Does connected components of a group scheme form a group scheme?












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$begingroup$


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.










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$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Jan 10 at 18:02







zzy

















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


















  • $begingroup$
    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










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