Lie Group has Euler characteristic zero. [closed]
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I need to find a proof that the Euler characteristic of a Lie group is zero that uses differential topology, instead of algebraic topology.
The $chi(X)$ defined as the index sum of a vector field $v$ in $X$. (Which is a topological invariant, i.e. does not depend on the choice of $v$).
Thanks in advance.
differential-topology
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
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closed as off-topic by onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy Dec 16 '18 at 0:41
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need to find a proof that the Euler characteristic of a Lie group is zero that uses differential topology, instead of algebraic topology.
The $chi(X)$ defined as the index sum of a vector field $v$ in $X$. (Which is a topological invariant, i.e. does not depend on the choice of $v$).
Thanks in advance.
differential-topology
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
$endgroup$
closed as off-topic by onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy Dec 16 '18 at 0:41
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need to find a proof that the Euler characteristic of a Lie group is zero that uses differential topology, instead of algebraic topology.
The $chi(X)$ defined as the index sum of a vector field $v$ in $X$. (Which is a topological invariant, i.e. does not depend on the choice of $v$).
Thanks in advance.
differential-topology
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
$endgroup$
I need to find a proof that the Euler characteristic of a Lie group is zero that uses differential topology, instead of algebraic topology.
The $chi(X)$ defined as the index sum of a vector field $v$ in $X$. (Which is a topological invariant, i.e. does not depend on the choice of $v$).
Thanks in advance.
differential-topology
differential-topology
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
asked Dec 15 '18 at 15:38
Bajo FondoBajo Fondo
410315
410315
closed as off-topic by onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy Dec 16 '18 at 0:41
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy Dec 16 '18 at 0:41
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – onurcanbektas, Paul Frost, Rebellos, Eevee Trainer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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$begingroup$
The space of left-invariant vector fields on a Lie group $G$ is isomorphic to the Lie algebra of $G$. A left-invariant vector space corresponding to a nonzero Lie
algebra element is nonzero everywhere, so $G$ has Euler characteristic zero.
(This of course fails when $G$ is zero-dimensional).
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
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– Bajo Fondo
Dec 15 '18 at 15:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The space of left-invariant vector fields on a Lie group $G$ is isomorphic to the Lie algebra of $G$. A left-invariant vector space corresponding to a nonzero Lie
algebra element is nonzero everywhere, so $G$ has Euler characteristic zero.
(This of course fails when $G$ is zero-dimensional).
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
$endgroup$
– Bajo Fondo
Dec 15 '18 at 15:55
add a comment |
$begingroup$
The space of left-invariant vector fields on a Lie group $G$ is isomorphic to the Lie algebra of $G$. A left-invariant vector space corresponding to a nonzero Lie
algebra element is nonzero everywhere, so $G$ has Euler characteristic zero.
(This of course fails when $G$ is zero-dimensional).
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
$endgroup$
– Bajo Fondo
Dec 15 '18 at 15:55
add a comment |
$begingroup$
The space of left-invariant vector fields on a Lie group $G$ is isomorphic to the Lie algebra of $G$. A left-invariant vector space corresponding to a nonzero Lie
algebra element is nonzero everywhere, so $G$ has Euler characteristic zero.
(This of course fails when $G$ is zero-dimensional).
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
The space of left-invariant vector fields on a Lie group $G$ is isomorphic to the Lie algebra of $G$. A left-invariant vector space corresponding to a nonzero Lie
algebra element is nonzero everywhere, so $G$ has Euler characteristic zero.
(This of course fails when $G$ is zero-dimensional).
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 15 '18 at 15:41
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
$endgroup$
– Bajo Fondo
Dec 15 '18 at 15:55
add a comment |
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
$endgroup$
– Bajo Fondo
Dec 15 '18 at 15:55
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
$endgroup$
– Bajo Fondo
Dec 15 '18 at 15:55
$begingroup$
Ok..Take $X$, lef-invariant with isolated zeros. If $X(x)=0$, then $d_xLg(X(x))=0 forall g in G$, then $X(gx)=0 forall g in G$, hence $X=0$ which has no isolated zeros, which contradicts the hypothesis. Thanks.
$endgroup$
– Bajo Fondo
Dec 15 '18 at 15:55
add a comment |
