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












2












$begingroup$


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 $mathfrak{so}(1,3)$ 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 $mathfrak{g}$ differentiating at the identity. Here one is doing the reverse!



In practice what is doing is: find a representation of $mathfrak{so}(1,3)$ 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 $D : mathfrak{so}(1,3)to operatorname{End}(V)$ be the representation of the algebra, define $mathscr{D} : SO_e^+(1,3)to GL(V)$



$$mathscr{D}(exp theta X)=exp theta D(X).$$



Now, this seems to be very subtle.



In general the exponential $exp : mathfrak{g}to G$ 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 $exp(mathfrak{g})$ 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 $mathfrak{g}$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrak{g})?



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










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    2












    $begingroup$


    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 $mathfrak{so}(1,3)$ 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 $mathfrak{g}$ differentiating at the identity. Here one is doing the reverse!



    In practice what is doing is: find a representation of $mathfrak{so}(1,3)$ 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 $D : mathfrak{so}(1,3)to operatorname{End}(V)$ be the representation of the algebra, define $mathscr{D} : SO_e^+(1,3)to GL(V)$



    $$mathscr{D}(exp theta X)=exp theta D(X).$$



    Now, this seems to be very subtle.



    In general the exponential $exp : mathfrak{g}to G$ 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 $exp(mathfrak{g})$ 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 $mathfrak{g}$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrak{g})?



    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









    $endgroup$















      2












      2








      2





      $begingroup$


      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 $mathfrak{so}(1,3)$ 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 $mathfrak{g}$ differentiating at the identity. Here one is doing the reverse!



      In practice what is doing is: find a representation of $mathfrak{so}(1,3)$ 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 $D : mathfrak{so}(1,3)to operatorname{End}(V)$ be the representation of the algebra, define $mathscr{D} : SO_e^+(1,3)to GL(V)$



      $$mathscr{D}(exp theta X)=exp theta D(X).$$



      Now, this seems to be very subtle.



      In general the exponential $exp : mathfrak{g}to G$ 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 $exp(mathfrak{g})$ 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 $mathfrak{g}$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrak{g})?



      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 $mathfrak{so}(1,3)$ 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 $mathfrak{g}$ differentiating at the identity. Here one is doing the reverse!



      In practice what is doing is: find a representation of $mathfrak{so}(1,3)$ 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 $D : mathfrak{so}(1,3)to operatorname{End}(V)$ be the representation of the algebra, define $mathscr{D} : SO_e^+(1,3)to GL(V)$



      $$mathscr{D}(exp theta X)=exp theta D(X).$$



      Now, this seems to be very subtle.



      In general the exponential $exp : mathfrak{g}to G$ 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 $exp(mathfrak{g})$ 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 $mathfrak{g}$ by exponentiation" can be done, or it really just gives representations of $exp(mathfrak{g})?



      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






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      asked 2 hours ago









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          $begingroup$

          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 $mathfrak{g}$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrak{g}$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbb{C})$. This means that not all representations of $mathfrak{so}(1, 3)$ 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
            24 mins ago












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          $begingroup$

          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 $mathfrak{g}$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrak{g}$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbb{C})$. This means that not all representations of $mathfrak{so}(1, 3)$ 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
            24 mins ago
















          4












          $begingroup$

          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 $mathfrak{g}$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrak{g}$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbb{C})$. This means that not all representations of $mathfrak{so}(1, 3)$ 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
            24 mins ago














          4












          4








          4





          $begingroup$

          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 $mathfrak{g}$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrak{g}$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbb{C})$. This means that not all representations of $mathfrak{so}(1, 3)$ 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 $mathfrak{g}$ and representations of the unique simply connected Lie group $G$ with Lie algebra $mathfrak{g}$. The proper orthochronous Lorentz group is not simply connected; its universal cover is $SL_2(mathbb{C})$. This means that not all representations of $mathfrak{so}(1, 3)$ 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
            24 mins 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
            24 mins 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
          24 mins 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
          24 mins ago


















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