I am reading Milnor's book "Topology from the Differentiable Viewpoint" and have a question of regular values. So, he defines a regular point as follows.

Let $f :M \to N $ be a smooth map between smooth manifolds of the same dimension. A point $p \in M$ is a $regular$ $point$ of $f$ if the derivative $df_{p}$ is nonsingular. Now, my question is the following. I am trying to find the regular points of the mapping

$f:S^{2} \to \mathbb{R}^{2}$ defined by $f(x,y,z)=(x,y).$

I wish to show that the set $\{(x,y,z) \in \mathbb{R}^{3}:z \neq 0\}$ is the set of regular points, and moreover points on $S^{1}$ are the critical points. Any helpful hints are greatly appreciated, thank you.

  • 3
    $\begingroup$ What is missing? $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '15 at 7:57
  • $\begingroup$ I can use the definition of limit, and I get that $df_{x}(h)=(x,y)$. This then implies that $df_{x}$ is onto and, moreover, as $df: \mathbb{R}^{2} \to \mathbb{R}^{2}$, then $df_{x}$ is one-to-one, namely; nonsingular. But I want to find the set of regular points. $\endgroup$ – Rene Cabrera Feb 18 '15 at 8:05
  • $\begingroup$ If you use the definition of limit, you should get $df_{\bf x}(h) = (h_x,h_y)$, where $h_x$ and $h_y$ are the $x-$ and $y-$components of $h$, respectively. This map is certainly singular at some points $\bf{x} \in S^2$ - namely, those points with $z-$coordinate equal to zero. Why? $\endgroup$ – user98602 Feb 18 '15 at 8:22
  • $\begingroup$ @MikeMiller, because any tangent space tangent to the equator of $S^{2}$, namely; $S^{1}$, there is a projection from $\mathbb{R}^{2}$ onto the line $\mathbb{R}$, making the derivative $df$ non-invertible. $\endgroup$ – Rene Cabrera Feb 18 '15 at 8:25
  • $\begingroup$ @MikeMiller, that formula for $df$ is not really true: the actual expression depends on what $\mathrm x$ is. $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '15 at 8:29

Recall the manner in which Milnor defines the differential of a map $f: X \to Y$ at a point $p$. To Milnor, all manifolds are embedded as subsets of some Euclidean space $\Bbb R^n$; we pick a smooth extension of $f$ to an open neighborhood of $X$. For us it will suffice to pick $f(x,y,z) = (x,y)$, where $f$ is now defined on the whole of $\Bbb R^3$. You should verify, using Milnor's definition, that the tangent space at a point $p \in S^2$ is precisely $T_pS^2 = \{h \in \Bbb R^3 | h \cdot p = 0\}.$ Now, if $h \in T_pS^2$, write for convenience $h = (h_x,h_y,h_z)$. By definition, $$df_p(h) = \lim_{t \to 0} \frac{f(p+th)-f(p)}{h} = \lim_{t \to 0}\frac{(th_x,th_y)}{h} = (h_x,h_y).$$ Because the codomain of $df_p$ is, as you mentioned, 2-dimensional, to find where $df_p$ is not surjective you need precisely to find what $p$ it's not injective for.

But $df_p(h_x,h_y,h_z) = 0$ iff $(h_x,h_y) = 0$; and the vectors $p$ such that $(0,0,1) \in T_pS^2$ are precisely those of the form $(x,y,0) \subset S^2$. These are the points in $S^2$ for which $f$ is not regular.

The intuition you should have for what's going on here is that in this situation, where we have a map between manifolds of the same dimension, that map $f$ has $p$ as a regular point precisely when $f$ is a diffeomorphism when restricted to a sufficiently small neighborhood of $p$. For points in the upper and lower hemispheres, this is true; but $f$ is not even injective on any neighborhood of a points on the equator.

  • $\begingroup$ where is your linear mapping $df_{p}$ defined on. By Milnor's definition applied to our question: for any smooth map $f: S^{2} \to \mathbb{R}^{2}$ with $f(x,y,z)=(x,y)$, the derivative $df_{p}$ at a point $p$ of $f$ is defined as follows. As $f:S^{2}\to \mathbb{R}^{2}$ is smooth, there exists an open set $U \subset \mathbb{R}^{3}$ containing $p$ and a smooth map $F:U \to \mathbb{R}^{2}$ that coincides with $f$ on the intersection $U \cap S^{2}$. Then $df_{p}(v)=dF_{p}(v)$ for all $v \in \mathbb{R}^{3}$. $\endgroup$ – Rene Cabrera Feb 19 '15 at 6:16
  • $\begingroup$ It's defined on the tangent space of the sphere $S^2$ at the point $p$. The definition in Milnor is on page 4 - it's not the same as when the domain is an open subset of $\Bbb R^n$. The definition is different when the domain is a submanifold. $\endgroup$ – user98602 Feb 19 '15 at 6:47
  • $\begingroup$ Okay, I am a bit confused. The tangent space of the sphere $S^{2}$ at a point $p$ is a plane $\mathbb{R}^{2}$, up to isomorphism. So then $T_{p}S^{2}=\mathbb{R}^{2}$. $\endgroup$ – Rene Cabrera Feb 19 '15 at 7:14
  • $\begingroup$ Also, does $T_{p}S^{2}=\{h \in \mathbb{R}^{3}:h \cdot p=0\}$ mean that at any point $p$ on $T_{p}S^{2}$, a vector $h$ is orthogonal to $p$? But $T_{p}S^{2}$ is equal to the image $df(\mathbb{R}^{2})$ $\endgroup$ – Rene Cabrera Feb 19 '15 at 7:48
  • $\begingroup$ may you please be more clear and specific on your hints/partial solution. Thank you. $\endgroup$ – Rene Cabrera Feb 19 '15 at 7:56

The smooth map $f:\mathbb S^2\to\mathbb R^2$ has a critical point at $p\in M$ iff the derivative $d_p:T_p\mathbb S^2\to T_{f(p)}\mathbb R^2$ is not surjective or equivalently here, is not injective. We are given that $f=F|\mathbb S^2$, with $F$ the linear projection. Thus $d_pf=d_pF|T_p\mathbb S^2$, hence $\ker(d_pf)=\ker(d_pF)\cap T_p\mathbb S^2$. Now: (i) $F$ is the linear projection, so that $F=d_pF$ and $\ker(d_pF)$ is generated by $(0,0,1)$, and (ii) $T_p\mathbb S^2$ is the plane perpendicular to $p$. Summing up, $p$ is singular if and only $(0,0,1)\in T_p\mathbb S^2$, that is $(0,0,1)$ is perpendicular to $p$, or in other words, $p\in\mathbb S^1$.


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