A homeomorphism $T$ from extended complex plane to itself preserving cross ratio is a Mobius map.












1












$begingroup$


Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
If we assume $T$ fixes infinity, can we prove $T$ is affine?






complex-analysis mobius-transformation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
    If we assume $T$ fixes infinity, can we prove $T$ is affine?






    complex-analysis mobius-transformation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
      If we assume $T$ fixes infinity, can we prove $T$ is affine?






      complex-analysis mobius-transformation







      share|cite|improve this question











      $endgroup$




      Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
      If we assume $T$ fixes infinity, can we prove $T$ is affine?






      complex-analysis mobius-transformation




      complex-analysis mobius-transformation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 13 at 18:17









      Thomas Shelby

      4,7362727




      4,7362727










      asked Jan 31 at 20:08









      SaikatSaikat

      61




      61






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          My answer needs two facts about Möbius transforms:




          • They preserve cross ratios (this can be checked by a direct calculation).

          • A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).


          You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.



          However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.



          We can express
          $$M(z) = frac{alpha z + beta}{gamma z + delta}$$
          Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.






          share|cite|improve this answer









          $endgroup$














            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',
            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
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095381%2fa-homeomorphism-t-from-extended-complex-plane-to-itself-preserving-cross-ratio%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









            1












            $begingroup$

            My answer needs two facts about Möbius transforms:




            • They preserve cross ratios (this can be checked by a direct calculation).

            • A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).


            You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.



            However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.



            We can express
            $$M(z) = frac{alpha z + beta}{gamma z + delta}$$
            Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              My answer needs two facts about Möbius transforms:




              • They preserve cross ratios (this can be checked by a direct calculation).

              • A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).


              You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.



              However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.



              We can express
              $$M(z) = frac{alpha z + beta}{gamma z + delta}$$
              Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                My answer needs two facts about Möbius transforms:




                • They preserve cross ratios (this can be checked by a direct calculation).

                • A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).


                You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.



                However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.



                We can express
                $$M(z) = frac{alpha z + beta}{gamma z + delta}$$
                Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.






                share|cite|improve this answer









                $endgroup$



                My answer needs two facts about Möbius transforms:




                • They preserve cross ratios (this can be checked by a direct calculation).

                • A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).


                You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.



                However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.



                We can express
                $$M(z) = frac{alpha z + beta}{gamma z + delta}$$
                Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 at 20:35









                0x5390x539

                1,450518




                1,450518






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • 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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095381%2fa-homeomorphism-t-from-extended-complex-plane-to-itself-preserving-cross-ratio%23new-answer', 'question_page');
                    }
                    );

                    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







                    Popular posts from this blog

                    Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                    Image
                    28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                    Read more

                    Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                    Image
                    up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                    Read more

                    A Topological Invariant for $pi_3(U(n))$

                    Image
                    3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                    Read more