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I know that in conformal field theories conformal group acts not by pushforwards but (e.g. for scalar field $\phi$ with conformal dimension $\Delta$) $$ \phi(x) \mapsto \phi'(x') = \left| \frac{\partial x^{' \mu}}{\partial x^\nu} \right|^{-\frac{\Delta}{d}}(x) \phi(x)\,. $$

On the other hand, I know how conformal algebra acts on fields at the origin. Indeed, finite dimensionality of the ''index space'', Schur's lemma and conformal algebra structure fix action to $$ \mathcal K_\mu \phi(0) = 0\,,\quad \mathcal D \phi(0) = \Delta \phi(0)\,,\quad \mathcal M_{\mu \nu} \phi(0) = R_{\mu \nu} \phi(0)\, $$ where $R_{\mu \nu} = r(\mathcal M_{\mu \nu})$ for some irrep $r$ of $\mathfrak{so}$. It is also natural to demand $\mathcal P_\mu \phi(x) = \partial_\mu \phi(x)$.

Here I denoted conformal Killing vectors by calligraphic letters: \begin{gather} \mathcal P_\mu = \partial_\mu,\;\; \mathcal M_{\mu \nu} = x_\mu \partial_\nu - x_\nu \partial_\mu,\\ \mathcal D = x^\mu \partial_\mu,\;\; \mathcal K_\mu = 2 x_\mu x^\nu \partial_\nu - x^2 \partial_\mu. \end{gather} I want to derive the first formula by exponentiating Lie algebra but I get some nonsense even for dilations. Using Hausdorff formula I get $$ \mathcal D \phi(x) = (\Delta + x \cdot \partial) \phi(x)\,. $$ I expect that $e^{t \mathcal D}$ correspond to dilatation $x \mapsto x' = e^{t} x$. So I should have $$\phi(x) \mapsto \phi'(x') = e^{- t \Delta} \phi(x).$$ But instead I have \begin{multline} e^{t \mathcal D} \phi(x) = e^{t \Delta} e^{t x \cdot \partial} \phi(x) = e^{t \Delta} \lim_{n \to \infty} \left( 1 + \frac{t x \cdot \partial}{n} \right)^n \phi(x)\\ = e^{t \Delta} \lim_{n \to \infty} \phi\left( \left( 1 + \frac{t}{n} \right)^n x \right) = e^{t \Delta} \phi\left( e^{t} x \right) = e^{t \Delta} \phi(x'). \end{multline} And if the overall multiple can be fixed by some minus, the primed argument is the issue.

Situation becomes even worse in quantum theory. In QFT we consider the complexification of conformal algebra with basis \begin{equation} P_\mu = - i \mathcal P_\mu \,, \quad M_{\mu \nu} = i \mathcal M_{\mu \nu} \,, \quad D = - i \mathcal D \,, \quad K_\mu = - i \mathcal K_\mu \,. \end{equation} Their commutators are written down in https://en.wikipedia.org/wiki/Conformal_symmetry#Commutation_relations

Conformal algebra acts on the space of operators by commutators such that (consider primary operator) \begin{equation} [K_\mu, \mathcal O(0)] = 0,\quad [D, \mathcal O(0)] = \Delta \mathcal O(0),\quad [M_{\mu \nu}, \mathcal O(0)] = R_{\mu \nu} O(0). \end{equation} We still want $[\mathcal P_\mu, \mathcal O(x)] = \partial_\mu \mathcal O(x)$ so $P_\mu$ acts like momentum operator in coordinate representation: $[P_\mu, \mathcal O(x)] = - i \partial_\mu \mathcal O(x)$.

Analogously, using Hausdorff formula we get $\mathcal O(x) = e^{x \cdot \mathcal P} \mathcal O(0) e^{-x \cdot \mathcal P}$ and $$ [D, \mathcal O(x)] = (\Delta + i x \cdot \partial) \mathcal O(x)\,. $$ Now we have $i$ and I don't see such nice trick as above with ''putting derivatives as arguments inside function'' (please don't say: ''Don't mind: analytical continuation, Wick rotation...'').

To summarize. How can finite form of conformal transformations can be derived from the infinitesimal one? There are some words in Simmons-Duffin: https://arxiv.org/abs/1602.07982 but I find them completely unconvincing. Di Franchesco says that we can do it and not doing it :)

Conformal Killing vectors commutation relations: \begin{align} [\mathcal M_{\mu \nu}, \mathcal M_{\rho \lambda}] &=\eta_{\mu \lambda} \mathcal M_{\nu \rho} + \eta_{\nu \rho} \mathcal M_{\mu \lambda} - \eta_{\mu \eta} \mathcal M_{\nu \lambda} - \eta_{\nu \rho} \mathcal M_{\mu \rho}\,,\\ [\mathcal M_{\mu \nu}, \mathcal P_\lambda] &= \eta_{\nu \lambda} \mathcal P_\mu - \eta_{\mu \lambda} \mathcal P_\nu\,,\\ [\mathcal D, \mathcal P_\mu ] &= - \mathcal P_\mu\,,\\ [\mathcal D, \mathcal K_\mu ] &= \mathcal K_\mu\,,\\ [\mathcal M_{\mu \nu}, \mathcal K_\lambda] &= \eta_{\nu \lambda} \mathcal K_\mu - \eta_{\mu \lambda} \mathcal K_\nu\,,\\ [\mathcal P_\mu, \mathcal K_\nu ] &= 2 \eta_{\mu \nu} \mathcal D + 2 \mathcal M_{\mu \nu}\,. \end{align}

Edit. It seems that I actually performed Simmons-Duffin argument for dilations. And it seems that he used another definition of action of conformal group: for $x \mapsto x'$ $$ \phi(x) \mapsto \phi'(x') = \left| \frac{\partial x^{' \mu}}{\partial x^\nu} \right|^{\frac{\Delta}{d}}(x) \phi(x') $$ (according to formula (55), which is just the answer to (kind of) my question, without derivation (there is an argument about factorization of infinitesimal Jacobi matrix (formula 25) but I don't see how to use it here)). But there is an other nice text: https://arxiv.org/abs/1805.04405 which, on one hand, relies on Simmons-Duffin in that, but on the other hand they use another representation (formula (18))...

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  • $\begingroup$ The aggressive tone in your answer is unnecessary. Instead of trying to understand pedagogical texts written by serious people, you have decided that their arguments are "completely unconvincing" but that your own formulas must be correct. Good luck with that... $\endgroup$ – Hans Moleman Jun 18 at 17:58
  • $\begingroup$ @Hans All I meant is that I was not convinced with that. That's shame on me. English is not my native language. Sorry if it sounded aggressively. I didn't mean any. $\endgroup$ – vanger Jun 18 at 18:06
  • $\begingroup$ Your first problem is that you are conflating a bunch different conventions together. You should understand that different sources define things in different ways, and you should not mix equations. Furthermore, I advise you to not use "$\phi(x)\mapsto ...$" to describe symmetry transformations. This is not an equation, so it is very easy to confuse yourself, there are many questions on this site caused by precisely this. You should always write $A=B$ and understand precisely what is in the left hand side and what is in the right... $\endgroup$ – Peter Kravchuk Jun 20 at 23:16
  • $\begingroup$ ... For example, what is $\mathcal{D}\phi(x)$? What is $\mathcal{D}$? What, mathematically, does it mean to write it to the left of $\phi(x)$? Why do you care what this is equal to? Why you write in some places $\mathcal{D}\phi(x)$ and $[D,\phi(x)]$ in others? Is $D=-i\mathcal{D}$ an operator on Hilbert space? Why then you write $\mathcal{D}=x_\mu \partial^\mu$? $\endgroup$ – Peter Kravchuk Jun 20 at 23:19
  • $\begingroup$ I suggest to read carefully Simmons-Duffin lectures, because he is careful in defining these things. If you are then confused by his derivation of finite transformation, you can ask a question about that. $\endgroup$ – Peter Kravchuk Jun 20 at 23:21

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