# $D^m\cup_h D^m$, joining $D^m \amalg D^m$ along the boundary $\partial D^m$

Given an orientation-preserving diffeomorphism $h: \partial D^m \to \partial D^m$, we can glue two copies of the closed unit disk $D^m$ along the boundary by identifying $x \sim h(x)$ to form the quotient space $$\Sigma(h) := (D^m \amalg D^m)/\sim$$ Now we can give this quotient a smooth structure such that the obvious inclusions $D^m \hookrightarrow \Sigma(h)$ are smooth embeddings and in fact it turns out that for any two smooth structures, there exists a diffeomorphism between them. So $\Sigma(h)$ is a unique manifold up to diffeomorphism.

So far, so good. Now, in Kosinski's 'Differential Manifolds', there is the following Lemma:

Lemma: $\Sigma(h)$ is diffeomorphic to $S^m$ if and only if $h$ extends over $D^m$. Moreover, $\Sigma(gh) = \Sigma(h)\# \Sigma(g)$.

Here $M\# N$ denotes the connected sum of two manifolds as usual.

The proof of this is left as an exercise for the reader, but I'm unsure, how one might construct an extension of $h$ to $D^m$, given that $\Sigma(h)$ is diffeomorphic to $S^m$?

I know that in this case $h(\partial D^m)$ necessarily separates $\Sigma(h) = S^m$ into two components, and since $h(\partial D^m)$ is an embedded compact $(m-1)$-manifold (which is smooth), I can also prove that $h(\partial D^m)$ is the boundary of both connected components of its complement.

But at this point I get lost. Is it clear that these two components are diffeomorphic to disks? Where might I find a proof of this?

I'm fine with the other parts of this Lemma, but I just don't see how to extend $h$ given $\Sigma(h) = S^m$. If you could help me out, this would be very much appreciated. Thank you for your help!

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– john mangual Oct 29 '12 at 19:27

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– john mangual Oct 29 '12 at 19:49
@johnmangual: Thank you for these links! Interesting stuff. – Sam Oct 29 '12 at 20:01
Finally, I think your components are m-disks by construction Since $h(\partial \mathbb{D}^m) = \partial \mathbb{D}^m$ you have $S^m \backslash h(\partial \mathbb{D}^m) \simeq S^m \backslash \partial \mathbb{D}^m$. So it's pretty clear that splits into two disks. I think it's possible to address all the parts of your question now. – john mangual Oct 29 '12 at 20:05
@john mangual: Sam's question is one of the key steps in setting up the bijective correspondence between homotopy-spheres (up to diffeomorphism) and the mapping class group of lower-dimensional (but standard) spheres. This is the beginning of the Smale-Milnor-Kervaire machine for computing the group of homotopy spheres in dimensions $n \geq 5$. – Ryan Budney Nov 3 '12 at 19:25

## 2 Answers

If your sphere $\Sigma(h)$ is diffeomorphic to a standard sphere, consider a diffeomorphism $\Sigma(h) \to S^m$. The two discs in $\Sigma(h)$ are smooth discs, so after applying the diffeomorphism, they're smooth discs in $S^m$. But smooth discs are tubular neighbourhoods of their centres. So they're unique up to embedded isotopy. In particular, this means that by the isotopy extension theorem you can isotope your diffeomorphism $\Sigma(h) \to S^m$ so that it sends the `bottom' disc $D^m$ in $\Sigma(h)$ to the lower hemi-sphere of $S^m$, moreover, you can ensure your diffeo $\Sigma(h) \to S^m$ is a standard diffeomorphism between the lower $D^m$ and the lower hemi-sphere. But now the upper $D^m$ in $\Sigma(h)$ is identified (via a diffeomorphism) with the upper hemi-sphere in $S^m$. So compose that map with a standard diffeomorphism between the upper-hemisphere and a $D^m$. Provided you choose it appropriately on the boundary, this is by design your extension of $h : \partial D^m \to \partial D^m$ to a diffeomorphism $\overline{h} : D^m \to D^m$.

So above, when I talk about `standard' diffeomorphisms between $D^m$ and the lower/upper hemi-spheres of $S^m$ what I mean is that $\partial D^m \times \{0\} = \partial H$ (set equality) where $H \subset S^m$ is either the upper or lower hemi-sphere in $S^m$. So to be standard I mean the diffeo must be the identity on the boundary in this sense.

Cerf went further than this, his pseudo-isotopy theorem now says that $h$ is isotopic to the identity on $\partial D^m$, provided $m \geq 6$.

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This is a great explanation! Very clear. What I was missing was the key step of moving one of the two hemi-spheres by an isotopy to make it 'nice' again. Thank you ever so much for writing it up for me and explaining it so well! – Sam Nov 3 '12 at 22:21

Topologically $\mathbb{D}^m \amalg \mathbb{D}^m / \sim$ is the m-sphere. Let's see if we can get this identification to agree with the smooth structures. Kosinski's book helped.

If $h:\partial\mathbb{D}^m \to \partial \mathbb{D}^m$ extends to a diffeomorphism $h:\mathbb{D}^m \to \mathbb{D}^m$, then there is a diffeomorphism

$$\mathbb{D}^m \amalg_h \mathbb{D}^m \simeq \mathbb{D}^m \amalg_{id} \mathbb{D}^m \simeq S^m$$

fixing the first disk and letting $h$ act on the second disc and its boundary. This map is a homeomorphism and differentiable.

If $\mathbb{D}^m \amalg_h \mathbb{D}^m \simeq S^m$ how do we get an extension? Using whatever diffeomorphism we have, $$S^m \backslash h(\partial \mathbb{D}^m) \simeq S^m \backslash \partial \mathbb{D}^m = \mathbb{D}^m \amalg \mathbb{D}^m$$ so we get a disjoint union of two disks.
Every diffeomorphism of $S^1 \to S^1$ can be extended to the disk. Such a map as a Fourier expansion: $$\sum a_n e^{i n \theta} \to \sum a_n r^n e^{i n \theta} \text{ with }|r|<1$$ so we have a map from $\mathbb{D} \to \mathbb{D}$ (by Cauchy-Schwartz).

Diffeomorphisms from $S^2 \to S^2$ extend to $\mathbb{D}^3$ this was proven by Munkres and by Smale.

In the comments of Tim Gower's blog, Greg Kuperberg explains that not all maps $S^6 \to S^6$ extend to $\mathbb{D}^7$. This has to do with the exotic structures on the 7-sphere, by Milnor.

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Thank you for your answer. I'm still not sure as to how exactly you get the extension of $h$. Could you maybe write out the details? – Sam Oct 30 '12 at 13:27
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