0
$\begingroup$

I am little confused about some basic symplectic geometry about Hamiltonian actions on sphere. I appreciate your comments.

Consider sphere $S^2 = \mathbb{C}P^1$ with its standard symplectic (Kaehler) form $\omega$ and usual action of $S^1$ on $\mathbb{C}P^1$ and its usual moment map $\phi: \mathbb{C}P^1 \to \mathbb{R}$ with image $[-1, 1]$.

Also consider the "n-tuple embedding" $\mathbb{C}P^1 \to \mathbb{C}P^n$ by $$(z:w) \mapsto (z^n : z^{n-1}w : \cdots : w^n),$$

and consider the Kaehler form $\omega_n$ on $\mathbb{C}P^1$ which is the pull-back of the Kaehler form on $\mathbb{C}P^n$ (Fubini-Study form). Let $\phi_n$ be the moment map for the Hamiltonian action of $S^1$ with respect to $\omega_n$ (that is the pull-back of $S^1$-moment map on $\mathbb{C}P^n$ to $\mathbb{C}P^1$). Here $t \in S^1$ acts on $\mathbb{C}P^1$ and $\mathbb{C}P^n$ respectively by: $$ t \cdot (z : w) = (z : tw),$$ $$t \cdot (z_0 : \cdots : z_n) = (z_0: tz_1: \cdots t^nz_n),$$

I am pretty sure that $\omega_n$ is a constant multiple of $\omega$ and in fact: $$\omega_n = n \omega. \quad (*)$$

From (*) it should follow that $n\phi(x)$ is a moment map for the $S^1$ action with respect to the symplectic form $\omega_n$. On the other hand moment map should be unique up to a constant and hence $\phi_n = n \phi + const.$ but this seem not to be the case (direct calculation).

What am I missing?

Here are more detail of computing the moment maps:

The moment map of the usual action of $(S^1)^{n+1}$ on $\mathbb{C}P^n$ is given by: $$\Phi(z_0: \cdots :z_n) = (\frac{|z_0|^2}{\sum_{i=0}^n |z_i|^2}, \ldots,\frac{|z_n|^2}{\sum_{i=0}^n |z_i|^2}).$$ Now as in above, under the $n$-uple embedding circle $S^1$ embeds in $(S^1)^{n+1}$ by: $$t \mapsto (1, t, \ldots, t^n).$$ The corresponding map between dual Lie algebras $\mathbb{R}^{n+1}=Lie((S^1)^{n+1})^* \to \mathbb{R}=Lie(S^1)^*$ is: $$(t_0, \ldots, t_n) \mapsto t_1 + 2t_2 + \cdots + nt_n.$$Thus the moment map of $S^1$ acting on $\mathbb{C}P^n$ is: $$\Phi(z_0 : \cdots : z_n) = \sum_{i=0}^n \frac{i|z_i|^2}{\sum_{i=0}^n |z_i|^2}.$$ Restricting to the image of $\mathbb{C}P^1$ in $\mathbb{C}P^n$ we get: $$\phi_n(1:w) =\sum_{i=0}^n \frac{i|w|^{2i}}{\sum_{i=0}^n |w|^{2i}}.$$ For n=1 we have the usual moment map of $\mathbb{C}P^1$ which is: $$\phi(1:w) = \frac{|w|^2}{1 + |w|^2}.$$ $\phi_n$ does not seem to be equal to $n\phi$ ...

$\endgroup$
2
  • 1
    $\begingroup$ You just multiply by n. Your direct calculation is wrong. I can't tell you anything more, since you don't reproduce it. $\endgroup$
    – Ben Webster
    Apr 12, 2014 at 7:02
  • $\begingroup$ @BenWebster I added details of (the standard) moment map calculations. $\endgroup$
    – user43696
    Apr 12, 2014 at 15:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.