Wightman axiom 2: what kind of representation? Both in Wikipedia and on page 98 of Streater, Wightman, PCT, Spin and Statistics and all that, the second axiom postulates that a field must transform according to a representation of the Poincaré group.
I am a mathematician and I wonder if there are implicit assumptions there. Is any representation of the Poincaré group acceptable? Should such a representation be real-valued (that is, if $\rho$ is such a representation, for any element $g$ of the Poincaré group, is $\rho(g)$ a real-valued matrix)? Should it be orthogonal, unitary (that is, should the aforementioned matrices be orthogonal, unitary)?
EDIT: Let me rephrase my question.
As far as I understand, axiomatically,

*

*a QFT should come with a (strictly speaking, projective, but I'm not sure it's relevant here) unitary representation $U$, that is, a continuous morphism from the Poincaré group to the group of unitaries of the Hilbert space;

*a $n$-dimensional vector-valued field is described by an $n$-tuple of maps $\phi := (\phi_1,\cdots,\phi_n)$ from the Minkowski space to the set of operators on the Hilbert space (I also knew that, strictly speaking, we should consider distributions instead of maps but I don't think it is relevant here);

*now, under the action of a symmetry (that is, under conjugation by a unitary -from the unitary representation $U$) each coordinate of $\phi$ becomes a linear combination of all the coordinates, and the coefficients are stored in a matrix that Streater and Wightman call $S$ (equation 3-4 in Streater-Wightman, page 99).

This $S$ is a morphism from the Poincaré group to the group of square invertible complex matrices of size $n$, so, as a mathematician, I also call $S$ a (finite-dimensional) representation of the Poincaré group.
My question is: is there any implicit assumption on $S$?
I think my question is motivated by my fear of coordinates (I don't like the idea that a field $\phi$ should be implemented as a tuple; it looks that we are making an arbitrary choice of coordinates).
 A: The value of the classical field at a point in spacetime $\phi(x)$ may transform under any finite-dimensional representation, not necessarily unitary or orthogonal etc.
But the quantum field as an operator-valued distribution transforms under an infinite-dimensional unitary representation that acts on the Hilbert space of the QFT.
Wightman axioms relate the two representations, postulating that
$$ U(\Lambda) \phi(f) U(\Lambda)^{\dagger} = P(\Lambda) \phi (S(\Lambda^{-1}) f). $$
Here $U(\Lambda)$ is the infinite-dimensional unitary representation, $P(\Lambda)$ is the finite-dimensional representation that acts on the classical field's value at a point, and $S(\Lambda)$ is the natural representation that acts on test functions over spacetime.
A: Well the answer here is that the author is talking about Irreducible representations (Irreps).
But @Plop you have asked me a great question in the comments, "why physicists like so much irreducible representations?". So I done my best to answer it.
TLDR: Physicists primarily deal with Lie Groups (Poincare group is a Lie Group) and there is a theorem which states that "If the Lie  group representation isn't already irreducible than it can be "completely reduced" into a collection of irreducible representations."
So by the nature of our math tools we MUST be using Irreps
$$----------------- \text{Long Answer} -----------------$$
I studied the Burau representation in undergrad (https://en.wikipedia.org/wiki/Burau_representation) and then quantuum chromodynamics in grad school, so I'll try to connect express how I have connected those two experiences.

*

*What is representation theory?
A represenation of a group $G$ is a homomorphic mapping of the group $G$ onto a non-singular group of $d \times d$ matrjces $\Gamma(T)$, where matrix multiplication is the group's multiplicative operation. (The group of matrices $\Gamma(T)$ forms a $d$-dimensional representation $\Gamma$ of group $G$)

Example I think that the unreduced Burau representation of the Braid Group $B_n$ (https://www.youtube.com/watch?v=uMMxD0Ak4lg) gives a great visual interpretation!
This representation maps the act of crossing two strands of hair (left over right) $\sigma_i$ onto the matrix,
$$ 
\begin{align}
(0) && \Gamma(\sigma_i) = 
\left[
\begin{array}{c|cc|c}
I_{i-1} & 0   & 0 & 0  \\ \hline
0       & 1-t & t & 0  \\
0       & 1   & 0 & 0  \\ \hline
0       & 0   & 0 & I_{n-i-1}
\end{array}
\right]
\end{align}
$$
By definition the representation connects to uncrossing two strands of hair (right over left) $\sigma_i^{-1}$ with matrix $\Gamma(\sigma_i)^{-1}=\Gamma(\sigma^{-1})$.
$$-------------------------------------------------------$$


*What are (ir)reducible representations?
Reducible representations have the form
$$
\begin{align}  \label{eq_reducible}
(1) && \Gamma(T) = \begin{bmatrix} \Gamma_{11}(T) & \Gamma_{12}(T) \\ 0 & \Gamma_{22}(T)  \end{bmatrix} 
\end{align}
$$.
Such that
$$
 \Gamma(T_1) \Gamma(T_2) =  \left[
\begin{array}{c|c}
\Gamma_{11}(T_1) \Gamma_{11}(T_2) & \Gamma_{11}(T_1) \Gamma_{12}(T_2) + \Gamma_{12}(T_1) \Gamma_{22}(T_2)   \\ \hline
0 & \Gamma_{22}(T_1) \Gamma_{22}(T_2)  
\end{array}
\right]
$$
We have arrived at the homomorphic property which defined a representation.
$$\Gamma_{11}(T_1 T_2)  = \Gamma_{11}(T_1) \Gamma_{11}(T_2) $$
$$\Gamma_{22}(T_1 T_2)  = \Gamma_{22}(T_1) \Gamma_{22}(T_2) $$
Therefore the representation $\Gamma$ contains representations $\Gamma_{11}$ and $\Gamma_{22}$ which are of a smaller dimension than $Gamma$.
Since these smaller representations exist within $\Gamma$, gamma is reducible.
An irreducible  group contains no such smaller dimensional representations.

Example
Fortunately for us, the Reducible Burau representation Eq.0 has this structure! Consider  $\Gamma(\sigma_1)$ and $\Gamma(\sigma_3)$ in $B_4$
$$ \Gamma(\sigma_1) \Gamma(\sigma_3) = 
\left[
\begin{array}{cc|cc}
1-t & t & 0 & 0  \\ 
1   & 0 & 0 & 0  \\ \hline
0   & 0 & 1 & 0  \\ 
0   & 0 & 0 & 1
\end{array}
\right]
\left[
\begin{array}{cc|cc}
1 & 0 & 0   & 0  \\
0 & 1 & 0   & 0  \\ \hline
0 & 0 & 1-t & t  \\ 
0 & 0 & 1   & 0
\end{array}
\right]
= 
\left[
\begin{array}{cc|cc}
1-t & t & 0   & 0  \\
1   & 0 & 0   & 0  \\ \hline
0   & 0 & 1-t & t  \\ 
0   & 0 & 1   & 0
\end{array}
\right]
$$
This is great and it indicates that the Unreduced Burau representation can be reduced! (https://en.wikipedia.org/wiki/Burau_representation#Explicit_matrices)
$$-------------------------------------------------------$$


*How does the process of Reduction Happen?
Suppose that $\Gamma_{11}(T)$ is itself reducible and be expressed in the same for as Eq.1. Fortunately you can always similarity transform $\Gamma(T)$ such that
$$S^{-1} \Gamma S = \Gamma' = 
\left[
\begin{array}{cccc}
\Gamma'_{11}(T) & \cdots & \cdots & \Gamma'_{1n}(T) \\ 
0               & \ddots & \cdots & \Gamma'_{2n}(T_2) \\  
\vdots          & 0      & \ddots & \vdots \\
0               & \cdots & 0      & \Gamma'_{nn}(T)
\end{array}
\right]$$
If this upper triangular representation can be "completely reduced" to a block diagonal  using another similarity transform, then $\Gamma(T)$ is a completely reducible representation.
There is a theorem which states that "if a Lie group is reducible it is completely reducible." - "Group Theory in Physics" Ch4-S4 J.F. Cornwell (1984).
$$\text{CITATION: https://www.sciencedirect.com/topics/mathematics/reducible-representation}$$
