# In what sense are fields representations of the Poincare group?

As far as I know, a representation is a homomorphism from the group to a vector space $$V$$ which preserves the group multiplication, i.e., if $$(\pi,V)$$ is a representation of the group $$G$$, then whenever $$g_1,g_2\in G$$ and $$g_1\cdot g_2 = g_3$$ then

$$$$\pi(g_1)\pi(g_2) = \pi(g_1\cdot g_2).$$$$

In Matthew Schwartz's QFT and the Standard Model, chapter 8 titled Spin 1 and Gauge Invariance, he says (page 110) that fields $$\phi(x)$$ form a representation of the group of translations because $$\phi(x) \to \phi(x+a)$$. I did not understand how this works and how one would define the group multiplication law.

After some scrounging and stumbling across Peter Woit's book, I found the following definition of a representation on a functional space (page 7, eq. (1.3)): given a group action of $$G$$ on a space $$M$$ (given as a set of points), functions on $$M$$ come with an action of $$G$$ by linear transformations $$\pi$$ given by $$$$(\pi(g))f(x) = f(g^{-1} \cdot x).$$$$ This made me conclude that it is better to think of the fields as being the 'vector' space on which the linear transformations associated with the group acts. Therefore since this 'vector space' is infinite-dimensional, I would think that it has infinite degrees of freedom. But what Schwartz says the problem is that the degrees of freedom are the indices that the fields carry (i.e. a vector field $$A_\mu(x)$$ has 4 d.o.f., tensor field $$T_{\mu\nu}$$ have $$4\times 4 = 16$$ and so on) and the problem is to embed the proper spin $$J$$ d.o.f. which are $$2J+1$$ in number into these fields. I do not understand how the indices are actually the d.o.f. but not the fields themselves. Can someone explain?

• Is this of any help? physics.stackexchange.com/q/349166/226902 "How are quantum fields related to representations of the Lorentz group?" See also physics.stackexchange.com/q/127989/226902 Commented Jan 31 at 13:11
• A first problem is that your definition of representation is sloppy, and I don't say this to play smart pants, but wondering whether the confusion you have arises from a misconception. Fields are not representations, but rather form a vector space $V$, over which the Poincaré group $G$ is represented through $\rho: G\to \operatorname{Aut}V$. Fields must live in a representation of the group by physical assumption: in flat spacetime, physics must remain unchanged under Poincaré transformations, so the fields describing this physics must transform under a definite representation of that group. Commented Jan 31 at 13:41
• @Albert I said in later in my post that '...fields as being the 'vector' space on which...' Commented Jan 31 at 13:57
• My confusion was why are the spacetime indices considered to be the d.o.f. instead of the fields themselves. Commented Jan 31 at 13:57
• You are welcome, @QFTheorist. It is important to remember that fields are operators acting on the Fock space where the (multi-)particle states live; ideally, and put simply, one tries to conceive a QFT in such a way that free particle states are created by the action of the field operators on the vacuum state. Commented Jan 31 at 15:24

Schwartz is trying to explain how $$e.g.$$ the $$3$$ irreducible d.o.f. of a spin-1 particle can be embedded into the 4 reducible d.o.f. of a spin-1 field or $$e.g.$$ the 1 irreducible d.o.f. of a spin-0 particle can be embedded into the 1 reducible d.o.f. of a spin-0 field.