1
$\begingroup$

I'm working through McDuff and Salamon's `Introduction to Symplectic Topology' (highly recommend). In example 3.1.2, we are given the example of a symplectic manifold: the 2-sphere with its standard area form. Spcifically, $$S^2 = \{(x_1,x_2,x_3)\in\mathbb{R}^{3}\:|\: x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\}$$ with induced area form $\omega_{x}(\xi,\eta) = \langle x,\xi\times\eta\rangle$ for $\xi,\eta\in T_{x}S^{2}$. The authors then comment that the total area is $4\pi$.

So my question is: what is the easiest way to see that the total area is $4\pi$? I would like to know how I can integrate this 2-form $$\int_{S^{2}}\omega$$ possibly using Stoke's theorem. I would also have a preference for coordinate free solutions!

Thanks in advance for your help.

$\endgroup$

1 Answer 1

3
$\begingroup$

If you read further into the book you'll learn many coordinate-free ways of seeing this, but they're not "directly" calculating the integral and depend on some extra structure that the sphere has. To do it directly, you do need coordinates.

The easiest coordinates in this case are cylindrical ones, you can check that writing the form in those terms gives $dz\wedge d\theta$, now the integral is easy to take.

$\endgroup$
7
  • $\begingroup$ "To do it directly, you do need coordinates." Boo to coordinates. $\endgroup$
    – Fomalhaut
    Commented Nov 8, 2019 at 2:06
  • $\begingroup$ excuse my poor understanding, but could you at least make this integral easier by breaking the integral into upper and lower hemispheres and then just integrate curves along the boundary via Stokes? $\endgroup$
    – no_idea
    Commented Nov 8, 2019 at 2:53
  • $\begingroup$ You can do that, but I don't think it makes it particularly easier, and integrating along the circle still requires coordinates. $\endgroup$ Commented Nov 8, 2019 at 20:42
  • 1
    $\begingroup$ @BalancedTryteOperators: Integrals are defined locally and with respect to parametrizations, so in general you cannot magically produce a number without appealing to coordinates somewhere. If you want to avoid them, you can appeal to special circumstances of your situation, e.g the circle action on the sphere. $\endgroup$ Commented Nov 8, 2019 at 20:48
  • $\begingroup$ Thankyou @AnonymousMemer. How would I explicitly compute this using cylindrical coordinates? Here is my try: $S^{2} = f^{-1}(0)$ where $f(r,\theta,z) = r^{2}+z^{2}-1$. Thus $\nabla f = (2r,0,2z)$ and thus $\xi$ and $\eta$ would be in the span of $(0,1,0)$ and $(z,0,-r)$. I then compute $\langle x, \xi\times\eta\rangle$ to be $-r^{2}-z^{2} = -1$. So $\int_{S^{2}}\omega = \int_{S^{2}}-1 d\theta\wedge dz = \int_{S^{2}} dz\wedge d\theta = 4\pi$. I'm not happy with my working though. $\endgroup$
    – no_idea
    Commented Nov 11, 2019 at 3:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .