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Degree Bound in Bend and Break Lemmas
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I have read the sections on the Bend & Break Lemmas in Koll'ar-Mori and Debarre and have the following question. (See below for background and what I do know.)
Question: I would like to know if the following is true: if $X$ is a normal (projective) variety and $-K_X$ is $mathbbQ$-Cartier and ample, for the generic point $x in X$, can one find a rational curve $C$ such that $-K_X cdot C le dim X + 1$?
On a related note, if this is false, I would also like to know if it is true for such varieties $X$ with terminal singularities. The reason I wonder if it is true in this case is because terminal singularities appear on minimal models of smooth varieties and we know the statement is true in that case.
Background: On a (smooth) Fano variety $X$, through every point $xin X$, there is a rational curve $C$ such that $0 < -K_X cdot C le dim X + 1$. However, if $X$ is singular, the situation differs. Theorem 3.6 in Debarre's Higher-Dimensional Algebraic Geometry implies that, if $-K_X$ is ample and $X$ is normal, there exists a rational curve $C$ through every point $xin X$ such that $0 < -K_X cdot C le 2 dim X$. I would like to understand why the bound on the degree changes. In my mind, I could see it coming from singular points where $K_X$ is not Cartier, so one doesn't expect the same behavior, or from some finer difference that I do not understand.
So, what I would like to know is: if $x$ is contained in the smooth locus of $X$, can one use the same Bend-and-Break argument to reduce the degree and find a curve $C$ with $-K_X cdot C le dim X + 1$? We can still find curves containing that point in the smooth locus of $X$, so can produce a rational curve through that point, so it seems like we can use the same trick (passing to characteristic $p$ and increasing the degree with the Frobenius) to find curves of lower degree. Perhaps, though, the problem comes when one tries to produce a rational curve--if it passes through the singular locus of $X$, the same argument will not work. I do not have enough experience in this area to know.
ag.algebraic-geometry
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I have read the sections on the Bend & Break Lemmas in Koll'ar-Mori and Debarre and have the following question. (See below for background and what I do know.)
Question: I would like to know if the following is true: if $X$ is a normal (projective) variety and $-K_X$ is $mathbbQ$-Cartier and ample, for the generic point $x in X$, can one find a rational curve $C$ such that $-K_X cdot C le dim X + 1$?
On a related note, if this is false, I would also like to know if it is true for such varieties $X$ with terminal singularities. The reason I wonder if it is true in this case is because terminal singularities appear on minimal models of smooth varieties and we know the statement is true in that case.
Background: On a (smooth) Fano variety $X$, through every point $xin X$, there is a rational curve $C$ such that $0 < -K_X cdot C le dim X + 1$. However, if $X$ is singular, the situation differs. Theorem 3.6 in Debarre's Higher-Dimensional Algebraic Geometry implies that, if $-K_X$ is ample and $X$ is normal, there exists a rational curve $C$ through every point $xin X$ such that $0 < -K_X cdot C le 2 dim X$. I would like to understand why the bound on the degree changes. In my mind, I could see it coming from singular points where $K_X$ is not Cartier, so one doesn't expect the same behavior, or from some finer difference that I do not understand.
So, what I would like to know is: if $x$ is contained in the smooth locus of $X$, can one use the same Bend-and-Break argument to reduce the degree and find a curve $C$ with $-K_X cdot C le dim X + 1$? We can still find curves containing that point in the smooth locus of $X$, so can produce a rational curve through that point, so it seems like we can use the same trick (passing to characteristic $p$ and increasing the degree with the Frobenius) to find curves of lower degree. Perhaps, though, the problem comes when one tries to produce a rational curve--if it passes through the singular locus of $X$, the same argument will not work. I do not have enough experience in this area to know.
ag.algebraic-geometry
asked 5 hours ago
be928
263
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have read the sections on the Bend & Break Lemmas in Koll'ar-Mori and Debarre and have the following question. (See below for background and what I do know.)
Question: I would like to know if the following is true: if $X$ is a normal (projective) variety and $-K_X$ is $mathbbQ$-Cartier and ample, for the generic point $x in X$, can one find a rational curve $C$ such that $-K_X cdot C le dim X + 1$?
On a related note, if this is false, I would also like to know if it is true for such varieties $X$ with terminal singularities. The reason I wonder if it is true in this case is because terminal singularities appear on minimal models of smooth varieties and we know the statement is true in that case.
Background: On a (smooth) Fano variety $X$, through every point $xin X$, there is a rational curve $C$ such that $0 < -K_X cdot C le dim X + 1$. However, if $X$ is singular, the situation differs. Theorem 3.6 in Debarre's Higher-Dimensional Algebraic Geometry implies that, if $-K_X$ is ample and $X$ is normal, there exists a rational curve $C$ through every point $xin X$ such that $0 < -K_X cdot C le 2 dim X$. I would like to understand why the bound on the degree changes. In my mind, I could see it coming from singular points where $K_X$ is not Cartier, so one doesn't expect the same behavior, or from some finer difference that I do not understand.
So, what I would like to know is: if $x$ is contained in the smooth locus of $X$, can one use the same Bend-and-Break argument to reduce the degree and find a curve $C$ with $-K_X cdot C le dim X + 1$? We can still find curves containing that point in the smooth locus of $X$, so can produce a rational curve through that point, so it seems like we can use the same trick (passing to characteristic $p$ and increasing the degree with the Frobenius) to find curves of lower degree. Perhaps, though, the problem comes when one tries to produce a rational curve--if it passes through the singular locus of $X$, the same argument will not work. I do not have enough experience in this area to know.
ag.algebraic-geometry
asked 5 hours ago
be928
263
I have read the sections on the Bend & Break Lemmas in Koll'ar-Mori and Debarre and have the following question. (See below for background and what I do know.)
Question: I would like to know if the following is true: if $X$ is a normal (projective) variety and $-K_X$ is $mathbbQ$-Cartier and ample, for the generic point $x in X$, can one find a rational curve $C$ such that $-K_X cdot C le dim X + 1$?
On a related note, if this is false, I would also like to know if it is true for such varieties $X$ with terminal singularities. The reason I wonder if it is true in this case is because terminal singularities appear on minimal models of smooth varieties and we know the statement is true in that case.
Background: On a (smooth) Fano variety $X$, through every point $xin X$, there is a rational curve $C$ such that $0 < -K_X cdot C le dim X + 1$. However, if $X$ is singular, the situation differs. Theorem 3.6 in Debarre's Higher-Dimensional Algebraic Geometry implies that, if $-K_X$ is ample and $X$ is normal, there exists a rational curve $C$ through every point $xin X$ such that $0 < -K_X cdot C le 2 dim X$. I would like to understand why the bound on the degree changes. In my mind, I could see it coming from singular points where $K_X$ is not Cartier, so one doesn't expect the same behavior, or from some finer difference that I do not understand.
So, what I would like to know is: if $x$ is contained in the smooth locus of $X$, can one use the same Bend-and-Break argument to reduce the degree and find a curve $C$ with $-K_X cdot C le dim X + 1$? We can still find curves containing that point in the smooth locus of $X$, so can produce a rational curve through that point, so it seems like we can use the same trick (passing to characteristic $p$ and increasing the degree with the Frobenius) to find curves of lower degree. Perhaps, though, the problem comes when one tries to produce a rational curve--if it passes through the singular locus of $X$, the same argument will not work. I do not have enough experience in this area to know.
ag.algebraic-geometry
ag.algebraic-geometry
asked 5 hours ago
be928
263
asked 5 hours ago
be928
263
asked 5 hours ago
be928
263
asked 5 hours ago
be928
263
asked 5 hours ago
be928
263
263
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1 Answer
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The bound $-K_X cdot C le dim X + 1$ can be guaranteed if $X$ has local complete intersection singularities, and $f(C)$ intersects the smooth locus of $X$; see [Kollár 1996, Thm. 5.14 and Rem. 5.15]. The reason is that you need certain lower bounds on dimensions of deformation spaces; see [Kollár 1996, Thm. 1.3].
I don't know, however, if there have been improvements since then.
[Kollár 1996] J. Kollár. Rational curves on algebraic varieties. Ergeb. Math. Grenzgeb. (3), Vol. 32. Berlin: Springer-Verlag, 1996. doi: 10.1007/978-3-662-03276-3. mr: 1440180.
answered 2 hours ago
Takumi Murayama
3331213
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The bound $-K_X cdot C le dim X + 1$ can be guaranteed if $X$ has local complete intersection singularities, and $f(C)$ intersects the smooth locus of $X$; see [Kollár 1996, Thm. 5.14 and Rem. 5.15]. The reason is that you need certain lower bounds on dimensions of deformation spaces; see [Kollár 1996, Thm. 1.3].
I don't know, however, if there have been improvements since then.
[Kollár 1996] J. Kollár. Rational curves on algebraic varieties. Ergeb. Math. Grenzgeb. (3), Vol. 32. Berlin: Springer-Verlag, 1996. doi: 10.1007/978-3-662-03276-3. mr: 1440180.
answered 2 hours ago
Takumi Murayama
3331213
add a comment |Â
up vote
3
down vote
The bound $-K_X cdot C le dim X + 1$ can be guaranteed if $X$ has local complete intersection singularities, and $f(C)$ intersects the smooth locus of $X$; see [Kollár 1996, Thm. 5.14 and Rem. 5.15]. The reason is that you need certain lower bounds on dimensions of deformation spaces; see [Kollár 1996, Thm. 1.3].
I don't know, however, if there have been improvements since then.
[Kollár 1996] J. Kollár. Rational curves on algebraic varieties. Ergeb. Math. Grenzgeb. (3), Vol. 32. Berlin: Springer-Verlag, 1996. doi: 10.1007/978-3-662-03276-3. mr: 1440180.
answered 2 hours ago
Takumi Murayama
3331213
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The bound $-K_X cdot C le dim X + 1$ can be guaranteed if $X$ has local complete intersection singularities, and $f(C)$ intersects the smooth locus of $X$; see [Kollár 1996, Thm. 5.14 and Rem. 5.15]. The reason is that you need certain lower bounds on dimensions of deformation spaces; see [Kollár 1996, Thm. 1.3].
I don't know, however, if there have been improvements since then.
[Kollár 1996] J. Kollár. Rational curves on algebraic varieties. Ergeb. Math. Grenzgeb. (3), Vol. 32. Berlin: Springer-Verlag, 1996. doi: 10.1007/978-3-662-03276-3. mr: 1440180.
answered 2 hours ago
Takumi Murayama
3331213
The bound $-K_X cdot C le dim X + 1$ can be guaranteed if $X$ has local complete intersection singularities, and $f(C)$ intersects the smooth locus of $X$; see [Kollár 1996, Thm. 5.14 and Rem. 5.15]. The reason is that you need certain lower bounds on dimensions of deformation spaces; see [Kollár 1996, Thm. 1.3].
I don't know, however, if there have been improvements since then.
[Kollár 1996] J. Kollár. Rational curves on algebraic varieties. Ergeb. Math. Grenzgeb. (3), Vol. 32. Berlin: Springer-Verlag, 1996. doi: 10.1007/978-3-662-03276-3. mr: 1440180.
answered 2 hours ago
Takumi Murayama
3331213
answered 2 hours ago
Takumi Murayama
3331213
answered 2 hours ago
Takumi Murayama
3331213
answered 2 hours ago
Takumi Murayama
3331213
3331213
add a comment |Â
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