mathbb{H},$ the Hamilton quaternions. Your isomorphism would then be
$$
\operatorname{U}(1,\mathbb{H})\cong \operatorname{SU}(2,\mathbb{C})\,.
$$