Getting representations of the Lie group out of representations of its Lie algebra Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Geometric algebra approach to Lorentz group representationsIsomorphisms of the Lorentz group and algebraIrreducible representations of the Lorentz Lie algebraRepresentation of Lie groups as exponentiations of algebra representations.Reference for rigorous treatment of the representation theory of the Lorentz groupClassification of representations of the lie algebra $mathfraku(2)$.Relation between representations of Lie Group and Lie AlgebraCorrespondence between representations of a Lie group and Lie algebra.Representations of $sl(2,C)$ as a real Lie algebraDifference between infinitesimal parameters of Lie algebra and group generators of Lie group

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Getting representations of the Lie group out of representations of its Lie algebra



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Geometric algebra approach to Lorentz group representationsIsomorphisms of the Lorentz group and algebraIrreducible representations of the Lorentz Lie algebraRepresentation of Lie groups as exponentiations of algebra representations.Reference for rigorous treatment of the representation theory of the Lorentz groupClassification of representations of the lie algebra $mathfraku(2)$.Relation between representations of Lie Group and Lie AlgebraCorrespondence between representations of a Lie group and Lie algebra.Representations of $sl(2,C)$ as a real Lie algebraDifference between infinitesimal parameters of Lie algebra and group generators of Lie group










3












$begingroup$


This is something that is usually done in QFT and that bothers me a lot because it seems to be done without much caution.



In QFT when classifying fields one looks for the irreducible representations of the proper orthochronous Lorentz group $SO_e^+(1,3)$.



But to do so what one does in practice is: look for representations of the Lie algebra $mathfrakso(1,3)$ and then exponentiate.



For instance, in Peskin's QFT book:




It is generally true that one can find matrix representations of a continuous group by finding matrix representations of the generators of the group, then exponentiating these infinitesimal transformations.




The same thing is done in countless other books.



Now I do agree that if we have a representation of $G$ we can get one of $mathfrakg$ differentiating at the identity. Here one is doing the reverse!



In practice what is doing is: find a representation of $mathfrakso(1,3)$ on a vector space $V$, then exponentiate it to get a representation of $SO_e^+(1,3)$. I think one way to write it would be as follows, let $D : mathfrakso(1,3)to operatornameEnd(V)$ be the representation of the algebra, define $mathscrD : SO_e^+(1,3)to GL(V)$



$$mathscrD(exp theta X)=exp theta D(X).$$



Now, this seems to be very subtle.



In general the exponential $exp : mathfrakgto G$ is not surjective. Even if it is, I think it need not be injective.



Also I've heard there is one very important and very subtle connection between $exp(mathfrakg)$ and the universal cover of $G$.



My question here is: how to understand this procedure Physicists do more rigorously? In general this process of "getting representations of $G$ out of representations of $mathfrakg$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrakg)?



Or in the end physicists are allowed to do this just because very luckilly in this case $exp$ is surjective onto $SO_e^+(1,3)$?










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    This is something that is usually done in QFT and that bothers me a lot because it seems to be done without much caution.



    In QFT when classifying fields one looks for the irreducible representations of the proper orthochronous Lorentz group $SO_e^+(1,3)$.



    But to do so what one does in practice is: look for representations of the Lie algebra $mathfrakso(1,3)$ and then exponentiate.



    For instance, in Peskin's QFT book:




    It is generally true that one can find matrix representations of a continuous group by finding matrix representations of the generators of the group, then exponentiating these infinitesimal transformations.




    The same thing is done in countless other books.



    Now I do agree that if we have a representation of $G$ we can get one of $mathfrakg$ differentiating at the identity. Here one is doing the reverse!



    In practice what is doing is: find a representation of $mathfrakso(1,3)$ on a vector space $V$, then exponentiate it to get a representation of $SO_e^+(1,3)$. I think one way to write it would be as follows, let $D : mathfrakso(1,3)to operatornameEnd(V)$ be the representation of the algebra, define $mathscrD : SO_e^+(1,3)to GL(V)$



    $$mathscrD(exp theta X)=exp theta D(X).$$



    Now, this seems to be very subtle.



    In general the exponential $exp : mathfrakgto G$ is not surjective. Even if it is, I think it need not be injective.



    Also I've heard there is one very important and very subtle connection between $exp(mathfrakg)$ and the universal cover of $G$.



    My question here is: how to understand this procedure Physicists do more rigorously? In general this process of "getting representations of $G$ out of representations of $mathfrakg$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrakg)?



    Or in the end physicists are allowed to do this just because very luckilly in this case $exp$ is surjective onto $SO_e^+(1,3)$?










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      This is something that is usually done in QFT and that bothers me a lot because it seems to be done without much caution.



      In QFT when classifying fields one looks for the irreducible representations of the proper orthochronous Lorentz group $SO_e^+(1,3)$.



      But to do so what one does in practice is: look for representations of the Lie algebra $mathfrakso(1,3)$ and then exponentiate.



      For instance, in Peskin's QFT book:




      It is generally true that one can find matrix representations of a continuous group by finding matrix representations of the generators of the group, then exponentiating these infinitesimal transformations.




      The same thing is done in countless other books.



      Now I do agree that if we have a representation of $G$ we can get one of $mathfrakg$ differentiating at the identity. Here one is doing the reverse!



      In practice what is doing is: find a representation of $mathfrakso(1,3)$ on a vector space $V$, then exponentiate it to get a representation of $SO_e^+(1,3)$. I think one way to write it would be as follows, let $D : mathfrakso(1,3)to operatornameEnd(V)$ be the representation of the algebra, define $mathscrD : SO_e^+(1,3)to GL(V)$



      $$mathscrD(exp theta X)=exp theta D(X).$$



      Now, this seems to be very subtle.



      In general the exponential $exp : mathfrakgto G$ is not surjective. Even if it is, I think it need not be injective.



      Also I've heard there is one very important and very subtle connection between $exp(mathfrakg)$ and the universal cover of $G$.



      My question here is: how to understand this procedure Physicists do more rigorously? In general this process of "getting representations of $G$ out of representations of $mathfrakg$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrakg)?



      Or in the end physicists are allowed to do this just because very luckilly in this case $exp$ is surjective onto $SO_e^+(1,3)$?










      share|cite|improve this question









      $endgroup$




      This is something that is usually done in QFT and that bothers me a lot because it seems to be done without much caution.



      In QFT when classifying fields one looks for the irreducible representations of the proper orthochronous Lorentz group $SO_e^+(1,3)$.



      But to do so what one does in practice is: look for representations of the Lie algebra $mathfrakso(1,3)$ and then exponentiate.



      For instance, in Peskin's QFT book:




      It is generally true that one can find matrix representations of a continuous group by finding matrix representations of the generators of the group, then exponentiating these infinitesimal transformations.




      The same thing is done in countless other books.



      Now I do agree that if we have a representation of $G$ we can get one of $mathfrakg$ differentiating at the identity. Here one is doing the reverse!



      In practice what is doing is: find a representation of $mathfrakso(1,3)$ on a vector space $V$, then exponentiate it to get a representation of $SO_e^+(1,3)$. I think one way to write it would be as follows, let $D : mathfrakso(1,3)to operatornameEnd(V)$ be the representation of the algebra, define $mathscrD : SO_e^+(1,3)to GL(V)$



      $$mathscrD(exp theta X)=exp theta D(X).$$



      Now, this seems to be very subtle.



      In general the exponential $exp : mathfrakgto G$ is not surjective. Even if it is, I think it need not be injective.



      Also I've heard there is one very important and very subtle connection between $exp(mathfrakg)$ and the universal cover of $G$.



      My question here is: how to understand this procedure Physicists do more rigorously? In general this process of "getting representations of $G$ out of representations of $mathfrakg$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrakg)?



      Or in the end physicists are allowed to do this just because very luckilly in this case $exp$ is surjective onto $SO_e^+(1,3)$?







      representation-theory lie-groups lie-algebras mathematical-physics quantum-field-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      user1620696user1620696

      11.8k742119




      11.8k742119




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          The exponential map doesn't need to be surjective. If $G$ is connected the exponential map is surjective onto a neighborhood of the identity, and since a neighborhood of the identity of a connected topological group generates it, once you know what a representation does to a neighborhood of the identity, that determines what it does everywhere.



          However, in general $G$ needs to be simply connected. That is, exponential in general provides an equivalence between representations of a finite-dimensional Lie algebra $mathfrakg$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrakg$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbbC)$. This means that not all representations of $mathfrakso(1, 3)$ exponentiate to representations of the proper orthochronous Lorentz group; some exponentiate to projective representations. As far as I know this is mostly fine for quantum, and so physicists don't seem to worry much about the distinction in practice.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
            $endgroup$
            – paul garrett
            1 hour ago











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          1 Answer
          1






          active

          oldest

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          active

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          active

          oldest

          votes









          5












          $begingroup$

          The exponential map doesn't need to be surjective. If $G$ is connected the exponential map is surjective onto a neighborhood of the identity, and since a neighborhood of the identity of a connected topological group generates it, once you know what a representation does to a neighborhood of the identity, that determines what it does everywhere.



          However, in general $G$ needs to be simply connected. That is, exponential in general provides an equivalence between representations of a finite-dimensional Lie algebra $mathfrakg$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrakg$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbbC)$. This means that not all representations of $mathfrakso(1, 3)$ exponentiate to representations of the proper orthochronous Lorentz group; some exponentiate to projective representations. As far as I know this is mostly fine for quantum, and so physicists don't seem to worry much about the distinction in practice.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
            $endgroup$
            – paul garrett
            1 hour ago















          5












          $begingroup$

          The exponential map doesn't need to be surjective. If $G$ is connected the exponential map is surjective onto a neighborhood of the identity, and since a neighborhood of the identity of a connected topological group generates it, once you know what a representation does to a neighborhood of the identity, that determines what it does everywhere.



          However, in general $G$ needs to be simply connected. That is, exponential in general provides an equivalence between representations of a finite-dimensional Lie algebra $mathfrakg$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrakg$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbbC)$. This means that not all representations of $mathfrakso(1, 3)$ exponentiate to representations of the proper orthochronous Lorentz group; some exponentiate to projective representations. As far as I know this is mostly fine for quantum, and so physicists don't seem to worry much about the distinction in practice.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
            $endgroup$
            – paul garrett
            1 hour ago













          5












          5








          5





          $begingroup$

          The exponential map doesn't need to be surjective. If $G$ is connected the exponential map is surjective onto a neighborhood of the identity, and since a neighborhood of the identity of a connected topological group generates it, once you know what a representation does to a neighborhood of the identity, that determines what it does everywhere.



          However, in general $G$ needs to be simply connected. That is, exponential in general provides an equivalence between representations of a finite-dimensional Lie algebra $mathfrakg$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrakg$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbbC)$. This means that not all representations of $mathfrakso(1, 3)$ exponentiate to representations of the proper orthochronous Lorentz group; some exponentiate to projective representations. As far as I know this is mostly fine for quantum, and so physicists don't seem to worry much about the distinction in practice.






          share|cite|improve this answer









          $endgroup$



          The exponential map doesn't need to be surjective. If $G$ is connected the exponential map is surjective onto a neighborhood of the identity, and since a neighborhood of the identity of a connected topological group generates it, once you know what a representation does to a neighborhood of the identity, that determines what it does everywhere.



          However, in general $G$ needs to be simply connected. That is, exponential in general provides an equivalence between representations of a finite-dimensional Lie algebra $mathfrakg$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrakg$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbbC)$. This means that not all representations of $mathfrakso(1, 3)$ exponentiate to representations of the proper orthochronous Lorentz group; some exponentiate to projective representations. As far as I know this is mostly fine for quantum, and so physicists don't seem to worry much about the distinction in practice.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Qiaochu YuanQiaochu Yuan

          282k32599946




          282k32599946











          • $begingroup$
            There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
            $endgroup$
            – paul garrett
            1 hour ago
















          • $begingroup$
            There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
            $endgroup$
            – paul garrett
            1 hour ago















          $begingroup$
          There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
          $endgroup$
          – paul garrett
          1 hour ago




          $begingroup$
          There's certainly also the issue of not-finite-dimensional representations... Wallach's and Casselman's "globalization" functors show two opposite extremes of adjoints to the functor that takes $G$ repns $V$ to $mathfrak g,K$ modules of smooth vectors $V^infty$.
          $endgroup$
          – paul garrett
          1 hour ago

















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