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What are the simply-connected non-compact irreducible symmetric spaces?
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Would someone be able to list (or provide a reference to) the simply-connected non-compact irreducible symmetric spaces of rank $ge 1$(as quotients of Lie groups $G/H$)?
Any help would be appreciated!
lie-groups symmetric-spaces
asked Jul 25 at 23:36
Multivariablecalculus
495313
migrated from mathoverflow.net Jul 26 at 17:07
This question came from our site for professional mathematicians.
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up vote
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Would someone be able to list (or provide a reference to) the simply-connected non-compact irreducible symmetric spaces of rank $ge 1$(as quotients of Lie groups $G/H$)?
Any help would be appreciated!
lie-groups symmetric-spaces
asked Jul 25 at 23:36
Multivariablecalculus
495313
migrated from mathoverflow.net Jul 26 at 17:07
This question came from our site for professional mathematicians.
2
Wolf (2011), p. 286 sq..
– Francois Ziegler
Jul 26 at 0:25
@FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups?
– Multivariablecalculus
Jul 26 at 0:47
I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.)
– Francois Ziegler
Jul 26 at 1:37
@FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= mathbbR^ntimesprod M_i times prod M_j^* times prod N_k times prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality".
– Multivariablecalculus
Jul 26 at 1:49
1
Yes. Not “changed†in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)†it was in reference to the “must†your very first comment above. E.g. $smashmathbf R^2timesmathrm S^2$ is noncompact with a compact factor.
– Francois Ziegler
Jul 26 at 2:36
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Would someone be able to list (or provide a reference to) the simply-connected non-compact irreducible symmetric spaces of rank $ge 1$(as quotients of Lie groups $G/H$)?
Any help would be appreciated!
lie-groups symmetric-spaces
asked Jul 25 at 23:36
Multivariablecalculus
495313
Would someone be able to list (or provide a reference to) the simply-connected non-compact irreducible symmetric spaces of rank $ge 1$(as quotients of Lie groups $G/H$)?
Any help would be appreciated!
lie-groups symmetric-spaces
asked Jul 25 at 23:36
Multivariablecalculus
495313
asked Jul 25 at 23:36
Multivariablecalculus
495313
asked Jul 25 at 23:36
Multivariablecalculus
495313
asked Jul 25 at 23:36
Multivariablecalculus
495313
495313
migrated from mathoverflow.net Jul 26 at 17:07
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Jul 26 at 17:07
This question came from our site for professional mathematicians.
2
Wolf (2011), p. 286 sq..
– Francois Ziegler
Jul 26 at 0:25
@FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups?
– Multivariablecalculus
Jul 26 at 0:47
I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.)
– Francois Ziegler
Jul 26 at 1:37
@FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= mathbbR^ntimesprod M_i times prod M_j^* times prod N_k times prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality".
– Multivariablecalculus
Jul 26 at 1:49
1
Yes. Not “changed†in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)†it was in reference to the “must†your very first comment above. E.g. $smashmathbf R^2timesmathrm S^2$ is noncompact with a compact factor.
– Francois Ziegler
Jul 26 at 2:36
 |Â
show 2 more comments
2
Wolf (2011), p. 286 sq..
– Francois Ziegler
Jul 26 at 0:25
@FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups?
– Multivariablecalculus
Jul 26 at 0:47
I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.)
– Francois Ziegler
Jul 26 at 1:37
@FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= mathbbR^ntimesprod M_i times prod M_j^* times prod N_k times prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality".
– Multivariablecalculus
Jul 26 at 1:49
1
Yes. Not “changed†in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)†it was in reference to the “must†your very first comment above. E.g. $smashmathbf R^2timesmathrm S^2$ is noncompact with a compact factor.
– Francois Ziegler
Jul 26 at 2:36
2
2
Wolf (2011), p. 286 sq..
– Francois Ziegler
Jul 26 at 0:25
Wolf (2011), p. 286 sq..
– Francois Ziegler
Jul 26 at 0:25
@FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups?
– Multivariablecalculus
Jul 26 at 0:47
@FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups?
– Multivariablecalculus
Jul 26 at 0:47
I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.)
– Francois Ziegler
Jul 26 at 1:37
I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.)
– Francois Ziegler
Jul 26 at 1:37
@FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= mathbbR^ntimesprod M_i times prod M_j^* times prod N_k times prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality".
– Multivariablecalculus
Jul 26 at 1:49
@FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= mathbbR^ntimesprod M_i times prod M_j^* times prod N_k times prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality".
– Multivariablecalculus
Jul 26 at 1:49
1
1
Yes. Not “changed†in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)†it was in reference to the “must†your very first comment above. E.g. $smashmathbf R^2timesmathrm S^2$ is noncompact with a compact factor.
– Francois Ziegler
Jul 26 at 2:36
Yes. Not “changed†in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)†it was in reference to the “must†your very first comment above. E.g. $smashmathbf R^2timesmathrm S^2$ is noncompact with a compact factor.
– Francois Ziegler
Jul 26 at 2:36
 |Â
show 2 more comments
1 Answer
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The reference from which I learned the lists of $G/K$'s of various sorts is S. Helgason's book "Differential Geometry and Symmetric Spaces".
(For my own mathematical purposes, explicit details about at least the "classical groups and domains" is very useful, so I do keep these things in my head.) Possibly dropping overt reference to the anomalous isogenies and edge cases and quibbles about notational conventions...
Type A: $SL_n(mathbb R)/SO(n,mathbb R)$, $SL_n(mathbb C)/SU(n)$, $SL_n(mathbb H)/Sp^*(n)$, $U(p,q)/U(p)times U(q)$
Types B,D: $O(p,q,mathbb R)/O(p)times O(q)$, $O(n,mathbb C)/SO(n,mathbb R)$, $O^*_2n/U(n)$.
Type C: $Sp_n(mathbb C)/Sp^*n$, $Sp_n(mathbb R)/U(n)$, $Sp^*_p,q/Sp^*_ptimes Sp^*_q$.
The not-well-known cases are: $Sp^*_p,q$ is (modeled by) the group of quaternion matrices preserving a quaternion hermitian form. Provably, these have signatures, much as Sylvester's inertia theorem for quadratic forms. And $O^*_2n$ is quaternion matrices preserving a skew-hermitian form.
Note that in all cases the three $mathbb R$-algebras $mathbb R, mathbb C, mathbb H$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.
answered Jul 26 at 21:48
paul garrett
30.8k360116
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The reference from which I learned the lists of $G/K$'s of various sorts is S. Helgason's book "Differential Geometry and Symmetric Spaces".
(For my own mathematical purposes, explicit details about at least the "classical groups and domains" is very useful, so I do keep these things in my head.) Possibly dropping overt reference to the anomalous isogenies and edge cases and quibbles about notational conventions...
Type A: $SL_n(mathbb R)/SO(n,mathbb R)$, $SL_n(mathbb C)/SU(n)$, $SL_n(mathbb H)/Sp^*(n)$, $U(p,q)/U(p)times U(q)$
Types B,D: $O(p,q,mathbb R)/O(p)times O(q)$, $O(n,mathbb C)/SO(n,mathbb R)$, $O^*_2n/U(n)$.
Type C: $Sp_n(mathbb C)/Sp^*n$, $Sp_n(mathbb R)/U(n)$, $Sp^*_p,q/Sp^*_ptimes Sp^*_q$.
The not-well-known cases are: $Sp^*_p,q$ is (modeled by) the group of quaternion matrices preserving a quaternion hermitian form. Provably, these have signatures, much as Sylvester's inertia theorem for quadratic forms. And $O^*_2n$ is quaternion matrices preserving a skew-hermitian form.
Note that in all cases the three $mathbb R$-algebras $mathbb R, mathbb C, mathbb H$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.
answered Jul 26 at 21:48
paul garrett
30.8k360116
add a comment |Â
up vote
2
down vote
accepted
The reference from which I learned the lists of $G/K$'s of various sorts is S. Helgason's book "Differential Geometry and Symmetric Spaces".
(For my own mathematical purposes, explicit details about at least the "classical groups and domains" is very useful, so I do keep these things in my head.) Possibly dropping overt reference to the anomalous isogenies and edge cases and quibbles about notational conventions...
Type A: $SL_n(mathbb R)/SO(n,mathbb R)$, $SL_n(mathbb C)/SU(n)$, $SL_n(mathbb H)/Sp^*(n)$, $U(p,q)/U(p)times U(q)$
Types B,D: $O(p,q,mathbb R)/O(p)times O(q)$, $O(n,mathbb C)/SO(n,mathbb R)$, $O^*_2n/U(n)$.
Type C: $Sp_n(mathbb C)/Sp^*n$, $Sp_n(mathbb R)/U(n)$, $Sp^*_p,q/Sp^*_ptimes Sp^*_q$.
The not-well-known cases are: $Sp^*_p,q$ is (modeled by) the group of quaternion matrices preserving a quaternion hermitian form. Provably, these have signatures, much as Sylvester's inertia theorem for quadratic forms. And $O^*_2n$ is quaternion matrices preserving a skew-hermitian form.
Note that in all cases the three $mathbb R$-algebras $mathbb R, mathbb C, mathbb H$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.
answered Jul 26 at 21:48
paul garrett
30.8k360116
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The reference from which I learned the lists of $G/K$'s of various sorts is S. Helgason's book "Differential Geometry and Symmetric Spaces".
(For my own mathematical purposes, explicit details about at least the "classical groups and domains" is very useful, so I do keep these things in my head.) Possibly dropping overt reference to the anomalous isogenies and edge cases and quibbles about notational conventions...
Type A: $SL_n(mathbb R)/SO(n,mathbb R)$, $SL_n(mathbb C)/SU(n)$, $SL_n(mathbb H)/Sp^*(n)$, $U(p,q)/U(p)times U(q)$
Types B,D: $O(p,q,mathbb R)/O(p)times O(q)$, $O(n,mathbb C)/SO(n,mathbb R)$, $O^*_2n/U(n)$.
Type C: $Sp_n(mathbb C)/Sp^*n$, $Sp_n(mathbb R)/U(n)$, $Sp^*_p,q/Sp^*_ptimes Sp^*_q$.
The not-well-known cases are: $Sp^*_p,q$ is (modeled by) the group of quaternion matrices preserving a quaternion hermitian form. Provably, these have signatures, much as Sylvester's inertia theorem for quadratic forms. And $O^*_2n$ is quaternion matrices preserving a skew-hermitian form.
Note that in all cases the three $mathbb R$-algebras $mathbb R, mathbb C, mathbb H$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.
answered Jul 26 at 21:48
paul garrett
30.8k360116
The reference from which I learned the lists of $G/K$'s of various sorts is S. Helgason's book "Differential Geometry and Symmetric Spaces".
(For my own mathematical purposes, explicit details about at least the "classical groups and domains" is very useful, so I do keep these things in my head.) Possibly dropping overt reference to the anomalous isogenies and edge cases and quibbles about notational conventions...
Type A: $SL_n(mathbb R)/SO(n,mathbb R)$, $SL_n(mathbb C)/SU(n)$, $SL_n(mathbb H)/Sp^*(n)$, $U(p,q)/U(p)times U(q)$
Types B,D: $O(p,q,mathbb R)/O(p)times O(q)$, $O(n,mathbb C)/SO(n,mathbb R)$, $O^*_2n/U(n)$.
Type C: $Sp_n(mathbb C)/Sp^*n$, $Sp_n(mathbb R)/U(n)$, $Sp^*_p,q/Sp^*_ptimes Sp^*_q$.
The not-well-known cases are: $Sp^*_p,q$ is (modeled by) the group of quaternion matrices preserving a quaternion hermitian form. Provably, these have signatures, much as Sylvester's inertia theorem for quadratic forms. And $O^*_2n$ is quaternion matrices preserving a skew-hermitian form.
Note that in all cases the three $mathbb R$-algebras $mathbb R, mathbb C, mathbb H$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.
answered Jul 26 at 21:48
paul garrett
30.8k360116
answered Jul 26 at 21:48
paul garrett
30.8k360116
answered Jul 26 at 21:48
paul garrett
30.8k360116
answered Jul 26 at 21:48
paul garrett
30.8k360116
30.8k360116
add a comment |Â
add a comment |Â
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2
Wolf (2011), p. 286 sq..
– Francois Ziegler
Jul 26 at 0:25
@FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups?
– Multivariablecalculus
Jul 26 at 0:47
I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.)
– Francois Ziegler
Jul 26 at 1:37
@FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= mathbbR^ntimesprod M_i times prod M_j^* times prod N_k times prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality".
– Multivariablecalculus
Jul 26 at 1:49
1
Yes. Not “changed†in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)†it was in reference to the “must†your very first comment above. E.g. $smashmathbf R^2timesmathrm S^2$ is noncompact with a compact factor.
– Francois Ziegler
Jul 26 at 2:36