Effective Bertini
Clash Royale CLAN TAG#URR8PPP
up vote
6
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
$$
V_q:= ; textrms. t. ; (X,frac1qD) ; textrm is klt
$$
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
1,037516
add a comment |
up vote
6
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
$$
V_q:= ; textrms. t. ; (X,frac1qD) ; textrm is klt
$$
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
1,037516
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcalO(1)$ 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 $X=mathbbP^2$, and $L=mathcalO(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcalO(2q-1)|$ 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
6
down vote
favorite
up vote
6
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
$$
V_q:= ; textrms. t. ; (X,frac1qD) ; textrm is klt
$$
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
1,037516
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
$$
V_q:= ; textrms. t. ; (X,frac1qD) ; textrm is klt
$$
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
1,037516
asked Nov 20 at 14:29
Giulio
1,037516
asked Nov 20 at 14:29
Giulio
1,037516
asked Nov 20 at 14:29
Giulio
1,037516
asked Nov 20 at 14:29
Giulio
1,037516
1,037516
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcalO(1)$ 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 $X=mathbbP^2$, and $L=mathcalO(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcalO(2q-1)|$ 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 $mathcalO(1)$ 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 $X=mathbbP^2$, and $L=mathcalO(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcalO(2q-1)|$ 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 $mathcalO(1)$ 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 $mathcalO(1)$ 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 $X=mathbbP^2$, and $L=mathcalO(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcalO(2q-1)|$ 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 $X=mathbbP^2$, and $L=mathcalO(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcalO(2q-1)|$ 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 $X = mathbbP^1 times mathbbP^2$ and let $L= mathcalO(1,1)$. 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
$$beginbmatrix
a & b& c \ d & e & f \ endbmatrix$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbbP^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbbP^2$ and a $mathbbP^1 times mathbbP^1$.) 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
105k8270531
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 $X = mathbbP^1 times mathbbP^2$ and let $L= mathcalO(1,1)$. 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
$$beginbmatrix
a & b& c \ d & e & f \ endbmatrix$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbbP^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbbP^2$ and a $mathbbP^1 times mathbbP^1$.) 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
105k8270531
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 $X = mathbbP^1 times mathbbP^2$ and let $L= mathcalO(1,1)$. 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
$$beginbmatrix
a & b& c \ d & e & f \ endbmatrix$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbbP^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbbP^2$ and a $mathbbP^1 times mathbbP^1$.) 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
105k8270531
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 $X = mathbbP^1 times mathbbP^2$ and let $L= mathcalO(1,1)$. 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
$$beginbmatrix
a & b& c \ d & e & f \ endbmatrix$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbbP^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbbP^2$ and a $mathbbP^1 times mathbbP^1$.) 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
105k8270531
Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = mathbbP^1 times mathbbP^2$ and let $L= mathcalO(1,1)$. 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
$$beginbmatrix
a & b& c \ d & e & f \ endbmatrix$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $mathbbP^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $mathbbP^2$ and a $mathbbP^1 times mathbbP^1$.) 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
105k8270531
answered Nov 20 at 14:47
David E Speyer
105k8270531
answered Nov 20 at 14:47
David E Speyer
105k8270531
answered Nov 20 at 14:47
David E Speyer
105k8270531
105k8270531
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 |
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315783%2feffective-bertini%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
For all $qgeq q_0$, the complement of $U_q$ is a divisor whose degree is known, cf. mathoverflow.net/questions/165672/… Replace $mathcalO(1)$ 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 $X=mathbbP^2$, and $L=mathcalO(3)$. Fix a line $H$, multiplying by $(q+1)H$ we can embed $| mathcalO(2q-1)|$ 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