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Varieties of dimension $n$ with an $n$-parameter family of lines
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Let $X$ be a projective variety of dimension $n$ with an $n$-parameter family of lines. Is $Xcong mathbbP^n$? I know this is true for $n=2$, but I'm curious about generalizations.
algebraic-geometry
asked Aug 6 at 21:01
AGmath
111
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up vote
2
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Let $X$ be a projective variety of dimension $n$ with an $n$-parameter family of lines. Is $Xcong mathbbP^n$? I know this is true for $n=2$, but I'm curious about generalizations.
algebraic-geometry
asked Aug 6 at 21:01
AGmath
111
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 21:05
The space of lines in $mathbf P^n$ is the Grassmannian $G(2,n+1)$, which has dimension $2(n-1)$. So for $n>2$ the space of lines has dimension strictly greater than $n$. Does your question mean "a family of lines of dimension at least $n$?"
– Asal Beag Dubh
Aug 9 at 13:06
By the way, this MO question is relevant: mathoverflow.net/questions/147229/…
– Asal Beag Dubh
Aug 9 at 14:04
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ be a projective variety of dimension $n$ with an $n$-parameter family of lines. Is $Xcong mathbbP^n$? I know this is true for $n=2$, but I'm curious about generalizations.
algebraic-geometry
asked Aug 6 at 21:01
AGmath
111
Let $X$ be a projective variety of dimension $n$ with an $n$-parameter family of lines. Is $Xcong mathbbP^n$? I know this is true for $n=2$, but I'm curious about generalizations.
algebraic-geometry
asked Aug 6 at 21:01
AGmath
111
asked Aug 6 at 21:01
AGmath
111
asked Aug 6 at 21:01
AGmath
111
asked Aug 6 at 21:01
AGmath
111
111
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 21:05
The space of lines in $mathbf P^n$ is the Grassmannian $G(2,n+1)$, which has dimension $2(n-1)$. So for $n>2$ the space of lines has dimension strictly greater than $n$. Does your question mean "a family of lines of dimension at least $n$?"
– Asal Beag Dubh
Aug 9 at 13:06
By the way, this MO question is relevant: mathoverflow.net/questions/147229/…
– Asal Beag Dubh
Aug 9 at 14:04
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 21:05
The space of lines in $mathbf P^n$ is the Grassmannian $G(2,n+1)$, which has dimension $2(n-1)$. So for $n>2$ the space of lines has dimension strictly greater than $n$. Does your question mean "a family of lines of dimension at least $n$?"
– Asal Beag Dubh
Aug 9 at 13:06
By the way, this MO question is relevant: mathoverflow.net/questions/147229/…
– Asal Beag Dubh
Aug 9 at 14:04
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 21:05
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 21:05
The space of lines in $mathbf P^n$ is the Grassmannian $G(2,n+1)$, which has dimension $2(n-1)$. So for $n>2$ the space of lines has dimension strictly greater than $n$. Does your question mean "a family of lines of dimension at least $n$?"
– Asal Beag Dubh
Aug 9 at 13:06
The space of lines in $mathbf P^n$ is the Grassmannian $G(2,n+1)$, which has dimension $2(n-1)$. So for $n>2$ the space of lines has dimension strictly greater than $n$. Does your question mean "a family of lines of dimension at least $n$?"
– Asal Beag Dubh
Aug 9 at 13:06
By the way, this MO question is relevant: mathoverflow.net/questions/147229/…
– Asal Beag Dubh
Aug 9 at 14:04
By the way, this MO question is relevant: mathoverflow.net/questions/147229/…
– Asal Beag Dubh
Aug 9 at 14:04
add a comment |Â
1 Answer
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There are other examples, for example hypersurfaces $X_dsubset mathbbP^n+1$ of dimension $n$ and low degree $d$ relative to $n$.
It suffices to show we can find $d$ so there is a 1-parameter family of lines through any point $pin X_d$. The lines through $p$ in $mathbbP^n+1$ are parameterized by a $mathbbP^n$. You can show (by expanding out the equation defining $X$ around $p$) that the family of lines through $p$ in $X_d$ is dimension at least $n-d$.
Therefore, it suffices to have $dleq n-1$.
More generally, if $Lsubset Xsubset mathbbP^N$ and the canonical divisor $K_X$ restricts to something of high enough degree on $L$ (I think $geq 3$), then we can wiggle the line in $X$ enough to get an $n$-dimensional family. Somebody who knows more about such varieties (and this story) should tell me more.
answered Aug 9 at 13:48
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
0
down vote
There are other examples, for example hypersurfaces $X_dsubset mathbbP^n+1$ of dimension $n$ and low degree $d$ relative to $n$.
It suffices to show we can find $d$ so there is a 1-parameter family of lines through any point $pin X_d$. The lines through $p$ in $mathbbP^n+1$ are parameterized by a $mathbbP^n$. You can show (by expanding out the equation defining $X$ around $p$) that the family of lines through $p$ in $X_d$ is dimension at least $n-d$.
Therefore, it suffices to have $dleq n-1$.
More generally, if $Lsubset Xsubset mathbbP^N$ and the canonical divisor $K_X$ restricts to something of high enough degree on $L$ (I think $geq 3$), then we can wiggle the line in $X$ enough to get an $n$-dimensional family. Somebody who knows more about such varieties (and this story) should tell me more.
answered Aug 9 at 13:48
Munchlax
1314
add a comment |Â
up vote
0
down vote
There are other examples, for example hypersurfaces $X_dsubset mathbbP^n+1$ of dimension $n$ and low degree $d$ relative to $n$.
It suffices to show we can find $d$ so there is a 1-parameter family of lines through any point $pin X_d$. The lines through $p$ in $mathbbP^n+1$ are parameterized by a $mathbbP^n$. You can show (by expanding out the equation defining $X$ around $p$) that the family of lines through $p$ in $X_d$ is dimension at least $n-d$.
Therefore, it suffices to have $dleq n-1$.
More generally, if $Lsubset Xsubset mathbbP^N$ and the canonical divisor $K_X$ restricts to something of high enough degree on $L$ (I think $geq 3$), then we can wiggle the line in $X$ enough to get an $n$-dimensional family. Somebody who knows more about such varieties (and this story) should tell me more.
answered Aug 9 at 13:48
Munchlax
1314
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There are other examples, for example hypersurfaces $X_dsubset mathbbP^n+1$ of dimension $n$ and low degree $d$ relative to $n$.
It suffices to show we can find $d$ so there is a 1-parameter family of lines through any point $pin X_d$. The lines through $p$ in $mathbbP^n+1$ are parameterized by a $mathbbP^n$. You can show (by expanding out the equation defining $X$ around $p$) that the family of lines through $p$ in $X_d$ is dimension at least $n-d$.
Therefore, it suffices to have $dleq n-1$.
More generally, if $Lsubset Xsubset mathbbP^N$ and the canonical divisor $K_X$ restricts to something of high enough degree on $L$ (I think $geq 3$), then we can wiggle the line in $X$ enough to get an $n$-dimensional family. Somebody who knows more about such varieties (and this story) should tell me more.
answered Aug 9 at 13:48
Munchlax
1314
There are other examples, for example hypersurfaces $X_dsubset mathbbP^n+1$ of dimension $n$ and low degree $d$ relative to $n$.
It suffices to show we can find $d$ so there is a 1-parameter family of lines through any point $pin X_d$. The lines through $p$ in $mathbbP^n+1$ are parameterized by a $mathbbP^n$. You can show (by expanding out the equation defining $X$ around $p$) that the family of lines through $p$ in $X_d$ is dimension at least $n-d$.
Therefore, it suffices to have $dleq n-1$.
More generally, if $Lsubset Xsubset mathbbP^N$ and the canonical divisor $K_X$ restricts to something of high enough degree on $L$ (I think $geq 3$), then we can wiggle the line in $X$ enough to get an $n$-dimensional family. Somebody who knows more about such varieties (and this story) should tell me more.
answered Aug 9 at 13:48
Munchlax
1314
answered Aug 9 at 13:48
Munchlax
1314
answered Aug 9 at 13:48
Munchlax
1314
answered Aug 9 at 13:48
Munchlax
1314
1314
add a comment |Â
add a comment |Â
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 21:05
The space of lines in $mathbf P^n$ is the Grassmannian $G(2,n+1)$, which has dimension $2(n-1)$. So for $n>2$ the space of lines has dimension strictly greater than $n$. Does your question mean "a family of lines of dimension at least $n$?"
– Asal Beag Dubh
Aug 9 at 13:06
By the way, this MO question is relevant: mathoverflow.net/questions/147229/…
– Asal Beag Dubh
Aug 9 at 14:04