Which representations of the (restricted) Lorentz group are real, and which are quaternionic? I know for compact semi-simple Lie groups, one can use the Frobenius-Schur indicator to classify the real and quaternionic representations. However this isn't possible for the Lorentz group, due to its non-compactness. Is there a catalogue or well known way to show which representations of the Lorentz group are real and which are quaternionic?
 A: I'll stick with Lie algebras. If somebody knows about a serious complication when going up to the group, please let me know.

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*The irreducible representations of $\mathfrak{so}(1,3)$ on complex vector spaces of finite dimension ("irreps") are parametrized by tensor products of pairs of irreps of $\mathfrak{sl}_2(\mathbb C)$. These, in turn, are well-known to be parametrized by integers (math convention) or half-integers (physics convention) $\ge 0$. Consequently, the irreps of $\mathfrak{so}(1,3)$ are parametrized by pairs of such (half)-integers $(m,n)$. Cf. answer to What is the relationship between the representations of ${\frak sl}(2;\Bbb C)$ when viewed as real Lie algebra or complex Lie algebra?


*All representations of $\mathfrak{so}(1,3)$, or any other real Lie algebra (on finite-dimensional complex vector spaces), fall under a trichotomy of "(truly) complex" versus "real" versus "quaternionic" representations. Cf. answer to Why can $(m,m)$ and $(m,n)\oplus(n,m)$ representations of $\mathfrak{so}(1,3)$ be represented over a real vector space? There are various equivalent ways to describe this trichotomy, including (as you ask in a comment), at least in the case of an irreducible representation, via the existence of a symmetric or alternating invariant bilinear form on it: see penultimate paragraph of part A of the answer in the last link. -- Especially in the case where our Lie algebra is compact, some theory has become standard in good sources. Unfortunately, the non-compact case (and as you correctly point out, $\mathfrak{so}(1,3)$ is an example for this) seems to have been treated badly, and there one finds confusion both in the terminology and in the results. Indeed, I link to two answers on this site there which seem to have wrongly confused the compact and the non-compact case. Cf. my attempt of a correction of one of those in https://math.stackexchange.com/a/3298058/96384, and also cf. $SO(p,q)$ Fundamental Weights?, which actually is a generalization of your question from $\mathfrak{so}(1,3)$ to general $\mathfrak{so}(p,q)$.


*To distinguish between "truly complex" and the other two cases, one has to check how complex conjugation acts on the weight lattice of the Lie algebra, in particular on the highest weight of the given representation. Roughly speaking, if it does not stabilize this highest weight, the representation will be "truly complex". If it does stabilize the highest weight, then the representation is either real or quaternionic: To determine which, one can check parity of a somewhat finer numerical invariant attached to that weight.


*In the case at hand, $\mathfrak{so}(1,3)$, notice that the pairs $(m,n)$ mentioned in 1 actually parametrize the possible highest weights. It turns out that here, complex conjugation acts via flipping $(m,n) \mapsto (n,m)$. Once one believes that, it is immediate that all irreps $(m,n)$ with $m \neq n$ are "truly complex". (By the way, I think what triggered the question  When will two isomorphic Lie algebras have the same representation?, and is among the first examples in my lengthy answer, is the innocuous fact that the inequivalent irreps $(1,0)$ and $(0,1)$ (math convention) of $\mathfrak{so}(1,3)$ are each other's complex conjugate, in particular truly complex; while when we look through the representations of the compact $\mathfrak{so}(4)$, which are parametrized by the very same pairs $(m,n)$, now each of $(1,0)$ and $(0,1)$ is self-conjugate, and turns out to be quaternionic. Note that the weight lattices of $\mathfrak{so}(1,3)$ and $\mathfrak{so}(4)$ are identical. Again, all the difference lies in how, depending on the real Lie algebra we look at, complex conjugation acts on the weights; in the case of $\mathfrak{so}(4)$, it happens to act trivially.)


*It remains to decide whether the irreps $(m,m)$, for $\mathfrak{so}(1,3)$, are real or quaternionic. In my answer to the first link in 2, I quote this as a special case of a ridiculously more general result by Jacques Tits. But as I say in part B of that answer, one should surely be able to decide it from whether a certain explicit map is a positive or a negative scalar (see, a parity distinction in disguise), and all bets are on "positive", which means that all these irreps $(m,m)$ are real, and the quaternionic case does not occur among them.
