In QFT, a representation of the Lorentz group is specified as follows:
$$
U^\dagger(\Lambda)\phi(x) U(\Lambda)= R(\Lambda)~\phi(\Lambda^{-1}x)
$$
Where $\Lambda$ is an element of the Lorentz group, $\phi(x)$ is a quantum field with possibly many components, $U$ is unitary, and $R$ an element in a representation of the Lorentz group.
We know that a representation is a map from a Lie group on to the group of linear operators on some vector space. My question is, for the representation specified as above, what is the vector space that the representation acts on?
Naively it may look like this representation act on the set of field operators, for $R$ maps some operator $\phi(x)$ to some other field operator $\phi(\Lambda^{-1}x)$, and if we loosely define field operators as things you get from canonically quantizing classical fields, we can possibly convince ourselves that this is indeed a vector space.
But then we recall that the dimension of a representation is simply the dimension of the space that it acts on. This means if we take $R$ to be in the $(1,1)$ singlet representation, this is a rep of dimension 1, hence its target space is of dimension 1. Then if we take the target space to the space of fields, this means $\phi(x)$ and $\phi(\Lambda x)$ are related by linear factors, which I am certainly not convinced of. EDIT: This can work if we view the set of all $\phi$ as a field over which we define the vector space, see the added section below.
I guess another way to state the question is the following: we all know that scalar fields and vector fields in QFT get their names from the fact that under Lorentz transformations, scalars transform as scalars, and vectors transform like vectors. I would like to state the statement "a scalar field transforms like a scalar" by precisely describing the target vector space of a scalar representation of the Lorentz group, how can this be done?
ADDED SECTION:
Let me give an explicit example of what I'm trying to get at:
Let's take the left handed spinor representation, $(2,1)$.
This is a 2 dimensional representation. We know that acts on things like $(\phi_1,\phi_2)$.
Let's call the space consisting of things of the form $(\phi_1,\phi_2)$ $V$. Is $V$ 2 dimensional?
Viewed as a classical field theory, yes, because each $\phi_i$ is just a scalar. As a quantum field theory, each $\phi_i$ is an operator.
We see that in order for $V$ space to be 2 dimensional after quantization, we need to able to view the scalar quantum fields as scalar multipliers of vectors in $V$. i.e. we need to view $V$ as a vector space defined over a (mathematical) field of (quantum) fields.
We therefore have to check whether the set of (quantum) fields satisfy (mathematical) field axioms. Can someone check this? Commutativity seem to hold if we, as in quantum mechanics, take fields and their complex conjugates to live in adjoint vector spaces, rather than the same one. Checking for closure under multiplication would require some axiomatic definition of what a quantum field is.
Best Answer
If $\mathcal H$ is the Hilbert space of the QFT, then \begin{align} U:\mathrm{SO}(3,1)\to \mathscr U(\mathcal H) \end{align} where $\mathscr U(\mathcal H)$ is the set of unitary operators on $\mathcal H$. In other words, $U$ is a unitary representation on the Hilbert space of the theory. If $V$ is the target space of the fields, then \begin{align} R:\mathrm{SO}(3,1)\to \mathrm{GL}(V), \end{align} where $\mathrm{GL}(V)$ is the vector space of invertible linear operators on $V$. In other words, $R$ is a representation on the target space of the fields in the theory. The mapping \begin{align} x\to \Lambda x \end{align} is the defining representation of $\mathrm{SO}(3,1)$ on $\mathbb R^{3,1}$. The statement that the field transforms in a particular way, namely the equation you wrote down, simply says that the action by conjugation of the Lorentz group on field operators by $U$ agrees with the composite of the representation on the target space and the inverse of the defining representaton.