Mixing
Is SL(2,C) unimodular
$begingroup$
My question is in the title, do you know if SL(2,C) is an unimodular group or not and how to prove it ?
Thank you.
lie-groups
asked Jan 28 at 16:58
théo jaminthéo jamin
1
$endgroup$
add a comment |
$begingroup$
My question is in the title, do you know if SL(2,C) is an unimodular group or not and how to prove it ?
Thank you.
lie-groups
asked Jan 28 at 16:58
théo jaminthéo jamin
1
$endgroup$
add a comment |
$begingroup$
My question is in the title, do you know if SL(2,C) is an unimodular group or not and how to prove it ?
Thank you.
lie-groups
asked Jan 28 at 16:58
théo jaminthéo jamin
1
$endgroup$
My question is in the title, do you know if SL(2,C) is an unimodular group or not and how to prove it ?
Thank you.
lie-groups
lie-groups
asked Jan 28 at 16:58
théo jaminthéo jamin
1
asked Jan 28 at 16:58
théo jaminthéo jamin
1
asked Jan 28 at 16:58
théo jaminthéo jamin
1
asked Jan 28 at 16:58
théo jaminthéo jamin
1
asked Jan 28 at 16:58
théo jaminthéo jamin
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
I found the answer in the article of JÖRG WINKELMANN : "Complex analytic geometry of complex parallelizable manifolds" :
Recall that a locally compact topological group $G$ is called unimodular iff a left-invariant Haar measure is also right-invariant.
Assume $G$ is a locally compact topological group.
Page $5$, J. Winkelmann proved that :
Let $Gamma$ be a cocompact discrete subgroup of $G$. Then $G/Gamma$ admits a $G$ invariant probability measure.
Let $Gamma$ be a discrete subgroup of $G$. Assume that there exists a $G$-invariant probability measure on $G/Gamma$ then $G$ is unimodular.
Finally, the existence of lattices in $SL_2(mathbb{C})$ is given by the existence of compact $3$-dimensional hyperbolic manifold and thus we know that $SL_2(mathbb{C})$ is unimodular.
I hope I did not make mistakes.
answered Jan 30 at 17:05
théo jaminthéo jamin
1
$endgroup$
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
add a comment |
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$begingroup$
I found the answer in the article of JÖRG WINKELMANN : "Complex analytic geometry of complex parallelizable manifolds" :
Recall that a locally compact topological group $G$ is called unimodular iff a left-invariant Haar measure is also right-invariant.
Assume $G$ is a locally compact topological group.
Page $5$, J. Winkelmann proved that :
Let $Gamma$ be a cocompact discrete subgroup of $G$. Then $G/Gamma$ admits a $G$ invariant probability measure.
Let $Gamma$ be a discrete subgroup of $G$. Assume that there exists a $G$-invariant probability measure on $G/Gamma$ then $G$ is unimodular.
Finally, the existence of lattices in $SL_2(mathbb{C})$ is given by the existence of compact $3$-dimensional hyperbolic manifold and thus we know that $SL_2(mathbb{C})$ is unimodular.
I hope I did not make mistakes.
answered Jan 30 at 17:05
théo jaminthéo jamin
1
$endgroup$
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
add a comment |
$begingroup$
I found the answer in the article of JÖRG WINKELMANN : "Complex analytic geometry of complex parallelizable manifolds" :
Recall that a locally compact topological group $G$ is called unimodular iff a left-invariant Haar measure is also right-invariant.
Assume $G$ is a locally compact topological group.
Page $5$, J. Winkelmann proved that :
Let $Gamma$ be a cocompact discrete subgroup of $G$. Then $G/Gamma$ admits a $G$ invariant probability measure.
Let $Gamma$ be a discrete subgroup of $G$. Assume that there exists a $G$-invariant probability measure on $G/Gamma$ then $G$ is unimodular.
Finally, the existence of lattices in $SL_2(mathbb{C})$ is given by the existence of compact $3$-dimensional hyperbolic manifold and thus we know that $SL_2(mathbb{C})$ is unimodular.
I hope I did not make mistakes.
answered Jan 30 at 17:05
théo jaminthéo jamin
1
$endgroup$
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
add a comment |
$begingroup$
I found the answer in the article of JÖRG WINKELMANN : "Complex analytic geometry of complex parallelizable manifolds" :
Recall that a locally compact topological group $G$ is called unimodular iff a left-invariant Haar measure is also right-invariant.
Assume $G$ is a locally compact topological group.
Page $5$, J. Winkelmann proved that :
Let $Gamma$ be a cocompact discrete subgroup of $G$. Then $G/Gamma$ admits a $G$ invariant probability measure.
Let $Gamma$ be a discrete subgroup of $G$. Assume that there exists a $G$-invariant probability measure on $G/Gamma$ then $G$ is unimodular.
Finally, the existence of lattices in $SL_2(mathbb{C})$ is given by the existence of compact $3$-dimensional hyperbolic manifold and thus we know that $SL_2(mathbb{C})$ is unimodular.
I hope I did not make mistakes.
answered Jan 30 at 17:05
théo jaminthéo jamin
1
$endgroup$
I found the answer in the article of JÖRG WINKELMANN : "Complex analytic geometry of complex parallelizable manifolds" :
Recall that a locally compact topological group $G$ is called unimodular iff a left-invariant Haar measure is also right-invariant.
Assume $G$ is a locally compact topological group.
Page $5$, J. Winkelmann proved that :
Let $Gamma$ be a cocompact discrete subgroup of $G$. Then $G/Gamma$ admits a $G$ invariant probability measure.
Let $Gamma$ be a discrete subgroup of $G$. Assume that there exists a $G$-invariant probability measure on $G/Gamma$ then $G$ is unimodular.
Finally, the existence of lattices in $SL_2(mathbb{C})$ is given by the existence of compact $3$-dimensional hyperbolic manifold and thus we know that $SL_2(mathbb{C})$ is unimodular.
I hope I did not make mistakes.
answered Jan 30 at 17:05
théo jaminthéo jamin
1
answered Jan 30 at 17:05
théo jaminthéo jamin
1
answered Jan 30 at 17:05
théo jaminthéo jamin
1
answered Jan 30 at 17:05
théo jaminthéo jamin
1
1
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
add a comment |
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
$begingroup$
This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant.
$endgroup$
– Moishe Kohan
Feb 1 at 18:41
add a comment |
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