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Is there any plane in a quintic threefold?
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Sorry to bother if this question is trivial: For a general smooth quintic threefold $V$ in $mathbbP^4$ over an algebraic closed field $k$ of characteristic $0$. Is there a plane in $V$?
If Yes, how do I find it(them)? If no, under what condition there is a plane?
Thanks for any hints and comments.
geometry algebraic-geometry
asked Jul 21 at 23:32
Leo D
686
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up vote
5
down vote
favorite
Sorry to bother if this question is trivial: For a general smooth quintic threefold $V$ in $mathbbP^4$ over an algebraic closed field $k$ of characteristic $0$. Is there a plane in $V$?
If Yes, how do I find it(them)? If no, under what condition there is a plane?
Thanks for any hints and comments.
geometry algebraic-geometry
asked Jul 21 at 23:32
Leo D
686
Any smooth hypersurface in $mathbbP^n, ngeq 4$, the Picard group is $mathbbZ$, genertaed by the hyperplane section. So, there can not be any plane in $V$ as in your case (no generality assumption is necessary).
– Mohan
Jul 21 at 23:51
Thanks @Mohan, just want to make sure, is that because of the Lefschetz hyperplane theorem?
– Leo D
Jul 22 at 0:43
Yes, or Grothendieck-Lefshetz for arbitrary characteristic.
– Mohan
Jul 22 at 1:06
Also, if you know of the $2875$ lines on a general quintic threefold, there'd be lines without bound if a plane was in there. Not a proof, just a mnemonic.
– Jan-Magnus Økland
Jul 22 at 14:10
Thanks @Jan-MagnusØkland, it's a good point.
– Leo D
Jul 23 at 17:28
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Sorry to bother if this question is trivial: For a general smooth quintic threefold $V$ in $mathbbP^4$ over an algebraic closed field $k$ of characteristic $0$. Is there a plane in $V$?
If Yes, how do I find it(them)? If no, under what condition there is a plane?
Thanks for any hints and comments.
geometry algebraic-geometry
asked Jul 21 at 23:32
Leo D
686
Sorry to bother if this question is trivial: For a general smooth quintic threefold $V$ in $mathbbP^4$ over an algebraic closed field $k$ of characteristic $0$. Is there a plane in $V$?
If Yes, how do I find it(them)? If no, under what condition there is a plane?
Thanks for any hints and comments.
geometry algebraic-geometry
asked Jul 21 at 23:32
Leo D
686
asked Jul 21 at 23:32
Leo D
686
asked Jul 21 at 23:32
Leo D
686
asked Jul 21 at 23:32
Leo D
686
686
Any smooth hypersurface in $mathbbP^n, ngeq 4$, the Picard group is $mathbbZ$, genertaed by the hyperplane section. So, there can not be any plane in $V$ as in your case (no generality assumption is necessary).
– Mohan
Jul 21 at 23:51
Thanks @Mohan, just want to make sure, is that because of the Lefschetz hyperplane theorem?
– Leo D
Jul 22 at 0:43
Yes, or Grothendieck-Lefshetz for arbitrary characteristic.
– Mohan
Jul 22 at 1:06
Also, if you know of the $2875$ lines on a general quintic threefold, there'd be lines without bound if a plane was in there. Not a proof, just a mnemonic.
– Jan-Magnus Økland
Jul 22 at 14:10
Thanks @Jan-MagnusØkland, it's a good point.
– Leo D
Jul 23 at 17:28
add a comment |Â
Any smooth hypersurface in $mathbbP^n, ngeq 4$, the Picard group is $mathbbZ$, genertaed by the hyperplane section. So, there can not be any plane in $V$ as in your case (no generality assumption is necessary).
– Mohan
Jul 21 at 23:51
Thanks @Mohan, just want to make sure, is that because of the Lefschetz hyperplane theorem?
– Leo D
Jul 22 at 0:43
Yes, or Grothendieck-Lefshetz for arbitrary characteristic.
– Mohan
Jul 22 at 1:06
Also, if you know of the $2875$ lines on a general quintic threefold, there'd be lines without bound if a plane was in there. Not a proof, just a mnemonic.
– Jan-Magnus Økland
Jul 22 at 14:10
Thanks @Jan-MagnusØkland, it's a good point.
– Leo D
Jul 23 at 17:28
Any smooth hypersurface in $mathbbP^n, ngeq 4$, the Picard group is $mathbbZ$, genertaed by the hyperplane section. So, there can not be any plane in $V$ as in your case (no generality assumption is necessary).
– Mohan
Jul 21 at 23:51
Any smooth hypersurface in $mathbbP^n, ngeq 4$, the Picard group is $mathbbZ$, genertaed by the hyperplane section. So, there can not be any plane in $V$ as in your case (no generality assumption is necessary).
– Mohan
Jul 21 at 23:51
Thanks @Mohan, just want to make sure, is that because of the Lefschetz hyperplane theorem?
– Leo D
Jul 22 at 0:43
Thanks @Mohan, just want to make sure, is that because of the Lefschetz hyperplane theorem?
– Leo D
Jul 22 at 0:43
Yes, or Grothendieck-Lefshetz for arbitrary characteristic.
– Mohan
Jul 22 at 1:06
Yes, or Grothendieck-Lefshetz for arbitrary characteristic.
– Mohan
Jul 22 at 1:06
Also, if you know of the $2875$ lines on a general quintic threefold, there'd be lines without bound if a plane was in there. Not a proof, just a mnemonic.
– Jan-Magnus Økland
Jul 22 at 14:10
Also, if you know of the $2875$ lines on a general quintic threefold, there'd be lines without bound if a plane was in there. Not a proof, just a mnemonic.
– Jan-Magnus Økland
Jul 22 at 14:10
Thanks @Jan-MagnusØkland, it's a good point.
– Leo D
Jul 23 at 17:28
Thanks @Jan-MagnusØkland, it's a good point.
– Leo D
Jul 23 at 17:28
add a comment |Â
1 Answer
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There is also an elementary argument (I first saw it in http://www.math.stonybrook.edu/~jstarr/papers/appendix3.pdf):
Let $X=V(F)$ and suppose $X$ contains $P=X_0=X_1=0subsetmathbbP^4$. Then, we can write
$$F=F_0X_0+F_1X_1,$$
as $Fin langle X_0,X_1rangle$ by definition. Now,
beginalign*
partial_0 F|_P &= F_0\
partial_1 F|_P &= F_1\
partial_2 F|_P &= 0\
partial_3 F|_P &= 0.
endalign*
Now, we notice that $F_0,F_1$ have a common zero when restricted to $P$, so $X=V(F)$ is not smooth.
In general, this shows we cannot have any planes in a hypersurface exceeding half the dimension of the hypersurface.
answered Jul 22 at 2:17
Munchlax
1314
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
There is also an elementary argument (I first saw it in http://www.math.stonybrook.edu/~jstarr/papers/appendix3.pdf):
Let $X=V(F)$ and suppose $X$ contains $P=X_0=X_1=0subsetmathbbP^4$. Then, we can write
$$F=F_0X_0+F_1X_1,$$
as $Fin langle X_0,X_1rangle$ by definition. Now,
beginalign*
partial_0 F|_P &= F_0\
partial_1 F|_P &= F_1\
partial_2 F|_P &= 0\
partial_3 F|_P &= 0.
endalign*
Now, we notice that $F_0,F_1$ have a common zero when restricted to $P$, so $X=V(F)$ is not smooth.
In general, this shows we cannot have any planes in a hypersurface exceeding half the dimension of the hypersurface.
answered Jul 22 at 2:17
Munchlax
1314
add a comment |Â
up vote
2
down vote
There is also an elementary argument (I first saw it in http://www.math.stonybrook.edu/~jstarr/papers/appendix3.pdf):
Let $X=V(F)$ and suppose $X$ contains $P=X_0=X_1=0subsetmathbbP^4$. Then, we can write
$$F=F_0X_0+F_1X_1,$$
as $Fin langle X_0,X_1rangle$ by definition. Now,
beginalign*
partial_0 F|_P &= F_0\
partial_1 F|_P &= F_1\
partial_2 F|_P &= 0\
partial_3 F|_P &= 0.
endalign*
Now, we notice that $F_0,F_1$ have a common zero when restricted to $P$, so $X=V(F)$ is not smooth.
In general, this shows we cannot have any planes in a hypersurface exceeding half the dimension of the hypersurface.
answered Jul 22 at 2:17
Munchlax
1314
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There is also an elementary argument (I first saw it in http://www.math.stonybrook.edu/~jstarr/papers/appendix3.pdf):
Let $X=V(F)$ and suppose $X$ contains $P=X_0=X_1=0subsetmathbbP^4$. Then, we can write
$$F=F_0X_0+F_1X_1,$$
as $Fin langle X_0,X_1rangle$ by definition. Now,
beginalign*
partial_0 F|_P &= F_0\
partial_1 F|_P &= F_1\
partial_2 F|_P &= 0\
partial_3 F|_P &= 0.
endalign*
Now, we notice that $F_0,F_1$ have a common zero when restricted to $P$, so $X=V(F)$ is not smooth.
In general, this shows we cannot have any planes in a hypersurface exceeding half the dimension of the hypersurface.
answered Jul 22 at 2:17
Munchlax
1314
There is also an elementary argument (I first saw it in http://www.math.stonybrook.edu/~jstarr/papers/appendix3.pdf):
Let $X=V(F)$ and suppose $X$ contains $P=X_0=X_1=0subsetmathbbP^4$. Then, we can write
$$F=F_0X_0+F_1X_1,$$
as $Fin langle X_0,X_1rangle$ by definition. Now,
beginalign*
partial_0 F|_P &= F_0\
partial_1 F|_P &= F_1\
partial_2 F|_P &= 0\
partial_3 F|_P &= 0.
endalign*
Now, we notice that $F_0,F_1$ have a common zero when restricted to $P$, so $X=V(F)$ is not smooth.
In general, this shows we cannot have any planes in a hypersurface exceeding half the dimension of the hypersurface.
answered Jul 22 at 2:17
Munchlax
1314
answered Jul 22 at 2:17
Munchlax
1314
answered Jul 22 at 2:17
Munchlax
1314
answered Jul 22 at 2:17
Munchlax
1314
1314
add a comment |Â
add a comment |Â
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Any smooth hypersurface in $mathbbP^n, ngeq 4$, the Picard group is $mathbbZ$, genertaed by the hyperplane section. So, there can not be any plane in $V$ as in your case (no generality assumption is necessary).
– Mohan
Jul 21 at 23:51
Thanks @Mohan, just want to make sure, is that because of the Lefschetz hyperplane theorem?
– Leo D
Jul 22 at 0:43
Yes, or Grothendieck-Lefshetz for arbitrary characteristic.
– Mohan
Jul 22 at 1:06
Also, if you know of the $2875$ lines on a general quintic threefold, there'd be lines without bound if a plane was in there. Not a proof, just a mnemonic.
– Jan-Magnus Økland
Jul 22 at 14:10
Thanks @Jan-MagnusØkland, it's a good point.
– Leo D
Jul 23 at 17:28