$n-2$-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 21 hours ago
Hans
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 21 hours ago
Hans
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 21 hours ago
Hans
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 21 hours ago
Hans
1,648516
asked 21 hours ago
Hans
1,648516
asked 21 hours ago
Hans
1,648516
asked 21 hours ago
Hans
1,648516
asked 21 hours ago
Hans
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 20 hours ago
Sasha
4,383139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
@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
17 hours ago
add a comment |
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1 Answer
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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 20 hours ago
Sasha
4,383139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
@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
17 hours ago
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 20 hours ago
Sasha
4,383139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
@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
17 hours ago
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 20 hours ago
Sasha
4,383139
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 20 hours ago
Sasha
4,383139
answered 20 hours ago
Sasha
4,383139
answered 20 hours ago
Sasha
4,383139
answered 20 hours ago
Sasha
4,383139
4,383139
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
@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
17 hours ago
add a comment |
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
@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
17 hours ago
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
I presume you meant $ngeq 4$ and not $ngeq 3$ above.
– Mohan
20 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
@Mohan: Indeed, $n$ is the dimension of a hypersurface.
– Sasha
18 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
In the question $n$ is the dimension of the projective space, at least in the first line.
– Mohan
17 hours ago
@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
17 hours ago
@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
17 hours ago
add a comment |
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