If infinitesimal transformations commute why don't the generators of the Lorentz group commute?












2












$begingroup$


If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?






special-relativity group-theory lorentz-symmetry commutator lie-algebra







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    $begingroup$


    If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?






    special-relativity group-theory lorentz-symmetry commutator lie-algebra







    share|cite|improve this question











    $endgroup$















      2












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      2


      1



      $begingroup$


      If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?






      special-relativity group-theory lorentz-symmetry commutator lie-algebra







      share|cite|improve this question











      $endgroup$




      If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?






      special-relativity group-theory lorentz-symmetry commutator lie-algebra




      special-relativity group-theory lorentz-symmetry commutator lie-algebra






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      edited yesterday









      Qmechanic♦

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          $begingroup$

          I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.



          Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.



          When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.






          share|cite|improve this answer









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            1












            $begingroup$


            1. Illustrative example: It is straightforward to prove that the Lie group
              $$ SO(3)~:=~{ Min {rm Mat}_{3times 3}(mathbb{R}) mid M^tM=mathbb{1}_{3times 3}, ~det(M)=1} $$
              of 3D rotations is generated by the corresponding Lie algebra
              $$ so(3)~:=~{ min {rm Mat}_{3times 3}(mathbb{R}) mid m^t=-m} $$
              of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


            2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!







            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              While
              begin{align}
              e^{epsilon A} e^{epsilon B}&=(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots)(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)tag{1}, ,\
              &= 1+ epsilon (A+B)+ frac{1}{2}
              epsilon^2 (A^2+AB +B^2)+ldots
              end{align}
              we have
              begin{align}
              e^{epsilon B} e^{epsilon A}&=(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots), ,tag{2}\
              &= 1+ epsilon (B+A)+ frac{1}{2}
              epsilon^2 (A^2+BA +B^2)+ldots
              end{align}
              so:




              1. To order $epsilon$, the transformations are the same,

              2. To order $epsilon^2$ they are different.


              Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






              share|cite|improve this answer









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                $begingroup$

                I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.



                Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.



                When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.






                share|cite|improve this answer









                $endgroup$


















                  8












                  $begingroup$

                  I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.



                  Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.



                  When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.






                  share|cite|improve this answer









                  $endgroup$
















                    8












                    8








                    8





                    $begingroup$

                    I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.



                    Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.



                    When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.






                    share|cite|improve this answer









                    $endgroup$



                    I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.



                    Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $exp(theta G_x)$ is a rotation through angle $theta$ about the $x$-axis, and $exp(phi G_y)$ is a rotation through angle $phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.



                    When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $theta$ and $phi$ both have infinitesimal magnitude $epsilonll 1$, then the rotations $exp(theta G_x)$ and $exp(phi G_y)$ commute with each other to first order in $epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $exp(theta G_x)exp(phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $epsilon^2llepsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $epsilon$.







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                    share|cite|improve this answer










                    answered 2 days ago









                    Chiral AnomalyChiral Anomaly

                    12.6k21542




                    12.6k21542























                        1












                        $begingroup$


                        1. Illustrative example: It is straightforward to prove that the Lie group
                          $$ SO(3)~:=~{ Min {rm Mat}_{3times 3}(mathbb{R}) mid M^tM=mathbb{1}_{3times 3}, ~det(M)=1} $$
                          of 3D rotations is generated by the corresponding Lie algebra
                          $$ so(3)~:=~{ min {rm Mat}_{3times 3}(mathbb{R}) mid m^t=-m} $$
                          of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                        2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!







                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$


                          1. Illustrative example: It is straightforward to prove that the Lie group
                            $$ SO(3)~:=~{ Min {rm Mat}_{3times 3}(mathbb{R}) mid M^tM=mathbb{1}_{3times 3}, ~det(M)=1} $$
                            of 3D rotations is generated by the corresponding Lie algebra
                            $$ so(3)~:=~{ min {rm Mat}_{3times 3}(mathbb{R}) mid m^t=-m} $$
                            of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                          2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!







                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$


                            1. Illustrative example: It is straightforward to prove that the Lie group
                              $$ SO(3)~:=~{ Min {rm Mat}_{3times 3}(mathbb{R}) mid M^tM=mathbb{1}_{3times 3}, ~det(M)=1} $$
                              of 3D rotations is generated by the corresponding Lie algebra
                              $$ so(3)~:=~{ min {rm Mat}_{3times 3}(mathbb{R}) mid m^t=-m} $$
                              of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                            2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!







                            share|cite|improve this answer









                            $endgroup$




                            1. Illustrative example: It is straightforward to prove that the Lie group
                              $$ SO(3)~:=~{ Min {rm Mat}_{3times 3}(mathbb{R}) mid M^tM=mathbb{1}_{3times 3}, ~det(M)=1} $$
                              of 3D rotations is generated by the corresponding Lie algebra
                              $$ so(3)~:=~{ min {rm Mat}_{3times 3}(mathbb{R}) mid m^t=-m} $$
                              of real $3times 3$ antisymmetric matrices, which clearly do not all commute.


                            2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!








                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            Qmechanic♦Qmechanic

                            106k121971227




                            106k121971227























                                1












                                $begingroup$

                                While
                                begin{align}
                                e^{epsilon A} e^{epsilon B}&=(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots)(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)tag{1}, ,\
                                &= 1+ epsilon (A+B)+ frac{1}{2}
                                epsilon^2 (A^2+AB +B^2)+ldots
                                end{align}
                                we have
                                begin{align}
                                e^{epsilon B} e^{epsilon A}&=(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots), ,tag{2}\
                                &= 1+ epsilon (B+A)+ frac{1}{2}
                                epsilon^2 (A^2+BA +B^2)+ldots
                                end{align}
                                so:




                                1. To order $epsilon$, the transformations are the same,

                                2. To order $epsilon^2$ they are different.


                                Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  While
                                  begin{align}
                                  e^{epsilon A} e^{epsilon B}&=(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots)(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)tag{1}, ,\
                                  &= 1+ epsilon (A+B)+ frac{1}{2}
                                  epsilon^2 (A^2+AB +B^2)+ldots
                                  end{align}
                                  we have
                                  begin{align}
                                  e^{epsilon B} e^{epsilon A}&=(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots), ,tag{2}\
                                  &= 1+ epsilon (B+A)+ frac{1}{2}
                                  epsilon^2 (A^2+BA +B^2)+ldots
                                  end{align}
                                  so:




                                  1. To order $epsilon$, the transformations are the same,

                                  2. To order $epsilon^2$ they are different.


                                  Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    While
                                    begin{align}
                                    e^{epsilon A} e^{epsilon B}&=(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots)(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)tag{1}, ,\
                                    &= 1+ epsilon (A+B)+ frac{1}{2}
                                    epsilon^2 (A^2+AB +B^2)+ldots
                                    end{align}
                                    we have
                                    begin{align}
                                    e^{epsilon B} e^{epsilon A}&=(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots), ,tag{2}\
                                    &= 1+ epsilon (B+A)+ frac{1}{2}
                                    epsilon^2 (A^2+BA +B^2)+ldots
                                    end{align}
                                    so:




                                    1. To order $epsilon$, the transformations are the same,

                                    2. To order $epsilon^2$ they are different.


                                    Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.






                                    share|cite|improve this answer









                                    $endgroup$



                                    While
                                    begin{align}
                                    e^{epsilon A} e^{epsilon B}&=(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots)(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)tag{1}, ,\
                                    &= 1+ epsilon (A+B)+ frac{1}{2}
                                    epsilon^2 (A^2+AB +B^2)+ldots
                                    end{align}
                                    we have
                                    begin{align}
                                    e^{epsilon B} e^{epsilon A}&=(1+epsilon B+frac{1}{2}epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac{1}{2}epsilon^2 A^2+ldots), ,tag{2}\
                                    &= 1+ epsilon (B+A)+ frac{1}{2}
                                    epsilon^2 (A^2+BA +B^2)+ldots
                                    end{align}
                                    so:




                                    1. To order $epsilon$, the transformations are the same,

                                    2. To order $epsilon^2$ they are different.


                                    Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $epsilon^2$ in (1) and (2) to see that they differ by a commutator.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered yesterday









                                    ZeroTheHeroZeroTheHero

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