Quantum Fields living in finite dimensional non-unitary irreducible representations of the Lorentz group In Non-unitary, finite dimensional representations of the Lorentz group it got clarified that the finite dimensional non-unitary reps of the Lorentz group are completely reducible. In physics, we use them to describe the representations of fields. Without going in the details as of why, since the post would be long, I claim that the irreps can be labelled by $(m,n)$, where $m,n$ are (half)integers since the Lorentz algebra is isomorphic to $su(2)\oplus su(2)$.
Now, I have a mathematical question. The above, namely that irreps can be labelled by $(m,n)$ would imply the representation spaces are finite dimensional. This is kind of counterintuitive to me and hard to understand, even if I see how It works using the rep theory argument. My concern is:A representation mathematically is a group homomorphism:$SO(3,1)\to GL(V)$, where $V$ is some vector space. Using the above, $V$ must be finite dimensional. But: up to my understanding in $V$ we have a list of ($\phi_i$), where $\phi$ are the field operators, which are mathematically operator valued distributions. So the main question is how can one mathematically precisely put a finite dimensional vector space structure on a space, whose elements are operator valued distributions, noting that the operators, where the distribution takes value could be possibly unbounded?
Edit: I am aware of a similar question being discussed on other posts on SE, but the explanations/questions there were not focusing on the mathematical aspect, rather the physical one.
 A: The finite-dimensional representation is on the target space of the classical fields, not on the space of fields itself. Of course, the "target space" of an operator-valued distribution is a bit annoying to formulate, so lets just do this differently:
Formally, there is some space of operator-valued distributions, let's call it $\mathcal{O}$. The quantum fields now should simply be thought of to take values in $V\otimes \mathcal{O}$, where $V$ is the finite-dimensional Lorentz representation from the question. Then in terms of a basis $v_i$ of $V$ we have that
$$ \phi = \sum_i v_i \otimes \phi_i,$$
where now the $\phi_i\in \mathcal{O}$ are "the components of the field in the basis $v_i$". Due to the tensor product, the finite-dimensional representation on $V$ extends to the infinite-dimensional $V\otimes \mathcal{O}$.
Note that this is the same logic we use for wavefunctions with spin in ordinary quantum mechanics - the state space of spinless objects is $L^2(\mathbb{R}^n)$ and formally making the wavefunction into a spinor with components is just tensoring that space with the corresponding spin representation.
