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).

  • 1
    $\begingroup$ I have heard much about this book, but not yet had the chance to work through it. As a physicist we mostly deal with irreducible representations, so that's likely the implicit label missing here. bohr.physics.berkeley.edu/classes/221/1011/notes/wigeck.pdf $\endgroup$
    – ThomasTuna
    Feb 3, 2021 at 20:48
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    $\begingroup$ I think a mathematician could describe the representations of the field as representations induced from finite-dimensional irreducible representations of the Lorentz subgroup. Basically you want your field to be $\phi^a(x)$, where $x\in \mathbb{R}^d$ and $a$ is in index in a finite-dimensional Lorentz irrep. Translations translate $x$ and Lorentz transformations rotate $x$ and $a$. There are no fundamental restrictions on the finite-dimensional Lorentz irrep in which $a$ lives, but some occur more often than others (scalar=trivial, traceless-symmetric tensors, 2-forms). $\endgroup$ Feb 3, 2021 at 21:29
  • $\begingroup$ The words that I said about induced representations are perhaps not exactly correct, so don't take them too far. The latter part of the above comment is what you're looking for. The $\phi^a(x)$ ends up being an operator-valued distribution, and the action of Poincare group on it is defined as I described. What particular representation this ends up being -- this is another question. $\endgroup$ Feb 3, 2021 at 21:32
  • $\begingroup$ Thank you Peter for your comment! I think I am beginning to understand a lot of things at once. $\endgroup$
    – Plop
    Feb 3, 2021 at 23:10
  • $\begingroup$ @ThomasTuna, why physicists like so much irreducible representations will be a future question, hahaha! $\endgroup$
    – Plop
    Feb 3, 2021 at 23:11

2 Answers 2


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.

  • $\begingroup$ So the finite-dimensional $P$ representation can be complex-valued? $\endgroup$
    – Plop
    Feb 4, 2021 at 10:06
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    $\begingroup$ @Plop if your theory is a theory of a complex-valued field, yes. $\endgroup$ Feb 4, 2021 at 11:56

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.

  1. 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 $\Gamma(\sigma_i)^{-1}=\Gamma(\sigma^{-1})$.


  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)


  1. 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).

$$\text{CITATION: https://www.sciencedirect.com/topics/mathematics/reducible-representation}$$

  • $\begingroup$ Good answer, but not the answer to the question that op asked. Consider asking a separate question and answering your own question (which is permitted on this site) $\endgroup$ Feb 5, 2021 at 3:27
  • $\begingroup$ haha I actually thought that you had given a good answer, but not o the one he had asked. Funny how that works. $\endgroup$
    – ThomasTuna
    Feb 5, 2021 at 3:38

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