Subgroup of Möbius transformations which are isometries with respect to the standard metric on the Riemann...












2














I'm trying to find which subgroup of Mobius transformations are isometries with respect to the standard metric on the Riemann sphere (the one induced from the Euclidean metric on $mathbb{R}^3$).



The question hints that the distance on the Riemann sphere corresponds to the distance function on $mathbb{P}^1$ given by



$$d(L_1,L_2) = 2sqrt{1 - frac{|langle v,w rangle|^2}{||v||^2||w||^2}}$$



where $ v in L_1backslash{0}$ and $w in L_2backslash{0}$, and that a Mobius map corresponds to the action of a matrix $A in GL_2(mathbb{C})$ on lines in $mathbb{C}^2$, and asks me to consider which $2x2$ matrices automatically preserve this expression for $d$.



I honestly have no idea where to start with this question, so I'd really appreciate whatever help you might be able to give.






complex-analysis metric-spaces isometry mobius-transformation stereographic-projections







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  • From the expression of $d$, it is clear that a Mobius transformation $A$ preserves the metric $d$ if $A$ preserves the inner product $left<cdot,cdotright>$, ie, $A$ is a rotation matrix. If you adopt the well-known fact that the isometry group of a sphere is the rotation group, then it follows that the subgroup is just the rotation group.
    – cjackal
    Jan 4 at 1:22










  • @cjackal note that one has to mod out $mathbb C^*$ in the $GL(2, mathbb C)$ action, so the subgroup in $GL(2, mathbb C)$ is definitely not the rotation group. On the other hand, there is no point to do this exercise, if one assume the knowledge about the isometry group of the sphere.
    – Arctic Char
    Jan 4 at 1:26












  • @ArcticChar Well we need to consider the matrices projectively, so $A$ is considered as an element in $PSL(2,mathbb{C})$ and my comment is just rephrasing the isomorphism $PSU(2,mathbb{C})cong SO(3)$. But it is true that my first comment is confusing in not distinguishing the complex inner product and real inner product. One may want to use an explicit formula converting a complex 2x2 matrix to a real 4x4 matrix to see the isomorphism above.
    – cjackal
    Jan 4 at 1:42
















2














I'm trying to find which subgroup of Mobius transformations are isometries with respect to the standard metric on the Riemann sphere (the one induced from the Euclidean metric on $mathbb{R}^3$).



The question hints that the distance on the Riemann sphere corresponds to the distance function on $mathbb{P}^1$ given by



$$d(L_1,L_2) = 2sqrt{1 - frac{|langle v,w rangle|^2}{||v||^2||w||^2}}$$



where $ v in L_1backslash{0}$ and $w in L_2backslash{0}$, and that a Mobius map corresponds to the action of a matrix $A in GL_2(mathbb{C})$ on lines in $mathbb{C}^2$, and asks me to consider which $2x2$ matrices automatically preserve this expression for $d$.



I honestly have no idea where to start with this question, so I'd really appreciate whatever help you might be able to give.






complex-analysis metric-spaces isometry mobius-transformation stereographic-projections







share|cite|improve this question







New contributor




jsnow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • From the expression of $d$, it is clear that a Mobius transformation $A$ preserves the metric $d$ if $A$ preserves the inner product $left<cdot,cdotright>$, ie, $A$ is a rotation matrix. If you adopt the well-known fact that the isometry group of a sphere is the rotation group, then it follows that the subgroup is just the rotation group.
    – cjackal
    Jan 4 at 1:22










  • @cjackal note that one has to mod out $mathbb C^*$ in the $GL(2, mathbb C)$ action, so the subgroup in $GL(2, mathbb C)$ is definitely not the rotation group. On the other hand, there is no point to do this exercise, if one assume the knowledge about the isometry group of the sphere.
    – Arctic Char
    Jan 4 at 1:26












  • @ArcticChar Well we need to consider the matrices projectively, so $A$ is considered as an element in $PSL(2,mathbb{C})$ and my comment is just rephrasing the isomorphism $PSU(2,mathbb{C})cong SO(3)$. But it is true that my first comment is confusing in not distinguishing the complex inner product and real inner product. One may want to use an explicit formula converting a complex 2x2 matrix to a real 4x4 matrix to see the isomorphism above.
    – cjackal
    Jan 4 at 1:42














2












2








2







I'm trying to find which subgroup of Mobius transformations are isometries with respect to the standard metric on the Riemann sphere (the one induced from the Euclidean metric on $mathbb{R}^3$).



The question hints that the distance on the Riemann sphere corresponds to the distance function on $mathbb{P}^1$ given by



$$d(L_1,L_2) = 2sqrt{1 - frac{|langle v,w rangle|^2}{||v||^2||w||^2}}$$



where $ v in L_1backslash{0}$ and $w in L_2backslash{0}$, and that a Mobius map corresponds to the action of a matrix $A in GL_2(mathbb{C})$ on lines in $mathbb{C}^2$, and asks me to consider which $2x2$ matrices automatically preserve this expression for $d$.



I honestly have no idea where to start with this question, so I'd really appreciate whatever help you might be able to give.






complex-analysis metric-spaces isometry mobius-transformation stereographic-projections







share|cite|improve this question







New contributor




jsnow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I'm trying to find which subgroup of Mobius transformations are isometries with respect to the standard metric on the Riemann sphere (the one induced from the Euclidean metric on $mathbb{R}^3$).



The question hints that the distance on the Riemann sphere corresponds to the distance function on $mathbb{P}^1$ given by



$$d(L_1,L_2) = 2sqrt{1 - frac{|langle v,w rangle|^2}{||v||^2||w||^2}}$$



where $ v in L_1backslash{0}$ and $w in L_2backslash{0}$, and that a Mobius map corresponds to the action of a matrix $A in GL_2(mathbb{C})$ on lines in $mathbb{C}^2$, and asks me to consider which $2x2$ matrices automatically preserve this expression for $d$.



I honestly have no idea where to start with this question, so I'd really appreciate whatever help you might be able to give.






complex-analysis metric-spaces isometry mobius-transformation stereographic-projections




complex-analysis metric-spaces isometry mobius-transformation stereographic-projections






share|cite|improve this question







New contributor




jsnow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




jsnow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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asked Jan 4 at 0:56









jsnow

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New contributor





jsnow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






jsnow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • From the expression of $d$, it is clear that a Mobius transformation $A$ preserves the metric $d$ if $A$ preserves the inner product $left<cdot,cdotright>$, ie, $A$ is a rotation matrix. If you adopt the well-known fact that the isometry group of a sphere is the rotation group, then it follows that the subgroup is just the rotation group.
    – cjackal
    Jan 4 at 1:22










  • @cjackal note that one has to mod out $mathbb C^*$ in the $GL(2, mathbb C)$ action, so the subgroup in $GL(2, mathbb C)$ is definitely not the rotation group. On the other hand, there is no point to do this exercise, if one assume the knowledge about the isometry group of the sphere.
    – Arctic Char
    Jan 4 at 1:26












  • @ArcticChar Well we need to consider the matrices projectively, so $A$ is considered as an element in $PSL(2,mathbb{C})$ and my comment is just rephrasing the isomorphism $PSU(2,mathbb{C})cong SO(3)$. But it is true that my first comment is confusing in not distinguishing the complex inner product and real inner product. One may want to use an explicit formula converting a complex 2x2 matrix to a real 4x4 matrix to see the isomorphism above.
    – cjackal
    Jan 4 at 1:42


















  • From the expression of $d$, it is clear that a Mobius transformation $A$ preserves the metric $d$ if $A$ preserves the inner product $left<cdot,cdotright>$, ie, $A$ is a rotation matrix. If you adopt the well-known fact that the isometry group of a sphere is the rotation group, then it follows that the subgroup is just the rotation group.
    – cjackal
    Jan 4 at 1:22










  • @cjackal note that one has to mod out $mathbb C^*$ in the $GL(2, mathbb C)$ action, so the subgroup in $GL(2, mathbb C)$ is definitely not the rotation group. On the other hand, there is no point to do this exercise, if one assume the knowledge about the isometry group of the sphere.
    – Arctic Char
    Jan 4 at 1:26












  • @ArcticChar Well we need to consider the matrices projectively, so $A$ is considered as an element in $PSL(2,mathbb{C})$ and my comment is just rephrasing the isomorphism $PSU(2,mathbb{C})cong SO(3)$. But it is true that my first comment is confusing in not distinguishing the complex inner product and real inner product. One may want to use an explicit formula converting a complex 2x2 matrix to a real 4x4 matrix to see the isomorphism above.
    – cjackal
    Jan 4 at 1:42
















From the expression of $d$, it is clear that a Mobius transformation $A$ preserves the metric $d$ if $A$ preserves the inner product $left<cdot,cdotright>$, ie, $A$ is a rotation matrix. If you adopt the well-known fact that the isometry group of a sphere is the rotation group, then it follows that the subgroup is just the rotation group.
– cjackal
Jan 4 at 1:22




From the expression of $d$, it is clear that a Mobius transformation $A$ preserves the metric $d$ if $A$ preserves the inner product $left<cdot,cdotright>$, ie, $A$ is a rotation matrix. If you adopt the well-known fact that the isometry group of a sphere is the rotation group, then it follows that the subgroup is just the rotation group.
– cjackal
Jan 4 at 1:22












@cjackal note that one has to mod out $mathbb C^*$ in the $GL(2, mathbb C)$ action, so the subgroup in $GL(2, mathbb C)$ is definitely not the rotation group. On the other hand, there is no point to do this exercise, if one assume the knowledge about the isometry group of the sphere.
– Arctic Char
Jan 4 at 1:26






@cjackal note that one has to mod out $mathbb C^*$ in the $GL(2, mathbb C)$ action, so the subgroup in $GL(2, mathbb C)$ is definitely not the rotation group. On the other hand, there is no point to do this exercise, if one assume the knowledge about the isometry group of the sphere.
– Arctic Char
Jan 4 at 1:26














@ArcticChar Well we need to consider the matrices projectively, so $A$ is considered as an element in $PSL(2,mathbb{C})$ and my comment is just rephrasing the isomorphism $PSU(2,mathbb{C})cong SO(3)$. But it is true that my first comment is confusing in not distinguishing the complex inner product and real inner product. One may want to use an explicit formula converting a complex 2x2 matrix to a real 4x4 matrix to see the isomorphism above.
– cjackal
Jan 4 at 1:42




@ArcticChar Well we need to consider the matrices projectively, so $A$ is considered as an element in $PSL(2,mathbb{C})$ and my comment is just rephrasing the isomorphism $PSU(2,mathbb{C})cong SO(3)$. But it is true that my first comment is confusing in not distinguishing the complex inner product and real inner product. One may want to use an explicit formula converting a complex 2x2 matrix to a real 4x4 matrix to see the isomorphism above.
– cjackal
Jan 4 at 1:42










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Note this could be all wrong.



It seems to me the isometries of the unit sphere in $Bbb R^3$ are rotations. Hence it seems very clear that $R_t$ leaves that metric invariant, if $$R_t(z)=e^{it}z$$for some $tinBbb R$.



I bet $phicirc R_tcircphi^{-1}$ is also an isometry; not willing to conjecture whether those are the only ones. (Of course there's also the question of giving a more intrinsic description. I bet that it's not hard to see that the $phicirc R_tcircphi^{-1}$ are precisely the Mobius transformatiions that fix exactly two points... (that would follow if the only transformations fixing $0$ and $infty$ were the $R_t$.))






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    Note this could be all wrong.



    It seems to me the isometries of the unit sphere in $Bbb R^3$ are rotations. Hence it seems very clear that $R_t$ leaves that metric invariant, if $$R_t(z)=e^{it}z$$for some $tinBbb R$.



    I bet $phicirc R_tcircphi^{-1}$ is also an isometry; not willing to conjecture whether those are the only ones. (Of course there's also the question of giving a more intrinsic description. I bet that it's not hard to see that the $phicirc R_tcircphi^{-1}$ are precisely the Mobius transformatiions that fix exactly two points... (that would follow if the only transformations fixing $0$ and $infty$ were the $R_t$.))






    share|cite|improve this answer


























      1














      Note this could be all wrong.



      It seems to me the isometries of the unit sphere in $Bbb R^3$ are rotations. Hence it seems very clear that $R_t$ leaves that metric invariant, if $$R_t(z)=e^{it}z$$for some $tinBbb R$.



      I bet $phicirc R_tcircphi^{-1}$ is also an isometry; not willing to conjecture whether those are the only ones. (Of course there's also the question of giving a more intrinsic description. I bet that it's not hard to see that the $phicirc R_tcircphi^{-1}$ are precisely the Mobius transformatiions that fix exactly two points... (that would follow if the only transformations fixing $0$ and $infty$ were the $R_t$.))






      share|cite|improve this answer
























        1












        1








        1






        Note this could be all wrong.



        It seems to me the isometries of the unit sphere in $Bbb R^3$ are rotations. Hence it seems very clear that $R_t$ leaves that metric invariant, if $$R_t(z)=e^{it}z$$for some $tinBbb R$.



        I bet $phicirc R_tcircphi^{-1}$ is also an isometry; not willing to conjecture whether those are the only ones. (Of course there's also the question of giving a more intrinsic description. I bet that it's not hard to see that the $phicirc R_tcircphi^{-1}$ are precisely the Mobius transformatiions that fix exactly two points... (that would follow if the only transformations fixing $0$ and $infty$ were the $R_t$.))






        share|cite|improve this answer












        Note this could be all wrong.



        It seems to me the isometries of the unit sphere in $Bbb R^3$ are rotations. Hence it seems very clear that $R_t$ leaves that metric invariant, if $$R_t(z)=e^{it}z$$for some $tinBbb R$.



        I bet $phicirc R_tcircphi^{-1}$ is also an isometry; not willing to conjecture whether those are the only ones. (Of course there's also the question of giving a more intrinsic description. I bet that it's not hard to see that the $phicirc R_tcircphi^{-1}$ are precisely the Mobius transformatiions that fix exactly two points... (that would follow if the only transformations fixing $0$ and $infty$ were the $R_t$.))







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 1:15









        David C. Ullrich

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