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

[cianfa72]

TL;DR Summary

About SL(n,R) as Lie group with the smooth structure as regular submanifold of GL(n,R)

Hi,
consider the group $SL(n,\mathbb R)$. It is a subgroup of $GL(n,\mathbb R)$. To show it is a Lie group we must assign a differential structure turning it into a differential manifold, proving further that multiplication and taking the inverse are actually smooth maps.

With the indentification $\mathbb M_n \cong \mathbb R^{2n}$ one can see that $GL(n,\mathbb R)$ is an open submanifold of $\mathbb M_n$ (it is the preimage of an open set through the determinant map $d$ that is continuous). Then by mean of Regular level set theorem, $SL(n,\mathbb R)$ is actually a regular submanifold of codimension 1.

Btw, if one restricts the domain of the determinant map $d$ to the subset $GL(n,\mathbb R)$, it is continuous as well. Therefore the preimage of the closed set $\{1\}$ under the "restricted determinant map" gives a closed set in $GL(n,\mathbb R)$. It isn't in general a closed subset of $\mathbb M_n \cong \mathbb R^{2n}$.

Is the above correct ? Thanks.

 

[fresh_42]

TL;DR Summary: About SL(n,R) as Lie group with the smooth structure as regular submanifold of GL(n,R)

Hi,
consider the group $SL(n,\mathbb R)$. It is a subgroup of $GL(n,\mathbb R)$. To show it is a Lie group we must assign a differential structure turning it into a differential manifold, proving further that multiplication and taking the inverse are actually smooth maps.

Which is easy. We have $n^2$ real indeterminates $x_{ij}$ and multiplication, as well as inversion are differential polynomials in these variables:
$$
(x_{ij})_{ij}\cdot (y_{ij})_{ij} = \left(\sum_{p=1}^n x_{ip}y_{pj}\right)_{ij}
$$
$$
(x_{ij})^{-1}_{ij}= \dfrac{\operatorname{adj (x_{ij})_{ij}}}{\det (x_{ij})_{ij}}
$$
where the adjugate matrix is again a matrix of determinants in the variables $x_{ij}.$

The defining polynomial $\det (x_{ij})_{ij} =1$ is also smooth. $\operatorname{GL}(n,\mathbb{R})\subseteq \mathbb{M}(n,\mathbb{R})$ is a dense subset, so that their tangential spaces
$$
T_1\operatorname{GL}(n,\mathbb{R})=\mathfrak{gl}(n,\mathbb{R})=\mathbb{M}(n,\mathbb{R})=T_0\mathbb{M}(n,\mathbb{R})
$$
are identical vector spaces.

With the indentification $\mathbb M_n \cong \mathbb R^{2n}$ ...

... assuming you meant $\mathbb{R}^{n^2}$ ...

... one can see that $GL(n,\mathbb R)$ is an open submanifold of $\mathbb M_n$ (it is the preimage of an open set through the determinant map $d$ that is continuous).

Yes, $\operatorname{GL}(n,\mathbb{R})=\det^{-1}(\mathbb{R}-\{0\}).$

Then by mean of Regular level set theorem, $SL(n,\mathbb R)$ is actually a regular submanifold of codimension 1.

Yes, $\dim \operatorname{SL}(n,\mathbb{R})=\dim \mathfrak{sl}(n,\mathbb{R}) = n^2-1.$ Another way to see it is, that $\mathfrak{sl}(n,\mathbb{R})$ has $n^2$ free parameters and one linear equation $\operatorname{trace} = x_{11}+\ldots+x_{nn}=0$ and the dimension of a Lie group is defined as the dimension of their Lie algebra, their tangent space at $1.$

Btw, if one restricts the domain of the determinant map $d$ to the subset $GL(n,\mathbb R)$, it is continuous as well. Therefore the preimage of the closed set $\{1\}$ under the "restricted determinant map" gives a closed set in $GL(n,\mathbb R)$.

Yes.

It isn't in general a closed subset of $\mathbb M_n \cong \mathbb R^{2n}$.

What is it? $\operatorname{SL}(n,\mathbb{R}),$ I assume.
$n^2,$ I asssume.
Now, I ran out of guesses.

$\operatorname{SL}(n,\mathbb{R}) \subseteq \mathbb{M}(n,\mathbb{R})$ is Zariski-closed.

But I have the feeling that you meant another topology. Which one is it? You embedded $\operatorname{SL}(n,\mathbb{R})$ in $\mathbb{M}(n,\mathbb{R})$ so I assume we have the subspace topology here. That leaves me to guess which topology you prefer on the vector space on the right. Induced by a matrix norm, maybe?

Is the above correct?

Depends on what you meant by "in general" and all the other things you left unexplained.

 

[cianfa72]

Which is easy. We have $n^2$ real indeterminates $x_{ij}$ and multiplication, as well as inversion are differential polynomials in these variables:
$$
(x_{ij})_{ij}\cdot (y_{ij})_{ij} = \left(\sum_{p=1}^n x_{ip}y_{pj}\right)_{ij}
$$ $$
(x_{ij})^{-1}_{ij}= \dfrac{\operatorname{adj (x_{ij})_{ij}}}{\det (x_{ij})_{ij}}
$$ where the adjugate matrix is again a matrix of determinants in the variables $x_{ij}.$

Sorry, what do you mean by the notation $(x_{ij})_{ij}$ ? Is $(x_{ij})$ just a notation for the $X$ matrix itself and the subscripts $i,j$ outside the brackets () actually select which indices ?

The defining polynomial $\det (x_{ij})_{ij} =1$ is also smooth. $\operatorname{GL}(n,\mathbb{R})\subseteq \mathbb{M}(n,\mathbb{R})$ is a dense subset, so that their tangential spaces
$$
T_1\operatorname{GL}(n,\mathbb{R})=\mathfrak{gl}(n,\mathbb{R})=\mathbb{M}(n,\mathbb{R})=T_0\mathbb{M}(n,\mathbb{R})
$$ are identical vector spaces.

You mean the tangent space at the Identity $I$ or $\mathbb 1$ of $\operatorname{GL}(n,\mathbb{R})$ is or defines its "associated" Lie algebra $\mathfrak{gl}(n,\mathbb{R})$.

... assuming you meant $\mathbb{R}^{n^2}$ ...

Yes, of course.

What is it? $\operatorname{SL}(n,\mathbb{R}),$ I assume.
$n^2,$ I assume.

$\operatorname{SL}(n,\mathbb{R}) \subseteq \mathbb{M}(n,\mathbb{R})$ is Zariski-closed.

But I have the feeling that you meant another topology. Which one is it? You embedded $\operatorname{SL}(n,\mathbb{R})$ in $\mathbb{M}(n,\mathbb{R})$ so I assume we have the subspace topology here. That leaves me to guess which topology you prefer on the vector space on the right. Induced by a matrix norm, maybe?

My point is that $\operatorname{SL}(n,\mathbb{R})$ is closed in $\operatorname{GL}(n,\mathbb{R})$ endowed with the subspace topology from $\mathbb{M}(n,\mathbb{R}) \cong \mathbb R^{n^2}$. However $\operatorname{GL}(n,\mathbb{R})$ is not closed inside $\mathbb{M}(n,\mathbb{R})$ then it does not follow that $\operatorname{SL}(n,\mathbb{R})$ isn't closed in $\mathbb{M}(n,\mathbb{R})$ as well (the latter endowed with the standard topology). As you pointed out $\operatorname{SL}(n,\mathbb{R})$ is actually Zariski-closed in $\mathbb{M}(n,\mathbb{R})$.

Depends on what you meant by "in general" and all the other things you left unexplained.

Sorry, I was sloppy. My "in general" actually referred to the above statement about the lack of closure of $\operatorname{SL}(n,\mathbb{R})$ inside $\mathbb{M}(n,\mathbb{R})$.

 

[fresh_42]

Sorry, what do you mean by the notation $(x_{ij})_{ij}$ ? Is $(x_{ij})$ just a notation for the $X$ matrix itself and the subscripts $i,j$ outside the brackets () actually select which indices ?

$x_{ij}$ are matrix entries, in this case, real variables, coordinates. $(x_{ij})$ is a matrix with these entries. $(x_{ij})_{ij}$ is short for $(x_{ij})_{\substack{1\leq i\leq n\\1\leq i\leq n}}$ which is the full description of a matrix where the range of the indices is also noted. The external term ${ij}$ indicates that these numbers vary. The internal term ${ij}$ indicates a specific position at row $i$ and column $j$.

You mean the tangent space at the Identity $I$ or $\mathbb 1$ of $\operatorname{GL}(n,\mathbb{R})$ is or defines its "associated" Lie algebra $\mathfrak{gl}(n,\mathbb{R})$.

Yes.

Sorry, I was sloppy. My "in general" actually referred to the above statement about the lack of closure of $\operatorname{SL}(n,\mathbb{R})$ inside $\mathbb{M}(n,\mathbb{R})$.

And my point was that $\operatorname{SL}(n,\mathbb{R})\subseteq \mathbb{M}(n,\mathbb{R})$ is closed in the Zariski topology of $\mathbb{M}(n,\mathbb{R}).$

 

[martinbn]

And my point was that $\operatorname{SL}(n,\mathbb{R})\subseteq \mathbb{M}(n,\mathbb{R})$ is closed in the Zariski topology of $\mathbb{M}(n,\mathbb{R}).$

And in the usual topology.

 

[cianfa72]

And in the usual topology.

Why ? How can one show that $\operatorname{SL}(n,\mathbb{R})$ is closed in the standard (euclidean) topology of $\mathbb M(n, \mathbb R) \cong \mathbb R^{n^2}$ ?

 

[cianfa72]

$x_{ij}$ are matrix entries, in this case, real variables, coordinates. $(x_{ij})$ is a matrix with these entries. $(x_{ij})_{ij}$ is short for $(x_{ij})_{\substack{1\leq i\leq n\\1\leq i\leq n}}$ which is the full description of a matrix where the range of the indices is also noted. The external term ${ij}$ indicates that these numbers vary. The internal term ${ij}$ indicates a specific position at row $i$ and column $j$.

Ah ok, so the external subscripts ${i,j}$ basically indicate the range of each index (from 1 to n). Therefore the RHS of $$(x_{ij})_{ij}\cdot (y_{ij})_{ij} = \left(\sum_{p=1}^n x_{ip}y_{pj}\right)_{ij}$$ is actually the definition of the product at LHS.

 

[martinbn]

Why ? How can one show that $\operatorname{SL}(n,\mathbb{R})$ is closed in the standard (euclidean) topology of $\mathbb M(n, \mathbb R) \cong \mathbb R^{n^2}$ ?

$det$ is continuous.

 

[cianfa72]

$det$ is continuous.

Ah ok, $\operatorname{SL}(n,\mathbb{R})$ is the preimage in $\mathbb M(n, \mathbb R) \cong \mathbb R^{n^2}$ of the closed set $\{1\}$ under the continuous function $det$ (w.r.t. standard topology).

 

[fresh_42]

Ah ok, $\operatorname{SL}(n,\mathbb{R})$ is the preimage in $\mathbb M(n, \mathbb R) \cong \mathbb R^{n^2}$ of the closed set $\{1\}$ under the continuous function $det$ (w.r.t. standard topology).

Yes. It is the same argument that you already stated ...

Btw, if one restricts the domain of the determinant map $d$ to the subset $GL(n,\mathbb R)$, it is continuous as well. Therefore the preimage of the closed set $\{1\}$ under the "restricted determinant map" gives a closed set in $GL(n,\mathbb R)$. It isn't in general a closed subset of $\mathbb M_n \cong \mathbb R^{2n}$.

... only without restricting the determinant.

 

[cianfa72]

Yes. It is the same argument that you already stated .... only without restricting the determinant.

Ah yes, the fact that $GL(n,\mathbb R)$ isn't closed in $\mathbb M \cong \mathbb R^{n^2}$ doesn't mean that $SL(n,\mathbb R)$ may not be closed in $\mathbb M$ (as it is).

 

[fresh_42]

Ah yes, the fact that $GL(n,\mathbb R)$ isn't closed in $\mathbb M \cong \mathbb R^{n^2}$ doesn't mean that $SL(n,\mathbb R)$ may not be closed in $\mathbb M$ (as it is).

Yes $\neq \{0\}$ is open and $=\{1\}$ is closed.

 

[cianfa72]

I found these interesting lectures. In particular here Lie groups at minute 19:30 he defines Lie group as

Connected portion of continuous group with analytic group composition function (including taking the inverse of any element).

So my point is: since for instance $O(3)$ isn't connected then $O(3)$ is not itself a Lie group (but just $SO(3)$ is since it is the connected component the includues the Identity)?

 

[martinbn]

I found these interesting lectures. In particular here Lie groups at minute 19:30 he defines Lie group as

So my point is: since for instance $O(3)$ isn't connected then $O(3)$ is not itself a Lie group (but just $SO(3)$ is since it is the connected component the includues the Identity)?

The usual definition does not insisy on conectedness. With this definition even $GL(n,\mathbb R)$ is not a Lie group.

 

[cianfa72]

The usual definition does not insisy on conectedness. With this definition even $GL(n,\mathbb R)$ is not a Lie group.

Btw, how is $O(n)$ or $GL(n,\mathbb R)$ topologized ? Their topology is assumed to be the subspace topology from $\mathbb R^{n^2}$ standard topology ?

In that (subspace) topology the group multiplication operation and taking the inverse are actually continuous maps.

 

[martinbn]

Btw, how is $O(n)$ or $GL(n,\mathbb R)$ topologized ? Their topology is assumed to be the subspace topology from $\mathbb R^{n^2}$ standard topology ?

In that (subspace) topology the group multiplication operation and taking the inverse are actually continuous maps.

Not just continuius, there are smooth. They are rational functions.

 

[cianfa72]

Not just continuius, there are smooth. They are rational functions.

Ok yes, I was thinking about the following.

Suppose to take the discrete topology for the set $\mathbb R^{n^2}$. By definition of discrete topology any subset is open. Therefore $O(n)$ and $GL(n,\mathbb R)$ are both open in it hence, if we assume the subspace topology from above as their topologies, then the topological spaces we get can no longer be turned in a (topological) manifold. Right?

 

[fresh_42]

Suppose to take the discrete topology for the set $\mathbb R^{n^2}$. By definition of discrete topology any subset is open. Therefore $O(n)$ and $GL(n,\mathbb R)$ are both open in it hence, if we assume the subspace topology from above as their topologies, then the topological spaces we get can no longer be turned in a (topological) manifold. Right?

Wrong. It is still a topological manifold because multiplication and inversion are trivially continuous. It will be hard to establish an analytical structure, but the topological comes for free. It is just not very interesting.

 

[cianfa72]

Wrong. It is still a topological manifold because multiplication and inversion are trivially continuous. It will be hard to establish an analytical structure, but the topological comes for free. It is just not very interesting.

Sorry, my definition of topological manifold requires locally euclidean. However, since the subspace topology is the same as the discrete topology (it follows from the fact that any subset of $\mathbb R^{n^2}$ is open in the discrete topology), then it can't be locally Euclidean.

 

[martinbn]

Sorry, my definition of topological manifold requires locally euclidean. However, since the subspace topology is the same as the discrete topology (it follows from the fact that any subset of $\mathbb R^{n^2}$ is open in the discrete topology), then it can't be locally Euclidean.

It can, but it will be zero dimensional.

For a ropological group it is not needed that it is a manifold.

 

[cianfa72]

It can, but it will be zero dimensional.

Ah, you mean it would be locally homeomorphic to a zero dimensional Euclidean space.

For a topological group it is not needed that it is a manifold.

In those lectures, he emphasizes the fact that a Lie group is a topological manifold.

 

[martinbn]

Ah, you mean it would be locally homeomorphic to a zero dimensional Euclidean space.

Yes, but it would be uncountble amd some exclude that.

In those lectures, he emphasizes the fact that a Lie group is a topological manifold.

A Lie group yes, it is even a differentiable manifold. My comment was about topological groups.

 

[jbergman]

I found these interesting lectures. In particular here Lie groups at minute 19:30 he defines Lie group as

So my point is: since for instance $O(3)$ isn't connected then $O(3)$ is not itself a Lie group (but just $SO(3)$ is since it is the connected component the includues the Identity)?

That's an incorrect definition.

 

[cianfa72]

Yes, but it would be uncountable and some exclude that.

Ah ok, so every singleton in such topological space would be homeomorphic to $\mathbb R^{0}$.

By uncountable I believe you mean that the topological space is no longer second countable (and as you pointed out some authors exclude that in their definitions).

 

[martinbn]

That's an incorrect definition.

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

 

[jbergman]

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

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

 

[cianfa72]

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

As reference for lectures on this subject, he uses the book "Lie groups, Lie algebras, and some of their applications" by R. Gilmore.

Therefore he takes from there the relevant definitions of structures involved.

 

[jbergman]

As reference for lectures on this subject, he uses the book "Lie groups, Lie algebras, and some of their applications" by R. Gilmore.

Therefore he takes from there the relevant definitions of structures involved.

On what page does that book say a Lie Group is a connected component of the identity. I didn't find that on a quick scan. I thik the YouTuber lost something in translation. I've watched that channel before, and my impression has been that it lacks rigor and precision.

 

[cianfa72]

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) ?

 

[martinbn]

On what page does that book say a Lie Group is a connected component of the identity. I didn't find that on a quick scan. I thik the YouTuber lost something in translation. I've watched that channel before, and my impression has been that it lacks rigor and precision.

Page 77.

 

[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 $\mathbb{Z}_2.$


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