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

How can I fix this gap between bookcases I made?

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

Is ipsum/ipsa/ipse a third person pronoun, or can it serve other functions?

Landlord wants to switch my lease to a "Land contract" to "get back at the city"

How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?

Doomsday-clock for my fantasy planet

What does "enim et" mean?

Prime joint compound before latex paint?

Email Account under attack (really) - anything I can do?

How is it possible for user's password to be changed after storage was encrypted? (on OS X, Android)

Does a dangling wire really electrocute me if I'm standing in water?

Are cabin dividers used to "hide" the flex of the airplane?

aging parents with no investments

Extreme, but not acceptable situation and I can't start the work tomorrow morning

Is Social Media Science Fiction?

Ideas for 3rd eye abilities

How to answer pointed "are you quitting" questioning when I don't want them to suspect

What is it called when oen voice type sings a 'solo?'

Crop image to path created in TikZ?

What to wear for invited talk in Canada

Hosting Wordpress in a EC2 Load Balanced Instance

What causes the sudden spool-up sound from an F-16 when enabling afterburner?

Patience, young "Padovan"

What do the Banks children have against barley water?



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


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 $GL(n,mathbb{K})$ is Lie group.factoring a neighborhood of identity in a compact connected Lie group with a closed Lie subgroupLattice 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 $mathbb R^2/S^1$?










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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
    23 hours ago


















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 $mathbb R^2/S^1$?










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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
    23 hours ago
















3












3








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 $mathbb R^2/S^1$?










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 17 hours ago









YuiTo Cheng

2,3184937




2,3184937










asked yesterday









Amrat AAmrat A

345111




345111








  • 3




    $begingroup$
    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
    23 hours ago
















  • 3




    $begingroup$
    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
    23 hours ago










3




3




$begingroup$
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
23 hours ago






$begingroup$
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
23 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

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 $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ 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. commutator) 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











$endgroup$













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












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










  • $begingroup$
    Oh yes, I just did. Thanks again!
    $endgroup$
    – Amrat A
    yesterday






  • 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
    yesterday






  • 1




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



















6












$begingroup$

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









$endgroup$









  • 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
    23 hours ago












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179002%2fwhat-is-an-example-of-an-abelian-lie-group-g-and-a-closed-subgroup-h-such-th%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ 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. commutator) 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











$endgroup$













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












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










  • $begingroup$
    Oh yes, I just did. Thanks again!
    $endgroup$
    – Amrat A
    yesterday






  • 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
    yesterday






  • 1




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
















4












$begingroup$

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 $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ 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. commutator) 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











$endgroup$













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












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










  • $begingroup$
    Oh yes, I just did. Thanks again!
    $endgroup$
    – Amrat A
    yesterday






  • 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
    yesterday






  • 1




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














4












4








4





$begingroup$

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 $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ 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. commutator) 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











$endgroup$



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 $mathbb{R}^mtimes (S^1)^n$ for some $m$ and $n$. In fact, given $G$ you can read off $m$ and $n$ 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. commutator) 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














share|cite|improve this answer



share|cite|improve this answer








edited 18 hours ago

























answered yesterday









Alex YoucisAlex Youcis

36.1k775115




36.1k775115












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












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










  • $begingroup$
    Oh yes, I just did. Thanks again!
    $endgroup$
    – Amrat A
    yesterday






  • 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
    yesterday






  • 1




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


















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












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










  • $begingroup$
    Oh yes, I just did. Thanks again!
    $endgroup$
    – Amrat A
    yesterday






  • 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
    yesterday






  • 1




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
















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






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














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




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












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




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




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
yesterday




$begingroup$
@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
yesterday




1




1




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




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











6












$begingroup$

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









$endgroup$









  • 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
    23 hours ago
















6












$begingroup$

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









$endgroup$









  • 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
    23 hours ago














6












6








6





$begingroup$

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









$endgroup$



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












share|cite|improve this answer



share|cite|improve this answer










answered 23 hours ago









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
    23 hours ago














  • 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
    23 hours ago








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
23 hours ago




$begingroup$
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
23 hours ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179002%2fwhat-is-an-example-of-an-abelian-lie-group-g-and-a-closed-subgroup-h-such-th%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

Castillo d'Acher Características Menú de navegación

Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...