Let $D^2$ denote the closed unit disk in $\mathbb{R}^2$. Let $\omega := dx \wedge dy$ denote the standard area form on $\mathbb{R}^2$ (and on $D^2$ by restriction). Let $\phi$ be a diffeomorphism of $D^2$ which is equal to the identity in a neighborhood of $\partial D^2$, and which preserves area; i.e. $\phi^\* \omega = \omega$. There is a $1$-form $\alpha$ with $d\alpha = \omega$. We have $\phi^\*\alpha - \alpha$ is exact, and equal to $df$ for some smooth function $f$ which vanishes on $\partial D$. The *Calabi invariant* of $\phi$ is the integral$$C(\phi) = \int\_{D^2} f\omega.$$What is an example where the Calabi invariant is nontrivial?



Example where Calabi invariant is nontrivial?
There are equivalent definitions of the Calabi invariant that allow to build easily such examples.


1) Given a Hamiltonian $H:[0,1]\times \mathbb{R^2}\to \mathbb{R}$, the Calabi invariant of its time-one map is given by
$$\int\_0^1\int H(t,x)\,\omega dt.$$
(See the book "Introduction to symplectic topology" by McDuff and Salamon.)
So any non negative and non identically zero Hamiltonian yields a positive Calabi invariant.


2) There is a dynamical interpretation of the Calabi invariant: Intuitively, it measures how pair of points wind around each other along an isotopy. This is due to Fathi, and two different proofs can be found in the paper of Gambaudo and Ghys "Enlacements asymptotiques". To be more precise, given a Hamiltonian isotopy $f^t$, consider for every pair $x\neq y$, the map 
$t\to Ang(\overrightarrow{f^t(x)f^t(y)},\overrightarrow{xy})$
from $[0,1]$ to $\mathbb{S}^1$, and lift it to a map $A^t(x,y):[0,1]\to\mathbb R$. Then the Calabi invariant of $f^1$ is 
$$\int\_{\mathbb{R}^2}A^1(x,y)\,dx\,dy.$$
So to build an example, take an isotopy that preserves all the circles with center in 0 and with non negative rotation on each of these circles. If your map is not the identity, you'll get a positive Calabi invariant. 


