3

Suppose I have 2-dimensional manifold embedded in $\mathbb{R}^3$. It's clear that the most natural Riemannian metric is the one induced by the usual inner product.

What about symplectic forms? Is there a canonical symplectic form I can put on this manifold? If not, then I ask for something weaker, an example of a typical form that one would expect to see on such a manifold, and can do explicit computations with.

user908123
  • 41
  • 2
    As the manifold is orientable, you have define on each tangent plane the anticlockwise rotation to 90 degree (called $J$). Then $w(X, Y) = g(X, JY)$, where $J$ is the rotation, will be a symplectic form. –  Dec 21 '13 at 06:45
  • 1
    If you choose a unit normal vector field $N$, then the form that John describes can also be defined by $w(X,Y) = (X\times Y)\cdot N$, where $\times$ is the cross product in $\mathbb{R}^3$. – Jim Belk Dec 21 '13 at 06:49

1 Answers1

1

If the 2-dimensional manifold is non-orientable (e.g. Klein bottle), then it will never admit a symplectic structure since the symplectic structure would be a volume form and a manifold is orientable iff it has a volume form.

Gilbert Easton
  • 575