I am teaching myself differential geometry on manifolds with some notes a professor gave me. As an initial calculation to prove that the Levi Civita connection is invariant under isometries, the notes threw the following result \eqref{push} leaving it as an exercise which I have been scratching my head about:
Let $F:M \to M'$ be a diffeomorphism and $Z$ be a vector field over $M$. The field $F_*Z$ over $M'$ is defined putting $(F_*Z)_{F(p)} = F_{*p}Z_p$. For example, if $(U,x)$ is a chart of $M$ we have that $x_* \partial_i = D_i$, partial derivative with respect to the variable. More in general, if the transformed chart $(F(U),x\circ F^{-1})$ of $M'$ has coordinate fields $\partial_i'$ we have that $F_*\partial_i=\partial_i'$ and \begin{equation}\label{push} \tag{$\sharp$} F_*Z=F_*(Z^i\partial_i)=(Z^i\circ F^{-1})\partial_i'. \end{equation}
Therefore Z is differentiable , because $F_*Z$ is differentiable
Note: $x_*$ is the pushforward map
My try:
I know the pushforward of a differentiable function $f$ between manifolds, $f_*$ is supposed to map tangent vector of one space to tangent vectors of the other one and it is defined as $(f_*\nu)h = \nu (h \circ f)$ when $\nu$ is a derivation and $h$ a smooth function, I don't know if that is useful but it took me nowhere, so I was trying to do something like :
$(F_*Z)h =Z (h \circ F)=Z^i \partial_i (h \circ F) $
and then I was hoping to find "(the expresion I want)h", so no idea where that $F^{-1}$ in \eqref{push} is coming from.
Can someone please show me the proof of \eqref{push}? Please be as detailed as posible, I am new at manipulating the pushforward map.
I think you started on the right path: work from $(F_*Z)(h)$ using the definition.
First, some notation: fix a system of local coordinates ${\bf x}=(x^1,\dots,x^n):U\to\mathbb{R}^n$ on the open set $U\subset M$, and let instead ${\bf y}={\bf x}\circ F^{-1}$ be the coordinates on the open set $U'=F(U)\subset M'$.
By definition, for every $h\in\mathcal{C}^\infty(M',\mathbb{R})$ and every $p\in M$ we have \begin{equation}\label{def} \tag{1} (F_*Z)_{F(p)}(h)=Z_p(h\circ F). \end{equation} We use this definition to prove the formula you are interested in. In our local coordinate systems, \eqref{def} reads, for $q=F(p)$ \begin{equation}\label{def_coords} \tag{2} (F_*Z)^i(q)\,\partial_{y^i}h(q)=Z^j(p)\,\partial_{x^j}(h\circ F)(p). \end{equation} Apply the chain rule to compute the derivative on the RHS of \eqref{def_coords}: \begin{equation} Z^j(p)\,\partial_{x^j}(h\circ F)(p)=Z^j(p)\,\partial_{y^k}h(q)\,\partial_{x^j}F^k(p) \end{equation} where $F^k(p)$ is the $k$th component of $F(p)$ in the coordinate system ${\bf y}$, i.e. $y^k(F(p))$. By definition of $y$, the $k$th component of ${\bf y}(F(p))$ is the $k$th component of ${\bf x}(p)$, so $\partial_{x^j}F^k(p)=\delta_j^k$.
Then, we can rewrite \eqref{def_coords} as \begin{equation} \begin{gathered} (F_*Z)^i(q)\,\partial_{y^i}h(q)=Z^j(p)\,\partial_{y^k}h(q)\,\partial_{x^j}F^k(p)=\\=Z^j(p)\,\partial_{y^j}h(q)=Z^j(F^{-1}(q))\,\partial_{y^j}h(q) \end{gathered} \end{equation} which is precisely what we were looking for: as $h$ was arbitrary \begin{equation} F_*Z=Z^j\circ F^{-1}\,\partial_{y^j}. \end{equation}