I having some problem for computing the differential of the function $f : \mathbb{T}^2 \rightarrow \mathbb{S}^2$ defined as the quotient of the function $F$ from $\mathbb{R}^2$ to $\mathbb{S}^2$ by :

$F(x,y) = (\cos(2\pi x)\cos(2\pi y), \cos(2\pi x)\sin(2\pi y), \sin(2\pi x))$

Actually, I wanted to considerer the function on the charts. I first tried to consider the charts on $\mathbb{S}^2$ given by the stereographic projection, but I had the impression that it was not the best choice. So, I considered the charts on $\mathbb{S}^2$ given by : $(U_i^{+/-}, g_i^{+/-})$ where $U_i^{+} = \{ (x_1,x_2,x_3) \in \mathbb{S}^2 \; | x_i > \;0 \}$, and $g_i^{+} : U_i^{+} \rightarrow \{ (x,y) \in \mathbb{R}^2 \; | \; x^2+y^2 <1\}$ defined by remove the $i$-th coordinates.

On $\mathbb{T}^2$, I would consider the Atlas given by the open sets $B((x_1, x_2), \frac{1}{4})$, where $x_i \in \{ 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \}$, and the projection $p : \mathbb{R}^2 \rightarrow \mathbb{T}^2$ restricted (and co-restricted) to the open ball $B((x_1, x_2), \frac{1}{4})$.

Thus, I would like to consider $F$ in the charts, but I had to distinguished the cases to put it in the right charts at the beginning and at the end, and it gave me a long list of cases to handle.

Finally, I was wondering if it was the good way to do in order to calculate the differential of this function, because it seems very long, and maybe it's not this way that I have to proceed ?

Thank you for your help !

  • $\begingroup$ I have a few clarifying questions for you. 1. When you write $S^2$, you're clearly thinking about it as a subset of $\Bbb R^3$. Are you doing the same for $T^2$ ("it's a donut!") or are you considering it as, for instance, a subset of $\Bbb R^4$? 2. Either way, can you say what you'd consider the differential to be? I mean, what would constitute an answer? How, exactly, do you want to express an element of the tangent space $T_p T^2$ at a point $p$ of the torus (which would be an argument to $df(p)$)? $\endgroup$ – John Hughes Apr 12 at 13:37
  • $\begingroup$ For $T^2$, I see it as the manifold $\mathbb{R}^2 / \mathbb{Z}^2$, for example induced by the action of $\mathbb{Z}^2$ on $\mathbb{R}^2$, but not as a subset of $\mathbb{R}^3$. For the differential, strictily, it's : $DF(p) : T_pT^2 \rightarrow T_{F(p)}S^2$ which send a path $[c]$ to $[F \circ c]$, but we identified this application with the differential of $F$ saw in the charts. $\endgroup$ – ChocoSavour Apr 12 at 14:29

OK, so there's a really nice parameterization of $T^2$ by $\Bbb R^2$, namely "quotient by $2 \pi$ in each coordinate." Calling that map $H$, we can identify $$ T_{H(p)} T^2 \sim T_p \Bbb R^2. $$ So I can write a tangent vector at the point $p = (\theta, \phi) \in T^2$ just using coordinates, which I'll write vertically: $\pmatrix{a\\b}$, and where this is a proxy for the vector $dH_{(\theta, \phi)}\pmatrix{a\\b}$. Now we'd like to figure out the following: what is $$ df(\theta, \phi)\pmatrix{a\\b}? $$ It should be a vector in $T_{f(\theta, \phi)} S^2 \subset T_{f(\theta, \phi)} \Bbb R^3$. Let's give the name $q$ to $f(\theta, \phi)$, so I don't have to keep writing it.

OK. Let's look at $g = f \circ H$; that's a map from $\Bbb R^2$ to $\Bbb R^3$. Its Jacobian matrix at $p$ is \begin{align} Dg(p) &= Dg(\theta, \phi) \\ &= 2\pi \pmatrix{ -\sin(2\pi \theta) \cos(2\pi \phi) & -\cos(2\pi \theta) \sin(2\pi \phi) \\ -\sin(2\pi \theta) \cos (2 \pi \phi) & \cos(2\pi \theta) cos(2\pi \phi) \\ \cos(2 \pi \theta) & 0}. \end{align}

Now knowing the chain rule, we know that $$ dg = df \cdot dh $$ and because, in our current coordinates, $dh$ is just the identity, we end up with $dg = df$. In short,

$$ df(\theta, \phi) \pmatrix{a, b} = 2\pi \pmatrix{ -\sin(2\pi \theta) \cos(2\pi \phi) & -\cos(2\pi \theta) \sin(2\pi \phi) \\ -\sin(2\pi \theta) \cos (2 \pi \phi) & \cos(2\pi \theta) cos(2\pi \phi) \\ \cos(2 \pi \theta) & 0}\pmatrix{a\\b}. $$


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