Is inverse map of a Lie group a Lie algebra homomorphism?

Question

I have been doing a problem (number 6.8) from a differential geometry book by Victor Prasolov:

Prove that the right commutator differs from the left commutator in sign i.e. $$[X^L, Y^L]_e=-[X^R,Y^R]_e$$ where $X^L,Y^L$ denote the left invariant fields defined by vectors $X,Y$ at $e$ and similarly $X^R,Y^R$ denote the right invariant vector fields. In the solution of the problem, an identity $Inv_*[X^L, Y^L] = [Inv_*X^L, Inv_*Y^L]$ is used, where $Inv_*$ is the differential of the inverse map. However, I couldn't prove this identity and, in fact, it seems that it isn't true. Take a function $f$, then we have $$ \partial_{Inv_*[X^L, Y^L]}f = \partial_{[X^L, Y^L]}f\circ Inv = (\partial_{X^L}\partial_{Y^L} - \partial_{Y^L}\partial_{X^L})f\circ Inv = (\partial_{X^L}\partial_{Inv_*Y^L} - \partial_{Y^L}\partial_{Inv_*X^L})f$$ which doesn't seem to equal $$\partial_{[Inv_*X^L, Inv_*Y^L]}f=(\partial_{Inv_*X^L}\partial_{Inv_*Y^L} - \partial_{Inv_*Y^L}\partial_{Inv_*X^L})f$$ If the fields were not left invariant, it would be almost hopeless to believe that the identity is true. But in the special case of left invariant fields is it true that $Inv_*[X^L, Y^L] = [Inv_*X^L, Inv_*Y^L]$ and is it possible to prove it without going into deep theory, as this problem appears in the beginning of the chapter about the Lie groups?

Answer 1

What is going on here is actually much more general than Lie groups. Specifically, we have the following proposition:

Suppose $f:M\rightarrow N$ is a diffeomorphism between smooth manifolds $M$ and $N$. If $X$ and $Y$ are vector fields on $N$, then $f_\ast([X,Y]) = [f_\ast X, f_\ast Y]$.

A proof of this is given here, but I would like to add that one has to be a bit careful about $f$. Specifically, if e.g., $f$ is not injective, then $f_\ast X$ need not even be a vector field on $N$. Of course, if $f$ is a diffeomorphism then these issues vanish, and that is sufficient in this situation, as $Inv$ is a diffeomorphism.

In terms of the calculation supplied, the error is that $\partial_{Y^L}(f\circ Inv)$ is equal to $(Inv_\ast Y^L)(f)\circ Inv,$ rather than being equal to $(Inv_\ast Y^L)(f)$.

Answer 2

I will correct the mistake in the calculation as pointed out by Jason DeVito. What happened is I didn't pay close attention to the point at which the function and its differential are being evaluated. In order to understand it more clearly it helped me to think in terms of the more general setting: for a diffeomorphism $\varphi$ $$\partial_{\varphi_*X}f = (\partial_X f\circ \varphi)\circ \varphi^{-1}$$ Specifically in this case, we have $$\partial_{Inv_*[X^L, Y^L]}f = ((\partial_{X^L}\partial_{Y^L} - \partial_{Y^L}\partial_{X^L})f\circ Inv)\circ Inv = (\partial_{X^L}((\partial_{Inv_*Y^L}f)\circ Inv) - \partial_{Y^L}((\partial_{Inv_*X^L}f)\circ Inv)\circ Inv = (\partial_{Inv_*X^L}\partial_{Inv_*Y^L}f\circ Inv - \partial_{Inv_*Y^L}\partial_{Inv_*X^L}f\circ Inv)\circ Inv = (\partial_{Inv_*X^L}\partial_{Inv_*Y^L}f- \partial_{Inv_*Y^L}\partial_{Inv_*X^L})f = \partial_{[Inv_*X^L,Inv_*Y^L]}f$$ Note that we did not use the fact that $X^L$ and $Y^L$ are left invariant. But again, all this is just a special case of the general proposition stated in Jason DeVito's answer.