Effective Bertini
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Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_qsubset |qL|$ such that every divisor $D$ in $U_q$ is smooth.
Warm up question: is the complement of $U_q$ always a divisor?
We can define a bigger open subset $V_qsubset |qL|$ as
$$
V_q:={Din |qL| ; textrm{s. t.} ; (X,frac{1}{q}D) ; textrm{ is klt} }
$$
My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? )
ag.algebraic-geometry
asked Nov 20 at 14:29
Giulio
1,042516
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up vote
7
down vote
favorite
Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_qsubset |qL|$ such that every divisor $D$ in $U_q$ is smooth.
Warm up question: is the complement of $U_q$ always a divisor?
We can define a bigger open subset $V_qsubset |qL|$ as
$$
V_q:={Din |qL| ; textrm{s. t.} ; (X,frac{1}{q}D) ; textrm{ is klt} }
$$
My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? )
ag.algebraic-geometry
asked Nov 20 at 14:29
Giulio
1,042516
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$.
– Jason Starr
Nov 20 at 14:41
Thanks! My main issue is with $V_q$.
– Giulio
Nov 20 at 14:43
Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=mathbb{P}^2$, and $L=mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment).
– Giulio
Nov 20 at 20:37
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_qsubset |qL|$ such that every divisor $D$ in $U_q$ is smooth.
Warm up question: is the complement of $U_q$ always a divisor?
We can define a bigger open subset $V_qsubset |qL|$ as
$$
V_q:={Din |qL| ; textrm{s. t.} ; (X,frac{1}{q}D) ; textrm{ is klt} }
$$
My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? )
ag.algebraic-geometry
asked Nov 20 at 14:29
Giulio
1,042516
Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_qsubset |qL|$ such that every divisor $D$ in $U_q$ is smooth.
Warm up question: is the complement of $U_q$ always a divisor?
We can define a bigger open subset $V_qsubset |qL|$ as
$$
V_q:={Din |qL| ; textrm{s. t.} ; (X,frac{1}{q}D) ; textrm{ is klt} }
$$
My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? )
ag.algebraic-geometry
ag.algebraic-geometry
asked Nov 20 at 14:29
Giulio
1,042516
asked Nov 20 at 14:29
Giulio
1,042516
asked Nov 20 at 14:29
Giulio
1,042516
asked Nov 20 at 14:29
Giulio
1,042516
asked Nov 20 at 14:29
Giulio
1,042516
1,042516
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$.
– Jason Starr
Nov 20 at 14:41
Thanks! My main issue is with $V_q$.
– Giulio
Nov 20 at 14:43
Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=mathbb{P}^2$, and $L=mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment).
– Giulio
Nov 20 at 20:37
add a comment |
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$.
– Jason Starr
Nov 20 at 14:41
Thanks! My main issue is with $V_q$.
– Giulio
Nov 20 at 14:43
Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=mathbb{P}^2$, and $L=mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment).
– Giulio
Nov 20 at 20:37
3
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$.
– Jason Starr
Nov 20 at 14:41
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$.
– Jason Starr
Nov 20 at 14:41
Thanks! My main issue is with $V_q$.
– Giulio
Nov 20 at 14:43
Thanks! My main issue is with $V_q$.
– Giulio
Nov 20 at 14:43
Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=mathbb{P}^2$, and $L=mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment).
– Giulio
Nov 20 at 20:37
Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=mathbb{P}^2$, and $L=mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment).
– Giulio
Nov 20 at 20:37
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = mathbb{P}^1 times mathbb{P}^2$ and let $L= mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form
$$u (ax+by+cz) + v (dx+ey+fz)=0.$$
$D$ is singular if and only if the matrix
$$begin{bmatrix}
a & b& c \ d & e & f \ end{bmatrix}$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbb{P}^2$ and a $mathbb{P}^1 times mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$.
The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
answered Nov 20 at 14:47
David E Speyer
105k8271532
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = mathbb{P}^1 times mathbb{P}^2$ and let $L= mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form
$$u (ax+by+cz) + v (dx+ey+fz)=0.$$
$D$ is singular if and only if the matrix
$$begin{bmatrix}
a & b& c \ d & e & f \ end{bmatrix}$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbb{P}^2$ and a $mathbb{P}^1 times mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$.
The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
answered Nov 20 at 14:47
David E Speyer
105k8271532
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
add a comment |
up vote
6
down vote
Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = mathbb{P}^1 times mathbb{P}^2$ and let $L= mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form
$$u (ax+by+cz) + v (dx+ey+fz)=0.$$
$D$ is singular if and only if the matrix
$$begin{bmatrix}
a & b& c \ d & e & f \ end{bmatrix}$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbb{P}^2$ and a $mathbb{P}^1 times mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$.
The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
answered Nov 20 at 14:47
David E Speyer
105k8271532
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
add a comment |
up vote
6
down vote
up vote
6
down vote
Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = mathbb{P}^1 times mathbb{P}^2$ and let $L= mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form
$$u (ax+by+cz) + v (dx+ey+fz)=0.$$
$D$ is singular if and only if the matrix
$$begin{bmatrix}
a & b& c \ d & e & f \ end{bmatrix}$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbb{P}^2$ and a $mathbb{P}^1 times mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$.
The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
answered Nov 20 at 14:47
David E Speyer
105k8271532
Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = mathbb{P}^1 times mathbb{P}^2$ and let $L= mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form
$$u (ax+by+cz) + v (dx+ey+fz)=0.$$
$D$ is singular if and only if the matrix
$$begin{bmatrix}
a & b& c \ d & e & f \ end{bmatrix}$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbb{P}^2$ and a $mathbb{P}^1 times mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$.
The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
answered Nov 20 at 14:47
David E Speyer
105k8271532
answered Nov 20 at 14:47
David E Speyer
105k8271532
answered Nov 20 at 14:47
David E Speyer
105k8271532
answered Nov 20 at 14:47
David E Speyer
105k8271532
105k8271532
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
add a comment |
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
Thanks!! In general, I am more interested in the asymptotic values of $q$ (actually, for $q=1$, $L$ might even have no sections at all!!). And of course, I think $V_q$ is the hard part. Thanks!!
– Giulio
Nov 20 at 20:33
add a comment |
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For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcal{O}(1)$ in that answer by an appropriately positive tensor power of $L$, and use the asymptotic formula to find $q_0$.
– Jason Starr
Nov 20 at 14:41
Thanks! My main issue is with $V_q$.
– Giulio
Nov 20 at 14:43
Just to make things more explicit, I am interested in the case $q>>0$. Moreover, let me explain a little computation to justify my last guess about the complement of $V_q$. Take $X=mathbb{P}^2$, and $L=mathcal{O}(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcal{O}(2q-1)|$ into the complement of $V_q$. My guess is that this is kind of the biggest thing that can be outside $V_q$ (but I have no proof for the moment).
– Giulio
Nov 20 at 20:37