I've been trying to understand how we get the $(m,n)$-labelled irreps of $SO^+(1,3)$ by reading posts such as this, this, this, this and links within, as well as the Wikipedia article on the matter, and remain rather confused about how some of the final steps in the process work.

We want the irreps of $SO^+(1,3)$, so we look at the corresponding algebra, $\mathfrak{so}^+(1,3)$. The six generators are labelled $J_a$ and $K_a$, $a \in \{1,2,3\}$, corresponding to rotations and boosts.

We then complexify the algebra, yielding $\mathfrak{so}^+(1,3)_\mathbb{C}$, and use a change of basis $A^\pm_a=(J_a \pm iK_a)/2.$ The new sets of generators $\{A^+_a\}$ and $\{A^-_a\}$ commute, and each generate an algebra isomorphic to $\mathfrak{su}(2)_\mathbb{C}$, which is the complexification of $\mathfrak{su}(2)$. Thus we have the following isomorphism:

$$\mathfrak{so}^+(1,3)_\mathbb{C} ~\cong~ \mathfrak{su}(2)_\mathbb{C} \oplus \mathfrak{su}(2)_\mathbb{C}$$

It is really the steps after this that I don't follow. We want the irreps of $\mathfrak{so}^+(1,3)$, not its complexification. From this post, I see that the above result does not mean that $\mathfrak{so}^+(1,3)$ is isomorphic to $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$. So how do we actually get around this issue to finish with being able to label the irreps of $\mathfrak{so}^+(1,3)$ using the labels for two irreps of $\mathfrak{su}(2)$?


1 Answer 1

  1. Any representation of a Lie algebra $\mathfrak{g}_{\mathbb{C}}$ trivially forms a restricted representation wrt. to a Lie subalgebra $\mathfrak{g}$.

  2. The interesting direction is the other way, cf. e.g. this & this related Phys.SE posts.

  • 1
    $\begingroup$ Ah, I think I see now. We can study $\mathfrak{so}^+(1,3)_\mathbb{C}$ to get the irreps of $\mathfrak{so}^+(1,3)$, so we use the labels of irreps of $(\mathfrak{su}(2) \oplus \mathfrak{su}(2))_\mathbb{C}$ to label the irreps of $\mathfrak{so}^+(1,3)$, and the Lie subalgebra that we use is just the de-complexified one generated by $\{A^+_a\}$ and $\{A^-_a\}$? $\endgroup$ Dec 11, 2021 at 18:22
  • $\begingroup$ Then, given the comments from SRS under this answer (physics.stackexchange.com/a/293606/101126) to the post you linked, there is a one-to-one correspondence, via a "change of basis", between the representations of $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$ used for classification, and the representations of the real restricted Lorentz algebra that we ultimately want. $\endgroup$ Dec 11, 2021 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.