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


Why do we use the complexification of the Lorentz group?Difference Between Algebra of Infinitesimal Conformal Transformations & Conformal AlgebraSubgroup of Lorentz Group Generated by BoostsLorentz Group Generators: Two MethodsMeaning of Lorentz GeneratorsFinite lorentz transform for 4-vectors in terms of the generatorsOn the generators of the Lorentz groupLie group compactness from generatorsRelation between the Dirac Algebra and the Lorentz groupCommutation relations of the generators of the Lorentz group













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?










share|cite|improve this question











$endgroup$
















    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?










    share|cite|improve this question











    $endgroup$














      2












      2








      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?










      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






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      edited 11 hours ago









      Qmechanic

      106k121961227




      106k121961227










      asked yesterday









      KALLE THE BAWSMANKALLE THE BAWSMAN

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          3 Answers
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          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









          $endgroup$




















            1












            $begingroup$

            1. Illustrative example: It is straightforward to prove that the Lie group
              $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
              of 3D rotations is generated by the corresponding Lie algebra
              $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
              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









            $endgroup$




















              0












              $begingroup$

              While
              beginalign
              e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
              &= 1+ epsilon (A+B)+ frac12
              epsilon^2 (A^2+AB +B^2)+ldots
              endalign

              we have
              beginalign
              e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
              &= 1+ epsilon (B+A)+ frac12
              epsilon^2 (A^2+BA +B^2)+ldots
              endalign

              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









              $endgroup$












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                3 Answers
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                3 Answers
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                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












                  $begingroup$

                  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 $epsilon$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    Chiral AnomalyChiral Anomaly

                    12.5k21542




                    12.5k21542





















                        1












                        $begingroup$

                        1. Illustrative example: It is straightforward to prove that the Lie group
                          $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                          of 3D rotations is generated by the corresponding Lie algebra
                          $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                          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
                            $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                            of 3D rotations is generated by the corresponding Lie algebra
                            $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                            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
                              $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                              of 3D rotations is generated by the corresponding Lie algebra
                              $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                              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
                              $$ SO(3)~:=~ Min rm Mat_3times 3(mathbbR) mid M^tM=mathbb1_3times 3, ~det(M)=1 $$
                              of 3D rotations is generated by the corresponding Lie algebra
                              $$ so(3)~:=~ min rm Mat_3times 3(mathbbR) mid m^t=-m $$
                              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 11 hours ago









                            QmechanicQmechanic

                            106k121961227




                            106k121961227





















                                0












                                $begingroup$

                                While
                                beginalign
                                e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                &= 1+ epsilon (A+B)+ frac12
                                epsilon^2 (A^2+AB +B^2)+ldots
                                endalign

                                we have
                                beginalign
                                e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                &= 1+ epsilon (B+A)+ frac12
                                epsilon^2 (A^2+BA +B^2)+ldots
                                endalign

                                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









                                $endgroup$

















                                  0












                                  $begingroup$

                                  While
                                  beginalign
                                  e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                  &= 1+ epsilon (A+B)+ frac12
                                  epsilon^2 (A^2+AB +B^2)+ldots
                                  endalign

                                  we have
                                  beginalign
                                  e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                  &= 1+ epsilon (B+A)+ frac12
                                  epsilon^2 (A^2+BA +B^2)+ldots
                                  endalign

                                  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









                                  $endgroup$















                                    0












                                    0








                                    0





                                    $begingroup$

                                    While
                                    beginalign
                                    e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                    &= 1+ epsilon (A+B)+ frac12
                                    epsilon^2 (A^2+AB +B^2)+ldots
                                    endalign

                                    we have
                                    beginalign
                                    e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                    &= 1+ epsilon (B+A)+ frac12
                                    epsilon^2 (A^2+BA +B^2)+ldots
                                    endalign

                                    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
                                    beginalign
                                    e^epsilon A e^epsilon B&=(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots)(1+epsilon B+frac12epsilon^2 B^2+ldots)tag1, ,\
                                    &= 1+ epsilon (A+B)+ frac12
                                    epsilon^2 (A^2+AB +B^2)+ldots
                                    endalign

                                    we have
                                    beginalign
                                    e^epsilon B e^epsilon A&=(1+epsilon B+frac12epsilon^2 B^2+ldots)(1+epsilon A+textstylefrac12epsilon^2 A^2+ldots), ,tag2\
                                    &= 1+ epsilon (B+A)+ frac12
                                    epsilon^2 (A^2+BA +B^2)+ldots
                                    endalign

                                    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 11 hours ago









                                    ZeroTheHeroZeroTheHero

                                    21.1k53364




                                    21.1k53364



























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