Thursday, 24 November 2011

groupoids - How difficult is Morse theory on stacks?

The title is a little tongue-in-cheek, since I have a very particular question, but I don't know how to condense it into a pithy title. If you have suggestions, let me know.



Suppose I have a Lie groupoid GrightrightarrowsG0, by which I mean the following data:



  • two finite-dimensional (everything is smooth) manifolds G,G0,

  • two surjective submersions l,r:GtoG0,

  • an embedding e:G0hookrightarrowG that is a section of both the maps l,r,

  • a composition law m:GtimesG0GtoG, where the fiber product is the pull back of GoversetrtoG0oversetlleftarrowG, intertwining the projections l,r to G0.

  • Such that m is associative, by which I mean the two obvious maps GtimesG0GtimesG0GtoG agree,

  • m(e(l(g)),g)=g=m(g,e(r(g))) for all ginG,

  • and there is a map i:GtoG, with icircl=r and icirci=textid and m(i(g),g)=e(r(g)) and m(g,i(g))=e(l(g)).

Then it makes sense to talk about smooth functors of Lie groupoids, smooth natural transformations of functors, etc. In particular, we can talk about whether two Lie groupoids are "equivalent", and I believe that a warm-up notion for "smooth stack" is "Lie groupoid up to equivalence". Actually, I believe that the experts prefer some generalizations of this — (certain) bibundles rather than functors, for example. But I digress.



Other than that we know what equivalences of Lie groupoids are, I'd like to point out that we can work also in small neighborhoods. Indeed, if U0 is an open neighborhood in G0, then I think I can let U=l1(U0)capr1(U0), and then UrightrightarrowsU0 is another Lie groupoid.



Oh, let me also recall the notion of tangent Lie algebroid AtoG0 to a Lie groupoid. The definition I'll write down doesn't look very symmetric in lleftrightarrowr, but the final object is. The fibers of the vector bundle AtoG0 are Ay=rmTe(y)(r1(y)), the tangent space along e(G0) to the r-fibers, and l:r1(y)toG0 determines a God-given anchor map alpha=dl:AtormTG0, and because e is a section of both l,r, this map intertwines the projections, and so is a vector bundle map. In fact, the composition m determines a Lie bracket on sections of A, and alpha is a Lie algebra homomorphism to vector fields on G0.



Suppose that I have a smooth function f:G0tomathbbR that is constant on G-orbits of G0, i.e. f(l(g))=f(r(g)) for all ginG. I'd like to think of f as a Morse function on "the stack G0//G". So, suppose [y]subseteqG0 is a critical orbit, by which I mean: it is an orbit of the G action on G0, and each yin[y] is a critical point of f. (Since f is G-invariant, critical points necessarily come in orbits.) If y is a critical point of f, then it makes sense to talk about the Hessian, which is a symmetric pairing (rmTyG0)otimes2tomathbbR, but I'll think of it as a map f(2)y:rmTyG0to(rmTyG0). In general, this map will not be injective, but rather the kernel will include alphay(Ay)subseteqrmTyG0. Let's say that the critical orbit [y] is nondegenerate if kerf(2)y=alphay(Ay), i.e. if the Hessian is nondegenerate as a pairing on rmTyG0/alphay(Ay). I'm pretty sure that this is a condition of the orbit, not of the individual point.



Nondegeneracy rules out some singular behavior of [y], like the irrational line in the torus.



Anyway, my question is as follows:




Suppose I have a Lie groupoid GrightrightarrowsG0 and a G-invariant smooth function f:G0tomathbbR and a nondegenerate critical orbit [y] of f. Can I find a G-invariant neighborhood U0supseteq[y] so that the corresponding Lie groupoid UrightrightarrowsU0 is equivalent to a groupoid VrightrightarrowsV0 in which [y] corresponds to a single point baryinV0? I.e. push/pull the function f over to V0 along the equivalence; then can I make [y] into an honestly-nondegenerate critical point baryinV0?




I'm assuming, in the second phrasing of the question, that f push/pulls along the equivalence to a V-invariant function barf on V0. I'm also assuming, so if I'm wrong I hope I'm set right, that rmTbaryV0congrmTyG0/alphay(Ay) canonically, so that e.g. barf(2)bary=f(2)y.

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