Automorphism group of union of varieties

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A projective hypersurface $mathcalV(F)$, given by a homogeneous polynomial $F$, can always be expressed as the union of its affine components $mathcalV(F_i)$, where $F_i=F(x_1,ldots,x_i-1,1,x_i+1,ldots,x_n)$ is the $i$-th dehomogenisation. That is we have
$$mathcalV(F)=bigcup_i=1^nmathcalV(F_i).$$
My question is; is there any relationship between the automorphism group of the projective hypersurface, $mathrmAut(mathcalV(F))$, and the affine automorphism groups, $mathrmAut(mathcalV(F_i))$?
To me it seems that each automorphism of $mathcalV(F)$ should act either as an isomorphism between pairs of its affine components $mathcalV(F_i)$ and $mathcalV(F_j)$, $ineq j$, or as an automorphism on each $mathcalV(F_i)$.
So then it should be the case that each $mathrmAut(mathcalV(F_i))$ is a subgroup of $mathrmAut(mathcalV(F))$?
algebraic-geometry projective-geometry automorphism-group
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A projective hypersurface $mathcalV(F)$, given by a homogeneous polynomial $F$, can always be expressed as the union of its affine components $mathcalV(F_i)$, where $F_i=F(x_1,ldots,x_i-1,1,x_i+1,ldots,x_n)$ is the $i$-th dehomogenisation. That is we have
$$mathcalV(F)=bigcup_i=1^nmathcalV(F_i).$$
My question is; is there any relationship between the automorphism group of the projective hypersurface, $mathrmAut(mathcalV(F))$, and the affine automorphism groups, $mathrmAut(mathcalV(F_i))$?
To me it seems that each automorphism of $mathcalV(F)$ should act either as an isomorphism between pairs of its affine components $mathcalV(F_i)$ and $mathcalV(F_j)$, $ineq j$, or as an automorphism on each $mathcalV(F_i)$.
So then it should be the case that each $mathrmAut(mathcalV(F_i))$ is a subgroup of $mathrmAut(mathcalV(F))$?
algebraic-geometry projective-geometry automorphism-group
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A projective hypersurface $mathcalV(F)$, given by a homogeneous polynomial $F$, can always be expressed as the union of its affine components $mathcalV(F_i)$, where $F_i=F(x_1,ldots,x_i-1,1,x_i+1,ldots,x_n)$ is the $i$-th dehomogenisation. That is we have
$$mathcalV(F)=bigcup_i=1^nmathcalV(F_i).$$
My question is; is there any relationship between the automorphism group of the projective hypersurface, $mathrmAut(mathcalV(F))$, and the affine automorphism groups, $mathrmAut(mathcalV(F_i))$?
To me it seems that each automorphism of $mathcalV(F)$ should act either as an isomorphism between pairs of its affine components $mathcalV(F_i)$ and $mathcalV(F_j)$, $ineq j$, or as an automorphism on each $mathcalV(F_i)$.
So then it should be the case that each $mathrmAut(mathcalV(F_i))$ is a subgroup of $mathrmAut(mathcalV(F))$?
algebraic-geometry projective-geometry automorphism-group
A projective hypersurface $mathcalV(F)$, given by a homogeneous polynomial $F$, can always be expressed as the union of its affine components $mathcalV(F_i)$, where $F_i=F(x_1,ldots,x_i-1,1,x_i+1,ldots,x_n)$ is the $i$-th dehomogenisation. That is we have
$$mathcalV(F)=bigcup_i=1^nmathcalV(F_i).$$
My question is; is there any relationship between the automorphism group of the projective hypersurface, $mathrmAut(mathcalV(F))$, and the affine automorphism groups, $mathrmAut(mathcalV(F_i))$?
To me it seems that each automorphism of $mathcalV(F)$ should act either as an isomorphism between pairs of its affine components $mathcalV(F_i)$ and $mathcalV(F_j)$, $ineq j$, or as an automorphism on each $mathcalV(F_i)$.
So then it should be the case that each $mathrmAut(mathcalV(F_i))$ is a subgroup of $mathrmAut(mathcalV(F))$?
algebraic-geometry projective-geometry automorphism-group
edited Aug 29 at 5:38
asked Aug 29 at 5:32
user551642
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For generic $F$, $mboxAut(mathcalV(F))< mboxAut(mathbbP^n-1)$ and there is a subgroup of $mboxAut(mathcalV(F))$ that acts on $mathcalV(F_i)$, namely the stabilizer of $x_i=0$.
If you can find coordinates such that every atutomorphism in $mboxAut(mathcalV(F))$ permutes the $mathcalV(F_i)$, then $mboxAut(mathcalV(F))$ is called an imprimitive subgroup of $mboxAut(mathbbP^n-1$). It can defenitely happen but not always. (For curves it happens for the Fermat curves but not for the Klein quartic)
The problem with $mboxAut(mathcalV(F_i)) < mboxAut(mathcalV(F))$ is what you want to the automorphism to preserve. You could have a birational self-map $T$ of $mathcalV(F)$ preserving $mathcalV(F_i)$ and this imply that $T in mboxAut(mathcalV(F_i))$ but $Tnotin mboxAut(mathcalV(F))$.
Let me know if it helps you.
Thanks for the answer.
â user551642
Aug 29 at 22:26
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
For generic $F$, $mboxAut(mathcalV(F))< mboxAut(mathbbP^n-1)$ and there is a subgroup of $mboxAut(mathcalV(F))$ that acts on $mathcalV(F_i)$, namely the stabilizer of $x_i=0$.
If you can find coordinates such that every atutomorphism in $mboxAut(mathcalV(F))$ permutes the $mathcalV(F_i)$, then $mboxAut(mathcalV(F))$ is called an imprimitive subgroup of $mboxAut(mathbbP^n-1$). It can defenitely happen but not always. (For curves it happens for the Fermat curves but not for the Klein quartic)
The problem with $mboxAut(mathcalV(F_i)) < mboxAut(mathcalV(F))$ is what you want to the automorphism to preserve. You could have a birational self-map $T$ of $mathcalV(F)$ preserving $mathcalV(F_i)$ and this imply that $T in mboxAut(mathcalV(F_i))$ but $Tnotin mboxAut(mathcalV(F))$.
Let me know if it helps you.
Thanks for the answer.
â user551642
Aug 29 at 22:26
add a comment |Â
up vote
2
down vote
accepted
For generic $F$, $mboxAut(mathcalV(F))< mboxAut(mathbbP^n-1)$ and there is a subgroup of $mboxAut(mathcalV(F))$ that acts on $mathcalV(F_i)$, namely the stabilizer of $x_i=0$.
If you can find coordinates such that every atutomorphism in $mboxAut(mathcalV(F))$ permutes the $mathcalV(F_i)$, then $mboxAut(mathcalV(F))$ is called an imprimitive subgroup of $mboxAut(mathbbP^n-1$). It can defenitely happen but not always. (For curves it happens for the Fermat curves but not for the Klein quartic)
The problem with $mboxAut(mathcalV(F_i)) < mboxAut(mathcalV(F))$ is what you want to the automorphism to preserve. You could have a birational self-map $T$ of $mathcalV(F)$ preserving $mathcalV(F_i)$ and this imply that $T in mboxAut(mathcalV(F_i))$ but $Tnotin mboxAut(mathcalV(F))$.
Let me know if it helps you.
Thanks for the answer.
â user551642
Aug 29 at 22:26
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For generic $F$, $mboxAut(mathcalV(F))< mboxAut(mathbbP^n-1)$ and there is a subgroup of $mboxAut(mathcalV(F))$ that acts on $mathcalV(F_i)$, namely the stabilizer of $x_i=0$.
If you can find coordinates such that every atutomorphism in $mboxAut(mathcalV(F))$ permutes the $mathcalV(F_i)$, then $mboxAut(mathcalV(F))$ is called an imprimitive subgroup of $mboxAut(mathbbP^n-1$). It can defenitely happen but not always. (For curves it happens for the Fermat curves but not for the Klein quartic)
The problem with $mboxAut(mathcalV(F_i)) < mboxAut(mathcalV(F))$ is what you want to the automorphism to preserve. You could have a birational self-map $T$ of $mathcalV(F)$ preserving $mathcalV(F_i)$ and this imply that $T in mboxAut(mathcalV(F_i))$ but $Tnotin mboxAut(mathcalV(F))$.
Let me know if it helps you.
For generic $F$, $mboxAut(mathcalV(F))< mboxAut(mathbbP^n-1)$ and there is a subgroup of $mboxAut(mathcalV(F))$ that acts on $mathcalV(F_i)$, namely the stabilizer of $x_i=0$.
If you can find coordinates such that every atutomorphism in $mboxAut(mathcalV(F))$ permutes the $mathcalV(F_i)$, then $mboxAut(mathcalV(F))$ is called an imprimitive subgroup of $mboxAut(mathbbP^n-1$). It can defenitely happen but not always. (For curves it happens for the Fermat curves but not for the Klein quartic)
The problem with $mboxAut(mathcalV(F_i)) < mboxAut(mathcalV(F))$ is what you want to the automorphism to preserve. You could have a birational self-map $T$ of $mathcalV(F)$ preserving $mathcalV(F_i)$ and this imply that $T in mboxAut(mathcalV(F_i))$ but $Tnotin mboxAut(mathcalV(F))$.
Let me know if it helps you.
answered Aug 29 at 15:12
Alan Muniz
1,411622
1,411622
Thanks for the answer.
â user551642
Aug 29 at 22:26
add a comment |Â
Thanks for the answer.
â user551642
Aug 29 at 22:26
Thanks for the answer.
â user551642
Aug 29 at 22:26
Thanks for the answer.
â user551642
Aug 29 at 22:26
add a comment |Â
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