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

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? ## 1 Answer 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. • 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\$. – paul garrett Apr 22 at 1:52
• 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! – user1620696 Apr 22 at 3:18