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.


1 Answer 1


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.

  • $\begingroup$ what would formally be the space $\mathcal{O}$? How do we put a vector (Hilbert) space structure on it, to be able to tensor product it with V, which is a finite dimensional vector space (the rep space)? $\endgroup$
    – ProphetX
    Commented Jun 29, 2022 at 18:07
  • 1
    $\begingroup$ @ProphetX I was very deliberately vague about that because physics is never really careful about that, so that depends on your specific formal approach. It's not really relevant for your question, though: Whatever the space of spinless fields is, tensoring it with the finite-dimensional representation of spin $(s_1,s_2)$ yields the space of fields of spin $(s_1,s_2)$. Asking what legitimate choices for $\mathcal{O}$ are is a different question ;) $\endgroup$
    – ACuriousMind
    Commented Jun 29, 2022 at 18:21
  • $\begingroup$ So to be really strict, when physics books say 'fields live in a finite-dimensional rep space of the Lorentz group', they really do mean only the specific spin part, and forget (the possible infinite dimensional) space it is tensored with, which you mentioned, right? My point is one of the confusions I had/have is why is the rep space finite dimensional per se. I think it can't be. Same as in QM, the only systems that have finite dim Hilbert space are pure spin systems. They would be the only ones in QFT too? $\endgroup$
    – ProphetX
    Commented Jun 29, 2022 at 19:47
  • $\begingroup$ @ProphetX Again, the space that is finite-dimensional is the target space of the fields, not the state space. Classically, a field that "transforms in the finite-dimensional rep $V$" is a function $\mathbb{R}^n\to V$. $\endgroup$
    – ACuriousMind
    Commented Jun 29, 2022 at 19:56

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