On the notion of projective isomorphism
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Assume that $X,Y$ are closed subschemes of some projective space $P$. I call $X,Y$ projectively isomorphic if there exists an isomorphism $f:Xrightarrow Y$ with $f^ast O_Y(1)=O_X(1)$. Here $O_X(1)$ is the pullback/restriction of $O_P(1)$. Does this imply that $f$ comes from an automorphism of $P$? I can easily prove the converse.
algebraic-geometry
asked Sep 5 at 10:30
Michel de Nostredame
249310
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Assume that $X,Y$ are closed subschemes of some projective space $P$. I call $X,Y$ projectively isomorphic if there exists an isomorphism $f:Xrightarrow Y$ with $f^ast O_Y(1)=O_X(1)$. Here $O_X(1)$ is the pullback/restriction of $O_P(1)$. Does this imply that $f$ comes from an automorphism of $P$? I can easily prove the converse.
algebraic-geometry
asked Sep 5 at 10:30
Michel de Nostredame
249310
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume that $X,Y$ are closed subschemes of some projective space $P$. I call $X,Y$ projectively isomorphic if there exists an isomorphism $f:Xrightarrow Y$ with $f^ast O_Y(1)=O_X(1)$. Here $O_X(1)$ is the pullback/restriction of $O_P(1)$. Does this imply that $f$ comes from an automorphism of $P$? I can easily prove the converse.
algebraic-geometry
asked Sep 5 at 10:30
Michel de Nostredame
249310
Assume that $X,Y$ are closed subschemes of some projective space $P$. I call $X,Y$ projectively isomorphic if there exists an isomorphism $f:Xrightarrow Y$ with $f^ast O_Y(1)=O_X(1)$. Here $O_X(1)$ is the pullback/restriction of $O_P(1)$. Does this imply that $f$ comes from an automorphism of $P$? I can easily prove the converse.
algebraic-geometry
algebraic-geometry
asked Sep 5 at 10:30
Michel de Nostredame
249310
asked Sep 5 at 10:30
Michel de Nostredame
249310
asked Sep 5 at 10:30
Michel de Nostredame
249310
asked Sep 5 at 10:30
Michel de Nostredame
249310
asked Sep 5 at 10:30
Michel de Nostredame
249310
249310
add a comment |Â
add a comment |Â
1 Answer
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I can take $X,Y$ to be 4 points in $mathbbP^1$, e.g. $X=0,1,infty,a$ and $Y=0,1,infty,b$, with $ane b$. Then clearly there is an isomorphism $f:Xrightarrow Y$ with $f^*mathcalO_Y(1) = mathcalO_X(1)$ but this does not come from an automorphism of $mathbbP^1$, since any such automorphism is determined by 3 points.
answered Sep 5 at 13:19
loch
1,623189
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
2
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I can take $X,Y$ to be 4 points in $mathbbP^1$, e.g. $X=0,1,infty,a$ and $Y=0,1,infty,b$, with $ane b$. Then clearly there is an isomorphism $f:Xrightarrow Y$ with $f^*mathcalO_Y(1) = mathcalO_X(1)$ but this does not come from an automorphism of $mathbbP^1$, since any such automorphism is determined by 3 points.
answered Sep 5 at 13:19
loch
1,623189
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
2
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
add a comment |Â
up vote
3
down vote
accepted
I can take $X,Y$ to be 4 points in $mathbbP^1$, e.g. $X=0,1,infty,a$ and $Y=0,1,infty,b$, with $ane b$. Then clearly there is an isomorphism $f:Xrightarrow Y$ with $f^*mathcalO_Y(1) = mathcalO_X(1)$ but this does not come from an automorphism of $mathbbP^1$, since any such automorphism is determined by 3 points.
answered Sep 5 at 13:19
loch
1,623189
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
2
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I can take $X,Y$ to be 4 points in $mathbbP^1$, e.g. $X=0,1,infty,a$ and $Y=0,1,infty,b$, with $ane b$. Then clearly there is an isomorphism $f:Xrightarrow Y$ with $f^*mathcalO_Y(1) = mathcalO_X(1)$ but this does not come from an automorphism of $mathbbP^1$, since any such automorphism is determined by 3 points.
answered Sep 5 at 13:19
loch
1,623189
I can take $X,Y$ to be 4 points in $mathbbP^1$, e.g. $X=0,1,infty,a$ and $Y=0,1,infty,b$, with $ane b$. Then clearly there is an isomorphism $f:Xrightarrow Y$ with $f^*mathcalO_Y(1) = mathcalO_X(1)$ but this does not come from an automorphism of $mathbbP^1$, since any such automorphism is determined by 3 points.
answered Sep 5 at 13:19
loch
1,623189
answered Sep 5 at 13:19
loch
1,623189
answered Sep 5 at 13:19
loch
1,623189
answered Sep 5 at 13:19
loch
1,623189
1,623189
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
2
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
add a comment |Â
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
2
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
+1 Interesting. Do you also know an example if $X$ and $Y$ are irreducible?
– Michel de Nostredame
Sep 5 at 21:46
2
2
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
@MicheldeNostredame Consider $X$ embedded in a $mathbbP^2 hookrightarrow mathbbP^3$ as a cubic curve and $Y$ to be the twisted cubic in $mathbbP^3$. Any automorphism of $mathbbP^3$ necessarily maps a $mathbbP^2$ to another $mathbbP^2$, but the twisted cubic is nondegenerate.
– loch
Sep 6 at 0:50
add a comment |Â
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