Effective Bertini
$begingroup$
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 '18 at 14:29
GiulioGiulio
1,047517
$endgroup$
add a comment |
$begingroup$
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 '18 at 14:29
GiulioGiulio
1,047517
$endgroup$
3
$begingroup$
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 $q_0$.
$endgroup$
– Jason Starr
Nov 20 '18 at 14:41
$begingroup$
Thanks! My main issue is with $V_q$.
$endgroup$
– Giulio
Nov 20 '18 at 14:43
$begingroup$
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).
$endgroup$
– Giulio
Nov 20 '18 at 20:37
add a comment |
$begingroup$
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 '18 at 14:29
GiulioGiulio
1,047517
$endgroup$
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 '18 at 14:29
GiulioGiulio
1,047517
asked Nov 20 '18 at 14:29
GiulioGiulio
1,047517
asked Nov 20 '18 at 14:29
GiulioGiulio
1,047517
asked Nov 20 '18 at 14:29
GiulioGiulio
1,047517
asked Nov 20 '18 at 14:29
GiulioGiulio
1,047517
1,047517
3
$begingroup$
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 $q_0$.
$endgroup$
– Jason Starr
Nov 20 '18 at 14:41
$begingroup$
Thanks! My main issue is with $V_q$.
$endgroup$
– Giulio
Nov 20 '18 at 14:43
$begingroup$
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).
$endgroup$
– Giulio
Nov 20 '18 at 20:37
add a comment |
3
$begingroup$
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 $q_0$.
$endgroup$
– Jason Starr
Nov 20 '18 at 14:41
$begingroup$
Thanks! My main issue is with $V_q$.
$endgroup$
– Giulio
Nov 20 '18 at 14:43
$begingroup$
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).
$endgroup$
– Giulio
Nov 20 '18 at 20:37
3
3
$begingroup$
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 $q_0$.
$endgroup$
– Jason Starr
Nov 20 '18 at 14:41
$begingroup$
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 $q_0$.
$endgroup$
– Jason Starr
Nov 20 '18 at 14:41
$begingroup$
Thanks! My main issue is with $V_q$.
$endgroup$
– Giulio
Nov 20 '18 at 14:43
$begingroup$
Thanks! My main issue is with $V_q$.
$endgroup$
– Giulio
Nov 20 '18 at 14:43
$begingroup$
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).
$endgroup$
– Giulio
Nov 20 '18 at 20:37
$begingroup$
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).
$endgroup$
– Giulio
Nov 20 '18 at 20:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
$endgroup$
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
$endgroup$
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
add a comment |
$begingroup$
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 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
$endgroup$
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
add a comment |
$begingroup$
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 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
$endgroup$
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 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
answered Nov 20 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
answered Nov 20 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
answered Nov 20 '18 at 14:47
David E SpeyerDavid E Speyer
107k8278537
107k8278537
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
add a comment |
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
$begingroup$
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!!
$endgroup$
– Giulio
Nov 20 '18 at 20:33
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
$begingroup$
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 $q_0$.
$endgroup$
– Jason Starr
Nov 20 '18 at 14:41
$begingroup$
Thanks! My main issue is with $V_q$.
$endgroup$
– Giulio
Nov 20 '18 at 14:43
$begingroup$
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).
$endgroup$
– Giulio
Nov 20 '18 at 20:37