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In [1], the successive finite Lorentz transformations ($\Lambda_1$ after $\Lambda_2$) on a classical scalar field is defined as

\begin{align*} \phi(x) \overbrace{\longrightarrow}^{\Lambda_2} \phi'(x) := U(\Lambda_2) \phi(x) \overbrace{\longrightarrow}^{\Lambda_1} \phi''(x) := U(\Lambda_2)U(\Lambda_1) \phi'(x) = U(\Lambda_2) \phi(\Lambda_1 x) = \phi(\Lambda_1 \Lambda_2 x)\tag{1.43} \end{align*}

Here the convention is unfamiliar with me, but it is assumed that the symmetry transformations act on fields directly. On the other hand, if a set of coordinate $x$ is transformed under $\Lambda_2$ and $\Lambda_1$, $x$ should be $x \rightarrow x' = \Lambda_1 \Lambda_2 x$. Hence the most right hand side should be $\phi(\Lambda_1 \Lambda_2 x)$ as written.

With this unfamiliar (at least to me) convention, there is an exercise (Exercise 1.7) whose statement is (equivalent to)

... Specifically, use the definition $$U(\Lambda) =e^{-\frac{1}{2} \lambda^{\rho \sigma} L_{[\rho \sigma]}}\tag{1.40}$$ to show that the order $\lambda_1 \lambda_2$ terms in the product $U(\Lambda_2) U(\Lambda_1) \phi$ of differential operators acting on $\phi$ agrees with therms of the same order in $\phi(\Lambda_1 \Lambda_2 x)$. ...

Here $L_{[\rho \sigma]} = x_\rho \partial_\sigma - x_\sigma \partial_\rho$ is a generator and $\lambda^{\rho \sigma}$ is an real anti-symmetric parameter.

However I have no idea for this problem. Indeed, the product $U(\Lambda_2) U(\Lambda_1)$ is expanded as

\begin{align*} U(\Lambda_2) U(\Lambda_1) &= \left(1 - \frac{1}{2} \lambda_2^{\rho \sigma} L_{\rho\sigma} + \frac{1}{4} \lambda_2^{\mu \nu} L_{\mu \nu} \lambda_2^{\rho \sigma} L_{\rho \sigma} + {\cal O}(\lambda^3)\right) (\lambda_2 \leftrightarrow \lambda_1)\\ &\supset x_\rho {{\lambda_2}^\rho}_{\mu} {{\lambda_1}^\mu}_{\nu} \partial^\nu \phi (x) + x_\rho x_\mu {\lambda_2}^{\rho \sigma} {\lambda_1}^{\mu \nu} \partial_\sigma \partial_\nu \phi(x) \end{align*}

hence this is inconsistent with the taylor expansion of $\phi(\Lambda_1 \Lambda_2 x)$ where the order is $\lambda_1 \lambda_2$.

It is obvious that I don't understand the convention correctly, but where is a mistake?

Also, if there is any advantage to following this convention, it would be helpful if you could clarify it.

[1]: D. Z. Freedman and A. Van Proeyen, "Supergravity", https://www.cambridge.org/core/books/supergravity/B7EEC3E37A39AB6E6625850857B96AA7.

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  • $\begingroup$ Your second term $x_\rho x_\mu \lambda_2^{\rho\sigma} \lambda_1^{\mu\nu} \partial_\sigma \partial_\nu \phi$ cannot be used to say anything about the order because it goes to itself under $1 \leftrightarrow 2$. The first term looks wrong because of the free indices. $\endgroup$ Apr 6, 2022 at 11:10
  • $\begingroup$ @ConnorBehan Sorry, corrected the free indice. The second term says nothing exactly.. actually this term vanishes under a commutation relation $[U(\Lambda_1), U(\Lambda_2)]$, and this should be another $U(\Lambda_3)$ because of the underlying Lie algebra. However I cannot figure out the correct answer itself. $\endgroup$
    – Keyflux
    Apr 6, 2022 at 11:43

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