# Extending diffeomorphisms of some faces of the standard $n$- simplex to a diffeomorphism of the $n$-simplex

I would be grateful if you could help me with the following question:

Let $n\in \mathbb{N}$ and $k\in \{1, \ldots, n+1\}$. For every $i=1, ..., k$ let $f_i : \partial_i \Delta^n \rightarrow \partial_i \Delta^n$ be an orientation preserving diffeomorphism of the i-th face of the standard $n$-simplex $\Delta^n$. Suppose $f_i = f_j$ on $\partial_i \Delta^n \cap \partial_j \Delta^n$ for all $i,j$. Does there exist a diffeomorphism $f: \Delta^n \rightarrow \Delta^n$ restricting to $f_i$ on $\partial_i \Delta^n$ for every $i$ ? Simplices are considered as manifolds with corners.

I think I can solve it for $k=n+1$ when $f_i$ are time one flows of some vector fields by reducing to the problem of extending a smooth function from a closed set. However, I am interested in the case when $f_i$ are arbitrary diffeomorphisms and $k=n$ so that there should not be any topological obstructions for the extension. I do not know how to start proving it except for trying to write down an explicit formula for the extension.

The relatively simple issue is that even if $f_i$ and $f_j$ agree on their common boundary, it may be impossible to extend them even locally to a smooth map in a neighbourhood of those boundary points. This is because the map may have a "corner" in its graph. The issue is the relative growth rate in the directions transverse to the common boundary. You can deal with this by pre-supposing local extensibility, or fixing a local normal growth rate, or a variety of fairly standard ways.
Once you've fixed that, there are more fundamental problems. In the $k=n+1$ case, the answer is generally no, it is impossible. This goes back to the work of Milnor-Kervaire. In dimensions $n \geq 6$ all homotopy-spheres can be represented as the gluing-together of two discs via a diffeomorphism of their boundary. In particular the orientation-preserving mapping class group of $S^{n-1}$ is isomorphic to the group of (oriented) homotopy-spheres in dimension $n$. You can represent such elements as a union of maps of your $f_i$ kind in the $k=n+1$ case. Extensibility in the level of generality you assert would be equivalent to the statement that there are no homotopy spheres.
When $k<n+1$ the answer is yes, it's possible to extend. For this you basically want to think of the $n$-disc as $D^n = D^{n-1} \times [0,1]$. You construct the extension by taking the product with the identity map. You'll have to be careful with how you identify your simplex with the disc to ensure you dont introduce corners into your map, but it can be done, with patience.
• Thank you very much for your helpful answer. I will keep thinking about the extension when $k < n + 1$ then. Btw. I do not understand the issue with smoothness at the corners: For instance, if $f, g: [0,1) \rightarrow \mathbb{R}$ are smooth functions with $f(0) = g(0)$ then $h(x,y) = f(x) + g(y) - f(0)$ is a smooth function $h: [0,1)^2 \rightarrow \mathbb{R}$ extending $f$ and $g$ from the boundary, right ? I can write similarly looking extensions in higher dimensions as well. – Pavel Jul 8 '14 at 15:26
• I am currently dealing with a different (but similar) issue. Here is a fact that is stated in Lee's book "Introduction to smooth manifolds". Lemma 16.8 says that, if $f\colon \partial\Delta \to M$ is a continuous map whose restriction to every face of $\Delta$ is smooth, then $f$ is smooth when considered as a map of the whole boundary $\partial \Delta$ into $M$. This statement does not deal with diffeomorphisms, but still ensures a smooth local extension of the $f_i$'s to a neighbourhood of $\partial \Delta$. – Roberto Frigerio Jul 10 '14 at 13:44
• Thank you very much for the reference. In fact, if $f: \partial \mathbb{R}^n_+ \rightarrow \mathbb{R}$ is a function smooth on $\partial_i \mathbb{R}^n_+ =\{x\in R^n_+ : x^i = 0\}$ for every $i$ then one can use the smooth extension $E(f)=\sum_{k=1}^n (-1)^{k+1}\sum_{\substack{I \subset \{1,\ldots,n\}}} f \circ \pi_I$ where $\pi_I$ is the obvious linear projection to $\partial_I \mathbb{R}^n_+ = \{x\in \mathbb{R}^n_+ : x^I = 0\}$. I guess this is what arises when one extends the Lee's inductive argument. – Pavel Jul 11 '14 at 8:45
• In addition, $\partial_i E(f) = E(\partial_i f)$. It follows, if $f: \partial\mathbb{R}^n_+ \rightarrow \partial\mathbb{R}^n_+$ restricts to or. pres. diffeos of $\partial_i \mathbb{R}^n_+$ for all $i$ then $\mathrm{d}E(f)(x)$ is inverible whenever $x$ lies in at least two $\partial_i\mathbb{R}^n_+$. Here, $E$ is applied componentwisely. The inverse function theorem for mfds with corners asserts $E(f)$ is a diffeo of nbhds of $x$ and $f(x)$ in $\mathbb{R}^n_+$. If $x$ lies in precisely one $\partial_i\mathbb{R}^n_+$ it is easy to write down a diffeo loc. extending $f$. Hence, locally it works. – Pavel Jul 11 '14 at 9:09