# Why the Lorentz group has complex generators in QFT treatments? [duplicate]

In Schwartz' and Peskin's QFT books, when trying to deal with representations of the Lorentz group the authors study the representations of the Lie algebra of such group.

By definition, if $SO(1,3)$ is the Lorentz group, the Lie algebra is $\mathfrak{so}(1,3)$ defined as the set of all left-invariant vector fields on $SO(1,3)$ which in turns is equivalent to the tangent space at the identity of $SO(1,3)$.

Now, $SO(1,3)$ is a real manifold. Hence its tangent space at the origin is a real vector space.

Anyway, the books say that there are elements on this Lie algebra called generators defined by some complex matrices $J_i$ and $K_i$ such that any group element is

$$\Lambda = \exp(i\theta^i J_i+i\beta^i K_i)$$

and such that

$$[J_i,J_j]=i\epsilon_{ijk}J_k$$

$$[K_i,K_j]=-i\epsilon_{ijk}J_k$$

$$[J_i,K_j]=i\epsilon_{ijk}K_k$$

Now there's something quite wrong here. There are two main points I've noticed:

1. How can the elements of $\mathfrak{so}(1,3)$ be complex matrices if this is a real vector space? The matrices certainly need to be real. Unless one is hidden a complexification somewhere, but this isn't made clear in the books. If that's the case, where and why does one use a complexification?

2. It is not true that all elements of the group can be recovered by exponentiation, if I'm not mistaken. I really don't remember this quite well, but claiming that all elements have that form seems wrong. Furthermore, the exponentiation I know about is the map $\exp : \mathfrak{so}(1,3)\to SO(1,3)$ defined by

$$\exp(A)=\phi^{X^A}_1(e),$$

where $X^A$ is the associated left-invariant vector field, $\phi^{X^A}_t$ is its flow and $e\in SO(1,3)$ is the identity. If I'm not mistaken this exponential map isn't surjective. How the author is justified to say that any group element is of this form?

In summary how to connect the physicst's approach the author presents to the usual Lie group / Lie algebra theory?

## marked as duplicate by ACuriousMind♦Jun 15 '17 at 18:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• For your first question. Are you sure you are not talking about the $SL(2,\mathbb{C})$ representation of the Lorentz group? As for your second question, the exponential maps the Lie algebra onto the connected component of the identity. True for any Lie group. – user154997 Jun 15 '17 at 18:18
• @LucJ.Bourhis well the authors say that's for $SO(1,3)$ and that's the reason for my confusion, since the Lie algebra should be real. I believe there's a complexification being done, about which the authors are not talking about. I'm just unsure why and how this is used for the problem at hand. – user1620696 Jun 15 '17 at 18:29
• ok, just wanted to make sure. See the references given by AccidentalFourierTransform as a starting point then. – user154997 Jun 15 '17 at 18:35
• Note that the exponential map is not always surjective even for connected Lie groups (it is for compact connected ones). The "onto" in Luc's first answer might give some people the wrong idea about surjectivity (this is a common misconception among physicists, as the exponential map turns out to be surjective for most if not all even non-compact groups used in physics). – Adomas Baliuka Jun 15 '17 at 18:43

## 1 Answer

A partial answer to (2): The more precise statement is that any group element which is continuously connected to the identity can be written in this way. I'm pretty sure we define "continuously connected" in such a way that the previous statement is a tautology, but hey that's physics!

There are important group elements like P (parity) and T (time reversal) which cannot be written as the exponentiation of generators, and they will play a big role later in QFT.