Finite lorentz transform for 4-vectors in terms of the generators One or two sets of notes (one of them by Timo Weigand) on QFT that I have come across state explicitly that a finite lorentz transformation for 4-vectors can be written in terms of the generators $J^{\rho\sigma}$ as:
$$\Lambda^{\mu}_{\,\nu}(\Omega)=\lim_{N\rightarrow \infty}(\delta^{\mu}_{\,\nu}-\frac{i}{2N}\Omega_{\rho\sigma}(J^{\rho\sigma})^{\mu}_{\,\nu})^{N}=\exp{(-\frac{i}{2}\Omega_{\rho\sigma}(J^{\rho\sigma}))^{\mu}_{\,\nu}}$$
Now, my question is that, isn't it required of the generators of a Lie group to commute(or equivalently for the group to be abelian) in order to be able to represent it as an exponential of its generators in the finite case? This follows from the condition that $e^{A+B}=e^{A}e^{B}$ only when $[A,B]=0$ (in this case, aren't we simply constructing the finite exponential by multiplying the infinitesimal exponential terms together and then assuming that the generators commute?). If so, than why can the above be written in the way it is given that the generators(J) do not commute. What am I missing?
 A: No, it's not required. You do need the generators to commute if you want to split the exponential as a product of exponentials, but there's no need for the latter. The linear combination $\Omega_{\rho\sigma}J^{\rho\sigma}$ is a perfectly valid four-by-four matrix, no matter what is coefficients or the commutation properties of its internal components, and it's perfectly OK to explore its matrix exponential. 
A: The extreme RHS exponential is correct.  This is exactly the way to write a Lorentz transformation for any size parameters $\Omega$.  It is also true for making the matrices of other representations, not just the 4x4 matrices which rotate 4-vectors. For the M x M representation, each of the 16 generators $J^{\rho \sigma}$ would be a M x M matrix.  Actually there are not 16 but only 6 generators for the Lorentz Group because the array of parameters is antisymmetric $\Omega _{\rho \sigma}=-\Omega_{\sigma \rho}$ and picks out only the antisymmetric set of generators $J^{\rho \sigma}=-J^{\sigma \rho}$  .
The limit expression is not written correctly. It should be written
$$\Lambda^{\mu}_{\,\nu}(\Omega)=[\lim_{N\rightarrow \infty}(I-\frac{i}{2N}\Omega_{\rho\sigma}J^{\rho\sigma})^{N}]^{\mu}_{\,\nu}=\exp{(-\frac{i}{2}\Omega_{\rho\sigma}J^{\rho\sigma})^{\mu}_{\,\nu}}$$
where I is the 4x4 identity matrix diag(1,1,1,1).  Now matrices are being multiplied together and not just the $\mu \nu $ elements.
Notice that each of the N terms in the limit product is exactly the same, and therefore commutes with all the other terms and your worry is resolved.
$$\Lambda^{\mu}_{\,\nu}(\Omega)=[\lim_{N\rightarrow \infty}(e^{-\frac{i}{2N}\Omega_{\rho\sigma}J^{\rho\sigma}})^{N}]^{\mu}_{\,\nu}=\exp{(-\frac{i}{2}\Omega_{\rho\sigma}J^{\rho\sigma})^{\mu}_{\,\nu}}$$
A: The equation that you have used is based on the simple limit
$$\lim_{n\rightarrow\infty}\bigg(1+\dfrac{x}{n}\bigg)^n=e^x.$$
This limit has nothing to do with the details of what $x$ is. To answer your question, usually any operator that is connected to the identity operator (${\delta^{\,\mu}}_{\nu}$ in this case) can be written as an exponential in terms of its generators because the above limit can be used. 
