2
$\begingroup$

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?

$\endgroup$
1
+50
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.