# $\{(x,y)\!\in\!\mathbb{B}^n; -\varepsilon\leq-\|x\|^2\!+\!\|y\|^2\leq\varepsilon\}\approx\mathbb{B}^k\!\times\!\mathbb{B}^{n-k}$

The question is motivated by the notion of handle attachment, Morse theory, critical points of index $k$, Morse lemma, sublevel sets, etc.

For $0\!\leq\!k\!\leq\!n$ and $0\!<\!\varepsilon\!<\!1$, how can I find a homeomorphism $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\{x\!\in\!\mathbb{B}^n;\, -\varepsilon\leq-\sum_{i=1}^k\!x_i^2+\!\sum_{i=k+1}^n\!x_i^2\leq\varepsilon\}\;\;\approx\;\;\mathbb{B}^k\!\times\!\mathbb{B}^{n-k}\;\;\;\text{ that sends}$$ $$\{x\!\in\!\mathbb{B}^n;\, -\varepsilon=-\sum_{i=1}^k\!x_i^2+\!\sum_{i=k+1}^n\!x_i^2\}\;\;\approx\;\;\mathbb{S}^{k-1}\!\times\!\mathbb{B}^{n-k}\;\;\;?$$

The case $n\!=\!2$, $k\!=\!1$:

The case $n\!=\!3$, $k\!=\!1$:

The case $n\!=\!3$, $k\!=\!2$:

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I have no idea why you ask for a homeomorphism. Your pairs of spaces are not homeomorphic. – Will Jagy Sep 7 '12 at 0:05
@WillJagy: Do you have a counter-example? – Dejan Govc Sep 7 '12 at 8:07
@DejanGovc, for the first two pairs, one is connected and the other not. In the third pair, one is contractible and the other has first homotopy group $\pi_1 = \mathbb Z.$ – Will Jagy Sep 7 '12 at 18:50
@WillJagy: Yes, but I believe the question is asking how to find a homeomorphism $h:X_{n,k}\to\Bbb B^k\times\Bbb B^{n-k}$ such that the restriction $h|_{Y_{n,k}}:Y_{n,k}\to\Bbb S^{k-1}\times\Bbb B^{n-k}$ is also a homeomorphism. (Here $Y_{n,k}\subseteq X_{n,k}$ are the pairs of spaces pictured above.) – Dejan Govc Sep 7 '12 at 21:02
@DejanGovc, in that case a different presentation of the question would have helped, such as the one in your latest comment. – Will Jagy Sep 7 '12 at 23:39

Denote $(x_1, \ldots, x_k)= Y$ and $(x_{k+1}, \ldots, x_n)= X$. Then the region is given by $-\epsilon \leq |X|^2-|Y|^2 \leq \epsilon$ and $|Y|^2+|X|^2 \leq 1$.

Here is an almost-solution, which unfortunately is not as straigthforwrd as I would like due to some technical issues, but the general idea should hopefully be clear and may be helpful.

We proceed in two parts. First part, is to say that the region in the problem is homeomorphic to the region $-\epsilon \leq |X|^2-|Y|^2 \leq \epsilon$ - without the other inequality - after compactifing it at infinity; this compactified region we shall call $R$.

The next step is to provide the needed homeomorphism from the new region $R$ to products of balls as desired (described below). Finally, one needs to see that the composition of the two homeomorphisms does the right thing on the relevant boundary.

For the homeomorpism from $R$ to the product of balls, the idea is to provide a diffeomorphism to $\mathbb{B}^1 \times \mathbb{B}^1$ for the case of $k=1$ $n=2$ and then use the $O(k) \times O(n-k)$ symmetry to extend it to the general case.

Let's start with the $k=1$ and $n=2$ case first. In this case $R$ is obtained by adding 4 "corner" points at infinity.

We have $-\epsilon \leq x^2-y^2 \leq \epsilon$ which we want to map to $\mathbb{B}^1$ in two "good" ways, so that we get a map to $\mathbb{B}^1 \times \mathbb{B}^1$. We are going to give one map, and the other will be obtained by switching $x$ and $y$.

Here by "good" we mean, for one, that the preimage of $0$ for the first map is the subset $y=0$, and the boundary $x^2-y^2 = -\epsilon$ is mapped to $\pm 1$. This suggests writing $y=0$ as $-y^2=0$ and interpolating between the two by $tx^2-y^2=-\epsilon t$, so that $t= \frac{y^2}{x^2+\epsilon}$ (here one would do well to draw a picture of the family of interpolating curves between $y=0$ and $x^2-y^2=-\epsilon$). This $t$ is between $0$ and $1$ and is a good definition for $y \geq 0$, but since $t$ is invariant under $y$ going to $-y$ it does not work well for $y<0$. To fix this we define $g_1(x,y)$ to be $t$ if $y\geq 0$ and $-t$ if $y<0$. Note that this is continuous and is $O(1)=\mathbb{Z}_2$ equivariant. We define $f_1$ as the restriction of $g_1$ to the region we started with, that is one with additional condition $x^2+y^2\leq 1$.

Further, we define $f_2(x,y)=f_1(y,x)$. We claim that $F=(f_1, f_2)$ is a homeomorphism to $\mathbb{B}^1 \times \mathbb{B}^1$. Indeed, recovering $(x,y)$ from $(t,s)$ requires solving a linear system for $x^2$ and $y^2$ which has determinant $\pm ts-1$ which is non-0 outside the 4 corners where $|t|=|s|=1$, and the corners get mapped to the 4 compactification points at infinity. One checks the continuity of this inverse map separately near the 4 corners and everywhere else.

Now, for the general case.

The compactification $R$ that is needed is the addition of the "corner stratum" $\mathbb{S}^{k-1} \times \mathbb{S}^{n-1}$, that is the $|X|^2=N+\epsilon$, $|Y|^2=N$ for large $N$ (more precisely, one would define $R$ as the original region disjoint union $\mathbb{S}^{k-1} \times \mathbb{S}^{n-1}$ and put the correct topology that makes $\mathbb{S}^{k-1} \times \mathbb{S}^{n-1}$ lie at infinity).

Let $H= (\frac{Y}{|Y|} f_1(|X|, |Y|), \frac{X}{|X|} f_2(|X|, |Y|))$. Note that this is a map to $\mathbb{B}^k \times \mathbb{B}^n$, that restricted to $|X|^2- |Y|^2=-\epsilon$ it maps to $\mathbb{S}^{k-1} \times \mathbb{B}^n$ (and that if $k=1$ and $n=2$ it gives back $F$). It is a homeomorphism, for essentially the same reasons $F$ was.

Note: All of this together does give the desired homomorphism. However, if I were to rewrite it from scratch, I would probably just restrict the map $H$ to the original domain $-\epsilon \leq |X|^2-|Y|^2 \leq \epsilon$ and $|Y|^2+|X|^2 \leq 1$, then claim that its a homeomorphism to its image inside $\mathbb{B}^k \times \mathbb{B}^n$ and then claim that the image is homeomorphic to all of $\mathbb{B}^k \times \mathbb{B}^n$.

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I was about to write a similar explanation, but was hesitating b/c of this corner smoothing issue. I don't know if you've noticed, Leon and LVK, but most topology books waive their hands at smooth handle attaching. It's not hard (as reading Max's explanation will make clear), but writing down every detail becomes a headache. – Sam Lisi Sep 7 '12 at 18:57
@Max: I'm in a really really tight spot time-wise. I will check your answer in a week, if that's ok. I don't really know how these bounties work, I hope you won't be robbed of your prize, If I accept after some time. – Leon Sep 8 '12 at 14:24
@Leon Lampert: I don't know how these work either. I seem to have gotten it (probably by LVK's doing). It's not 100% polished answer in any case, so I would not feel "robbed" if there is no bounty - perhaps the points can attract someone else to write out more details, if LVK decides that this is something to do. Though I'd be glad to discuss any particular point that seems to be of interest. – Max Sep 8 '12 at 16:05
@LeonLampret Bounties are independent of acceptance. I awarded the bounty since Max's answer was obviously the best given within the timeframe. I know that writing down every detail of a such a diffeomorphism is a headache, and likely to be a reader's headache as well... – user31373 Sep 23 '12 at 2:28

I can find an answer for the special cases you listed here but I'm having a tough time generalizing to arbitrary $n$ and $k$. Hopefully this will at least help a bit. In what follows, let $\beta = \sqrt{(1+\epsilon)/2}$.

Let $\xi^{2} = \sum_{i=2}^{n} x_{i}^{2}$ and we will call the hyperboloid sheet $M$. Whenever $k = 1$ there is an obvious strategy to define a homeomorphism: You want to flatten out your hyperboloid sheets. The hyperboloid is given by the equation $x_{1}^{2} - \xi^{2} = \epsilon$, so re-arranging gives:

$x_{1} = \pm\sqrt{\epsilon + \xi^{2}}$

We can now define the homeomorphism and its inverse:

$\varphi: M\to S^{0}\times B^{n-1}$

$\varphi(x_{1},\dots,x_{n}) = (\mathrm{sgn}x_{1}, x_{2},\dots,x_{n})$

The last $n-2$ coordinates do indeed map to a ball since $\xi^{2} \leq \beta^{2}$. The inverse can be easily checked to be:

$\varphi^{-1}(\pm 1, x_{1},\dots,x_{n}) = (\pm \sqrt{\epsilon+\xi^{2}},x_{2},\dots,x_{n})$.

Now if $k = 2$ it is slightly trickier. I will do the case of $n=3$ for concreteness. I'll call my coordinates $(x,y,z)$ and to be in $M$ we must satisfy $x^{2} +y^{2} - z^{2} = \epsilon$. Let's find the intersection of $B^{3}$ with $M$: We must have $x^{2}+y^{2}+z^{2} = 1$ and $x^{2}+y^{2}-z^{2} = \epsilon$, so the intersection should satisfy $x^{2} + y^{2} = \beta^{2}$.

I would like to find a $c$ such that $c^{2}(x^{2}+y^{2}) = \beta^{2}$ for any $(x,y,z) \in M$. Notice that:

$c^{2}(x^{2}+y^{2}) = c^{2}(z^{2}+\epsilon) = \beta^{2} \Longrightarrow c = \sqrt{\frac{\beta}{z^{2}+\epsilon}}$

Now I can define my homeomorphism:

$\psi: M \to S^{1}\times B^{1}$

$\psi(x,y,z) = (cx,cy,z)$

Where $\vert z \vert < \sqrt{\frac{1-\epsilon}{2}}$ and $(cx)^{2} + (cy)^{2} = \beta^{2}$ so we are mapping to where we want. The inverse to this map is the obvious one.

Anyway hopefully this helps a bit!

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