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Is an open subscheme of a rationally connected variety, rationally connected?
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Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness?
By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve.
Any idea/reference is most welcome.
ag.algebraic-geometry birational-geometry
asked 3 hours ago
Chen
51439
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up vote
4
down vote
favorite
Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness?
By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve.
Any idea/reference is most welcome.
ag.algebraic-geometry birational-geometry
asked 3 hours ago
Chen
51439
3
When you write "rational curve" do you mean a dense open of $mathbbP^1$? (If not, $mathbbP^1 - S$ is not rationally connected when $Sneq emptyset$.)
– Ariyan Javanpeykar
3 hours ago
2
Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...).
– Jason Starr
3 hours ago
I guess it should be specified in the question (not just the title) that $X$ is rationally connected.
– Laurent Moret-Bailly
2 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness?
By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve.
Any idea/reference is most welcome.
ag.algebraic-geometry birational-geometry
asked 3 hours ago
Chen
51439
Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness?
By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve.
Any idea/reference is most welcome.
ag.algebraic-geometry birational-geometry
ag.algebraic-geometry birational-geometry
asked 3 hours ago
Chen
51439
asked 3 hours ago
Chen
51439
edited 1 hour ago
asked 3 hours ago
Chen
51439
asked 3 hours ago
Chen
51439
asked 3 hours ago
Chen
51439
51439
3
When you write "rational curve" do you mean a dense open of $mathbbP^1$? (If not, $mathbbP^1 - S$ is not rationally connected when $Sneq emptyset$.)
– Ariyan Javanpeykar
3 hours ago
2
Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...).
– Jason Starr
3 hours ago
I guess it should be specified in the question (not just the title) that $X$ is rationally connected.
– Laurent Moret-Bailly
2 hours ago
add a comment |Â
3
When you write "rational curve" do you mean a dense open of $mathbbP^1$? (If not, $mathbbP^1 - S$ is not rationally connected when $Sneq emptyset$.)
– Ariyan Javanpeykar
3 hours ago
2
Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...).
– Jason Starr
3 hours ago
I guess it should be specified in the question (not just the title) that $X$ is rationally connected.
– Laurent Moret-Bailly
2 hours ago
3
3
When you write "rational curve" do you mean a dense open of $mathbbP^1$? (If not, $mathbbP^1 - S$ is not rationally connected when $Sneq emptyset$.)
– Ariyan Javanpeykar
3 hours ago
When you write "rational curve" do you mean a dense open of $mathbbP^1$? (If not, $mathbbP^1 - S$ is not rationally connected when $Sneq emptyset$.)
– Ariyan Javanpeykar
3 hours ago
2
2
Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...).
– Jason Starr
3 hours ago
Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...).
– Jason Starr
3 hours ago
I guess it should be specified in the question (not just the title) that $X$ is rationally connected.
– Laurent Moret-Bailly
2 hours ago
I guess it should be specified in the question (not just the title) that $X$ is rationally connected.
– Laurent Moret-Bailly
2 hours ago
add a comment |Â
1 Answer
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If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $mathbbP^2$ and let $X subset mathbbP^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves.
However, let $Y = X setminus x_0 $. Let $pi : Y to E$ be the projection. For any rational curve $C subset Y$, the map $pi : C to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $pi$, and points in different fibers of $pi$ cannot be connected by a chain of rational curves.
Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.
answered 2 hours ago
David E Speyer
103k8264526
2
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $mathbbP^2$ and let $X subset mathbbP^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves.
However, let $Y = X setminus x_0 $. Let $pi : Y to E$ be the projection. For any rational curve $C subset Y$, the map $pi : C to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $pi$, and points in different fibers of $pi$ cannot be connected by a chain of rational curves.
Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.
answered 2 hours ago
David E Speyer
103k8264526
2
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
add a comment |Â
up vote
5
down vote
If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $mathbbP^2$ and let $X subset mathbbP^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves.
However, let $Y = X setminus x_0 $. Let $pi : Y to E$ be the projection. For any rational curve $C subset Y$, the map $pi : C to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $pi$, and points in different fibers of $pi$ cannot be connected by a chain of rational curves.
Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.
answered 2 hours ago
David E Speyer
103k8264526
2
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $mathbbP^2$ and let $X subset mathbbP^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves.
However, let $Y = X setminus x_0 $. Let $pi : Y to E$ be the projection. For any rational curve $C subset Y$, the map $pi : C to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $pi$, and points in different fibers of $pi$ cannot be connected by a chain of rational curves.
Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.
answered 2 hours ago
David E Speyer
103k8264526
If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $mathbbP^2$ and let $X subset mathbbP^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves.
However, let $Y = X setminus x_0 $. Let $pi : Y to E$ be the projection. For any rational curve $C subset Y$, the map $pi : C to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $pi$, and points in different fibers of $pi$ cannot be connected by a chain of rational curves.
Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.
answered 2 hours ago
David E Speyer
103k8264526
answered 2 hours ago
David E Speyer
103k8264526
answered 2 hours ago
David E Speyer
103k8264526
answered 2 hours ago
David E Speyer
103k8264526
103k8264526
2
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
add a comment |Â
2
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
2
2
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
You have the correct definitions, and your example illustrates that rational chain connectedness is fragile without some further hypothesis on the singularities. If the singularities are KLT (your singularities are log canonical, slightly worse than KLT), then it is expected that the smooth locus does inherit good rational connectedness properties of the projective variety. The main theorems in this direction are due to Keel-McKernan and to Chenyang Xu.
– Jason Starr
1 hour ago
add a comment |Â
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3
When you write "rational curve" do you mean a dense open of $mathbbP^1$? (If not, $mathbbP^1 - S$ is not rationally connected when $Sneq emptyset$.)
– Ariyan Javanpeykar
3 hours ago
2
Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...).
– Jason Starr
3 hours ago
I guess it should be specified in the question (not just the title) that $X$ is rationally connected.
– Laurent Moret-Bailly
2 hours ago