I am trying to figure out why the following equality is true :
$$f_*[X,Y]=[f_*X,f_*Y]$$
where $f:M\rightarrow N$ is a diffeomorphism, $M$, $N$ are smooth manifolds, $X$, $Y$ are smooth vector fields on $M$.
I have tried to write $$f_*[X,Y]=\dfrac{\partial f^i}{\partial x^j}\left( \chi^k \dfrac{\partial \psi^j}{\partial x^k}-\psi^k \dfrac{\partial \chi^j}{\partial x^k}\right)\dfrac {\partial}{\partial y^i}$$
where $$X=\chi^k \dfrac{\partial}{\partial x^k},Y=\psi^k \dfrac{\partial}{\partial x^k}, [X,Y]=\left( \chi^k \dfrac{\partial \psi^j}{\partial x^k}-\psi^k \dfrac{\partial \chi^j}{\partial x^k}\right)\dfrac {\partial}{\partial x^j}.$$
However, when it comes to write the second part of the equality:
$$[f_*X,f_*Y]=\left( (f_*X)^k \dfrac{\partial (f_*Y)^j}{\partial y^k}-(f_*Y)^k \dfrac{\partial (f_*X)^j}{\partial y^k}\right)\dfrac {\partial}{\partial y^j}$$
where $y^j$ is a coordinate basis of N.
The problem I face is that I cannot differentiate $f_*Y, f_*X$ with respect to the basis $y^j$, in the above expression. Any help would be appreciated.
( I would prefer an answer which is based on the definition of Lie Bracket with coordinates, as I worked above)
Best Answer
Well, you see it is much simpler in coordinate independent form. As for diffeomorphism $ f : M \rightarrow N $ you have $ f_* : \mathcal{X}(M) \rightarrow \mathcal{X}(N) $ and hence for $p\in M $ it maps tangent spaces $T_p(M)$ to $T_{f(p)}(N) $ given by for $ g \in C^\infty(N) $ you have $f_* (X)(g)(f(p)) = X(g\circ f)(p)$ hence $ f_*(X)(g)\circ f = X(g\circ f ) $ Thus for $X,Y \in \mathcal{X}(M) $ we have for all $ g \in C^\infty(N) $ \begin{align*} & f_*[X,Y]_{f(p)}(g) = [X,Y]_p(g\circ f) \\ & = X_p(Y(g\circ f))-Y_p(X(g\circ f)) \\ & = X_p(f_*(Y)(g)\circ f) - Y_p(f_*(X)(g)\circ f) \\ & = f_*(X)_{f(p)}(f_*(Y)(g))-f_*(Y)_{f(p)}(f_*(X)(g)) \\ & = [f_*(X),f_*(Y)]_{f(p)} (g) \end{align*} Hence $ f_*[X,Y] = [f_*(X),f_*(Y)] $