I’m reading ‘Bombay Lectures on Highest Weight Representations of Infinite Dimensional Lie Algebras’ by Kac and Raina. And there is a following paragraph in the first chapter.
The Lie algebra $\operatorname{Vect}(S^{1})$ can be considered as the Lie algebra of the group $G$ of orientation preserving diffeomorphisms of $S^{1}$. If $\zeta_{1}$, $\zeta_{2}$ are two elements of $G$ then their product is defined by composition: $$\zeta_{1}\cdot\zeta_{2} = \zeta_{1}\big(\zeta_{2}(z)\big)$$ for each $z = \exp (i\theta)$ on $S^{1}$. If $f(z)$ is an element of the vector space of smooth complex-valued functions on $S^{1}$, then $\gamma\in G$ acts on $f(z)$ by $$\pi(\gamma) f(z) = f\big(\gamma^{-1}(z)\big).$$ This clearly defines a representation of $G$.
Probably this is a notation subtlety I missed on Lie Groups representations, but I can’t see why $\pi$ is a representation, as it switches the order of multiples: \begin{multline} \forall\xi_{1},\xi_{2}\in G: \pi(\xi_{1}\cdot\xi_{2})f(z) = f\big((\xi_{1}\cdot\xi_{2})^{-1}(z)\big) = f\left(\left(\xi_{2}^{-1}\cdot\xi_{1}^{-1}\right)(z)\right) = \\ = f\left(\xi_{2}^{-1}\left(\xi_{1}^{-1}\left(z\right)\right)\right) = \pi(\xi_{2})f\left(\xi_{1}^{-1}(z)\right)= \big(\pi(\xi_{2})\cdot\pi(\xi_{1})\big)f(z) \end{multline} Therefore $\pi(\xi_{1}\cdot\xi_{2}) = \pi(\xi_{2})\cdot\pi(\xi_{1})$ so it kind of works like an ‘anti’-representation. However if we denote \begin{equation} (\zeta_{1}\cdot\zeta_{2})(z) = \zeta_{2}\big(\zeta_{1}(z)\big) \end{equation} everything works as it should \begin{multline} \forall \xi_{1},\xi_{2}\in G : \pi(\xi_{1}\cdot\xi_{2})f(z) = f\big((\xi_{1}\cdot\xi_{2})^{-1}(z)\big) = f\left(\left(\xi_{2}^{-1}\cdot\xi_{1}^{-1}\right)(z)\right) = \\ = f\left(\xi_{1}^{-1}\left( \xi_{2}^{-1}(z)\right)\right) = \pi(\xi_{1})f\left(\xi_{2}^{-1}(z)\right) = \left(\pi(\xi_{1})\cdot\pi(\xi_{2})\right)f(z) \end{multline} therefore $\pi(\xi_{1}\cdot\xi_{2}) = \pi(\xi_{1})\cdot\pi(\xi_{2})$.
I’m not sure if it is a typo as I saw similar notation in other places and I do recall that a similar discussion was dedicated to the order of multiples in the case of regular representations for finite groups. But there it was more obvious. If $f:G\to \mathbb{C}$ we had three representations \begin{gather} \lambda_{g}f(h) = f(gh)\\ \rho_{g}f(h) = f\left(hg^{-1}\right)\\ \pi_{g}f(h) = f\left(ghg^{-1}\right) \end{gather} and it is straightforward why all of them are well-defined.
The more I think in which order should I multiply $\xi_{i}$, $\xi_{i}^{-1}$ and $\pi(\xi_{i})$ the more clueless I get. Could anyone clarify this concept, please!
Introducing extra notation for the intermediate steps of the computation and worrying about the point evaluations at $z\in S^1$ separately might make things clearer.
Let $g:=\pi(\xi_2)f$. Applying the definition of the action, we obtain at each $z\in S^1$ $$\pi(\xi_1)g(z) = g(\xi_1^{-1}(z)).$$ On the other hand, by definition of the action we have at each point $w\in S^1$ $$g(w) = f(\xi_2^{-1}(w)).$$ Evaluating at $w=\xi_1^{-1}(z)$, we obtain $$\pi(\xi_1)(\pi(\xi_2)f)(z) = \pi(\xi_1)g(z) = g(\xi_1^{-1}(z)) = f(\xi_2^{-1}(\xi_1^{-1}(z))),$$ so we see that the order gets flipped correctly.
You can also avoid the point evaluations altogether by thinking in terms of composition of functions, so the group product $\xi_1\cdot\xi_2$ is composition $\xi_1\circ\xi_2$ and the action is $\pi(\gamma)f = f\circ\gamma^{-1}$. Then the required computation takes the form $$\pi(\xi_1\circ\xi_2)f = f\circ(\xi_1\circ\xi_2)^{-1} = f\circ\xi_2^{-1}\circ\xi_1^{-1} = (\pi(\xi_2)f)\circ \xi_1^{-1} = \pi(\xi_1)(\pi(\xi_2)f).$$