0
$\begingroup$

Spivak defines manifolds in this way:

A subset $M$ of $\mathbf R^n$ is called a $k$-dimensional manifold (in $\mathbf R^n$) if for every point $x\in M$ the following condition is satisfied:

$(M)$ There is an open set $U$ containing $x$, an open set $V\subset \mathbf R^n$, and a diffeomorphism $h:U\to V$ such that $$h(U\cap M)=V\cap(\mathbf R^k\times\{0\})=\{y\in V: y^{k+1}=\dots=y^n=0 \}.$$

And gives this theorem:

5-2 $\,\,\,$ Theorem. $\textit{A subset}$ $M$ $\textit{of}$ $\mathbf{R}^n$ $\textit{is a}$ $k\textit{-dimensional manifold}$ $\textit{if and}$ $\textit{only if}$ $\textit{for each point}$ $x\in M$ $\textit{the following}$ $\textit{“coordinate condition”}$ $\textit{is satisfied:}$

$\quad (C)$ $\textit{There is}$ $\textit{an open set}$ $U$ $\textit{containing}$ $x\textit{, an}$ $\textit{open set}$ $W\subset\mathbf{R}^k\textit{,}$ $\textit{and a}$ $1\textit{-}1$ $\textit{differentiable}$ $\textit{function}$ $f:W\to\mathbf{R}^n$ $\textit{such that}$ $$\begin{align}(1) \ & f(W)=M\cap U,\\ (2)\ & f'(y)\textit{ has rank }k\textit{ for each }y\in W,\\ (3) \ & f^{-1}:f(W)\to W\textit{ is continuous}.\end{align}$$

He then asks to make up a counterexample to Theorem 5-2 if condition $(3)$ is omitted and as a hint says "wrap an open interval into a figure six". But I don't have any idea on how to do that, nor do I have other ideas for counterexamples.

The only idea that I had is to translate Spivak's condition $(C)$ into more comprehensible (i.e., more comprehensible/familiar for/to me) language (see my other question) and then figure out what condition $(3)$ means in terms of the setting which I'm familiar with. But probably it's easier to do this problem directly, I don't know.

$\endgroup$
  • $\begingroup$ Can you see what his hint means pictorially even if you can't write down an explicit function? $\endgroup$ – Ted Shifrin Apr 17 '18 at 22:13
  • $\begingroup$ @TedShifrin: I can see that "6" (the closed figure 6) is not a 1-dimensional manifold (in the most general definition). $\endgroup$ – user531232 Apr 17 '18 at 22:17
  • $\begingroup$ So you need a mapping from an open interval $(a,b)$ to the plane so that the "junction point" of the 6 is $f(c)$ and $\lim\limits_{t\to b^-} f(t) = f(c)$? $\endgroup$ – Ted Shifrin Apr 17 '18 at 22:24
  • $\begingroup$ Well, if this condition holds, then $f$ will fail to be continuous at $b$ from the right, but I don't see why is that what we need. We need $f$ to have nonzero derivative on the whole $(a,b)$ and its inverse must be discontinuous at some point. $\endgroup$ – user531232 Apr 17 '18 at 22:47
1
$\begingroup$

See this question. The interval $(-\pi,\pi)$ is "wrapped" to form a figure eight via the map $(-\pi,\pi) \to \mathbb{R}^2$, $t \mapsto (\sin 2t, \sin t)$. This map is one-to-one and differentiable.

However, the inverse map from the image to $(-\pi,\pi)$ fails to be continuous for open neighborhoods of the "crossover point." This is easiest to think about in terms of limits. There are sequences in the image converging toward the crossover point, but the termwise preimage of the sequence coverges to $\pi$ or $-\pi$, not the preimage of the crossover point, which is zero.

EDIT: Answer to question in comment. Let $a_n$ be a sequence in $(-\pi,\pi)$ converging to $\pi$, for example $a_n = \pi - \frac{1}{n}$. Then $f(a_n)$ is a sequence in the figure-eight converging to the origin. Note that $f$ is injective, so it is bijective when we restrict the range to be the image of $f$. Also note that $f(0)$ is the origin. If $f^{-1}$ is continuous, then $$ \lim_{n \to \infty} f^{-1}(f(a_n) ) = f^{-1} \big(\lim_{n \to \infty} f(a_n)\big) $$ The LHS is $\lim a_n = \pi$, but the RHS is $f^{-1}(0,0) = 0$, so $f^{-1}$ can't be continuous.

$\endgroup$
  • $\begingroup$ How do I see that the preimage of such a sequence converges to $\pm \pi$? $\endgroup$ – user531232 Apr 17 '18 at 22:51
1
$\begingroup$

Consider$$\begin{array}{rccc}f\colon&(-1,1)&\longrightarrow&\mathbb{R}^2\\&t&\mapsto&\begin{cases}(t,0)&\text{ if }t<0\\\left(\cos\left(-\frac\pi2+2\pi t\right),1+\sin\left(-\frac\pi2+2\pi t\right)\right)&\text{ otherwise.}\end{cases}\end{array}$$Then $f\bigl((-1,1)\bigr)$ is the union of the segment joining $(-1,0)$ to $(0,0)$ together with the circle centered at $(0,1)$ with radius $1$. It is not a manifold because of the point $(0,0)$. If you remoe it, then, yes, it is a manifold.

$\endgroup$
  • $\begingroup$ But $(C)$ says that the map should be from an open subset of $\mathbf R^k$, so probably you need to take the open interval? $\endgroup$ – user531232 Apr 17 '18 at 22:24
  • $\begingroup$ @user531299 Sure. My mistake. I've edited my answer. $\endgroup$ – José Carlos Santos Apr 17 '18 at 22:26
  • $\begingroup$ Thanks. It seems to be a hassle to check that this piecewise defined function has a nonzero derivative everywhere... $\endgroup$ – user531232 Apr 17 '18 at 22:49
  • $\begingroup$ @user531299 Did you try it? It's really easy. $\endgroup$ – José Carlos Santos Apr 17 '18 at 22:54

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.