Why is the projective symmetry group a group?












3












$begingroup$


I am reading the paper from X. Wen about quantum orders and symmetric spin liquids. It can be found here: https://arxiv.org/abs/cond-mat/0107071



The Hamiltonian he is writing about looks like this:
begin{align}
H_{MF} = sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j + text{h.c.} + text{const.} + text{Lagrange-multiplier terms}
end{align}

where
begin{align}
Psi_i^dagger := (f_{iuparrow}^dagger,, f_{idownarrow} )
end{align}

and $U_{ij}$ is some complex $2times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $Psi_i to G(i)^daggerPsi(i)$, where each $G(i) in SU(2)$. A symmetry-transformation $T$ is of the form $Psi_i to Psi_{T^{-1}(i)}$, where (I guess) $T$ has to be a bijective function from the lattice to the lattice.



He introduces something called projective symmetry group (PSG). Its elements are pairs $(G_T,T)$ of a Gauge-transformation together with a symmetry such that the Hamilton operator is invariant under the transformation $G_T T$. This means $(G_T,T) in PSG$ has to satisfy
begin{align}
sum_{<ij>} G_TT(Psi_i)^dagger, U_{ij}, G_TT(Psi_j)&= sum_{<T(i)T(j)>} Psi_i^dagger, G_T(T(i))U_{T(i)T(j)} G_T(T(j))^dagger, Psi_j^dagger \&=^! sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j
end{align}

Thus,
begin{align}
G(T(i))U_{T(i)T(j)}G(T(j))^dagger = U_{ij}
end{align}

I do understand the definition. And I hope I got it right, how Gauge transformation and symmetries act on the basis. But I don't understand why it is calle a group.



So basically I am looking for the group operation
begin{align}
cdot: PSG times PSG to PSG, qquad (G_T,T) cdot (G_S, S) = ?!
end{align}

It is nowhere mentioned, and I searched a lot. It is probably very easy, but I just can not figure it out. I tried $(G_TG_S,TS)$ and $(G_TG_S, ST)$. Both are of the right form, Gauge and symmetry, but they don't seem to leave the Hamilton invariant. At least I can't see it. I know, that e.g. $G_TTG_SS$ will not change $H_{MF}$, but it is not of the form $(text{Gauge}, text{Symmetry})$. So what to do?










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I am reading the paper from X. Wen about quantum orders and symmetric spin liquids. It can be found here: https://arxiv.org/abs/cond-mat/0107071



    The Hamiltonian he is writing about looks like this:
    begin{align}
    H_{MF} = sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j + text{h.c.} + text{const.} + text{Lagrange-multiplier terms}
    end{align}

    where
    begin{align}
    Psi_i^dagger := (f_{iuparrow}^dagger,, f_{idownarrow} )
    end{align}

    and $U_{ij}$ is some complex $2times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $Psi_i to G(i)^daggerPsi(i)$, where each $G(i) in SU(2)$. A symmetry-transformation $T$ is of the form $Psi_i to Psi_{T^{-1}(i)}$, where (I guess) $T$ has to be a bijective function from the lattice to the lattice.



    He introduces something called projective symmetry group (PSG). Its elements are pairs $(G_T,T)$ of a Gauge-transformation together with a symmetry such that the Hamilton operator is invariant under the transformation $G_T T$. This means $(G_T,T) in PSG$ has to satisfy
    begin{align}
    sum_{<ij>} G_TT(Psi_i)^dagger, U_{ij}, G_TT(Psi_j)&= sum_{<T(i)T(j)>} Psi_i^dagger, G_T(T(i))U_{T(i)T(j)} G_T(T(j))^dagger, Psi_j^dagger \&=^! sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j
    end{align}

    Thus,
    begin{align}
    G(T(i))U_{T(i)T(j)}G(T(j))^dagger = U_{ij}
    end{align}

    I do understand the definition. And I hope I got it right, how Gauge transformation and symmetries act on the basis. But I don't understand why it is calle a group.



    So basically I am looking for the group operation
    begin{align}
    cdot: PSG times PSG to PSG, qquad (G_T,T) cdot (G_S, S) = ?!
    end{align}

    It is nowhere mentioned, and I searched a lot. It is probably very easy, but I just can not figure it out. I tried $(G_TG_S,TS)$ and $(G_TG_S, ST)$. Both are of the right form, Gauge and symmetry, but they don't seem to leave the Hamilton invariant. At least I can't see it. I know, that e.g. $G_TTG_SS$ will not change $H_{MF}$, but it is not of the form $(text{Gauge}, text{Symmetry})$. So what to do?










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I am reading the paper from X. Wen about quantum orders and symmetric spin liquids. It can be found here: https://arxiv.org/abs/cond-mat/0107071



      The Hamiltonian he is writing about looks like this:
      begin{align}
      H_{MF} = sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j + text{h.c.} + text{const.} + text{Lagrange-multiplier terms}
      end{align}

      where
      begin{align}
      Psi_i^dagger := (f_{iuparrow}^dagger,, f_{idownarrow} )
      end{align}

      and $U_{ij}$ is some complex $2times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $Psi_i to G(i)^daggerPsi(i)$, where each $G(i) in SU(2)$. A symmetry-transformation $T$ is of the form $Psi_i to Psi_{T^{-1}(i)}$, where (I guess) $T$ has to be a bijective function from the lattice to the lattice.



      He introduces something called projective symmetry group (PSG). Its elements are pairs $(G_T,T)$ of a Gauge-transformation together with a symmetry such that the Hamilton operator is invariant under the transformation $G_T T$. This means $(G_T,T) in PSG$ has to satisfy
      begin{align}
      sum_{<ij>} G_TT(Psi_i)^dagger, U_{ij}, G_TT(Psi_j)&= sum_{<T(i)T(j)>} Psi_i^dagger, G_T(T(i))U_{T(i)T(j)} G_T(T(j))^dagger, Psi_j^dagger \&=^! sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j
      end{align}

      Thus,
      begin{align}
      G(T(i))U_{T(i)T(j)}G(T(j))^dagger = U_{ij}
      end{align}

      I do understand the definition. And I hope I got it right, how Gauge transformation and symmetries act on the basis. But I don't understand why it is calle a group.



      So basically I am looking for the group operation
      begin{align}
      cdot: PSG times PSG to PSG, qquad (G_T,T) cdot (G_S, S) = ?!
      end{align}

      It is nowhere mentioned, and I searched a lot. It is probably very easy, but I just can not figure it out. I tried $(G_TG_S,TS)$ and $(G_TG_S, ST)$. Both are of the right form, Gauge and symmetry, but they don't seem to leave the Hamilton invariant. At least I can't see it. I know, that e.g. $G_TTG_SS$ will not change $H_{MF}$, but it is not of the form $(text{Gauge}, text{Symmetry})$. So what to do?










      share|cite|improve this question











      $endgroup$




      I am reading the paper from X. Wen about quantum orders and symmetric spin liquids. It can be found here: https://arxiv.org/abs/cond-mat/0107071



      The Hamiltonian he is writing about looks like this:
      begin{align}
      H_{MF} = sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j + text{h.c.} + text{const.} + text{Lagrange-multiplier terms}
      end{align}

      where
      begin{align}
      Psi_i^dagger := (f_{iuparrow}^dagger,, f_{idownarrow} )
      end{align}

      and $U_{ij}$ is some complex $2times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $Psi_i to G(i)^daggerPsi(i)$, where each $G(i) in SU(2)$. A symmetry-transformation $T$ is of the form $Psi_i to Psi_{T^{-1}(i)}$, where (I guess) $T$ has to be a bijective function from the lattice to the lattice.



      He introduces something called projective symmetry group (PSG). Its elements are pairs $(G_T,T)$ of a Gauge-transformation together with a symmetry such that the Hamilton operator is invariant under the transformation $G_T T$. This means $(G_T,T) in PSG$ has to satisfy
      begin{align}
      sum_{<ij>} G_TT(Psi_i)^dagger, U_{ij}, G_TT(Psi_j)&= sum_{<T(i)T(j)>} Psi_i^dagger, G_T(T(i))U_{T(i)T(j)} G_T(T(j))^dagger, Psi_j^dagger \&=^! sum_{<ij>} Psi_i^dagger , U_{ij} Psi_j
      end{align}

      Thus,
      begin{align}
      G(T(i))U_{T(i)T(j)}G(T(j))^dagger = U_{ij}
      end{align}

      I do understand the definition. And I hope I got it right, how Gauge transformation and symmetries act on the basis. But I don't understand why it is calle a group.



      So basically I am looking for the group operation
      begin{align}
      cdot: PSG times PSG to PSG, qquad (G_T,T) cdot (G_S, S) = ?!
      end{align}

      It is nowhere mentioned, and I searched a lot. It is probably very easy, but I just can not figure it out. I tried $(G_TG_S,TS)$ and $(G_TG_S, ST)$. Both are of the right form, Gauge and symmetry, but they don't seem to leave the Hamilton invariant. At least I can't see it. I know, that e.g. $G_TTG_SS$ will not change $H_{MF}$, but it is not of the form $(text{Gauge}, text{Symmetry})$. So what to do?







      symmetry gauge-invariance gauge spin-liquid






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 30 '18 at 23:30









      Glorfindel

      3231311




      3231311










      asked Dec 30 '18 at 17:23









      N.BeckN.Beck

      1656




      1656






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.



          begin{align}
          cdot : PSG times PSG to PSG, , qquad (G_T,T)cdot(G_S,S) := (G_TTG_ST^{-1},TS)
          end{align}



          This has the required form, since
          begin{align}
          G_TTG_ST^{-1} (Psi_i) = G_T(i)^dagger G_S(T(i))^dagger , Psi_i
          end{align}

          is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by
          begin{align}
          (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1})
          end{align}

          where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)






          share|cite|improve this answer









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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.



            begin{align}
            cdot : PSG times PSG to PSG, , qquad (G_T,T)cdot(G_S,S) := (G_TTG_ST^{-1},TS)
            end{align}



            This has the required form, since
            begin{align}
            G_TTG_ST^{-1} (Psi_i) = G_T(i)^dagger G_S(T(i))^dagger , Psi_i
            end{align}

            is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by
            begin{align}
            (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1})
            end{align}

            where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.



              begin{align}
              cdot : PSG times PSG to PSG, , qquad (G_T,T)cdot(G_S,S) := (G_TTG_ST^{-1},TS)
              end{align}



              This has the required form, since
              begin{align}
              G_TTG_ST^{-1} (Psi_i) = G_T(i)^dagger G_S(T(i))^dagger , Psi_i
              end{align}

              is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by
              begin{align}
              (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1})
              end{align}

              where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.



                begin{align}
                cdot : PSG times PSG to PSG, , qquad (G_T,T)cdot(G_S,S) := (G_TTG_ST^{-1},TS)
                end{align}



                This has the required form, since
                begin{align}
                G_TTG_ST^{-1} (Psi_i) = G_T(i)^dagger G_S(T(i))^dagger , Psi_i
                end{align}

                is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by
                begin{align}
                (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1})
                end{align}

                where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)






                share|cite|improve this answer









                $endgroup$



                Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.



                begin{align}
                cdot : PSG times PSG to PSG, , qquad (G_T,T)cdot(G_S,S) := (G_TTG_ST^{-1},TS)
                end{align}



                This has the required form, since
                begin{align}
                G_TTG_ST^{-1} (Psi_i) = G_T(i)^dagger G_S(T(i))^dagger , Psi_i
                end{align}

                is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by
                begin{align}
                (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1})
                end{align}

                where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 30 '18 at 18:29









                N.BeckN.Beck

                1656




                1656






























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