Kdjykuj

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 $SO_e^+(1,3)$?

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.