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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 $SO_e^+(1,3)$?

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 $\mathfrak{g}$ by differentiation. Not all representations of $\mathfrak{g}$ come from derivatives like this, however. These representations of $\mathfrak{g}$ come from derivatives of representations of the universal cover of $G$, though. Then when $G$ is simply connected, all representations of $\mathfrak{g}$ indeed come from $G$ as derivatives.

Now, if we know the representations of $\mathfrak{g}$ we can determine by exponentiation the representations of the universal cover $\tilde{G}$ of $G$ from which they are derived by exponentiation. This determines them in a neigbhorhood of the identity.

For the representations of $\mathfrak{g}$ 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 $\mathfrak{so}(1,3)$ which do not come from $SO_e^+(1,3)$ come from the universal cover $SL_2(\mathbb{C})$.

Is this the whole point?

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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.

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  • $\begingroup$ 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 Apr 22 at 1:52
  • $\begingroup$ 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 Apr 22 at 3:18

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