When $J(tau)inmathbbR$? ($J$ is Klein's $j$-invariant.)

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











up vote
2
down vote

favorite












I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.



The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.







share|cite|improve this question
























    up vote
    2
    down vote

    favorite












    I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
    Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.



    The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.







    share|cite|improve this question






















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
      Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.



      The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.







      share|cite|improve this question












      I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
      Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.



      The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 23 at 3:25









      user356126

      1106




      1106




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          The $j$-invariant is real on the fundamental domain, if and only if $tau$
          lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).



          To prove this, note that on the vertical lines with integer and half-integer real
          values, the $h$-function is real, and the unit semicircle is equivalent uunder
          $textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.



          Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
          lines, so takes every real value, by the IVT, on the stated set. Now since
          the $j$-function is bijective on the "strict" fundamental region, then elsewhere
          the $j$-function is non-real.






          share|cite|improve this answer




















          • Thank you for your answer. I understood the behavior of $J(tau)$ well.
            – user356126
            Aug 25 at 5:07










          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: "69"
          ;
          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: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2fmath.stackexchange.com%2fquestions%2f2891670%2fwhen-j-tau-in-mathbbr-j-is-kleins-j-invariant%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
          1
          down vote



          accepted










          The $j$-invariant is real on the fundamental domain, if and only if $tau$
          lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).



          To prove this, note that on the vertical lines with integer and half-integer real
          values, the $h$-function is real, and the unit semicircle is equivalent uunder
          $textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.



          Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
          lines, so takes every real value, by the IVT, on the stated set. Now since
          the $j$-function is bijective on the "strict" fundamental region, then elsewhere
          the $j$-function is non-real.






          share|cite|improve this answer




















          • Thank you for your answer. I understood the behavior of $J(tau)$ well.
            – user356126
            Aug 25 at 5:07














          up vote
          1
          down vote



          accepted










          The $j$-invariant is real on the fundamental domain, if and only if $tau$
          lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).



          To prove this, note that on the vertical lines with integer and half-integer real
          values, the $h$-function is real, and the unit semicircle is equivalent uunder
          $textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.



          Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
          lines, so takes every real value, by the IVT, on the stated set. Now since
          the $j$-function is bijective on the "strict" fundamental region, then elsewhere
          the $j$-function is non-real.






          share|cite|improve this answer




















          • Thank you for your answer. I understood the behavior of $J(tau)$ well.
            – user356126
            Aug 25 at 5:07












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The $j$-invariant is real on the fundamental domain, if and only if $tau$
          lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).



          To prove this, note that on the vertical lines with integer and half-integer real
          values, the $h$-function is real, and the unit semicircle is equivalent uunder
          $textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.



          Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
          lines, so takes every real value, by the IVT, on the stated set. Now since
          the $j$-function is bijective on the "strict" fundamental region, then elsewhere
          the $j$-function is non-real.






          share|cite|improve this answer












          The $j$-invariant is real on the fundamental domain, if and only if $tau$
          lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).



          To prove this, note that on the vertical lines with integer and half-integer real
          values, the $h$-function is real, and the unit semicircle is equivalent uunder
          $textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.



          Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
          lines, so takes every real value, by the IVT, on the stated set. Now since
          the $j$-function is bijective on the "strict" fundamental region, then elsewhere
          the $j$-function is non-real.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 23 at 6:21









          Lord Shark the Unknown

          88k954115




          88k954115











          • Thank you for your answer. I understood the behavior of $J(tau)$ well.
            – user356126
            Aug 25 at 5:07
















          • Thank you for your answer. I understood the behavior of $J(tau)$ well.
            – user356126
            Aug 25 at 5:07















          Thank you for your answer. I understood the behavior of $J(tau)$ well.
          – user356126
          Aug 25 at 5:07




          Thank you for your answer. I understood the behavior of $J(tau)$ well.
          – user356126
          Aug 25 at 5:07

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2891670%2fwhen-j-tau-in-mathbbr-j-is-kleins-j-invariant%23new-answer', 'question_page');

          );

          Post as a guest













































































          這個網誌中的熱門文章

          tkz-euclide: tkzDrawCircle[R] not working

          How to combine Bézier curves to a surface?

          1st Magritte Awards