Does connected components of a group scheme form a group scheme?
$begingroup$
Let $G$ be a group scheme locally finite type and smooth over a base scheme $S$, and assume $S$ is normal and integral.
Then does the set of (geometrical) connected components of a group scheme form a group? or even point of a group scheme over $S$?
If $S=Spec k $ where $k$ is a field, then this is true and we have a theory of $pi_0(G)$ using etale algebras over a field. I wonder what will happen for the generic case.
general-topology algebraic-geometry algebraic-groups
asked Jan 10 at 6:45
zzyzzy
2,4051419
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group scheme locally finite type and smooth over a base scheme $S$, and assume $S$ is normal and integral.
Then does the set of (geometrical) connected components of a group scheme form a group? or even point of a group scheme over $S$?
If $S=Spec k $ where $k$ is a field, then this is true and we have a theory of $pi_0(G)$ using etale algebras over a field. I wonder what will happen for the generic case.
general-topology algebraic-geometry algebraic-groups
asked Jan 10 at 6:45
zzyzzy
2,4051419
$endgroup$
$begingroup$
If $S$ is spectrum of a field and $G$ is a finite group, the connected components are not group schemes (at least not in a natural way, except the identity component).
$endgroup$
– Mohan
Jan 10 at 14:35
$begingroup$
@Mohan Milne has a section "The group of connected components of an algebraic group" at jmilne.org/math/CourseNotes/iAG200.pdf, and I think that works well over a field.
$endgroup$
– zzy
Jan 10 at 14:56
$begingroup$
I don't think it does. You have a group homomorphism $Gto pi_0(G)$ and the fiber over $e$ is a normal subgroup. The fiber over other points of $pi_0(G)$ are not subgroups.
$endgroup$
– Mohan
Jan 10 at 16:31
$begingroup$
@Mohan PROPOSITION 5.48. (a), I mean the set of connected components, you may misunderstand my question?
$endgroup$
– zzy
Jan 10 at 18:02
add a comment |
$begingroup$
Let $G$ be a group scheme locally finite type and smooth over a base scheme $S$, and assume $S$ is normal and integral.
Then does the set of (geometrical) connected components of a group scheme form a group? or even point of a group scheme over $S$?
If $S=Spec k $ where $k$ is a field, then this is true and we have a theory of $pi_0(G)$ using etale algebras over a field. I wonder what will happen for the generic case.
general-topology algebraic-geometry algebraic-groups
asked Jan 10 at 6:45
zzyzzy
2,4051419
$endgroup$
Let $G$ be a group scheme locally finite type and smooth over a base scheme $S$, and assume $S$ is normal and integral.
Then does the set of (geometrical) connected components of a group scheme form a group? or even point of a group scheme over $S$?
If $S=Spec k $ where $k$ is a field, then this is true and we have a theory of $pi_0(G)$ using etale algebras over a field. I wonder what will happen for the generic case.
general-topology algebraic-geometry algebraic-groups
general-topology algebraic-geometry algebraic-groups
asked Jan 10 at 6:45
zzyzzy
2,4051419
asked Jan 10 at 6:45
zzyzzy
2,4051419
edited Jan 10 at 18:02
zzy
asked Jan 10 at 6:45
zzyzzy
2,4051419
asked Jan 10 at 6:45
zzyzzy
2,4051419
asked Jan 10 at 6:45
zzyzzy
2,4051419
2,4051419
$begingroup$
If $S$ is spectrum of a field and $G$ is a finite group, the connected components are not group schemes (at least not in a natural way, except the identity component).
$endgroup$
– Mohan
Jan 10 at 14:35
$begingroup$
@Mohan Milne has a section "The group of connected components of an algebraic group" at jmilne.org/math/CourseNotes/iAG200.pdf, and I think that works well over a field.
$endgroup$
– zzy
Jan 10 at 14:56
$begingroup$
I don't think it does. You have a group homomorphism $Gto pi_0(G)$ and the fiber over $e$ is a normal subgroup. The fiber over other points of $pi_0(G)$ are not subgroups.
$endgroup$
– Mohan
Jan 10 at 16:31
$begingroup$
@Mohan PROPOSITION 5.48. (a), I mean the set of connected components, you may misunderstand my question?
$endgroup$
– zzy
Jan 10 at 18:02
add a comment |
$begingroup$
If $S$ is spectrum of a field and $G$ is a finite group, the connected components are not group schemes (at least not in a natural way, except the identity component).
$endgroup$
– Mohan
Jan 10 at 14:35
$begingroup$
@Mohan Milne has a section "The group of connected components of an algebraic group" at jmilne.org/math/CourseNotes/iAG200.pdf, and I think that works well over a field.
$endgroup$
– zzy
Jan 10 at 14:56
$begingroup$
I don't think it does. You have a group homomorphism $Gto pi_0(G)$ and the fiber over $e$ is a normal subgroup. The fiber over other points of $pi_0(G)$ are not subgroups.
$endgroup$
– Mohan
Jan 10 at 16:31
$begingroup$
@Mohan PROPOSITION 5.48. (a), I mean the set of connected components, you may misunderstand my question?
$endgroup$
– zzy
Jan 10 at 18:02
$begingroup$
If $S$ is spectrum of a field and $G$ is a finite group, the connected components are not group schemes (at least not in a natural way, except the identity component).
$endgroup$
– Mohan
Jan 10 at 14:35
$begingroup$
If $S$ is spectrum of a field and $G$ is a finite group, the connected components are not group schemes (at least not in a natural way, except the identity component).
$endgroup$
– Mohan
Jan 10 at 14:35
$begingroup$
@Mohan Milne has a section "The group of connected components of an algebraic group" at jmilne.org/math/CourseNotes/iAG200.pdf, and I think that works well over a field.
$endgroup$
– zzy
Jan 10 at 14:56
$begingroup$
@Mohan Milne has a section "The group of connected components of an algebraic group" at jmilne.org/math/CourseNotes/iAG200.pdf, and I think that works well over a field.
$endgroup$
– zzy
Jan 10 at 14:56
$begingroup$
I don't think it does. You have a group homomorphism $Gto pi_0(G)$ and the fiber over $e$ is a normal subgroup. The fiber over other points of $pi_0(G)$ are not subgroups.
$endgroup$
– Mohan
Jan 10 at 16:31
$begingroup$
I don't think it does. You have a group homomorphism $Gto pi_0(G)$ and the fiber over $e$ is a normal subgroup. The fiber over other points of $pi_0(G)$ are not subgroups.
$endgroup$
– Mohan
Jan 10 at 16:31
$begingroup$
@Mohan PROPOSITION 5.48. (a), I mean the set of connected components, you may misunderstand my question?
$endgroup$
– zzy
Jan 10 at 18:02
$begingroup$
@Mohan PROPOSITION 5.48. (a), I mean the set of connected components, you may misunderstand my question?
$endgroup$
– zzy
Jan 10 at 18:02
add a comment |
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$begingroup$
If $S$ is spectrum of a field and $G$ is a finite group, the connected components are not group schemes (at least not in a natural way, except the identity component).
$endgroup$
– Mohan
Jan 10 at 14:35
$begingroup$
@Mohan Milne has a section "The group of connected components of an algebraic group" at jmilne.org/math/CourseNotes/iAG200.pdf, and I think that works well over a field.
$endgroup$
– zzy
Jan 10 at 14:56
$begingroup$
I don't think it does. You have a group homomorphism $Gto pi_0(G)$ and the fiber over $e$ is a normal subgroup. The fiber over other points of $pi_0(G)$ are not subgroups.
$endgroup$
– Mohan
Jan 10 at 16:31
$begingroup$
@Mohan PROPOSITION 5.48. (a), I mean the set of connected components, you may misunderstand my question?
$endgroup$
– zzy
Jan 10 at 18:02