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Consider the mapping $f:S^2\rightarrow \mathbb{R}$, with $f(x,y,z)=z$, giving the height of a point on the sphere. I am asked to find the points for which the differential $(df)_p$ is surjective and i have failed so far. It's an exercise on an introductory course on smooth manifolds, so please try to be clear about what theorems you use, i'm just a beginner!

Thanks a lot in advance!

kleinmeinpouts
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2 Answers2

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Intuition

Draw a picture. At a tangent space, visualize tangent vectors and the result of projecting those tangent vectors to the $z$ axis. Do you see what are the (hint: only two) points such that the projection is not surjective?

Computation

We know that $f$ is actually the restriction of the projection map $F:\mathbb{R}^3\to \mathbb{R}$ taking $(x,y,z)\mapsto z$. Compute the differential of $F$, call it $dF$. Don't think too hard about this.

Now recall that the tangent space to $S^2$ at a point $p$ can be identified with the plane of vectors in $\mathbb{R}^3$ perpendicular to $p$. (Why?) Given a $p\in S^2$, compute a basis for $T_pS^2$.

Compute the matrix for $df_p$ by restricting $dF$ to the tangent space (which you know, as you have a basis).

What are the points where this map is not onto?

Neal
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  • Well, working with the usual atlas that divides the sphere in 6 hemispheres, i showed that $g$ is differentiable and found a local expression for it, but i don't really know if that's helpful. I tried drawing a picture of a sphere as you suggested, along with a vertical line on the side acting as a measurement for height. My intuition tells me that the points which have a problem are the poles, but i am pretty much stuck here. I cannot show it formally and it wasn't explained whether we are supposed to solve it formally or not. – kleinmeinpouts Sep 24 '16 at 21:16
  • @kleinmeinpouts First, you should show it formally. Second, It's a good exercise to work in atlases -- they will give you local coordinate expressions and from there it's a matter of row reducing. But, third, in this case, as I suggested in my answer, we have extra structure: the sphere is immersed in 3-space and your function is the restriction of a very natural function 3-space . Why not use that structure? – Neal Sep 24 '16 at 21:29
  • I said i'm a beginner, meaning on this course specifically! Our professor has not taught us at all about immersions, submanifolds etc. They were part of presentations by students (sigh) and he leaves pretty much everything as an exercise. So, if by 'immersion' you don't actually mean 'injective' or 'restriction', then i am pretty much hopeless at this point. I think i'm beginning to understand what you mean in your previous comments, though i would really appreciate a formal answer, just to be sure i get it right! Sorry for wasting your time! – kleinmeinpouts Sep 24 '16 at 21:48
  • So you mean that i should write $F:\mathbb{R}^3\rightarrow \mathbb{R}$, $(x,y,z) \mapsto z$, then $f=F|_{S^2}$. So if we take a vector $v\in \mathbb{R}^3$, then $(dF)_p(v)=\bigtriangledown f(p)\cdot v=(0,0,1)\cdot v=v\cdot e_3$ and if we take $v\in S^2\subseteq \mathbb{R}^3$ we have that $(df)_p(v)=(dF)_p(v)=v\cdot e_3$, since $T_pS^2\subseteq T_p\mathbb{R}^3$. Something like that? – kleinmeinpouts Sep 24 '16 at 22:00
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For $u=(x,y,z)\in R^3,v=(a,b,c)\in T_uR^3 df_u(v)=c$.Thus $df_u(v)=0$, i.e $c=0$, if $u\in S^2, T_uS^2\subset T_uS^3$ and $df_u(T_uS^2)$ i.e $T_uS^2=\{(a,b,0),a,b\in R\}$ i.e $u$ is the north pole or the south pole.

Tsemo Aristide
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