Degree Bound in Bend and Break Lemmas

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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.










share|cite|improve this question

























    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.










    share|cite|improve this question























      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.










      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      be928

      263




      263




















          1 Answer
          1






          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.






          share|cite|improve this answer




















            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',
            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
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f314861%2fdegree-bound-in-bend-and-break-lemmas%23new-answer', 'question_page');

            );

            Post as a guest






























            1 Answer
            1






            active

            oldest

            votes








            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.






            share|cite|improve this answer
























              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.






              share|cite|improve this answer






















                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Takumi Murayama

                3331213




                3331213



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f314861%2fdegree-bound-in-bend-and-break-lemmas%23new-answer', 'question_page');

                    );

                    Post as a guest













































































                    Popular posts from this blog

                    How to check contact read email or not when send email to Individual?

                    Displaying single band from multi-band raster using QGIS

                    How many registers does an x86_64 CPU actually have?