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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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA Updated.
  $endgroup$
  – Alex Youcis
  3 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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA Updated.
  $endgroup$
  – Alex Youcis
  3 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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmratA Updated.
  $endgroup$
  – Alex Youcis
  3 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