Non-unitary, finite dimensional representations of the Lorentz group On Wikipedia and often in physics books it is stated, that the Lorenz group's finite dimensional non-unitary representations are completely reducible. We heavily use this in physics, however I do not know of a direct proof or reference for this. Could anyone help in providing such? It is clear to me that the Lorentz algebra $so(3,1)$ is semisimple, thus completely reducible. However, I don't see how we go to the group level, cause the exponential map is not surjective, since $SO(3,1)$ is non-compact and disconnected.
In particular, this is of interest because fields transform in irreps of $SO(3,1)$, so complete reducibility would be useful to have.
 A: *

*To every Lie algebra $\mathfrak{g}$ there is an associated simply-connected Lie group $G_s$. By the general correspondence of Lie groups and algebras, every representation of $G_s$ induces one of $\mathfrak{g}$ and vice versa, regardless of surjectivity of the exponential map. Hence a representation of $G_s$ is completely reducible if and only if the corresponding representation of $\mathfrak{g}$ is.


*For any connected Lie group $G_c$ with Lie algebra $\mathfrak{g}$, this is still true: If we have a representation $V_r$ of $G_c$ that has a subrepresentation $W_r\subset V_r$ of $G_c$ and $\mathfrak{g}$ is semi-simple, then we get $V_r = W_r\oplus W'_r$ for some other representation $W'_r$ of $\mathfrak{g}$. But $W'_r$ must be a subrepresentation of $G_c$ also, since you can always use the exponential map locally to get the representation of $G_c$ from the representation of $\mathfrak{g}$ - very similar to the general proof of the algebra-group correspondence, surjectivity is not needed here either, only connectedness so that you can "shift" the exponential map along paths in the group.


*For a disconnected Lie group $G_d$ with Lie algebra $\mathfrak{g}$, we have that all connected components are diffeomorphic to the identity component $G_0$ and there is an exact sequence
$$ 1 \to G_0 \to G \to \pi_0(G)=G/G_0\to 1. \tag{1}$$
If this sequence splits, we have $G = \pi_0(G)\rtimes G_0$ where $\rtimes$ is a semi-direct product.  If $\pi_0(G)$ is additionally Abelian, Mackey's theory of induced representations tells us that every representation of $G$ comes from representations of the factors $\pi_0(G)$ and $G_0$ and every pair of irreducible representations of the factors defines an irreducible representation of $G$. Since Maschke's theorem tells us that finite group have only completely reducible finite-dimensional representations, this means that a Lie group for which (1) splits and for which $\pi_0(G)$ is Abelian has completely reducible representations if $\mathfrak{g}$ is semi-simple.
Applying this to the Lorentz group, we have $\pi_0(G) = \mathbb{Z}_2\times\mathbb{Z}_2$ as the group of space parity and time reversal. It is Abelian, and one can show that the Lorentz group is indeed the semi-direct product of this and its connected component by explicitly writing out its matrix structure. Therefore, all finite-dimensional representations of the Lorentz group are completely reducible, and its irreducible representations can be given by an irreducible representation of the proper orthochronous Lorentz group $(V,\rho)$ + an irreducible representation of time reversal (which under the tensor product of the induced representation becomes just some operator on $V$ that squares to the identity) + an irreducible representation of space parity (likewise).
