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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?

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  • $\begingroup$ What else do you know about Calabi's invariant? How does it behave under deformations of $\phi$? Have to tried to compute $\frac d{dt}C(\phi_t)$ where $\phi_t$ is the flow of a divergence free vector field? $\endgroup$ – Sebastian Goette Mar 11 '16 at 13:43
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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.

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  • $\begingroup$ I don't understand - maybe the formulas for the Calabi invariant are incompatible. In your rotational example, take $\alpha=\frac 12r^2\,d\varphi$ in polar coordinates (OP's notation). This is well-defined and gives $d\alpha=r\,dr\,d\varphi=\omega$. Then $\phi^*\alpha=\alpha$ because each circle $r=\mathrm{const}$ is rotated with speed depending on $r$ only. So the OP's version of the Calabi invariant vanishes. Or am I wrong here? $\endgroup$ – Sebastian Goette Mar 12 '16 at 21:48
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    $\begingroup$ Your example is not the identity near the boundary. All what I said applies for compactly supported maps, which is the usual setting for the Calabi invariant $\endgroup$ – Vincent H Mar 12 '16 at 22:10
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    $\begingroup$ I still don't understand - you say rotate each circle at positive speed, depending on the radius. So with speed $0$ close to the boundary, everything is fine - except that the Calabi invariant in the question above vanishes. I therefore believe that either the OP or @VincentH has the wrong definition. $\endgroup$ – Sebastian Goette Mar 13 '16 at 14:28
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    $\begingroup$ Sorry for my last comment which was not addressing the right point. I think that the equality $\phi^*\alpha=\alpha$ does not hold in the case we are considering: our $\phi$ can be written in polar coodinates $\phi(r,\theta)=(r,\theta+f(r))$ and computation gives $\phi^*d\theta=f'(r)\,dr+d\theta$, from which one can derive the correct formula for $\phi^*\alpha$. $\endgroup$ – Vincent H Mar 14 '16 at 8:58
  • $\begingroup$ I see, I made a stupid mistake. So in the end, the value is $\int f(r)\,\omega=2\pi\int_0^1r\,f(r)\,dr$ (with $f(1)=0$), which can well be different from $0$. $\endgroup$ – Sebastian Goette Mar 14 '16 at 16:25

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