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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.

Ook
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  • Here, you would need 2 coordinate functions $(y^1,y^2)$ which are defined on a certain open subset $V$ of $S^2$. The function $F^j$ is by definition $y^j\circ F$ (or to be really precise it is $y^j\circ (F|_{F^{-1}(V)\cap U})$, where $U$ is the domain of a coordinate chart on $S^3$). You seem to be making the grave sin of confusing coordinates on the ambient $\Bbb{R}^4,\Bbb{R}^3$ for coordinates on the spheres $S^3,S^2$. – peek-a-boo Sep 25 '23 at 16:36
  • If you want to compute in local coordinates then you must decide first and foremost, which local coordinates are you choosing on the domain and target? Note that there are several possible choices: common ones include graph coordinates, polar coordinates, sterographic projection coordinates etc. Each has advantages and disadvantages, but you need to decide before delving into computations. Btw, here is a somewhat related question where OP made the same mistake as you are right now. – peek-a-boo Sep 25 '23 at 16:40
  • As peek-a-boo pointed out already: $S^3$ is a three-dimensional manifold and has as basis only three vector fields $\frac{\partial}{\partial x^i},.$ You seem to think it has four but your Hopf map is expressed in coordinates of the ambient Euclidean space. Likewise $S^2$ has only two basis vector fields. You might have an easier life when you look at $M=N=\mathbb R^2$ and take as $F$ the coordinate transformation from Cartesian to polar coordinates. – Kurt G. Sep 25 '23 at 16:55
  • @peek-a-boo I think I understand, so I should be thinking of $F$ taking $S^3$ to $S^2$ as taking $(x^1,x^2,x^3)$ to $(y^1,y^2)$ right? Then use the atlases on those spheres (which was done using stereographic projection) to do thing in $\mathbb{R}^3$ and $\mathbb{R}^2$. Do you mind giving a basic example so that I can solidify this idea in my mind (maybe taking the cartesian to polar coordinates) as suggested by Kurt G. – Ook Sep 25 '23 at 17:04

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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 $F$).

Kurt G.
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  • Just as a technical question but will $\frac{\partial F^i}{\partial x^j}$ always be the Jacobian? My understanding is that $\partial$ in tangent spaces might not be the partial derivative that we are used to. – Ook Sep 25 '23 at 17:27
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    It will be the Jacobian of a function that you get once you have chosen coordinate charts. I do not want to elaborate on this further since this amounts to giving a lecture on differential geometry. Be patient, work with concrete examples and you will get there. The real Hopf map you should work with through your entire life. Have fun! – Kurt G. Sep 25 '23 at 17:31
  • just another question but if I am asked to find a formula for $F$ in local coordinates, would that be the same as finding $\psi\circ F\circ \phi^-1$ where $\psi$ is my coordinate chat from $N$ to $\mathbb{R}^n$ and $\phi$ is from $M$ to $\mathbb{R}^m$? Do local coordinates mean the coordinates in euclidean space or coordinates on the manifold? – Ook Sep 25 '23 at 18:57
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    With $\psi\circ F\circ \phi^{-1}$ you are getting close. That is a function from $\mathbb R^m$ to $\mathbb R^n$ and has a Jacobian. The coordinates are tuples in $\mathbb R^m,,$ resp. in $\mathbb R^n$ by which we label points in $M$ resp. in $N,.$ Coordinates on the manifold? Forget that. – Kurt G. Sep 25 '23 at 19:27
  • @ Kurt G. So is that why we always consider things in terms of neighborhoods around points. Because then we can essentially consider a point of a manifold being the same thing as a point in $\mathbb{R}^n$. So in the case of the hopf map we are really just computing the differential of $\psi\circ F\circ \phi^{-1}$ since euclidean space and the manifold are essentially the same? Am I sort of understanding. Also the $y^i$ depend on the $x^i$ right? – Ook Sep 25 '23 at 19:48
  • I do not want to say "right" here and let you move on with something else. The next fruitful exercise you may want to do is: choose coordinates for $S^2$ (most popular are spherical $(\theta,\varphi)$) and for $S^3$ (spherical or quaternions). Then calculate $dF$ for the real Hopf map. Getting into rabbit holes is healthy. Not asking questions like manifold and $\mathbb R^n$ are "essentially" the same. – Kurt G. Sep 26 '23 at 06:40