$n-2$-spaces on cubic hypersurfaces
Let $Xsubsetmathbb{P}^n$ be a smooth cubic hypersurface. Is there, as in the case of $n=3$, an $n-2$-dimensional linear subspace contained in $X$? More general, is there a good description of the Picard group of $X$ for some $ngeq 4$?
algebraic-geometry cubic-equations
asked Jan 4 at 15:50
HansHans
1,648516
add a comment |
Let $Xsubsetmathbb{P}^n$ be a smooth cubic hypersurface. Is there, as in the case of $n=3$, an $n-2$-dimensional linear subspace contained in $X$? More general, is there a good description of the Picard group of $X$ for some $ngeq 4$?
algebraic-geometry cubic-equations
asked Jan 4 at 15:50
HansHans
1,648516
add a comment |
Let $Xsubsetmathbb{P}^n$ be a smooth cubic hypersurface. Is there, as in the case of $n=3$, an $n-2$-dimensional linear subspace contained in $X$? More general, is there a good description of the Picard group of $X$ for some $ngeq 4$?
algebraic-geometry cubic-equations
asked Jan 4 at 15:50
HansHans
1,648516
Let $Xsubsetmathbb{P}^n$ be a smooth cubic hypersurface. Is there, as in the case of $n=3$, an $n-2$-dimensional linear subspace contained in $X$? More general, is there a good description of the Picard group of $X$ for some $ngeq 4$?
algebraic-geometry cubic-equations
algebraic-geometry cubic-equations
asked Jan 4 at 15:50
HansHans
1,648516
asked Jan 4 at 15:50
HansHans
1,648516
asked Jan 4 at 15:50
HansHans
1,648516
asked Jan 4 at 15:50
HansHans
1,648516
asked Jan 4 at 15:50
HansHans
1,648516
1,648516
add a comment |
add a comment |
1 Answer
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By Lefschetz theorem for $n ge 3$ the Picard group of a smooth hypersurface is generated by the class of a hyperplane section.
Similarly, for $n ge 5$ (again by Lefschetz theorem) the Chow group of codimension 2 cycles is generated by the class of a codimension 2 linear section; in particular a smooth hypersurface of dimension $n ge 5$ has no codimension 2 linear subspaces.
In dimension $n = 4$ the moduli space of smooth cubic hypersurfaces (which is 20-dimensional) contains an irreducible divisor of hypersurfaces containing a plane. Thus a general smooth cubic 4-fold has no planes, while some special smooth cubic 4-folds have.
answered Jan 4 at 16:50
SashaSasha
4,443139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
add a comment |
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1 Answer
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By Lefschetz theorem for $n ge 3$ the Picard group of a smooth hypersurface is generated by the class of a hyperplane section.
Similarly, for $n ge 5$ (again by Lefschetz theorem) the Chow group of codimension 2 cycles is generated by the class of a codimension 2 linear section; in particular a smooth hypersurface of dimension $n ge 5$ has no codimension 2 linear subspaces.
In dimension $n = 4$ the moduli space of smooth cubic hypersurfaces (which is 20-dimensional) contains an irreducible divisor of hypersurfaces containing a plane. Thus a general smooth cubic 4-fold has no planes, while some special smooth cubic 4-folds have.
answered Jan 4 at 16:50
SashaSasha
4,443139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
add a comment |
By Lefschetz theorem for $n ge 3$ the Picard group of a smooth hypersurface is generated by the class of a hyperplane section.
Similarly, for $n ge 5$ (again by Lefschetz theorem) the Chow group of codimension 2 cycles is generated by the class of a codimension 2 linear section; in particular a smooth hypersurface of dimension $n ge 5$ has no codimension 2 linear subspaces.
In dimension $n = 4$ the moduli space of smooth cubic hypersurfaces (which is 20-dimensional) contains an irreducible divisor of hypersurfaces containing a plane. Thus a general smooth cubic 4-fold has no planes, while some special smooth cubic 4-folds have.
answered Jan 4 at 16:50
SashaSasha
4,443139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
add a comment |
By Lefschetz theorem for $n ge 3$ the Picard group of a smooth hypersurface is generated by the class of a hyperplane section.
Similarly, for $n ge 5$ (again by Lefschetz theorem) the Chow group of codimension 2 cycles is generated by the class of a codimension 2 linear section; in particular a smooth hypersurface of dimension $n ge 5$ has no codimension 2 linear subspaces.
In dimension $n = 4$ the moduli space of smooth cubic hypersurfaces (which is 20-dimensional) contains an irreducible divisor of hypersurfaces containing a plane. Thus a general smooth cubic 4-fold has no planes, while some special smooth cubic 4-folds have.
answered Jan 4 at 16:50
SashaSasha
4,443139
By Lefschetz theorem for $n ge 3$ the Picard group of a smooth hypersurface is generated by the class of a hyperplane section.
Similarly, for $n ge 5$ (again by Lefschetz theorem) the Chow group of codimension 2 cycles is generated by the class of a codimension 2 linear section; in particular a smooth hypersurface of dimension $n ge 5$ has no codimension 2 linear subspaces.
In dimension $n = 4$ the moduli space of smooth cubic hypersurfaces (which is 20-dimensional) contains an irreducible divisor of hypersurfaces containing a plane. Thus a general smooth cubic 4-fold has no planes, while some special smooth cubic 4-folds have.
answered Jan 4 at 16:50
SashaSasha
4,443139
answered Jan 4 at 16:50
SashaSasha
4,443139
answered Jan 4 at 16:50
SashaSasha
4,443139
answered Jan 4 at 16:50
SashaSasha
4,443139
4,443139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
add a comment |
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
Jan 4 at 17:07
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
Jan 4 at 18:54
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
Jan 4 at 19:37
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
@Mohan: and in my answer $n$ is the dimension of a hypersurface. Thus it answers the question (which actually is about codimension 1 cycles) in the first paragraph, and gives more information about codimension 2 cycles in the last two.
– Sasha
Jan 4 at 19:50
add a comment |
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