A short background from Smooth Manifolds, by Lee.
Given $G$ a Lie group and a right $G$-action on a smooth manifold $M$, we have an infinitesimal $\mathfrak{g}:=Lie(G)$-action over $M$. That is, a Lie homomorphism $\mathfrak{g} \rightarrow \mathfrak{X}(M)$, where $\mathfrak{X}(M)$ is the Lie algebra of vector fields over $M$ with the natural bracket. Moreover, if we are considering a left $G$-action, we can still define an infinitesimal $\mathfrak{g}$-action in the same way, however the induced map $\mathfrak{g} \rightarrow \mathfrak{X}(M)$ is a Lie anti-homomorphism.
Now, my question.
I was reading about Hamiltonian actions of a Lie group $G$ over a manifold $M$ on Representation Theory and Complex Geometry, by Chriss and Ginzburg, p. 42, where you can read the following lines:
Are they assuming that $G$ acts on the right? Is there any convention in this field? Because before of this they made examples of natural (on the left) $G$-actions.
P.S. It doesn't seem to be a mistake because they use this fact in the next pages to define Hamiltonian actions for example. Unfortunately, I cannot find any pdf of the reference they give about this (would be great if anyone can tell me where to find it) which is Orbits, Symplectic structure and Representation Theory, by Kostant, 1966.
Best Answer
Let $\mathcal C(X)$ the algebra of functions on $X$. (I'm being a bit vague here, because Lee's book works with smooth functions, but the Chriss-Ginzburg book sometimes works with algebraic varieties or complex manifolds in which case you would consider algebraic or analytic functions.) If $G$ acts on $X$ on the left, then there is a natural linear action of $G$ on $\mathcal C(X)$ given by $$ g(f)(x) = f(g.x), \quad \forall g \in G, f \in \mathcal C(X), x \in X. $$ However, a simple calculation shows that this gives a right action of $G$ on $\mathcal C(X)$, or if you prefer an anti-homomorphism from $G$ to $\text{Aut}(X)$ the group of (smooth) automorphisms of $X$. If you differentiate this action you obtain the anti-homomorphism mentioned by the OP. One way to deal with this is simply to allow left and right linear actions as well as left and right group actions, but then one has to (repeatedly) specify which is intended.
Alternatively, one notes that groups and Lie algebras are equipped with a canonical anti-automorphism, so one can always switch between left and right actions using it. For groups this is the inversion map $\iota\colon G\to G$, given by $\iota(g)=g^{-1}$, while for Lie algebras, it is $a=d\iota(x)$ where $a(x) = -x$, for all $x \in \mathfrak g$. Thus, for example, if you take the induced action of $G$ on $\mathcal C(X)$ to be given by the formula $g(f)(x) = f(g^{-1}.x)$, then you obtain a left linear action (or left representation) of $G$ on $\mathcal C(X)$. If you then differentiate it you obtain a Lie algebra homomorphism $\mathfrak g \to \mathfrak X(X)$.
I expect that Chriss-Ginzburg simply failed to specify that they compose with the antiautomorphism $\iota$ or $a=d\iota$ in order to obtain a left-action. In general, it is a good idea to check the definition that is used when an action on a space is used to induce an action on functions on that space before doing any substantial calculations, because signs will change depending on the convention being used. (If the reference studies actions by nonabelian groups, it is likely that they will use the definition given in the formula above, but if all groups used in the reference are abelian (e.g. as may be the case when studying torus actions) then sometimes the definition $g(f)(x) = f(g(x))$.
A publicly available reference on this point can be found here . Meinrenken uses the same choice of signs in \S1.2 of this note, and has a brief discussion on the issue in Remark 1.8.