Let $M$ be a smooth manifold.

Let $p\in M$. A tangent vector at $p$ is a an equivalence class $[\gamma]$ of smooth curves $\gamma : (-\epsilon,\epsilon)\rightarrow M$, with $\gamma(0) = p$, where the equivalence relation is as follows: $$\gamma_1\sim\gamma_2 \iff \exists \text{ chart } (U,\phi) \text{ such that } (\phi\circ\gamma_1)'(0) = (\phi\circ\gamma_2)'(0)$$

One can now define the tangent space $T_pM$ as the set of all tangent vectors. And define the derivative as follows:

Let $f:M\rightarrow N$ be a differentiable map between manifolds, $p\in M$. The derivative of $f$ at $p$ is $$(f_*)_p:T_pM\rightarrow T_pN: [\gamma]\mapsto[f\circ\gamma] $$

I understand that the derivative at $f$ just changes the tangent vectors at $M$ to tangent vectors at $N$ such that it has nice properties. But I'm having a hard time understanding what this means practically of how to interpret these tangent vectors.

The exercise I am trying to solve defines $f:S^2\rightarrow \mathbb{R}:(x,y,z)\mapsto z^2$ and asks for which $p$ $(f_*)_p = 0$.

I have proven that if $F:M\rightarrow \mathbb{R}$ and $N$ a submanifold, that if $f = F\mid_N:N\rightarrow \mathbb{R}$ that for all $p\in N$: $(f_*)_p = 0\iff T_pN\subset ker(F_*)_p$.

It seems like this can be useful in some way, but then again I think I need to calculate $(F_*)_p$ which I don't know how to do.


The theorem you have indeed is useful, provided you know what the tangent space to $N$ looks like/ if you can easily figure it out, and provided you know what $F,M$ should be. In your particular case, it should be natural to choose $M= \Bbb{R}^3$, equipped with the maximal atlas containing the identity chart $(\Bbb{R}^3, id)$. Then, of course, we take $N = S^2$, and lastly, it should be reasonable to define $F: \Bbb{R}^3 \to \Bbb{R}$ by \begin{align} F(x,y,z) = z^2 \end{align}

Then, clearly, $f = F|_N$.

Now, let's side-track slightly and see how to think of tangent vectors as actual elements of your model space, so that we can see how to apply this to your question. Let $X$ be a smooth manifold (I avoid $M$ because I already used it above) of dimension $n$. Let $p \in X$, and let $(U, \alpha)$ be any chart of $X$ containing the point $p$ (i.e $p \in U$). Then, one can construct an isomorphism of $T_p(X)$ onto $\Bbb{R}^n$ as follows: define the map $\Phi_{\alpha, p}: T_pX \to \Bbb{R}^n$ by \begin{align} \Phi_{\alpha, p}\left( [c] \right) &:= (\alpha \circ c)'(0) \end{align} By definition of the equivalence relation, this map is well defined (in fact it is pretty much using this map that you use to get a vector space structure on $T_pX$).

Now, if $g: X \to Y$ is a smooth map between smooth manifolds, and $(U, \alpha)$ is a chart about a point $p \in X$, and $(V, \beta)$ is a chart for $Y$ about the point $g(p)$, then we have the following commutative diagram:

$\require{AMScd}$ \begin{CD} T_pX @>{g_{*p}}>> T_{g(p)}Y \\ @V{\Phi_{\alpha,p}}VV @VV{\Phi_{\beta, g(p)}}V \\ \Bbb{R}^n @>>{D(\beta \circ g \circ \alpha^{-1})_{\alpha(p)}}> \Bbb{R}^m \end{CD} Which I'll leave to you to verify that it is actually commutative. In words what this means is that to compute $g_{*,p}$, you can choose charts on the domain and the target, and instead consider the (familiar euclidean type) derivative of the chart-representative map $\beta \circ g \circ \alpha^{-1}$.

Now, a special case of interest is the following: we have $X$ as a submanifold of some $\Bbb{R}^l$, and we're given a point $p \in X$. How do we think of $T_pX$ in an intuitive way? Well, $p$ is an element of $X$ and hence $\Bbb{R}^l$, so it (trivially) lies in the identity chart $(\Bbb{R}^l, id)$. Thus, rather than thinking of the tangent space as $T_pX$, which set-theoretically consists of equivalence classes of curves (which is rather abstract and tough to compute with), consider instead the $\dim X$-dimensional subspace $\Phi_{id,p}\left( T_pX \right) \subset \Bbb{R}^l$. This is precisely the intuitive picture of tangent space you would have.

For example, if $X= S^2$ considered as a submanifold of $\Bbb{R}^3$, then for each $p \in X= S^2$, $\Phi_{id,p}(T_pS^2)$ will be the (translated) tangent plane $\{\xi \in \Bbb{R}^3: \langle \xi, p\rangle = 0 \}$ (which is precisely the usual intuitive picture you might have). To prove this in a rigorous manner, it will be much easier if you know that $S^2$ can be written as a level set, say $h^{-1}(\{1\})$, where $h: \Bbb{R}^3 \to \Bbb{R}$ is defined by $h(x,y,z) = x^2 + y^2 + z^2$, and that $T_pS^2 = \ker h_{*p}$, so that (by the above commutative diagram, and basic linear algebra), \begin{align} \Phi_{\text{id}_{\Bbb{R}^3}, p}(T_pS^2) = \ker D(\text{id}_{\Bbb{R}} \circ h\circ \text{id}_{\Bbb{R}^3}^{-1})_{\text{id}_{\Bbb{R}^3}(p)} = \ker(Dh_p) \end{align} where $Dh_p : \Bbb{R}^3 \to \Bbb{R}$ is the usual derivative.

So, now back to your question. You seek all $p \in S^2$ such that $f_{*,p} = 0$, or equivalently, by your theorem, those $p \in S^2$ such that $T_pS^2 \subset \ker F_{*,p}$. Or equivalently, those $p \in S^2$ such that \begin{align} \Phi_{\text{id}_{\Bbb{R}^3}, p}(T_pS^2) \subset \Phi_{\text{id}_{\Bbb{R}^3}, p} \left( \ker F_{*,p} \right) = \ker DF_p \end{align} So, to answer your question, you just have to compute $DF_{p}$ for all $p \in S^2$ (in the usual calculus sense), compute the kernel of this map, and then see whether the translated tangent plane to the sphere lies inside the kernel.

I believe that this final computational part isn't difficult so I'll leave this all to you. I felt it is more important to see the logic behind what kind of computation needs to be performed.


In your particular question, you've been aided by the fact that the theorem you stated gives a nice short proof (after a bit of practice, the reasoning I explained above in gory detail will become natural, so you'll be able to directly jump to my paragraph above... so this really is a short solution). However, suppose that you didn't know about that theorem. Then how would you go about finding the set of $p$ where $f_{*p} = 0$?

Well, the answer is very simple and straight forward (perhaps algebraically more tedious if you don't remember the charts). The sphere $S^2$ is a manifold, and as such, it has charts. The sphere is so nice that it can be covered by $2$-charts, (if you use stereographic projection).

Consider the stereographic projection from the north-pole: let $U_N = S^2 \setminus \{(0,0,1)\}$, and $\sigma_N : U_N \to \Bbb{R}^2$ \begin{align} \sigma_N(x,y,z) = \left( \dfrac{x}{1-z}, \dfrac{y}{1-z} \right) \end{align} Its inverse is $\sigma_N^{-1}: \Bbb{R}^2 \to U_N$ \begin{align} \sigma_N^{-1}(\xi,\eta) = \left( \dfrac{2\xi}{\xi^2 + \eta^2 +1}, \dfrac{2\eta}{\xi^2 + \eta^2 +1}, \dfrac{\xi^2 + \eta^2 - 1}{\xi^2 + \eta^2 +1} \right) \end{align}

This chart covers the whole $S^2$ except the north-pole $(0,0,1)$. Now, it should be easy enough to verify that for $p \in U_N$, $f_{*,p} = 0$ if and only if $D(f \circ \sigma_N^{-1})_{\sigma_N(p)} = 0$. Or said differently, $f_{*, \sigma_N^{-1}(\xi,\eta)} = 0$ if and only if $D(f \circ \sigma_N^{-1})_{(\xi,\eta)} = 0$. But \begin{align} f \circ \sigma_N^{-1}(\xi, \eta) = \left( \dfrac{\xi^2 + \eta^2 - 1}{\xi^2 + \eta^2 +1} \right)^2 \end{align} So, it should be easy to compute the standard derivative, and find where it vanishes. Then lastly, you just have to see if $f_{*, (0,0,1)} = 0$. To do this, you have to choose a chart which covers the north pole; you could use the stereographic projection from the south pole, or you could instead use the much simpler "graph chart" given by $V_{z,+} = \left\{(x,y,z) \in S^2| \, z>0 \right\}$ and $\psi_{z,+} : V_{z,+} \to \{(x,y)|\, x^2 + y^2 < 1\}$ given by \begin{align} \psi_{z,+}(x,y,z) = (x,y) \end{align} Note that I restricted the domain and target so that this is invertible, with inverse \begin{align} \psi_{z,+}^{-1}(x,y) = (x,y,\sqrt{1-x^2-y^2}) \end{align} (I chose postive square root because of the definition of $V_{z,+}$). Hence, in this case, \begin{align} (f \circ \psi_{z,+}^{-1})(x,y) = 1-x^2-y^2 \end{align} So, $f_{*, (0,0,1)} = 0$ if and only if $D(f \circ \psi_{z,+}^{-1})_{\psi_{z,+}(0,0,1)} = D(f \circ \psi_{z,+}^{-1})_{(0,0)} = 0$. Again, it should be easy enough to verify whether or not this condition is satisfied.

To recap: if you did not know that theorem, you just find an atlas for the manifold (and for computational purposes, find one with the fewest/simplest charts). Then, any property you wish to investigate about the push-forward $f_{*p}$ can be phrased equivalently in terms of the derivatives of the chart-representative maps, and solve the question in the chart (this is useful in general too).

For instance, if you hada slightly tougher question, say you're given some map $g: S^3 \to \Bbb{R}^4$, and you were asked to find all points where $g_{*p}$ had full rank, then I think a coordinate approach would be very mechanical and straight-forward. (As much as possible, it is a good idea to avoid charts, but it is also good to get used to them, because sometimes, they can provide a much quicker solution.)

  • $\begingroup$ Thanks for the thorough explanation, this clarifies a lot. One thing I'm not sure about though is the statement that if a submanifold $N$ can be written as a level set $h^{-1}(\{c\})$ that its tangent space at $p$ is equal to the kernel of $(h_*)_p$. Is this easy to see or does it require some work to prove? $\endgroup$ – Jarne Renders Nov 7 '19 at 13:44
  • $\begingroup$ @JarneRenders Its a standard, but usually technical theorem (usually called the preimage theorem/ regular-value theorem) which is proven in virtually any book on differential geometry. The difficulty comes in because it uses the inverse/implicit function theorems to prove that (if $c$ is a regular value of $h$ then ) $h^{-1}(\{c\})$ is actually a submanifold of the domain of $h$. It is easy to prove the claim on tangent spaces, as follows: let $[\gamma] \in T_p h^{-1}(\{c\})$. Then, $h_{*p}([\gamma]) = [h \circ \gamma] = [t \mapsto c]$, which is a constant curve, and hence the zero vector $\endgroup$ – peek-a-boo Nov 7 '19 at 13:52
  • $\begingroup$ Equivalence classes of constant curves are the zero vector of the tangent space because if you use the chart-induced isomorphism $\Phi$ I described in my answer, then you'll see that this curve gets mapped to $0 \in \Bbb{R}^n$. But if a linear isomorphism maps something to zero, then that something must have already been the zero of the vector space. What this argument shows is that $T_p h^{-1}(\{c\}) \subseteq \ker h_{*p}$. Now, when actually proving that $h^{-1}(\{c\})$ is a submanifold, the proof will tell you the dimension of this manifold. It turns out that $\endgroup$ – peek-a-boo Nov 7 '19 at 13:55
  • 1
    $\begingroup$ It turns out that the dimension of $h^{-1}(\{c\})$ as a manifold equals the dimension of $\ker(h_{*p})$ (as a vector space). Since a manifold and its tangent space have same dimension it follows that $\dim T_ph^{-1}(\{c\}) = \dim \ker h_{*p}$. Hence, we have equality of these vector spaces (not just inclusion). So... once again, the tough (depending on your background) part is proving the level set is a manifold. The rest is an easy corollary $\endgroup$ – peek-a-boo Nov 7 '19 at 13:58
  • $\begingroup$ @JarneRenders By the way, you might find this old answer of mine math.stackexchange.com/questions/3262559/… useful as well (I address a similar issue about computing the push-forward map) $\endgroup$ – peek-a-boo Nov 7 '19 at 14:03

Consider the curves on $S^2$ that go from "south" pole to "north" pole in such a way that $z$ changes montonically from $-1$ to $1$ along the curve. You could use $z$ to parameterise any one of these curves. How does $f(z)=z^2$ vary with $z$ along one of these curves ? Where is $\frac{df}{dz}$ equal to $0$ ?

  • $\begingroup$ $f(z)$ would by $1$ for $z = -1$ go to $0$ for $z = 0$ and back to $1$ for $z = 1$. $\dfrac{df}{dz}$ is $0$ for $z = 0$, so I guess for the set $\{(x,y,z)\in \mathbb{R}^2\mid x^2+y^2 = 1, z=0 \}$. But aren't there a lot more curves than this? I still feel I lack some intuition, these are some curves in $S^2$ but which of these are equivalent and is it enough to look only at these curves? $\endgroup$ – Jarne Renders Nov 7 '19 at 12:56

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