Guillemin Sternberg Conjecture(proved) says that for symplectic manifold $(M,\omega)$ with $[Q,R]=0$ condiction, with compact group action $G$, such that $\mu:M\to \mathfrak g^*$ is regular at $0$, and $G$ action freely on $\mu^{-1}(0)$. Then $Ind^G(D_M^L)=Ind(D^{L_G}_{M//G})$.

Is there an example?

Here is an idea.

Let $M=\mathbb P^2$ $L=\mathcal O(1)$ and $G=S^1$ action on $M$ on the first coordinate, i.e. $\theta\cdot[z_1,z_2,z_3]=[\theta z_1,z_2,z_3]$. Then, $M//G=\mathbb P^1$.

Q: I donot know what is $L_G$. $\mathcal O(1)$ or trivial bundle?

By my understanding, I think $L_G=\mathcal O(1)$, then $Ind(D^{\mathcal O(1)}_{\mathbb P^1})=2$. Is this correct?


1 Answer 1


Since you are looking for an official source, I recommend the book Symplectic Fibrations and Multiplicity Diagrams by Guillemin, Lerman and Sternberg. It has a lot to say about symplectic reduction on homogeneous manifolds. However, I don't have it with me right now and cannot point you directly to relevant examples there.

For your example at hand, let us fix the $S^1$-action on $L=\mathcal O(1)\to\mathbb P^k$ first. The bundle is generated by its holomorphic sections, which can be identified with linear maps from $\mathbb C^{k+1}$ to $\mathbb C$. I assume you want an action that multiplicies the first coordinate of a vector in $\mathbb C^{k+1}$ (an element of the tautological bundle) by $\theta\in S^1$. The corresponding dual action multiplies the first coordinate of a linear form by $\bar\theta$.

Now, here is number of claims that you should check step by step.

  1. Your $S^1$ action has a moment map that is a function of $|z_1|/|(z_1,z_2,z_3)|$. The regular level sets are diffeomorphic to $S^3$, and the quotient by the $S^1$ action is $\mathbb P^1$. A regular level set $\mu^{-1}(r)$ can be represented by $\{(z_1,z_2,z_3)\mid(z_2,z_3)\in S^3\}$ with fixed $z_1\in\mathbb R_+\subset\mathbb C$. And $\theta\in S^1$ acts by multiplying $(z_2,z_3)$ with $\bar\theta$. For $z_1\to 0$, the level sets collapse to a copy of $\mathbb P^1$ on which $S^1$ acts trivially.

  2. Your bundle $\mathcal O(1)$ is dual to the tautological bundle $L^*$, which I find easier to handle. It has a trivialisation on the level set $\mu^{-1}(r)$ as above, given by elements of the form $(1,z_2,z_3)$. The group $S^1$ still acts by multiplying the first coordinate with $\theta$. The bundle $(L^*)^{S^1}$ is the quotient of $L^*|_{\mu^{-1}(r)}$ by this action. One can check that this gives the tautological bundle on $\mathbb P^1$. Dual to this, therefore $L^{S^1}\cong\mathcal O(1)$.

  3. Finally, compute the indices. $D$ is the Dolbeault operator. Holomorphic sections of $\mathcal O(1)\to\mathbb P^k$ correspond to linear functions on $\mathbb C^{k+1}$, and to the best of my knowledge, there is no higher cohomology, so $\mathrm{ind}(D_{\mathbb P^1}^{\mathcal O(1)})$ would be $2$. On the other hand, the index on $\mathbb P^2$ would be the dual space of $\mathbb C^3$, where the $S^1$ action above fixes a two-dimensional subspace, so again, $\mathrm{ind}(D_{\mathbb P^2}^{\mathcal O(1)})^{S^1}=2$.

On the other hand, one is free to vary the $S^1$ action by multiplying the action above for $\theta\in S^1$ by any power of $\theta$. For example, if you replace the action $\rho$ on $\mathcal O(1)$ above by $\theta\rho(\theta)$, the action $\rho^*$ on the tautological bundle is replaced by $\bar\theta\rho^*(\theta)$. In this case, the bundles $(L^*)^{S^1}$ and $L^{S^1}$ over $\mathbb P^1$ are both trivial, and the index on $\mathbb P^1$ is $1$. Moreover, the action on $(\alpha_1,\alpha_2,\alpha_3)\colon\mathbb C^3\to\mathbb C$ now has a one-dimensional fixed subspace spanned by $(1,0,0)$, so we have $\mathrm{ind}(D_{\mathbb P^1}^{\mathcal O(1)})=\mathrm{ind}(D_{\mathbb P^2}^{\mathcal O(1)})^{S^1}=1$ instead.


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