$mathbb R$-points of semisimple real algebraic groups, connectivity, and Cartan involutions: some questionsA question about Lie algebras corresponding to Lie groups and algebraic groupsHow to classify all $theta$-stable Cartan sub algebras?Question on unitary representation of non-compact simple Lie groupsMaximal tori in Lie vs algebraic groupsScalar product on Lie algebra of compact Lie groupProve identity map of a Lie algebra is unique Cartan involution when Killing form is negative definiteMost general definition of Borel and parabolic Lie algebras?Is every connected complex semisimple Lie group the complexification of a compact Lie group?Semisimple linear algebraic groups in positive characteristicSymmetric space associated to a connected semisimple linear algebraic $mathbbQ$-group
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$mathbb R$-points of semisimple real algebraic groups, connectivity, and Cartan involutions: some questions
A question about Lie algebras corresponding to Lie groups and algebraic groupsHow to classify all $theta$-stable Cartan sub algebras?Question on unitary representation of non-compact simple Lie groupsMaximal tori in Lie vs algebraic groupsScalar product on Lie algebra of compact Lie groupProve identity map of a Lie algebra is unique Cartan involution when Killing form is negative definiteMost general definition of Borel and parabolic Lie algebras?Is every connected complex semisimple Lie group the complexification of a compact Lie group?Semisimple linear algebraic groups in positive characteristicSymmetric space associated to a connected semisimple linear algebraic $mathbbQ$-group
$begingroup$
I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.
Let $G = mathbf G(mathbb R)$, which is a real Lie group.
$G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?
A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?
Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?
Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?
Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?
algebraic-geometry representation-theory lie-groups lie-algebras algebraic-groups
$endgroup$
add a comment |
$begingroup$
I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.
Let $G = mathbf G(mathbb R)$, which is a real Lie group.
$G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?
A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?
Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?
Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?
Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?
algebraic-geometry representation-theory lie-groups lie-algebras algebraic-groups
$endgroup$
add a comment |
$begingroup$
I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.
Let $G = mathbf G(mathbb R)$, which is a real Lie group.
$G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?
A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?
Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?
Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?
Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?
algebraic-geometry representation-theory lie-groups lie-algebras algebraic-groups
$endgroup$
I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.
Let $G = mathbf G(mathbb R)$, which is a real Lie group.
$G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?
A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?
Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?
Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?
Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?
algebraic-geometry representation-theory lie-groups lie-algebras algebraic-groups
algebraic-geometry representation-theory lie-groups lie-algebras algebraic-groups
asked Mar 21 at 17:45
D_SD_S
14.2k61653
14.2k61653
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1 Answer
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$begingroup$
In a slightly different order:
For future visitors to this problem, here an answer about connectivity on MO.
Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.
In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)
One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.
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1 Answer
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$begingroup$
In a slightly different order:
For future visitors to this problem, here an answer about connectivity on MO.
Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.
In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)
One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.
$endgroup$
add a comment |
$begingroup$
In a slightly different order:
For future visitors to this problem, here an answer about connectivity on MO.
Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.
In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)
One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.
$endgroup$
add a comment |
$begingroup$
In a slightly different order:
For future visitors to this problem, here an answer about connectivity on MO.
Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.
In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)
One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.
$endgroup$
In a slightly different order:
For future visitors to this problem, here an answer about connectivity on MO.
Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.
In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)
One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.
answered Mar 29 at 4:48
Stefan DawydiakStefan Dawydiak
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