# Are the so-called representations of the Lorentz group actually all representations of it?

Fermionic fields change sign under a rotation by $$2\pi$$. However, in $$SO^+\left(1,3\right)$$ a rotation by $$2\pi$$ is the identity. For any representation $$R$$ of $$SO\left(1,3\right)$$ then we have $$R\left(\Lambda_1\right)R\left(\Lambda_2\right)=R\left(\Lambda_1\Lambda_2\right)$$ taking $$\Lambda_1=\Lambda_2=$$rotation by $$2\pi=$$ the identity, we have $$R\left(1\right)R\left(1\right)=R\left(1\right)$$ giving $$R\left(1\right)=1\neq-1$$. Hence, these fermionic fields do not transform under a representation of $$SO^+\left(1,3\right)$$. Of course, fermions do transform under a representation of the double cover of $$SO^+\left(1,3\right)$$, $$SL\left(2,\mathbb{C}\right)$$, but by the above argument these are not representations of the Lorentz group.

I feel like my professors talk as if this is not the case, calling them all representations of the Lorentz group. I just wanted to check that this is correct, and they are not really representations of the Lorentz group?

• Define "so-called representations." Do you mean projective representations? – Qmechanic Feb 6 '19 at 14:36

Before answering the question, I'll review the definition of "representation" so that the answer will be clear. If $$G$$ and $$H$$ are groups, then a homomorphism from $$G$$ to $$H$$ is a map that assigns each element $$g\in G$$ to some element $$\sigma(g)\in H$$ such that the basic group operations are respected, such as $$\sigma(g_1)\sigma(g_2)=\sigma(g_1 g_2)$$ and $$\sigma(1)=1$$ where $$1$$ is the identity element in $$G$$ or $$H$$, respectively.
A matrix representation of a group $$G$$ is a homomorphism $$\sigma:G\to H$$ where $$H$$ is a group of matrices (usually over $$\mathbb{C}$$) using matrix multiplication as the multiplication rule for the group. This answer will focus on matrix representations, as implied by the OP.
By this definition, the spin-$$1/2$$ "representation" of the rotation group $$SO(3)$$ is not a representation, for precisely the reason noted in the OP. It is a legitimate reprentation of the covering group $$SU(2)$$, again as noted in the OP. Here I'm using just the rotation group $$SO(3)$$ as an example instead of the Lorentz group $$SO(1,3)$$, but the idea is the same.
If $$F$$ is a covering group of $$G$$, then a representation of $$F$$ can be regarded as a projective representation of $$G$$. Intuitively, if $$F$$ reduces to $$G$$ when certain differences between elements of $$F$$ are ignored (technically, when $$G$$ is regarded as a quotient of $$F$$), then a representation of $$F$$ gives a "representation" (in scare quotes) of $$G$$ if certain differences bewteen representation-matrices are ignored. The word "projective" refers to ignoring these factors (like factors of $$-1$$ in the OP's example).
On the other hand, if $$G$$ is a Lie group, then $$G$$ has an associated Lie algebra, and we can also talk about representations of the Lie algebra. Since $$SO(3)$$ and $$SU(2)$$ have the same Lie algebra (just like $$SO(1,3)$$ and $$SL(2,\mathbb{C})$$ have the same Lie algebra), a representation of the Lie algebra of $$SU(2)$$ is also a legitimate representation of the Lie algebra of $$SO(3)$$.
So the OP's suspicion is correct: people often use the language carelessly (or at least in non-standard ways). People might say "representation of the group $$G$$" when they really mean "projective reprentation of the group $$G$$" or "representation of a covering group of $$G$$" or "representation of the Lie algebra corresponding to $$G$$."