In the book by Di Franceso, Mathieu, Senechal, equation (5.46) shows that (assuming $\bar\epsilon = 0$) $$ (*) \qquad \langle \delta_{\epsilon, 0} \mathcal{O}\rangle = - \oint_\infty \frac{dz}{2\pi i z}\epsilon(z) \langle T(z)\mathcal{O}\rangle, $$ where on the left is the variation of $\mathcal{O}$ under an arbitrary local conformal transformation with param $\epsilon$. If I understand correctly, this equation starts from the three Ward identities eq (5.32) for the global conformal symmetry, then through some tricks around eq (5.42), an arbitrary conformal transformation parameter $\epsilon$ is brought in to arrive at the above equation, which seems to be much powerful than the original 3 identities.

Now, in the note Conformal field theory on the plane by Ribault, equation (2.2.13) writes for any meromorphic $\epsilon(z)$ that has no poles outside $z_1, ..., z_N$, $$ (**) \qquad 0 = \oint_\infty dz \epsilon(z) \langle T(z) \mathcal{O}(z_1, ..., z_N)\rangle, \qquad \text{if } \epsilon(z \to \infty) = O(z^2) \ . $$

In the two Ward identities, the right hand side looks the same, but the left do not.

I wonder

(1) Is the $(*)$ equation more powerful than the original Ward identities eq (5.32) for global conformal symmetry?

(2) Are the two statements $(*), (**)$ the same, or complimentary, or one is the special case of another?

In other words, what is the most general $\epsilon$ such that $\langle\delta_\epsilon \mathcal{O}(z_1, ..., z_N) \rangle = 0$ and how to see that? To begin, $\epsilon(z) = a + bz + cz^2$ is a solution, since it corresponds to the global conformal transformations, the "true symmetry of the system". But from $(**)$ it seems to suggest that any things like $\epsilon(z) = \frac{1}{z - z_1} + \frac{3.14}{(z - z_N)^{2023}} + 1000 z^2$ will set $\delta_\epsilon \langle \mathcal{O}(z_1, ..., z_N)\rangle = 0$, but I don't quite understand why.

(3) In Ribault, also considering the two bullets point after $(**)$, it seems that only $\epsilon(z \to \infty) = O(z^2)$ and $\epsilon(z \to \infty) = O(1/z)$ are considered. What about $\epsilon(z \to \infty) = O(z^{1000})$? Is there any thing one can say about such $\epsilon$? (Below is a snapshot)

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1 Answer 1


Your first equation is more general, and can be obtained from the infinitesimal change of variables in the path integral. The second equation is a special case where the integrand has no poles, hence the infinite contour integral is 0. Thus, in my opinion, both are consistent. The second equation simply tells you that $\delta_\epsilon \langle O \rangle = 0$ when $\epsilon \sim z^2$ at $z\to \infty$.

The two equations are complementary. The first one is the important property, and the second one tells you that in some special case, it vanishes. But this is not a simple application of the first equation, because one needs to understand the pole structure of the integrand to claim such a property.

  • $\begingroup$ So you mean the condition $\epsilon(z \to \infty) \to z^2 $ is equivalent to saying that $\epsilon$ corresponds to a global conformal transformation, and therefore $\delta_\epsilon \langle \mathcal{O}\rangle = 0$ due to the global conformal invariance? Is it ok to also consider things like $\epsilon(z) = z^2 + 1/(z - z_1)$? thanks. $\endgroup$
    – user060606
    Commented Jul 16, 2023 at 22:10
  • $\begingroup$ What do you mean by global? Here I guess the conformal transformation is an infinitesimal local function of $z$ $\endgroup$ Commented Jul 16, 2023 at 22:18
  • $\begingroup$ Oh I mean the infinitesimal global conformal transformations (translation/rotation/dilatation/...) correspond to param $\epsilon = a + bz + cz^2$; like the bullet points below $(**)$ in the note. I want to know whether $\epsilon(z \to \infty) = O(z^2)$ is equivalent to "setting $\epsilon$ generate the 6 transformations". $\endgroup$
    – user060606
    Commented Jul 16, 2023 at 22:23
  • $\begingroup$ Sorry, I guess I misunderstand the symbol $O(z^2)$: how does $z^2 + 1/(z - z_1) \sim 1/z$ at infinity? To me, the $z^2$ term dominates, and should $= O(z^2)$. Correct me if I'm wrong. $\endgroup$
    – user060606
    Commented Jul 16, 2023 at 22:25
  • $\begingroup$ I think you're right, but it makes no sense to me, because $\epsilon(z)$ could take very negative powers of z, and be mulplied by the taylor expansion of the OPE between $T$ and the operator which in general never truncates, so the full integrand would have a non-zero pole yielding a non-zero (in general) result $\endgroup$ Commented Jul 16, 2023 at 22:38

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