I've just started working through "Supersymmetric Gauge Field Theory and String Theory" by D. Balin and A. Love and I got a question to something right at the beginning.

They are revisiting the Poincare-Group and give the following construction of the generators:

For an infinitesimal translation it is

$$x'^\mu = x^\mu + a^\mu = x^\mu-ia_\lambda(P^\lambda)^\mu$$ and from this they conclude that the generator for translations is given by $$(P^\lambda)_\mu = i \delta^\lambda_\mu.\tag 1 \label{1}$$ I'm a little confused by this. Aren't the finite translations given by $$x'^\mu = \text{exp}(-i a_\lambda (P^\lambda)^\mu_\nu) x^\nu$$ and shouldn't the generator therefore read $$(P^\lambda)^\mu_\nu = i \delta^\mu_\nu \partial^\lambda \tag{2}$$ In other words, why is the generator not an operator valued matrix (as it would clearly be in my case.)

For Lorentz transformations they give the derivation $$x'^\mu = x^\rho + \omega^\rho_\sigma x^\sigma =x^\rho-\frac{1}{2} \omega_{\mu\nu}(M^{\mu\nu})^\rho_\sigma x^\sigma$$ and conclude $$(M^{\mu \nu})^{\rho \sigma} = i (\eta^{\mu \rho}\eta^{\nu \sigma} - \eta^{\mu \sigma}\eta^{\nu \rho}),\tag 3$$ which I understand a little better, as it at least makes sense with my understanding of generators. But with those two definitions, there is no way to make the commutator relation $$[M^{\mu \nu}, P^{\lambda}]=i (\eta^{\nu \lambda} P^\mu -\eta^{\mu \lambda}P^\nu) \tag 4$$work out. How would you even define this commutator, given that the $P^\mu$ in (1) has only one internal index?

On the other hand, to fulfill the commutation relation with the operator in (2) I think you also need the spacetime representation of the Lorentz generator, which is the typical $$L^{\mu\nu} = i(x^\mu \partial^\nu - x^\nu \partial^\mu).\tag 5$$

So I guess, my question boils down to: if (2) and (5) actually fulfill (4), then what is the translation generator corresponding to (3)? Which one of those representation is actually the action of the Poincare group on spacetime? I kind of want to say both but I'm just not sure.



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