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Closed subgroups of abelian groups

Embedded Lie subgroups are closed.The injectivity of torus in the category of abelian Lie groupsCenter of compact lie group closed?Closedness of connected semisimple Lie subgroups of semisimple groupsIntersection of a family of closed Lie subgroupsExamples about maximal Abelian subgroup is not a maximal torus in compact connected Lie group $G$.A question on abelian Lie groups and maximal compact subgroupReal and complex nilpotent Lie groupsClosed Subgroups of Lie Groups are closed Lie Subgroups?Lattice and abelian Lie groups

3

$begingroup$

What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?

Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?

general-topology differential-geometry lie-groups lie-algebras

share|cite|improve this question

edited 4 hours ago

Clayton

19.6k33288

asked 4 hours ago

Amrat AAmrat A

345111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
  $endgroup$
  – Randall
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?

Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?

general-topology differential-geometry lie-groups lie-algebras

share|cite|improve this question

edited 4 hours ago

Clayton

19.6k33288

asked 4 hours ago

Amrat AAmrat A

345111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
  $endgroup$
  – Randall
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?

Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?

general-topology differential-geometry lie-groups lie-algebras

share|cite|improve this question

edited 4 hours ago

Clayton

19.6k33288

asked 4 hours ago

Amrat AAmrat A

345111

$endgroup$

share|cite|improve this question

edited 4 hours ago

Clayton

19.6k33288

asked 4 hours ago

Amrat AAmrat A

345111

share|cite|improve this question

edited 4 hours ago

Clayton

19.6k33288

asked 4 hours ago

Amrat AAmrat A

345111

edited 4 hours ago

Clayton

19.6k33288

edited 4 hours ago

Clayton

19.6k33288

asked 4 hours ago

Amrat AAmrat A

345111

asked 4 hours ago

Amrat AAmrat A

345111

2 Answers
2

active

oldest

votes

2

$begingroup$

EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.

Every connected real abelian Lie group $G$ is isomorphic to $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ and $m=dim G-n$.

Now, if you have a short exact sequence of abelian Lie groups

$$0to Hto Gto G/Hto 0$$

Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence

$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$

So, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired

EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that

$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$

and

$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$

The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:

$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$

---

(Below is for the non-abelian situation)
Here's a simple interesting example.

Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.

share|cite|improve this answer

edited 2 hours ago

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
  $endgroup$
  – Amrat A
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh yes, I just did. Thanks again!
  $endgroup$
  – Amrat A
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA Updated.
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  

 | 
show 5 more comments

4

$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).

share|cite|improve this answer

answered 2 hours ago

RandallRandall

10.7k11431

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
  $endgroup$
  – Alex Youcis
  2 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.

Every connected real abelian Lie group $G$ is isomorphic to $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ and $m=dim G-n$.

Now, if you have a short exact sequence of abelian Lie groups

$$0to Hto Gto G/Hto 0$$

Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence

$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$

So, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired

EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that

$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$

and

$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$

The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:

$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$

---

(Below is for the non-abelian situation)
Here's a simple interesting example.

Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.

share|cite|improve this answer

edited 2 hours ago

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
  $endgroup$
  – Amrat A
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh yes, I just did. Thanks again!
  $endgroup$
  – Amrat A
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA Updated.
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  

 | 
show 5 more comments

2

$begingroup$

EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.

Every connected real abelian Lie group $G$ is isomorphic to $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ and $m=dim G-n$.

Now, if you have a short exact sequence of abelian Lie groups

$$0to Hto Gto G/Hto 0$$

Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence

$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$

So, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired

EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that

$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$

and

$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$

The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:

$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$

---

(Below is for the non-abelian situation)
Here's a simple interesting example.

Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.

share|cite|improve this answer

edited 2 hours ago

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
  $endgroup$
  – Amrat A
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh yes, I just did. Thanks again!
  $endgroup$
  – Amrat A
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA Updated.
  $endgroup$
  – Alex Youcis
  4 hours ago
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.

Every connected real abelian Lie group $G$ is isomorphic to $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ and $m=dim G-n$.

Now, if you have a short exact sequence of abelian Lie groups

$$0to Hto Gto G/Hto 0$$

Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence

$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$

So, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired

EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that

$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$

and

$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$

The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:

$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$

---

(Below is for the non-abelian situation)
Here's a simple interesting example.

Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.

share|cite|improve this answer

edited 2 hours ago

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

answered 4 hours ago

Alex YoucisAlex Youcis

36.1k775115

4

$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).

share|cite|improve this answer

answered 2 hours ago

RandallRandall

10.7k11431

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
  $endgroup$
  – Alex Youcis
  2 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).

share|cite|improve this answer

answered 2 hours ago

RandallRandall

10.7k11431

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
  $endgroup$
  – Alex Youcis
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).

share|cite|improve this answer

answered 2 hours ago

RandallRandall

10.7k11431

$endgroup$

share|cite|improve this answer

answered 2 hours ago

RandallRandall

10.7k11431

answered 2 hours ago

RandallRandall

10.7k11431

answered 2 hours ago

RandallRandall

10.7k11431