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Tangent bundle of a group quotient [closed]
If $G$ is a Lie group and $Gamma$ is a discrete subgroup of $G$. What is the tangent bundle of the quotient $G/Gamma$. More specifically, since $T(G)$ is a trivial bundle, how to show that $T(G/Gamma)$ is also a trivial bundle?
differential-geometry algebraic-geometry lie-groups
edited Nov 21 '18 at 7:10
Travis
59.8k767146
asked Nov 21 '18 at 6:53
Ronald
1,7431921
closed as off-topic by KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher Nov 21 '18 at 15:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
If $G$ is a Lie group and $Gamma$ is a discrete subgroup of $G$. What is the tangent bundle of the quotient $G/Gamma$. More specifically, since $T(G)$ is a trivial bundle, how to show that $T(G/Gamma)$ is also a trivial bundle?
differential-geometry algebraic-geometry lie-groups
edited Nov 21 '18 at 7:10
Travis
59.8k767146
asked Nov 21 '18 at 6:53
Ronald
1,7431921
closed as off-topic by KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher Nov 21 '18 at 15:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
For a free proper, smooth action of a discrete group $Gamma$ on a manifold $M$ the orbit map $Mrightarrow M/Gamma$ induces a bundle isomorphism $(TM)/Gammacong T(M/Gamma)$.
– Tyrone
Nov 22 '18 at 9:28
add a comment |
If $G$ is a Lie group and $Gamma$ is a discrete subgroup of $G$. What is the tangent bundle of the quotient $G/Gamma$. More specifically, since $T(G)$ is a trivial bundle, how to show that $T(G/Gamma)$ is also a trivial bundle?
differential-geometry algebraic-geometry lie-groups
edited Nov 21 '18 at 7:10
Travis
59.8k767146
asked Nov 21 '18 at 6:53
Ronald
1,7431921
If $G$ is a Lie group and $Gamma$ is a discrete subgroup of $G$. What is the tangent bundle of the quotient $G/Gamma$. More specifically, since $T(G)$ is a trivial bundle, how to show that $T(G/Gamma)$ is also a trivial bundle?
differential-geometry algebraic-geometry lie-groups
differential-geometry algebraic-geometry lie-groups
edited Nov 21 '18 at 7:10
Travis
59.8k767146
asked Nov 21 '18 at 6:53
Ronald
1,7431921
edited Nov 21 '18 at 7:10
Travis
59.8k767146
asked Nov 21 '18 at 6:53
Ronald
1,7431921
edited Nov 21 '18 at 7:10
Travis
59.8k767146
edited Nov 21 '18 at 7:10
Travis
59.8k767146
edited Nov 21 '18 at 7:10
Travis
59.8k767146
59.8k767146
asked Nov 21 '18 at 6:53
Ronald
1,7431921
asked Nov 21 '18 at 6:53
Ronald
1,7431921
asked Nov 21 '18 at 6:53
Ronald
1,7431921
1,7431921
closed as off-topic by KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher Nov 21 '18 at 15:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher Nov 21 '18 at 15:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – KReiser, John B, José Carlos Santos, Davide Giraudo, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
For a free proper, smooth action of a discrete group $Gamma$ on a manifold $M$ the orbit map $Mrightarrow M/Gamma$ induces a bundle isomorphism $(TM)/Gammacong T(M/Gamma)$.
– Tyrone
Nov 22 '18 at 9:28
add a comment |
For a free proper, smooth action of a discrete group $Gamma$ on a manifold $M$ the orbit map $Mrightarrow M/Gamma$ induces a bundle isomorphism $(TM)/Gammacong T(M/Gamma)$.
– Tyrone
Nov 22 '18 at 9:28
For a free proper, smooth action of a discrete group $Gamma$ on a manifold $M$ the orbit map $Mrightarrow M/Gamma$ induces a bundle isomorphism $(TM)/Gammacong T(M/Gamma)$.
– Tyrone
Nov 22 '18 at 9:28
For a free proper, smooth action of a discrete group $Gamma$ on a manifold $M$ the orbit map $Mrightarrow M/Gamma$ induces a bundle isomorphism $(TM)/Gammacong T(M/Gamma)$.
– Tyrone
Nov 22 '18 at 9:28
add a comment |
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For a free proper, smooth action of a discrete group $Gamma$ on a manifold $M$ the orbit map $Mrightarrow M/Gamma$ induces a bundle isomorphism $(TM)/Gammacong T(M/Gamma)$.
– Tyrone
Nov 22 '18 at 9:28