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Let's say that I have finite chiral transform and I would like to show invariance of Dirac's Lagrangian when $m=0$ under it.

The chiral transform is defined as: $$\psi(x) \rightarrow \psi'(x) =e^{i \alpha \gamma_5} \psi(x)$$ where the Dirac's Lagrangian: $$L = \bar \psi (x) (i \hbar \gamma^{\mu} \partial_{\mu} - 0c) \psi(x)$$

If I consider infinitsmall transformation of above: $$\psi(x) \rightarrow \psi'(x) =(1 + i \alpha \gamma_5) \psi(x)$$ I obtain that Lagrangian transforms as: $$L \rightarrow L' = L + O(\alpha^2)$$ where $O(\alpha^2)$ is term with with order $\alpha^2$. Is it enough to say that Lagrangian is invariant under finite transformation?

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    $\begingroup$ Have a look at my answer here. The surprising thing is that you are not actually doing an approximation when you only consider the transformation up to first order, as you rigorously show invariance under a Lie algebra, which implies invariance under the corresponding Lie group. $\endgroup$ – ACuriousMind Mar 6 '15 at 21:11
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    $\begingroup$ @ACuriousMind - That is incorrect. Invariance under infinitesimal transformations only implies invariance under the connected part of the Lie group. $\endgroup$ – Prahar Mar 6 '15 at 21:32
  • $\begingroup$ @Prahar: Yes, correct. In the case where you speak of "infinitesimal" and "finite" transformations, though, it is implicit that the "finite transformations" are precisely those generated from the "infinitesimal transformations". (At least, it makes no sense to call a transformation a finite version of an infinitesimal one if it cannot reached by it) $\endgroup$ – ACuriousMind Mar 6 '15 at 21:42
  • $\begingroup$ @ACuriousMind - I'm not talking about that. Rather, I am speaking about "finite" v.s. "disconnected" transformations. For instance, parity and time reversal are elements of the Lorentz group that can never be generated from any amount of infinitesimal transformations. $\endgroup$ – Prahar Mar 6 '15 at 23:49
  • $\begingroup$ In fact, that is also incorrect in general. The exponential map taking the Lie algebra to the lie group need not be surjective for connected Lie groups. I think for most groups in physics it is surjective though. See this topic: math.stackexchange.com/questions/551548/… $\endgroup$ – Adomas Baliuka Dec 9 '16 at 21:51
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This is frequently good enough, but in your specific case its actually much easier to show this holds exactly in the finite case. I won't do this for you, but note that this is a global symmetry, i.e. alpha has no dependence on x. At most, you might also need to use the Baker-Campbell-Hausdorff formula.

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