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Why is hermitian symmetric domain simply connected?
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In the proof of thereom 1.9 in Milne's note of Shimura varieties, he uses prop 1.14 but does not give a reference for the proof that any hermitian symmetric domain is simply connected, where could I find one?
algebraic-topology lie-groups riemannian-geometry
asked Jul 22 at 6:33
zzy
2,048319
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In the proof of thereom 1.9 in Milne's note of Shimura varieties, he uses prop 1.14 but does not give a reference for the proof that any hermitian symmetric domain is simply connected, where could I find one?
algebraic-topology lie-groups riemannian-geometry
asked Jul 22 at 6:33
zzy
2,048319
This can’t be understood
– Elad
Jul 22 at 7:34
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the proof of thereom 1.9 in Milne's note of Shimura varieties, he uses prop 1.14 but does not give a reference for the proof that any hermitian symmetric domain is simply connected, where could I find one?
algebraic-topology lie-groups riemannian-geometry
asked Jul 22 at 6:33
zzy
2,048319
In the proof of thereom 1.9 in Milne's note of Shimura varieties, he uses prop 1.14 but does not give a reference for the proof that any hermitian symmetric domain is simply connected, where could I find one?
algebraic-topology lie-groups riemannian-geometry
asked Jul 22 at 6:33
zzy
2,048319
edited Jul 22 at 8:39
asked Jul 22 at 6:33
zzy
2,048319
asked Jul 22 at 6:33
zzy
2,048319
asked Jul 22 at 6:33
zzy
2,048319
2,048319
This can’t be understood
– Elad
Jul 22 at 7:34
add a comment |Â
This can’t be understood
– Elad
Jul 22 at 7:34
This can’t be understood
– Elad
Jul 22 at 7:34
This can’t be understood
– Elad
Jul 22 at 7:34
add a comment |Â
1 Answer
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In fact, hermitian symmetric spaces (simple, of non-compact type) have Harish-Chandra models which are convex open subsets of some $mathbb C^n$.
These can be made explicit and elementary (as Siegel did, also see Piatetski-Shapiro's book) for the classical groups and domains.
For compact type, I do not know any comparably simple blanket assertion, although in the classical cases one can do it case-by-case.
answered Jul 22 at 13:23
paul garrett
30.8k360116
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In fact, hermitian symmetric spaces (simple, of non-compact type) have Harish-Chandra models which are convex open subsets of some $mathbb C^n$.
These can be made explicit and elementary (as Siegel did, also see Piatetski-Shapiro's book) for the classical groups and domains.
For compact type, I do not know any comparably simple blanket assertion, although in the classical cases one can do it case-by-case.
answered Jul 22 at 13:23
paul garrett
30.8k360116
add a comment |Â
up vote
1
down vote
accepted
In fact, hermitian symmetric spaces (simple, of non-compact type) have Harish-Chandra models which are convex open subsets of some $mathbb C^n$.
These can be made explicit and elementary (as Siegel did, also see Piatetski-Shapiro's book) for the classical groups and domains.
For compact type, I do not know any comparably simple blanket assertion, although in the classical cases one can do it case-by-case.
answered Jul 22 at 13:23
paul garrett
30.8k360116
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In fact, hermitian symmetric spaces (simple, of non-compact type) have Harish-Chandra models which are convex open subsets of some $mathbb C^n$.
These can be made explicit and elementary (as Siegel did, also see Piatetski-Shapiro's book) for the classical groups and domains.
For compact type, I do not know any comparably simple blanket assertion, although in the classical cases one can do it case-by-case.
answered Jul 22 at 13:23
paul garrett
30.8k360116
In fact, hermitian symmetric spaces (simple, of non-compact type) have Harish-Chandra models which are convex open subsets of some $mathbb C^n$.
These can be made explicit and elementary (as Siegel did, also see Piatetski-Shapiro's book) for the classical groups and domains.
For compact type, I do not know any comparably simple blanket assertion, although in the classical cases one can do it case-by-case.
answered Jul 22 at 13:23
paul garrett
30.8k360116
answered Jul 22 at 13:23
paul garrett
30.8k360116
answered Jul 22 at 13:23
paul garrett
30.8k360116
answered Jul 22 at 13:23
paul garrett
30.8k360116
30.8k360116
add a comment |Â
add a comment |Â
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This can’t be understood
– Elad
Jul 22 at 7:34