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
$begingroup$
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 $SO_e^+(1,3)$?
Edit: I think I got, so I'm going to post a summary of what I understood to confirm it:
Let $G$ be a Lie group. All representations of $G$ give rise to representations of $mathfrakg$ by differentiation. Not all representations of $mathfrakg$ come from derivatives like this, however. These representations of $mathfrakg$ come from derivatives of representations of the universal cover of $G$, though. Then when $G$ is simply connected, all representations of $mathfrakg$ indeed come from $G$ as derivatives.
Now, if we know the representations of $mathfrakg$ we can determine by exponentiation the representations of the universal cover $tildeG$ of $G$ from which they are derived by exponentiation. This determines them in a neigbhorhood of the identity.
For the representations of $mathfrakg$ that indeed come from $G$, if $G$ is connected, then a neigbhorhood of the identity generates it, so that this is enough to reconstruct the representation everywhere.
Nevertheless, in the particular case of $SO_e^+(1,3)$ it so happens that this neighborhood of the identity reconstructed by the exponential is the whole group. Finally the representations of $mathfrakso(1,3)$ which do not come from $SO_e^+(1,3)$ come from the universal cover $SL_2(mathbbC)$.
Is this the whole point?
representation-theory lie-groups lie-algebras mathematical-physics quantum-field-theory
asked 3 hours ago
user1620696user1620696
11.8k742119
$endgroup$
add a comment |
$begingroup$
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 $SO_e^+(1,3)$?
Edit: I think I got, so I'm going to post a summary of what I understood to confirm it:
Let $G$ be a Lie group. All representations of $G$ give rise to representations of $mathfrakg$ by differentiation. Not all representations of $mathfrakg$ come from derivatives like this, however. These representations of $mathfrakg$ come from derivatives of representations of the universal cover of $G$, though. Then when $G$ is simply connected, all representations of $mathfrakg$ indeed come from $G$ as derivatives.
Now, if we know the representations of $mathfrakg$ we can determine by exponentiation the representations of the universal cover $tildeG$ of $G$ from which they are derived by exponentiation. This determines them in a neigbhorhood of the identity.
For the representations of $mathfrakg$ that indeed come from $G$, if $G$ is connected, then a neigbhorhood of the identity generates it, so that this is enough to reconstruct the representation everywhere.
Nevertheless, in the particular case of $SO_e^+(1,3)$ it so happens that this neighborhood of the identity reconstructed by the exponential is the whole group. Finally the representations of $mathfrakso(1,3)$ which do not come from $SO_e^+(1,3)$ come from the universal cover $SL_2(mathbbC)$.
Is this the whole point?
representation-theory lie-groups lie-algebras mathematical-physics quantum-field-theory
asked 3 hours ago
user1620696user1620696
11.8k742119
$endgroup$
add a comment |
$begingroup$
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 $SO_e^+(1,3)$?
Edit: I think I got, so I'm going to post a summary of what I understood to confirm it:
Let $G$ be a Lie group. All representations of $G$ give rise to representations of $mathfrakg$ by differentiation. Not all representations of $mathfrakg$ come from derivatives like this, however. These representations of $mathfrakg$ come from derivatives of representations of the universal cover of $G$, though. Then when $G$ is simply connected, all representations of $mathfrakg$ indeed come from $G$ as derivatives.
Now, if we know the representations of $mathfrakg$ we can determine by exponentiation the representations of the universal cover $tildeG$ of $G$ from which they are derived by exponentiation. This determines them in a neigbhorhood of the identity.
For the representations of $mathfrakg$ that indeed come from $G$, if $G$ is connected, then a neigbhorhood of the identity generates it, so that this is enough to reconstruct the representation everywhere.
Nevertheless, in the particular case of $SO_e^+(1,3)$ it so happens that this neighborhood of the identity reconstructed by the exponential is the whole group. Finally the representations of $mathfrakso(1,3)$ which do not come from $SO_e^+(1,3)$ come from the universal cover $SL_2(mathbbC)$.
Is this the whole point?
representation-theory lie-groups lie-algebras mathematical-physics quantum-field-theory
asked 3 hours ago
user1620696user1620696
11.8k742119
$endgroup$
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 $SO_e^+(1,3)$?
Edit: I think I got, so I'm going to post a summary of what I understood to confirm it:
Let $G$ be a Lie group. All representations of $G$ give rise to representations of $mathfrakg$ by differentiation. Not all representations of $mathfrakg$ come from derivatives like this, however. These representations of $mathfrakg$ come from derivatives of representations of the universal cover of $G$, though. Then when $G$ is simply connected, all representations of $mathfrakg$ indeed come from $G$ as derivatives.
Now, if we know the representations of $mathfrakg$ we can determine by exponentiation the representations of the universal cover $tildeG$ of $G$ from which they are derived by exponentiation. This determines them in a neigbhorhood of the identity.
For the representations of $mathfrakg$ that indeed come from $G$, if $G$ is connected, then a neigbhorhood of the identity generates it, so that this is enough to reconstruct the representation everywhere.
Nevertheless, in the particular case of $SO_e^+(1,3)$ it so happens that this neighborhood of the identity reconstructed by the exponential is the whole group. Finally the representations of $mathfrakso(1,3)$ which do not come from $SO_e^+(1,3)$ come from the universal cover $SL_2(mathbbC)$.
Is this the whole point?
representation-theory lie-groups lie-algebras mathematical-physics quantum-field-theory
representation-theory lie-groups lie-algebras mathematical-physics quantum-field-theory
asked 3 hours ago
user1620696user1620696
11.8k742119
asked 3 hours ago
user1620696user1620696
11.8k742119
edited 16 mins ago
user1620696
asked 3 hours ago
user1620696user1620696
11.8k742119
asked 3 hours ago
user1620696user1620696
11.8k742119
asked 3 hours ago
user1620696user1620696
11.8k742119
11.8k742119
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
$endgroup$
$begingroup$
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
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 mins ago
add a comment |
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1 Answer
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1 Answer
1
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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 $mathfrakg$ 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.
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
$endgroup$
$begingroup$
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
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 mins ago
add a comment |
$begingroup$
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.
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
$endgroup$
$begingroup$
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
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 mins ago
add a comment |
$begingroup$
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.
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
$endgroup$
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.
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
answered 2 hours ago
Qiaochu YuanQiaochu Yuan
282k32599946
answered 2 hours 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
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 mins ago
add a comment |
$begingroup$
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
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 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
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
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 mins ago
$begingroup$
Thanks very much @QiaochuYuan, I think I finally got it. I posted one edit with a summary of what I understood of this matter. Could you please tell me if I got it right or if I misunderstood something? Thanks very much again!
$endgroup$
– user1620696
14 mins ago
add a comment |
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