Pushforward of Lie Bracket

Question

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)

Answer 1

To do this computation in coordinates without using functions and points you have to adopt the physicist way of writing things which is messy and unplesant :-) However chain rule takes care of all the evaluation matters.

Let f map $x$ to $y$. We will denote the Jacobian and inverse Jacobian by $\frac{\partial y^j}{\partial x^i}, \frac{\partial x^j}{\partial y^i}$

We will write $\tilde{Z}=f^*Z$ and $\tilde{W}=f^*W$ so the "components change as"

$Z^j = \tilde{Z}^i\frac{\partial x^j}{\partial y^i}$

$W^j = \tilde{W}^i\frac{\partial x^j}{\partial y^i}$

(here we really see Z as a pushforward of $\tilde{Z}$ by the inverse map etc)

Consider $f^*[Z,W]$

$=((Z^i\frac{\partial}{\partial x^i}(W^j) - W^i\frac{\partial}{\partial x^i}(Z^j))\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}$

$=((\tilde{Z}^k\frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial x^i}(\tilde{W}^m\frac{\partial x^j}{\partial y^m}) - (\tilde{W}^k\frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial x^i}(\tilde{Z}^m\frac{\partial x^j}{\partial y^m}))\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}$

The mixed derivative term is of the form

$((\tilde{Z}^k(\tilde{W}^m\frac{\partial}{\partial y^k}\frac{\partial x^j}{\partial y^m})-(\tilde{W}^k(\tilde{Z}^m\frac{\partial}{\partial y^k}\frac{\partial x^j}{\partial y^m}))$ =0

and the remaining terms give

$=((\tilde{Z}^k\frac{\partial}{\partial y^k}(\tilde{W}^m) - (\tilde{W}^k\frac{\partial}{\partial y^k}(\tilde{Z}^m))\delta_{lm}\frac{\partial}{\partial y^l}$

which is $[f^*Z,f^*W]$. The trick is to use chain rule when ever you want the derivation to be compatible with the function you are applying it to. So the middle expressions might not be completely sensible (just formal expressions to see the steps) but derivations are. Why this only works for diffeomorphisms is the component change rules given above is basically the pushforward expression in coordinates and that expression allows usage of chain rule at certain parts.

Answer 2

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)] $

Answer 3

I would give a basis-independent proof: $f:M\rightarrow N$ and $g:N\rightarrow\mathbb{R}$ \begin{eqnarray} f_*[X,Y](g)&=&\{[X,Y](g\circ f)\}\circ f^{-1}\\ &=&\{(X\circ Y)(g\circ f)\}\circ f^{-1}-(X\leftrightarrow Y)\\ &=&\{X[Y(g\circ f)]\circ f^{-1}\}-(X\leftrightarrow Y)\\ &=&\{X[Y(g\circ f)\circ f^{-1}\circ f]\circ f^{-1}\}-(X\leftrightarrow Y)\\ &=&\{X[f_*Y(g)\circ f]\circ f^{-1}\}-(X\leftrightarrow Y)\\ &=&(f_*X\circ f_*Y)(g)-(X\leftrightarrow Y)\\ &=&[f_*X,f_*Y](g). \end{eqnarray} Here one should keep in mind that: $X$ or $Y$ is a MAPPING from differential functions (on $M$) to differential functions (on $M$).