Counterexample to Lie's second theorem for SO(3)












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$begingroup$


Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $Phi$ of its Lie algebra $mathfrak{g}$ lifts to an isomorphism $phi$ of $G$, i.e. such that $dphi_e = Phi$ where we identify $mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.



Now consider $G = mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $Phi$ of $mathfrak{so}(3)$ which is not the differential of any isomorphism $phi$ of $mathrm{SO}(3)$?



I feel like, if $(eta_1, eta_2, eta_3)$ is the usual basis of $mathfrak{so}(3)$, where $[eta_i, eta_j] = eta_k$ cyclically, then a map like $Phi(eta_1)=eta_2$, $Phi(eta_2)=eta_1$, $Phi(eta_3)=-eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $phi$. I've been trying to find $t_1, t_2, t_3 in mathbb{R}$ and a relation involving $(e^{t_1 eta_1}, e^{t_2 eta_2}, e^{t_3 eta_3})$ that is not satisfied by $(e^{t_1 eta_2}, e^{t_2 eta_1}, e^{-t_3 eta_3})$ but I can't come up with one.










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$endgroup$

















    6












    $begingroup$


    Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $Phi$ of its Lie algebra $mathfrak{g}$ lifts to an isomorphism $phi$ of $G$, i.e. such that $dphi_e = Phi$ where we identify $mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.



    Now consider $G = mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $Phi$ of $mathfrak{so}(3)$ which is not the differential of any isomorphism $phi$ of $mathrm{SO}(3)$?



    I feel like, if $(eta_1, eta_2, eta_3)$ is the usual basis of $mathfrak{so}(3)$, where $[eta_i, eta_j] = eta_k$ cyclically, then a map like $Phi(eta_1)=eta_2$, $Phi(eta_2)=eta_1$, $Phi(eta_3)=-eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $phi$. I've been trying to find $t_1, t_2, t_3 in mathbb{R}$ and a relation involving $(e^{t_1 eta_1}, e^{t_2 eta_2}, e^{t_3 eta_3})$ that is not satisfied by $(e^{t_1 eta_2}, e^{t_2 eta_1}, e^{-t_3 eta_3})$ but I can't come up with one.










    share|cite|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $Phi$ of its Lie algebra $mathfrak{g}$ lifts to an isomorphism $phi$ of $G$, i.e. such that $dphi_e = Phi$ where we identify $mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.



      Now consider $G = mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $Phi$ of $mathfrak{so}(3)$ which is not the differential of any isomorphism $phi$ of $mathrm{SO}(3)$?



      I feel like, if $(eta_1, eta_2, eta_3)$ is the usual basis of $mathfrak{so}(3)$, where $[eta_i, eta_j] = eta_k$ cyclically, then a map like $Phi(eta_1)=eta_2$, $Phi(eta_2)=eta_1$, $Phi(eta_3)=-eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $phi$. I've been trying to find $t_1, t_2, t_3 in mathbb{R}$ and a relation involving $(e^{t_1 eta_1}, e^{t_2 eta_2}, e^{t_3 eta_3})$ that is not satisfied by $(e^{t_1 eta_2}, e^{t_2 eta_1}, e^{-t_3 eta_3})$ but I can't come up with one.










      share|cite|improve this question











      $endgroup$




      Lie's second theorem says that if $G$ is a simply connected Lie group, then every isomorphism $Phi$ of its Lie algebra $mathfrak{g}$ lifts to an isomorphism $phi$ of $G$, i.e. such that $dphi_e = Phi$ where we identify $mathfrak{g}$ with $T_e G$, the tangent space to $G$ at the identity.



      Now consider $G = mathrm{SO}(3)$, which is not simply connected. Is there an explicit counterexample to Lie's second theorem in this case? That is, can we write down a Lie algebra isomorphism $Phi$ of $mathfrak{so}(3)$ which is not the differential of any isomorphism $phi$ of $mathrm{SO}(3)$?



      I feel like, if $(eta_1, eta_2, eta_3)$ is the usual basis of $mathfrak{so}(3)$, where $[eta_i, eta_j] = eta_k$ cyclically, then a map like $Phi(eta_1)=eta_2$, $Phi(eta_2)=eta_1$, $Phi(eta_3)=-eta_3$ ought to work, but I can't figure out how to prove there is no such group isomorphism $phi$. I've been trying to find $t_1, t_2, t_3 in mathbb{R}$ and a relation involving $(e^{t_1 eta_1}, e^{t_2 eta_2}, e^{t_3 eta_3})$ that is not satisfied by $(e^{t_1 eta_2}, e^{t_2 eta_1}, e^{-t_3 eta_3})$ but I can't come up with one.







      differential-geometry lie-groups lie-algebras






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      edited 6 hours ago









      José Carlos Santos

      160k22126232




      160k22126232










      asked 6 hours ago









      Nate EldredgeNate Eldredge

      63.2k682171




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          2 Answers
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          3












          $begingroup$

          There is an isomorphism $r: text{Aut}(SU(2)) to text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(rho) = text{Id}$, then necessarily $rho(g) = pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $rho$ is a homomorphism, the sign is 1, as desired.



          To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $text{Aut}(G) hookrightarrow text{Aut}(mathfrak g)$. By Lie's theorem, the map $text{Aut}(SU(2)) to text{Aut}(mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.



          In fact, $$text{Aut}(SU(2)) = text{Inn}(SU(2)) cong text{Inn}(SO(3)) = SO(3).$$



          José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
            $endgroup$
            – Torsten Schoeneberg
            4 hours ago






          • 1




            $begingroup$
            @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
            $endgroup$
            – Mike Miller
            3 hours ago





















          2












          $begingroup$

          See $SU(2)$ as the group of quaterninons with norm $1$. Let$$operatorname{Im}mathbb{H}={ai+bj+ckinmathbb{H},|,a,b,cinmathbb{R}}$$and, for each $qin SU(2)$, let $varphi(q)colonoperatorname{Im}mathbb{H}longrightarrowoperatorname{Im}mathbb{H}$ be the map defined by $varphi(q)(v)=q^{-1}vq$. Then $operatorname{Im}mathbb{H}simeqmathbb{R}^3$ and $varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,mathbb{R})$, whose kernel is $pm1$. Then $Dvarphi_1colonmathfrak{su}(2)longrightarrowmathfrak{so}(3,mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $mathfrak{so}(3,mathbb{R})$ in $mathbb{C}^2$. But you cannot lift it to $SO(3,mathbb{R})$, since $SU(2)$ and $SO(3,mathbb{R})$ are not isomorphic.






          share|cite|improve this answer











          $endgroup$













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            3












            $begingroup$

            There is an isomorphism $r: text{Aut}(SU(2)) to text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(rho) = text{Id}$, then necessarily $rho(g) = pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $rho$ is a homomorphism, the sign is 1, as desired.



            To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $text{Aut}(G) hookrightarrow text{Aut}(mathfrak g)$. By Lie's theorem, the map $text{Aut}(SU(2)) to text{Aut}(mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.



            In fact, $$text{Aut}(SU(2)) = text{Inn}(SU(2)) cong text{Inn}(SO(3)) = SO(3).$$



            José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
              $endgroup$
              – Torsten Schoeneberg
              4 hours ago






            • 1




              $begingroup$
              @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
              $endgroup$
              – Mike Miller
              3 hours ago


















            3












            $begingroup$

            There is an isomorphism $r: text{Aut}(SU(2)) to text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(rho) = text{Id}$, then necessarily $rho(g) = pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $rho$ is a homomorphism, the sign is 1, as desired.



            To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $text{Aut}(G) hookrightarrow text{Aut}(mathfrak g)$. By Lie's theorem, the map $text{Aut}(SU(2)) to text{Aut}(mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.



            In fact, $$text{Aut}(SU(2)) = text{Inn}(SU(2)) cong text{Inn}(SO(3)) = SO(3).$$



            José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
              $endgroup$
              – Torsten Schoeneberg
              4 hours ago






            • 1




              $begingroup$
              @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
              $endgroup$
              – Mike Miller
              3 hours ago
















            3












            3








            3





            $begingroup$

            There is an isomorphism $r: text{Aut}(SU(2)) to text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(rho) = text{Id}$, then necessarily $rho(g) = pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $rho$ is a homomorphism, the sign is 1, as desired.



            To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $text{Aut}(G) hookrightarrow text{Aut}(mathfrak g)$. By Lie's theorem, the map $text{Aut}(SU(2)) to text{Aut}(mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.



            In fact, $$text{Aut}(SU(2)) = text{Inn}(SU(2)) cong text{Inn}(SO(3)) = SO(3).$$



            José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.






            share|cite|improve this answer











            $endgroup$



            There is an isomorphism $r: text{Aut}(SU(2)) to text{Aut}(SO(3))$, because any automorphism of $SU(2)$ necessarily preserves the center and $SO(3) = SU(2)/Z(SU(2))$. To see that $r$ is injective, observe that if $r(rho) = text{Id}$, then necessarily $rho(g) = pm g$ for each $g$. By continuity, the sign is independent of $g$, and because $rho$ is a homomorphism, the sign is 1, as desired.



            To see that $r$ is surjective, observe that taking the derivative at the identity gives an embedding $text{Aut}(G) hookrightarrow text{Aut}(mathfrak g)$. By Lie's theorem, the map $text{Aut}(SU(2)) to text{Aut}(mathfrak{su}(2))$ is an isomorphism, and in particular we have factored this isomorphism as the composite of two injective maps. Both maps must therefore be isomorphisms.



            In fact, $$text{Aut}(SU(2)) = text{Inn}(SU(2)) cong text{Inn}(SO(3)) = SO(3).$$



            José's answer gives an example of a map of Lie algebras which may not be lifted to a map of Lie groups, but we see here that if you're looking to find an automorphism of $mathfrak{so}(3)$ which does not lift to an automorphism of $SO(3)$, you are out of luck. You'll have to try something more complicated.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 5 hours ago

























            answered 5 hours ago









            Mike MillerMike Miller

            37.2k472139




            37.2k472139








            • 1




              $begingroup$
              1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
              $endgroup$
              – Torsten Schoeneberg
              4 hours ago






            • 1




              $begingroup$
              @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
              $endgroup$
              – Mike Miller
              3 hours ago
















            • 1




              $begingroup$
              1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
              $endgroup$
              – Torsten Schoeneberg
              4 hours ago






            • 1




              $begingroup$
              @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
              $endgroup$
              – Mike Miller
              3 hours ago










            1




            1




            $begingroup$
            1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
            $endgroup$
            – Torsten Schoeneberg
            4 hours ago




            $begingroup$
            1. Does math.stackexchange.com/q/1020865/96384 imply more generally that the inner automorphisms are the same for every group "in between" the simply connected and the adjoint group, thus can never work as counterexample? 2. Would e.g. something like the triality morphism on $mathfrak{so}_8$ work, as lifting to $Spin(8)$ but not to $SO(8)$?
            $endgroup$
            – Torsten Schoeneberg
            4 hours ago




            1




            1




            $begingroup$
            @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
            $endgroup$
            – Mike Miller
            3 hours ago






            $begingroup$
            @TorstenSchoeneberg Your first sentence seems correct to me. I'm not sure about your second sentence: I think that there is an identification of $text{Out}(G)$ with automorphisms of the Dynkin diagram in more generality than simply connected Lie groups, but I'm not sure at what generality it is true. I vaguely recall that it is at least true for $SO(n)$. In general, I am not so comfortable with the high-dimensional stuff. I'd love to see an example worked out of a group where an automorphism of $mathfrak g$ doesn't lift to $G$.
            $endgroup$
            – Mike Miller
            3 hours ago













            2












            $begingroup$

            See $SU(2)$ as the group of quaterninons with norm $1$. Let$$operatorname{Im}mathbb{H}={ai+bj+ckinmathbb{H},|,a,b,cinmathbb{R}}$$and, for each $qin SU(2)$, let $varphi(q)colonoperatorname{Im}mathbb{H}longrightarrowoperatorname{Im}mathbb{H}$ be the map defined by $varphi(q)(v)=q^{-1}vq$. Then $operatorname{Im}mathbb{H}simeqmathbb{R}^3$ and $varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,mathbb{R})$, whose kernel is $pm1$. Then $Dvarphi_1colonmathfrak{su}(2)longrightarrowmathfrak{so}(3,mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $mathfrak{so}(3,mathbb{R})$ in $mathbb{C}^2$. But you cannot lift it to $SO(3,mathbb{R})$, since $SU(2)$ and $SO(3,mathbb{R})$ are not isomorphic.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              See $SU(2)$ as the group of quaterninons with norm $1$. Let$$operatorname{Im}mathbb{H}={ai+bj+ckinmathbb{H},|,a,b,cinmathbb{R}}$$and, for each $qin SU(2)$, let $varphi(q)colonoperatorname{Im}mathbb{H}longrightarrowoperatorname{Im}mathbb{H}$ be the map defined by $varphi(q)(v)=q^{-1}vq$. Then $operatorname{Im}mathbb{H}simeqmathbb{R}^3$ and $varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,mathbb{R})$, whose kernel is $pm1$. Then $Dvarphi_1colonmathfrak{su}(2)longrightarrowmathfrak{so}(3,mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $mathfrak{so}(3,mathbb{R})$ in $mathbb{C}^2$. But you cannot lift it to $SO(3,mathbb{R})$, since $SU(2)$ and $SO(3,mathbb{R})$ are not isomorphic.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                See $SU(2)$ as the group of quaterninons with norm $1$. Let$$operatorname{Im}mathbb{H}={ai+bj+ckinmathbb{H},|,a,b,cinmathbb{R}}$$and, for each $qin SU(2)$, let $varphi(q)colonoperatorname{Im}mathbb{H}longrightarrowoperatorname{Im}mathbb{H}$ be the map defined by $varphi(q)(v)=q^{-1}vq$. Then $operatorname{Im}mathbb{H}simeqmathbb{R}^3$ and $varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,mathbb{R})$, whose kernel is $pm1$. Then $Dvarphi_1colonmathfrak{su}(2)longrightarrowmathfrak{so}(3,mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $mathfrak{so}(3,mathbb{R})$ in $mathbb{C}^2$. But you cannot lift it to $SO(3,mathbb{R})$, since $SU(2)$ and $SO(3,mathbb{R})$ are not isomorphic.






                share|cite|improve this answer











                $endgroup$



                See $SU(2)$ as the group of quaterninons with norm $1$. Let$$operatorname{Im}mathbb{H}={ai+bj+ckinmathbb{H},|,a,b,cinmathbb{R}}$$and, for each $qin SU(2)$, let $varphi(q)colonoperatorname{Im}mathbb{H}longrightarrowoperatorname{Im}mathbb{H}$ be the map defined by $varphi(q)(v)=q^{-1}vq$. Then $operatorname{Im}mathbb{H}simeqmathbb{R}^3$ and $varphi$ is a surjective group homomorphism from $SU(2)$ onto $SO(3,mathbb{R})$, whose kernel is $pm1$. Then $Dvarphi_1colonmathfrak{su}(2)longrightarrowmathfrak{so}(3,mathbb{R})$ is a Lie algebra isomorphism and its inverse is a representation of $mathfrak{so}(3,mathbb{R})$ in $mathbb{C}^2$. But you cannot lift it to $SO(3,mathbb{R})$, since $SU(2)$ and $SO(3,mathbb{R})$ are not isomorphic.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 6 hours ago

























                answered 6 hours ago









                José Carlos SantosJosé Carlos Santos

                160k22126232




                160k22126232






























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