Hom scheme of algebraic groups
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Let $G,H$ be two smooth algebraic groups over a field $k$. Consider the functor $T mapsto operatorname{Hom}_T(H_T,G_T)$ from $text{Sch}/k$ to $text{Set}$, is it representable even when $G,H$ is not projective? When is it smooth or etale? (For example, it is etale if $H$ is projective and $G$ is affine). Here Hom means the set of algebraic group homomorphisms.
algebraic-geometry algebraic-groups
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
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add a comment |
$begingroup$
Let $G,H$ be two smooth algebraic groups over a field $k$. Consider the functor $T mapsto operatorname{Hom}_T(H_T,G_T)$ from $text{Sch}/k$ to $text{Set}$, is it representable even when $G,H$ is not projective? When is it smooth or etale? (For example, it is etale if $H$ is projective and $G$ is affine). Here Hom means the set of algebraic group homomorphisms.
algebraic-geometry algebraic-groups
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
$endgroup$
$begingroup$
It follows from here(mathoverflow.net/a/314738/58056) that the space $Hom(G,H)$ is always representable as an algebraic space. This might not be of any help. BTW, where can i find a proof of the for example part of the statement or is it easy to prove?
$endgroup$
– random123
Dec 15 '18 at 6:05
add a comment |
$begingroup$
Let $G,H$ be two smooth algebraic groups over a field $k$. Consider the functor $T mapsto operatorname{Hom}_T(H_T,G_T)$ from $text{Sch}/k$ to $text{Set}$, is it representable even when $G,H$ is not projective? When is it smooth or etale? (For example, it is etale if $H$ is projective and $G$ is affine). Here Hom means the set of algebraic group homomorphisms.
algebraic-geometry algebraic-groups
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
$endgroup$
Let $G,H$ be two smooth algebraic groups over a field $k$. Consider the functor $T mapsto operatorname{Hom}_T(H_T,G_T)$ from $text{Sch}/k$ to $text{Set}$, is it representable even when $G,H$ is not projective? When is it smooth or etale? (For example, it is etale if $H$ is projective and $G$ is affine). Here Hom means the set of algebraic group homomorphisms.
algebraic-geometry algebraic-groups
algebraic-geometry algebraic-groups
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
edited Dec 13 '18 at 19:41
zzy
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
asked Apr 3 '18 at 16:23
zzyzzy
2,5651420
2,5651420
$begingroup$
It follows from here(mathoverflow.net/a/314738/58056) that the space $Hom(G,H)$ is always representable as an algebraic space. This might not be of any help. BTW, where can i find a proof of the for example part of the statement or is it easy to prove?
$endgroup$
– random123
Dec 15 '18 at 6:05
add a comment |
$begingroup$
It follows from here(mathoverflow.net/a/314738/58056) that the space $Hom(G,H)$ is always representable as an algebraic space. This might not be of any help. BTW, where can i find a proof of the for example part of the statement or is it easy to prove?
$endgroup$
– random123
Dec 15 '18 at 6:05
$begingroup$
It follows from here(mathoverflow.net/a/314738/58056) that the space $Hom(G,H)$ is always representable as an algebraic space. This might not be of any help. BTW, where can i find a proof of the for example part of the statement or is it easy to prove?
$endgroup$
– random123
Dec 15 '18 at 6:05
$begingroup$
It follows from here(mathoverflow.net/a/314738/58056) that the space $Hom(G,H)$ is always representable as an algebraic space. This might not be of any help. BTW, where can i find a proof of the for example part of the statement or is it easy to prove?
$endgroup$
– random123
Dec 15 '18 at 6:05
add a comment |
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$begingroup$
It follows from here(mathoverflow.net/a/314738/58056) that the space $Hom(G,H)$ is always representable as an algebraic space. This might not be of any help. BTW, where can i find a proof of the for example part of the statement or is it easy to prove?
$endgroup$
– random123
Dec 15 '18 at 6:05