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I recently realized that I don't know any non-linear diffeomorphisms of the plane (or $\mathbb{R}^n$ in general) except for linear ones, so I want to ask rather broad questions hoping to be pointed to the appropriate literature.

1) Are there simple ways of constructing autodiffeomorphisms of $\mathbb{R}^n$ that can be expressed in closed form? UPD: Ok, there's e.g. $(x, y) \mapsto (x, y + f(x))$, where $f: \mathbb{R} \to \mathbb{R}$ is smooth, so I call this one back - kind of, if you know some exciting and unusual family, feel free to share :)
2) Is every autodiffeomorphism of $\mathbb{R}^n$ isotopic to a linear one? Obviously, every one is homotopic to any other due to $\mathbb{R}^n$ being contractible, but since $\mathrm{GL}(\mathbb{R}^n)$ is not connected, $\mathbb{R}^n$ is a k-space), and homotopies behave nicely under differentials at a point, even some linear autodiffeomorphisms are not isotopic, if I'm not mistaken.

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Maybe you should try harder to find examples! People will come and give you all sorts of different families... and that will ruin your finding one! – Mariano Suárez-Alvarez Nov 5 '11 at 18:57
(I never cease to marvel at the fact that people somehow manage to end up knowing what isotopies of differomorphisms are, without ever having seen a non-linear diffeo of the plane! :( ) – Mariano Suárez-Alvarez Nov 5 '11 at 18:59
@Henning, yes, I figured this kind out by myself when Mariano motivated me with public humiliation :) – Alexei Averchenko Nov 5 '11 at 19:14
@Alexei: please don't read my comment as a comment on you, but on the way we teach math. In any case, great you found those triangular examples! You may be interested in knowing that—if we restrict to polynomial diffeos with polynomial inverse— in the plane all examples are compositions of linear maps and triangular ones. It was very recently shown that in three dimensions this is not true. This is a very classical problem, of famous difficulty. – Mariano Suárez-Alvarez Nov 7 '11 at 13:17
up vote 4 down vote accepted

The answer to your question (2) is yes.

The proof goes like this. Let $f$ be a diffeomorphism. We find an isotopy from $f$ to a diffeomorphism $g$ with $g(0)=0$. The isotopy is

$$(x,t) \longmapsto f(x)-tf(0) $$

when $t=0$ this is $f$, when $t=1$ you get $f(x)-f(0)$.

Next, given a diffeo $g$ with $g(0)=0$ we isotope $g$ to a linear diffeomorphism. The map is this:

$$(x,t) \longmapsto \frac{g((1-t)x)}{1-t}$$

for $t \in [0,1)$ and at $t=1$ we have

$$(x,1) \longmapsto Dg_0(x)$$

You can check this map is continuous provided $g$ is $C^1$.

Regarding your 1st question, one of the most common techniques is to integrate vector fields.

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Consider $\mathbb{R}^2$ first. Let $f$ be a smooth function on $\mathbb{R}^2$. If we consider a matrix $$ \begin{bmatrix} \cos(f(x,y)) & -\sin(f(x,y))\\ \sin(f(x,y)) & \cos(f(x,y)) \end{bmatrix}$$ This is non linear map and it gives a diffeomorphism on $\mathbb{R}^2$. Using these $2\times2$ blocks we can build diffeomorphisms on $\mathbb{R}^n$ just like we build rotations from the $2\times2$ blocks. We can use hyperbolic cos and sine also. $$ \begin{bmatrix} \cosh(f(x,y)) & \sinh(f(x,y))\\ \sinh(f(x,y)) & \cosh(f(x,y)) \end{bmatrix}$$ Another set of non linear diffeomorphisms are given by, $$\begin{bmatrix} 1 & f(x,y)\\0 & 1\end{bmatrix}$$ we can choose any non zero constants on the diagonals instead of $1$ and can have lower triangular matrices also. Product of all these types also gives non linear diffeomorphisms.

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