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Consider the circle action on $\mathbb C^n$ given by $(e^{it},z)\to e^{it}z$. A moment map for this action is $J:\mathbb C^n\to\mathbb R:z\to -\frac{1}{2}|z|^2$. Let $M_l=J^{-1}(-\frac{l}{2})/U(1)$ for $l>0$.

How can we calculate the function $f(l)=\frac{\text{Vol}(M_l)}{\text{Vol}(M_1)}$, where $\text{Vol}(M_l)$ is the volume of $M_l$ given by the reduced symplectic form $\omega_l$?

Solution: $J^{-1}(-\frac{l}{2})$ is $S^{2n-1}$ the sphere with radius $\sqrt l$, and so $M_l=S^{2n-1}/U(1)$ must be diffeomorphic to $\mathbb CP^n$. Obviously, $(M_1,\omega_1)$ is $\mathbb CP^n$ with its Study-Fubinni form.

Indeed, the Study-Fubini form is the projection of $i^* \omega$, the restriction to the sphere of the canonical symplectic form on the complex vector space to the quotient and from Marsden-Weinstein-Meyer Theorem, i.e. $i^*\omega=\pi^*\omega_1$, where $i:J(-\frac{1}{2})=S^{2n-1}\to\mathbb C^n$ is the inclusion and $\pi:J(-\frac{1}{2})\to M_1$ is the projection.

Moreover, by Fubiny's Theorem of integral calculus $\text{Vol}(S^{2n+1})=\text{Vol}(S^1)\text{Vol}(\mathbb C P^n)$. Hence, $\text{Vol}(\mathbb C P^n)={1\over 2\pi}\text{Vol}(S^{2n+1})=\frac{\pi^{n}}{n!}$.

Problem: The last calculation is right up to a constant of proportionality. What the constant of proportionality relating the volume element of $\mathbb C P^n$ used above and the volume element $w_1^n$, the $n$'th power of the Fubini-Study form?

Continued: Analogously, consider spheres of radius $r$ which we denote by $S^{2n+1}(r)$. I use that $M_l=S^{2n+1}(\sqrt l)/S^1(\sqrt l)$ as a Riemannian manifold. Thus, $\text{Vol}(M_l)={1\over 2\pi\sqrt l}\text{Vol}(S^{2n+1}(\sqrt l))=\frac{\pi^{n}}{n!}\sqrt l^{2n}$.

Problem: Should I have really taken the quotient by $S^1(\sqrt l)$ instead of a quotient by the unit sphere? How does this choice reflect on the volume forms? Again, what is the constant of proportionality relating the volume element of $M_l$ used above and the volume element $w_l^n$?

Finally, the constants of proportionality should cancel in the quotient $f(l)=\frac{\text{Vol}(M_l)}{\text{Vol}(M_1)}=\sqrt l^{2n}$.

To sum up, does taking the quotient of metric spheres $S^{2n+1}(\sqrt l)/S^1(\sqrt l)$ really give the induced metric on $M_l$ and how much do the Riemannian and symplectic volumes disagree on $M_l$? Do I need consider the metric structures at all?

  • Possibly of interest: Diameter of the Grassmannian. If I understand what you're asking, the symplectic form of volume $\pi^{n}/n!$ gives (complex projective) lines constant curvature $1$, hence area $4\pi$, and is therefore $\sqrt{4\pi}$ times the generator of integral cohomology. – Andrew D. Hwang May 23 '16 at 18:51
  • No, I'm sorry. This is not what I'm asking, not to my undestanding. – superAnnoyingUser May 23 '16 at 18:52
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    My comment gives the relative scaling between the Riemannian and integral Fubini-Study forms, which I thought could be used to address all your concerns. On the other hand, I count seven question marks in the current wording. Could you please clarify what exactly you are asking? Thank you. – Andrew D. Hwang May 23 '16 at 19:00
  • I recognize this question. No you don't need to consider the metric structures. As you noted it is tedious to determine how they relate to the $\omega_{red}$ w.r.t. which we have to calculate the volume. – user2520938 May 24 '16 at 11:35
  • So how do we go about it without considering the metric structures, @user2520938? – superAnnoyingUser May 24 '16 at 20:17
  • The most important question is how to rescale the form $\omega_l$ so $\omega_l^n$ integrates to the same total volume as that of the sphere of radius $\sqrt l$ divided by the length of a great circle of the same radius. I could not make this out from your other answer, @AndrewD.Hwang. Did I get the right answer for $f(l)$? – superAnnoyingUser May 24 '16 at 20:18
  • Yes, $l^{n} = \sqrt{l}^{2n}$ is the correct scale factor for dividing a $(2n+1)$-sphere of radius $\sqrt{l}$ by the circle action coming from complex scalar multiplication (i.e., by a circle of radius $\sqrt{l}$). Under this scaling, the Fubini-Study form, which scales like area, is multiplied by $l$, so the volume of complex projective space scales by $l^{n}$. – Andrew D. Hwang May 24 '16 at 22:38

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