2
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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, $p\in\mathbb S^1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.