The set of real structures on $mathbb{C}^n$ is isomorphic to $GL(n,mathbb{C})/GL(n,mathbb{R})$ as a...












1












$begingroup$


In representation theory a real structure on a $G$-module $V$ (finite dimensional complex vector space, in which the group $G$ has a linear action on it) can be definide as a conjugate-linear $G$-map $J:Vto V$ (that is, $J(g.v)=g.J(v)$) s.t. $J^2=id$.



Let $X$ be the set of all real structures on $mathbb{C}^n$. I want to show that $X$ is isomorphic to $GL(n,mathbb{C})/GL(n,mathbb{R})$ as a $GL(n,mathbb{C})$-space.



For this porpose I know that $GL(n,mathbb{C})$ acts on $X$ by $(A,J)mapsto AJA^{-1}$, as $(AJA^{-1})^2=id$.



But I cannot go any further. For example, given a $Bin GL(n,mathbb{R})$ i cannot see why $BJ=id$ for every $Jin X$.






linear-algebra linear-transformations representation-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    In representation theory a real structure on a $G$-module $V$ (finite dimensional complex vector space, in which the group $G$ has a linear action on it) can be definide as a conjugate-linear $G$-map $J:Vto V$ (that is, $J(g.v)=g.J(v)$) s.t. $J^2=id$.



    Let $X$ be the set of all real structures on $mathbb{C}^n$. I want to show that $X$ is isomorphic to $GL(n,mathbb{C})/GL(n,mathbb{R})$ as a $GL(n,mathbb{C})$-space.



    For this porpose I know that $GL(n,mathbb{C})$ acts on $X$ by $(A,J)mapsto AJA^{-1}$, as $(AJA^{-1})^2=id$.



    But I cannot go any further. For example, given a $Bin GL(n,mathbb{R})$ i cannot see why $BJ=id$ for every $Jin X$.






    linear-algebra linear-transformations representation-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In representation theory a real structure on a $G$-module $V$ (finite dimensional complex vector space, in which the group $G$ has a linear action on it) can be definide as a conjugate-linear $G$-map $J:Vto V$ (that is, $J(g.v)=g.J(v)$) s.t. $J^2=id$.



      Let $X$ be the set of all real structures on $mathbb{C}^n$. I want to show that $X$ is isomorphic to $GL(n,mathbb{C})/GL(n,mathbb{R})$ as a $GL(n,mathbb{C})$-space.



      For this porpose I know that $GL(n,mathbb{C})$ acts on $X$ by $(A,J)mapsto AJA^{-1}$, as $(AJA^{-1})^2=id$.



      But I cannot go any further. For example, given a $Bin GL(n,mathbb{R})$ i cannot see why $BJ=id$ for every $Jin X$.






      linear-algebra linear-transformations representation-theory







      share|cite|improve this question









      $endgroup$




      In representation theory a real structure on a $G$-module $V$ (finite dimensional complex vector space, in which the group $G$ has a linear action on it) can be definide as a conjugate-linear $G$-map $J:Vto V$ (that is, $J(g.v)=g.J(v)$) s.t. $J^2=id$.



      Let $X$ be the set of all real structures on $mathbb{C}^n$. I want to show that $X$ is isomorphic to $GL(n,mathbb{C})/GL(n,mathbb{R})$ as a $GL(n,mathbb{C})$-space.



      For this porpose I know that $GL(n,mathbb{C})$ acts on $X$ by $(A,J)mapsto AJA^{-1}$, as $(AJA^{-1})^2=id$.



      But I cannot go any further. For example, given a $Bin GL(n,mathbb{R})$ i cannot see why $BJ=id$ for every $Jin X$.






      linear-algebra linear-transformations representation-theory




      linear-algebra linear-transformations representation-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 16 at 15:29









      Andre GomesAndre Gomes

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          $begingroup$

          You shouldn't expect $BJ=mathrm{Id}$ for all $Bin GL(n,mathbb{R})$.



          Try this: Let $V=mathbb{C}^n$ and $J:Vto V$ be the conjugate linear map $J(v)=bar{v}$, where
          $$v=begin{pmatrix}a_1\vdots\a_nend{pmatrix};Rightarrow;bar{v}=begin{pmatrix}bar{a_1}\vdots\bar{a_n}end{pmatrix}.$$ Define a map $$f:GL(n,mathbb{C})to X;;;mbox{by};;;f(A)=AJA^{-1}.$$
          You need to show two things:




          1) $f$ is surjective, and



          2) $f(A)=f(B)$ if, and only if, $B^{-1}Ain GL(n,mathbb{R})$.




          These two facts together imply that there is a well defined bijection
          $$bar{f}:GL(n,mathbb{C})/GL(n,mathbb{R})to X$$
          given by $overline{f}(A, GL(n,mathbb{R}))=f(A)$.



          To prove 2, note that $f(B)=J$ if, and only if $Bin GL(n,mathbb{R})$. Indeed,
          begin{align}
          (BJB^{-1})(v)=J(v);forall vin V&iff Bbar{B}^{-1}(bar{v})=bar{v};forall vin V\
          &iff Bbar{B}^{-1}=mathrm{Id}_V\
          &iff B=bar{B}\
          &iff Bin GL(n,mathbb{R}).
          end{align}
          Now,
          begin{align}
          f(A)=f(B)&iff AJA^{-1}=BJB^{-1}\
          &iff J=B^{-1}AJA^{-1}B=B^{-1}AJ(B^{-1}A)^{-1}\
          &iff B^{-1}Ain GL(n,mathbb{R}).
          end{align}
          This proves 2.



          To prove 1, let $Kin X$ and let $V^K={vin Vmid K(v)=v}$ be the set of $K$-fixed points. Let $W=mathrm{span}_{mathbb{C}}V^K$. We claim that $W=V$. If not, $V=Woplus U$ for some complementary subspace $U$. If $uin Ubackslash{0}$, then $K(u)=w+u'$ for some $win W$ and $u'in U$. Since $u+K(u)=w+(u+u')in W$ we must have $u'=-u$. Now, consider the element $2iu-iwin Vbackslash W$. We have
          $$K(2iu-iw)=-2i(w-u)+iw=2iu-iw.$$
          Hence $2iu-iwin W$, a contradiction. Hence $W=V$ as desired.



          Finally, let $beta={v_1,ldots,v_n}subset V^K$ be a basis for $V$. The map $A:Vto V$ given by $A(e_i)=v_i$ ($e_i$ the $i$th coordinate vector) satisfies $f(A)=AJA^{-1}=K$, proving 1.



          Now, $GL(n,mathbb{C})$ acts on $GL(n,mathbb{C}/GL(n,mathbb{R})$ by left multipication:
          $$A.(B,GL(n,mathbb{R})=(AB),GL(n,mathbb{R}),$$
          and $GL(n,mathbb{C})$ acts on $X$ by conjugation:
          $$A.K=AKA^{-1}.$$
          Now, observe that
          begin{align}
          overline{f}(A.(B,GL(n,mathbb{R}))&=overline{f}((AB),GL(n,mathbb{R}))\
          &=f(AB)\
          &=(AB)J(AB)^{-1}\
          &=A(BJB^{-1})A^{-1}\
          &=A.(BJB^{-1})\
          &=A.f(B)\
          &=A.overline{f}(B,GL(n,mathbb{R})).
          end{align}
          Therefore, $overline{f}$ is a map of $GL(n,mathbb{C})$-spaces.






          share|cite|improve this answer











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            $begingroup$

            You shouldn't expect $BJ=mathrm{Id}$ for all $Bin GL(n,mathbb{R})$.



            Try this: Let $V=mathbb{C}^n$ and $J:Vto V$ be the conjugate linear map $J(v)=bar{v}$, where
            $$v=begin{pmatrix}a_1\vdots\a_nend{pmatrix};Rightarrow;bar{v}=begin{pmatrix}bar{a_1}\vdots\bar{a_n}end{pmatrix}.$$ Define a map $$f:GL(n,mathbb{C})to X;;;mbox{by};;;f(A)=AJA^{-1}.$$
            You need to show two things:




            1) $f$ is surjective, and



            2) $f(A)=f(B)$ if, and only if, $B^{-1}Ain GL(n,mathbb{R})$.




            These two facts together imply that there is a well defined bijection
            $$bar{f}:GL(n,mathbb{C})/GL(n,mathbb{R})to X$$
            given by $overline{f}(A, GL(n,mathbb{R}))=f(A)$.



            To prove 2, note that $f(B)=J$ if, and only if $Bin GL(n,mathbb{R})$. Indeed,
            begin{align}
            (BJB^{-1})(v)=J(v);forall vin V&iff Bbar{B}^{-1}(bar{v})=bar{v};forall vin V\
            &iff Bbar{B}^{-1}=mathrm{Id}_V\
            &iff B=bar{B}\
            &iff Bin GL(n,mathbb{R}).
            end{align}
            Now,
            begin{align}
            f(A)=f(B)&iff AJA^{-1}=BJB^{-1}\
            &iff J=B^{-1}AJA^{-1}B=B^{-1}AJ(B^{-1}A)^{-1}\
            &iff B^{-1}Ain GL(n,mathbb{R}).
            end{align}
            This proves 2.



            To prove 1, let $Kin X$ and let $V^K={vin Vmid K(v)=v}$ be the set of $K$-fixed points. Let $W=mathrm{span}_{mathbb{C}}V^K$. We claim that $W=V$. If not, $V=Woplus U$ for some complementary subspace $U$. If $uin Ubackslash{0}$, then $K(u)=w+u'$ for some $win W$ and $u'in U$. Since $u+K(u)=w+(u+u')in W$ we must have $u'=-u$. Now, consider the element $2iu-iwin Vbackslash W$. We have
            $$K(2iu-iw)=-2i(w-u)+iw=2iu-iw.$$
            Hence $2iu-iwin W$, a contradiction. Hence $W=V$ as desired.



            Finally, let $beta={v_1,ldots,v_n}subset V^K$ be a basis for $V$. The map $A:Vto V$ given by $A(e_i)=v_i$ ($e_i$ the $i$th coordinate vector) satisfies $f(A)=AJA^{-1}=K$, proving 1.



            Now, $GL(n,mathbb{C})$ acts on $GL(n,mathbb{C}/GL(n,mathbb{R})$ by left multipication:
            $$A.(B,GL(n,mathbb{R})=(AB),GL(n,mathbb{R}),$$
            and $GL(n,mathbb{C})$ acts on $X$ by conjugation:
            $$A.K=AKA^{-1}.$$
            Now, observe that
            begin{align}
            overline{f}(A.(B,GL(n,mathbb{R}))&=overline{f}((AB),GL(n,mathbb{R}))\
            &=f(AB)\
            &=(AB)J(AB)^{-1}\
            &=A(BJB^{-1})A^{-1}\
            &=A.(BJB^{-1})\
            &=A.f(B)\
            &=A.overline{f}(B,GL(n,mathbb{R})).
            end{align}
            Therefore, $overline{f}$ is a map of $GL(n,mathbb{C})$-spaces.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              You shouldn't expect $BJ=mathrm{Id}$ for all $Bin GL(n,mathbb{R})$.



              Try this: Let $V=mathbb{C}^n$ and $J:Vto V$ be the conjugate linear map $J(v)=bar{v}$, where
              $$v=begin{pmatrix}a_1\vdots\a_nend{pmatrix};Rightarrow;bar{v}=begin{pmatrix}bar{a_1}\vdots\bar{a_n}end{pmatrix}.$$ Define a map $$f:GL(n,mathbb{C})to X;;;mbox{by};;;f(A)=AJA^{-1}.$$
              You need to show two things:




              1) $f$ is surjective, and



              2) $f(A)=f(B)$ if, and only if, $B^{-1}Ain GL(n,mathbb{R})$.




              These two facts together imply that there is a well defined bijection
              $$bar{f}:GL(n,mathbb{C})/GL(n,mathbb{R})to X$$
              given by $overline{f}(A, GL(n,mathbb{R}))=f(A)$.



              To prove 2, note that $f(B)=J$ if, and only if $Bin GL(n,mathbb{R})$. Indeed,
              begin{align}
              (BJB^{-1})(v)=J(v);forall vin V&iff Bbar{B}^{-1}(bar{v})=bar{v};forall vin V\
              &iff Bbar{B}^{-1}=mathrm{Id}_V\
              &iff B=bar{B}\
              &iff Bin GL(n,mathbb{R}).
              end{align}
              Now,
              begin{align}
              f(A)=f(B)&iff AJA^{-1}=BJB^{-1}\
              &iff J=B^{-1}AJA^{-1}B=B^{-1}AJ(B^{-1}A)^{-1}\
              &iff B^{-1}Ain GL(n,mathbb{R}).
              end{align}
              This proves 2.



              To prove 1, let $Kin X$ and let $V^K={vin Vmid K(v)=v}$ be the set of $K$-fixed points. Let $W=mathrm{span}_{mathbb{C}}V^K$. We claim that $W=V$. If not, $V=Woplus U$ for some complementary subspace $U$. If $uin Ubackslash{0}$, then $K(u)=w+u'$ for some $win W$ and $u'in U$. Since $u+K(u)=w+(u+u')in W$ we must have $u'=-u$. Now, consider the element $2iu-iwin Vbackslash W$. We have
              $$K(2iu-iw)=-2i(w-u)+iw=2iu-iw.$$
              Hence $2iu-iwin W$, a contradiction. Hence $W=V$ as desired.



              Finally, let $beta={v_1,ldots,v_n}subset V^K$ be a basis for $V$. The map $A:Vto V$ given by $A(e_i)=v_i$ ($e_i$ the $i$th coordinate vector) satisfies $f(A)=AJA^{-1}=K$, proving 1.



              Now, $GL(n,mathbb{C})$ acts on $GL(n,mathbb{C}/GL(n,mathbb{R})$ by left multipication:
              $$A.(B,GL(n,mathbb{R})=(AB),GL(n,mathbb{R}),$$
              and $GL(n,mathbb{C})$ acts on $X$ by conjugation:
              $$A.K=AKA^{-1}.$$
              Now, observe that
              begin{align}
              overline{f}(A.(B,GL(n,mathbb{R}))&=overline{f}((AB),GL(n,mathbb{R}))\
              &=f(AB)\
              &=(AB)J(AB)^{-1}\
              &=A(BJB^{-1})A^{-1}\
              &=A.(BJB^{-1})\
              &=A.f(B)\
              &=A.overline{f}(B,GL(n,mathbb{R})).
              end{align}
              Therefore, $overline{f}$ is a map of $GL(n,mathbb{C})$-spaces.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                You shouldn't expect $BJ=mathrm{Id}$ for all $Bin GL(n,mathbb{R})$.



                Try this: Let $V=mathbb{C}^n$ and $J:Vto V$ be the conjugate linear map $J(v)=bar{v}$, where
                $$v=begin{pmatrix}a_1\vdots\a_nend{pmatrix};Rightarrow;bar{v}=begin{pmatrix}bar{a_1}\vdots\bar{a_n}end{pmatrix}.$$ Define a map $$f:GL(n,mathbb{C})to X;;;mbox{by};;;f(A)=AJA^{-1}.$$
                You need to show two things:




                1) $f$ is surjective, and



                2) $f(A)=f(B)$ if, and only if, $B^{-1}Ain GL(n,mathbb{R})$.




                These two facts together imply that there is a well defined bijection
                $$bar{f}:GL(n,mathbb{C})/GL(n,mathbb{R})to X$$
                given by $overline{f}(A, GL(n,mathbb{R}))=f(A)$.



                To prove 2, note that $f(B)=J$ if, and only if $Bin GL(n,mathbb{R})$. Indeed,
                begin{align}
                (BJB^{-1})(v)=J(v);forall vin V&iff Bbar{B}^{-1}(bar{v})=bar{v};forall vin V\
                &iff Bbar{B}^{-1}=mathrm{Id}_V\
                &iff B=bar{B}\
                &iff Bin GL(n,mathbb{R}).
                end{align}
                Now,
                begin{align}
                f(A)=f(B)&iff AJA^{-1}=BJB^{-1}\
                &iff J=B^{-1}AJA^{-1}B=B^{-1}AJ(B^{-1}A)^{-1}\
                &iff B^{-1}Ain GL(n,mathbb{R}).
                end{align}
                This proves 2.



                To prove 1, let $Kin X$ and let $V^K={vin Vmid K(v)=v}$ be the set of $K$-fixed points. Let $W=mathrm{span}_{mathbb{C}}V^K$. We claim that $W=V$. If not, $V=Woplus U$ for some complementary subspace $U$. If $uin Ubackslash{0}$, then $K(u)=w+u'$ for some $win W$ and $u'in U$. Since $u+K(u)=w+(u+u')in W$ we must have $u'=-u$. Now, consider the element $2iu-iwin Vbackslash W$. We have
                $$K(2iu-iw)=-2i(w-u)+iw=2iu-iw.$$
                Hence $2iu-iwin W$, a contradiction. Hence $W=V$ as desired.



                Finally, let $beta={v_1,ldots,v_n}subset V^K$ be a basis for $V$. The map $A:Vto V$ given by $A(e_i)=v_i$ ($e_i$ the $i$th coordinate vector) satisfies $f(A)=AJA^{-1}=K$, proving 1.



                Now, $GL(n,mathbb{C})$ acts on $GL(n,mathbb{C}/GL(n,mathbb{R})$ by left multipication:
                $$A.(B,GL(n,mathbb{R})=(AB),GL(n,mathbb{R}),$$
                and $GL(n,mathbb{C})$ acts on $X$ by conjugation:
                $$A.K=AKA^{-1}.$$
                Now, observe that
                begin{align}
                overline{f}(A.(B,GL(n,mathbb{R}))&=overline{f}((AB),GL(n,mathbb{R}))\
                &=f(AB)\
                &=(AB)J(AB)^{-1}\
                &=A(BJB^{-1})A^{-1}\
                &=A.(BJB^{-1})\
                &=A.f(B)\
                &=A.overline{f}(B,GL(n,mathbb{R})).
                end{align}
                Therefore, $overline{f}$ is a map of $GL(n,mathbb{C})$-spaces.






                share|cite|improve this answer











                $endgroup$



                You shouldn't expect $BJ=mathrm{Id}$ for all $Bin GL(n,mathbb{R})$.



                Try this: Let $V=mathbb{C}^n$ and $J:Vto V$ be the conjugate linear map $J(v)=bar{v}$, where
                $$v=begin{pmatrix}a_1\vdots\a_nend{pmatrix};Rightarrow;bar{v}=begin{pmatrix}bar{a_1}\vdots\bar{a_n}end{pmatrix}.$$ Define a map $$f:GL(n,mathbb{C})to X;;;mbox{by};;;f(A)=AJA^{-1}.$$
                You need to show two things:




                1) $f$ is surjective, and



                2) $f(A)=f(B)$ if, and only if, $B^{-1}Ain GL(n,mathbb{R})$.




                These two facts together imply that there is a well defined bijection
                $$bar{f}:GL(n,mathbb{C})/GL(n,mathbb{R})to X$$
                given by $overline{f}(A, GL(n,mathbb{R}))=f(A)$.



                To prove 2, note that $f(B)=J$ if, and only if $Bin GL(n,mathbb{R})$. Indeed,
                begin{align}
                (BJB^{-1})(v)=J(v);forall vin V&iff Bbar{B}^{-1}(bar{v})=bar{v};forall vin V\
                &iff Bbar{B}^{-1}=mathrm{Id}_V\
                &iff B=bar{B}\
                &iff Bin GL(n,mathbb{R}).
                end{align}
                Now,
                begin{align}
                f(A)=f(B)&iff AJA^{-1}=BJB^{-1}\
                &iff J=B^{-1}AJA^{-1}B=B^{-1}AJ(B^{-1}A)^{-1}\
                &iff B^{-1}Ain GL(n,mathbb{R}).
                end{align}
                This proves 2.



                To prove 1, let $Kin X$ and let $V^K={vin Vmid K(v)=v}$ be the set of $K$-fixed points. Let $W=mathrm{span}_{mathbb{C}}V^K$. We claim that $W=V$. If not, $V=Woplus U$ for some complementary subspace $U$. If $uin Ubackslash{0}$, then $K(u)=w+u'$ for some $win W$ and $u'in U$. Since $u+K(u)=w+(u+u')in W$ we must have $u'=-u$. Now, consider the element $2iu-iwin Vbackslash W$. We have
                $$K(2iu-iw)=-2i(w-u)+iw=2iu-iw.$$
                Hence $2iu-iwin W$, a contradiction. Hence $W=V$ as desired.



                Finally, let $beta={v_1,ldots,v_n}subset V^K$ be a basis for $V$. The map $A:Vto V$ given by $A(e_i)=v_i$ ($e_i$ the $i$th coordinate vector) satisfies $f(A)=AJA^{-1}=K$, proving 1.



                Now, $GL(n,mathbb{C})$ acts on $GL(n,mathbb{C}/GL(n,mathbb{R})$ by left multipication:
                $$A.(B,GL(n,mathbb{R})=(AB),GL(n,mathbb{R}),$$
                and $GL(n,mathbb{C})$ acts on $X$ by conjugation:
                $$A.K=AKA^{-1}.$$
                Now, observe that
                begin{align}
                overline{f}(A.(B,GL(n,mathbb{R}))&=overline{f}((AB),GL(n,mathbb{R}))\
                &=f(AB)\
                &=(AB)J(AB)^{-1}\
                &=A(BJB^{-1})A^{-1}\
                &=A.(BJB^{-1})\
                &=A.f(B)\
                &=A.overline{f}(B,GL(n,mathbb{R})).
                end{align}
                Therefore, $overline{f}$ is a map of $GL(n,mathbb{C})$-spaces.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 18 at 16:07

























                answered Jan 18 at 0:00









                David HillDavid Hill

                9,1111619




                9,1111619






























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                    How to fix TextFormField cause rebuild widget in Flutter

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                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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