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Download Introduction to Foliations and Lie Groupoids by I. Moerdijk, J. Mrcun PDF

By I. Moerdijk, J. Mrcun

In keeping with a graduate direction taught at Utrecht collage, this booklet offers a brief creation to the idea of Foliations and Lie Groupoids to scholars who've already taken a primary direction in differential geometry. Ieke Moerdijk and Janez Mrcun contain distinct references to permit scholars to discover the needful historical past fabric within the study literature. The textual content gains many routines and labored examples.

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Extra resources for Introduction to Foliations and Lie Groupoids

Example text

5 Let (F, g) be a Riemannian foliation of M . Let L be a leaf of F, α a path in L, and let T and S be transversal sections of F with α(0) ∈ T and α(1) ∈ S. 2 Riemannian foliations 27 is the germ of an isometry. Proof We have to prove that h = holS,T (α) preserves the metric. By the definition of holonomy, we can assume that α is inside a surjective foliation chart ϕ = (x1 , . . , xn−q , y1 , . . , yq ): U → Rn−q × Rq of F and that T, S ⊂ U . We can also assume without loss of generality that ∂ |T form a frame for the ϕ(T ) ⊂ {0} × Rq , so that the vector fields ∂y i tangent bundle of T .

Since H acts on V by holonomy diffeomorphisms, the orbit lies in a leaf L of F. The intersection L ∩ S is discrete, by compactness of L. Since U , and hence V , is relatively compact in S, it follows that L ∩ V is finite. This implies that the orbit is finite as well. By taking the differential at x, or equivalently, by conjugating with expx , we faithfully represent the group H as a group of orthogonal transformations of Tx (S), with the property that any orbit of the action of H in Tx (S) is finite.

Then Tx = r−1 (x) defines a transversal section at x, so we have chosen transversal sections Tx ‘smoothly in x’. (ii) Choose a finite cover U of L by domains of foliation charts of F. We assume that for any U ∈ U we have U ⊂ N , while the corresponding foliation chart (ϕ : U → Rn−q × Rq ) has image of the form B × Rq for a simply connected open subset B of Rn−q and satisfies ϕ(r(x)) = (pr1 (ϕ(x)), 0) for any x ∈ U . In particular, each U ∈ U intersects L in only one simply connected plaque. We can further assume that the intersection U ∩ U ∩ L for any two elements U and U of U is either connected or empty.

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