Let $F:M\to N$ be a smooth map between manifolds and $p\in M$. I am having trouble understanding the differential $dF_p$. I understand how $\left\{\frac{\partial}{\partial x^1},..., \frac{\partial}{\partial x^n}\right\}$. is a basis of $T_pM$. So we only really care where $dF_p$ maps basis elements. After a long string of calculations we showed that $dF_p(\frac{\partial}{\partial x^i})=\frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial y^j}$. But where are the $F^j$ coming from? As far as I know, we only started with one smooth function $F$. I thought I would try an example so I tried it with the Hopf map $F:S^3\to S^2$ defined by $F(x^1,x^2,x^3,x^4)=(2(x^1x^2+x^3x^4),2(x^1x^4-x^2x^3),(x^1)^2+(x^3)^2-(x^2)^2-(x^4)^2)$. Now I want to compute $dF_p(v)$.
I think that the notation is confusing me, but I believe that $y^1(x^1,x^2,x^3,x^4)=2(x^1x^2+x^3x^4)$ and so on. Does $F^j$ represent the same thing as $y^j$? Because if that is the case then the computation would be pretty easy, I think that $\frac{\partial F^j}{\partial x^i}$ would just be the partial derivative but then what is $\frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial y^j}$? Also I believe that this is a sum right? The notation is weird but i think this is $\sum_j \frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial y^j}$. Would someone mind going through this example for me so I can explicitly see what each thing represents.
As a first technical exercise I think it is legit to view the map $$ F:\begin{pmatrix}x^1\\x^2\\x^3\\x^4\end{pmatrix}\mapsto\begin{pmatrix}(2(x^1x^2+x^3x^4)\\2(x^1x^4-x^2x^3)\\(x^1)^2+(x^3)^2-(x^2)^2-(x^4)^2\end{pmatrix} $$ not as the Hopf map but as a map from $\mathbb R^4$ to $\mathbb R^3$ each equipped with their standard coordinate bases $$ \Big\{ \frac{\partial}{\partial x^1}\,,\frac{\partial}{\partial x^2}\,,\frac{\partial}{\partial x^3}\,,\frac{\partial}{\partial x^4}\Big\}\,,\text{ resp. } \Big\{\frac{\partial}{\partial y^1}\,,\frac{\partial}{\partial y^2}\,,\frac{\partial}{\partial y^3}\Big\}\,. $$ To answer your questions briefly in this case: $$ \frac{\partial F^i}{\partial x^j} $$ is the Jacobian of $F\,.$ Please calculate it.
$$\tag{1} \frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial y^j} $$ is a sum $$\tag{2} \sum_{j=1}^3\frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial y^j}\,. $$ (1) and (2) is a vector field on $\mathbb R^3$ and denoted by $$ dF\Big(\frac{\partial}{\partial x^i}\Big) $$ and also by $$ F_*\Big(\frac{\partial}{\partial x^i}\Big) $$ (which is called push forward of $\frac{\partial }{\partial x^i}$ by $F$).