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What is the kernel of $i^*:H^*(\overline{\mathcal{M}}_{g,n},\mathbb{Q}) \to H^*(\overline{\mathcal{M}}_{g-1,n+2},\mathbb{Q})$?
@DanPetersen, yes, I have the same feeling. For instance, imagine you want to define a sort of "compact type COhFT" replacing the loop axiom with the fact that the classes vanish on $\delta_{irr}$. Then the natural modification of the Givental group simply sums over stable trees instead of any stable graph. This action commutes with multiplying by "\lambda_g", so you can't get out of that ideal. This is how I got to asking this question.
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What is the kernel of $i^*:H^*(\overline{\mathcal{M}}_{g,n},\mathbb{Q}) \to H^*(\overline{\mathcal{M}}_{g-1,n+2},\mathbb{Q})$?
@JasonStarr, oops did I write it in reverse? I guess Dan edited it. Thanks Dan.
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Relationship between virtual cohomological dimension and tautological rings for moduli spaces of curves
this question has been here since 2013 and i just noticed it. Dan, do you happen to have got an answer since then?
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Connection between integrable systems and group actions
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Why is every Hamiltonian system locally integrable?
thanks @Ben, yes you are right of course, I even edited the question.
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Why is every Hamiltonian system locally integrable?
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Why is every Hamiltonian system locally integrable?
one of the quotation symbols ">" ended up inside a math environment, becoming an undesired symbol $>$
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Why is every Hamiltonian system locally integrable?
yes, thank you! I had JUST found this theorem myself, precisely in the reference you give. I agree that it is a slight generalization of Darboux Theorem, when you look at the proof. However I am glad I actually learned it. Thanks again.
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Why is every Hamiltonian system locally integrable?
I fail to see how the Hamiltonian vector field and its symmetries take part in Darboux theorem. What I am looking for is the result that, locally, I can find a Lagrangian foliation tangent to the vector field. Why does this come trivially from the normal form of the symplectic structure?