I Came across this following example from previous year class notes: let $f:S^2\rightarrow\mathbb{R} f(x,y,z)=z$ so $f$ is the restriction of $F:\mathbb{R}^3\rightarrow\mathbb{R}$ given by $F(x,y,z)=z$ to the sphere $S^2$. Therefor $F$ is linear transformation and $$DF|_p(v_1 ,v_2, v_3)=F(v_1 ,v_2, v_3)=v_3$$ for every $p\in \mathbb{R}^3$, in other words the differential $DF|_p$ in the regular basis is $$DF|p=(0,0,1)$$ and therefor exists that $$df|_p(v_1 ,v_2, v_3)=DF|_p(v_1 ,v_2, v_3)=v_3$$ My questions:
- Why does the last equation holds? $df|_p(v_1 ,v_2, v_3)=DF|_p(v_1 ,v_2, v_3)=v_3$
- I think that there should be $p$'s s.t. $df|_p=0$ because the poles of the sphere are critical points for $f$ but on the other hand $DF|_p$ isn't the zero vector for every $p$
The last equation doesn't hold. $df_p$ is a function from the $2$-dimensional space $TS^2_p$ to the $1$-dimensional $T\mathbb{R}_{f(p)}$. For that reason, it is not clear what the notation $df_p(v_1,v_2,v_3)$ means.
There are indeed points where $df_p=0$. Assume that $p=(0,0,1)$ (as you said, the other point is $(0,0,-1)$). Then a tangent vector at $p$ is an equivalence class of smooth curves $g=(g_1,g_2,g_3):\mathbb{R}\to S^2$ such that $g(0)=p$. This means, in particular, that $g_3\leq1$ and $g_3(0)=1$. Then $df_p([g])=[f\circ g]_p=[g_3]_p$. Since $g_3$ has a maximum at $t=0$, the derivative of $g_3$ at $t=0$ is $0$. Therefore, $[g_3]_p=[1]_p$, which is the zero in the tangent space of $\mathbb{R}$ at $f(p)=1$.
So, $df_{p}$, for $p=(0,0,1)$ sends all tangent vectors at $p$ to the zero tangent vector to $\mathbb{R}$ at $1$.
On the other hand $dF_p$, for $p=(0,0,1)$ is not the zero linear map. Consider the class of $g(t)=(0,0,1+t)$ in $T\mathbb{R}^3_{p}$. Then $dF_p([g])=[F\circ g]=[1+t]$, which is not the same class as $[1]$ because the derivative of $1+t$ is $1$, which is different from the derivative of $1$, which is $0$. Therefore $dF_p$ is not the zero function.