What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The closed subgroup of Lie groupExample: Lie group compact, abelian and disconnected.Center of compact lie group closed?Lie Subgroup Example - Explanation?Examples about maximal Abelian subgroup is not a maximal torus in compact connected Lie group $G$.Maximal compact subgroup of abelian Lie groupA question on abelian Lie groups and maximal compact subgroupClosed Subgroup of $GL(n,mathbbK)$ is Lie group.factoring a neighborhood of identity in a compact connected Lie group with a closed Lie subgroupLattice and abelian Lie groups
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What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The closed subgroup of Lie groupExample: Lie group compact, abelian and disconnected.Center of compact lie group closed?Lie Subgroup Example - Explanation?Examples about maximal Abelian subgroup is not a maximal torus in compact connected Lie group $G$.Maximal compact subgroup of abelian Lie groupA question on abelian Lie groups and maximal compact subgroupClosed Subgroup of $GL(n,mathbbK)$ is Lie group.factoring a neighborhood of identity in a compact connected Lie group with a closed Lie subgroupLattice and abelian Lie groups
$begingroup$
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
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
asked Apr 8 at 1:15
Amrat AAmrat A
350111
$endgroup$
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
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
asked Apr 8 at 1:15
Amrat AAmrat A
350111
$endgroup$
3
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
Apr 8 at 3:14
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
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
asked Apr 8 at 1:15
Amrat AAmrat A
350111
$endgroup$
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
general-topology differential-geometry lie-groups lie-algebras
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
asked Apr 8 at 1:15
Amrat AAmrat A
350111
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
asked Apr 8 at 1:15
Amrat AAmrat A
350111
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
edited Apr 8 at 9:07
YuiTo Cheng
2,40641037
2,40641037
asked Apr 8 at 1:15
Amrat AAmrat A
350111
asked Apr 8 at 1:15
Amrat AAmrat A
350111
asked Apr 8 at 1:15
Amrat AAmrat A
350111
350111
3
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
Apr 8 at 3:14
add a comment |
3
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
Apr 8 at 3:14
3
3
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
Apr 8 at 3:14
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
Apr 8 at 3:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $mathbbR^mtimes (S^1)^n$ for some $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ as $n=mathrmrank(pi_1(G))$ 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, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. 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
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutator) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
$endgroup$
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
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
Apr 8 at 1:58
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
Apr 8 at 2:08
|
show 5 more comments
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. 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 $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered Apr 8 at 3:24
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
Apr 8 at 3:37
add a comment |
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2 Answers
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2 Answers
2
active
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votes
active
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oldest
votes
$begingroup$
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 $mathbbR^mtimes (S^1)^n$ for some $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ as $n=mathrmrank(pi_1(G))$ 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, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. 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
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutator) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
$endgroup$
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
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
Apr 8 at 1:58
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
Apr 8 at 2:08
|
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 $mathbbR^mtimes (S^1)^n$ for some $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ as $n=mathrmrank(pi_1(G))$ 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, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. 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
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutator) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
$endgroup$
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
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
Apr 8 at 1:58
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
Apr 8 at 2:08
|
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 $mathbbR^mtimes (S^1)^n$ for some $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ as $n=mathrmrank(pi_1(G))$ 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, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. 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
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutator) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
$endgroup$
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 $mathbbR^mtimes (S^1)^n$ for some $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ as $n=mathrmrank(pi_1(G))$ 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, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. 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
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutator) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
edited Apr 8 at 8:35
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
answered Apr 8 at 1:26
Alex YoucisAlex Youcis
36.4k775115
36.4k775115
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
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
Apr 8 at 1:58
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
Apr 8 at 2:08
|
show 5 more comments
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
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
Apr 8 at 1:58
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
Apr 8 at 2:08
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
Apr 8 at 1:34
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
Apr 8 at 1:36
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
Apr 8 at 1:37
1
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
Apr 8 at 1:58
$begingroup$
@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
Apr 8 at 1:58
1
1
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@AmratA Updated.
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– Alex Youcis
Apr 8 at 2:08
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
Apr 8 at 2:08
|
show 5 more comments
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Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. 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 $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered Apr 8 at 3:24
RandallRandall
10.7k11431
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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
Apr 8 at 3:37
add a comment |
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. 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 $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered Apr 8 at 3:24
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
Apr 8 at 3:37
add a comment |
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. 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 $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered Apr 8 at 3:24
RandallRandall
10.7k11431
$endgroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. 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 $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered Apr 8 at 3:24
RandallRandall
10.7k11431
answered Apr 8 at 3:24
RandallRandall
10.7k11431
answered Apr 8 at 3:24
RandallRandall
10.7k11431
answered Apr 8 at 3:24
RandallRandall
10.7k11431
10.7k11431
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
Apr 8 at 3:37
add a comment |
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
Apr 8 at 3:37
2
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
Apr 8 at 3:37
$begingroup$
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
Apr 8 at 3:37
add a comment |
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Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
Apr 8 at 3:14