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On local integration of Lie brackets

  • Alejandro Cabrera , Ioan Mărcuţ and María Amelia Salazar

Abstract

We give a direct, explicit and self-contained construction of a local Lie groupoid integrating a given Lie algebroid which only depends on the choice of a spray vector field lifting the underlying anchor map. This construction leads to a complete account of local Lie theory and, in particular, to a finite-dimensional proof of the fact that the category of germs of local Lie groupoids is equivalent to that of Lie algebroids.

Funding statement: Ioan Mărcuţ was supported by the NWO Veni grant 613.009.031 and the NSF grant DMS 14-05671. María Amelia Salazar was a Post-Doctorate at IMPA during part of this project, funded by CAPES-Brazil. Alejandro Cabrera would like to thank CNPq and FAPERJ for financial support.

A Differentiating integrated cochains

In this appendix, we consider the differentiation of the local Lie groupoid cochains obtained by integration through the map Ψ defined in Section 2.5. The main result is Lemma A.1 which is used in the proof of item (ii) of Proposition 2.10.

Following the notation of Section 2.5, let αΓ(pA*) be an algebroid p-cochain and let Ψ(α):G(p) be the corresponding local groupoid cochain. In order to apply the van Est map, we consider sections a1,,akΓ(A) and compute the iterated differentiation DapDa1Ψ(α). We do this inductively on the number p of derivatives.

Let us fix composable arrows g2,,gpG(p-1) in a sufficiently small neighborhood of M, and consider h1(ϵ)=ϵa1(τ(g2)) so that

Da1Ψ(α)(g2,,gp)
  =ddϵ|ϵ=0Ψ(α)(h1(ϵ),g2,,gp)
  =ddϵ|ϵ=0Ip(θ*α)(1γh1(ϵ),g2,,gp,,pγh1(ϵ),g2,,gp)𝑑t1𝑑tp.

By using the relations

(A.1)γϵg1,g2,,gp(t1,,tp)=γg1,,gp(ϵt1,t2,,tp),
γτ(g2),g2,,gp(t1,,tp)=γg2,,gp(t2,,tp),

where τ(g2)=σ(g1)=0g1, it follows that, for aΓ(A) and small (g2,,gn)G(n),

(A.2)1γϵa(τg2),g2,,gn(t1,,tn)=ϵ1γa(τ(g2)),g2,,gn(ϵt1,t2,,tn),
jγτ(g2),g2,,gn(t1,t2,,tn)=j-1γg2,,gn(t2,,tn),j>1.

Applying (A.2), we can compute ddϵ|ϵ=0 (equivalently, apply the substitution u=ϵt1 as in Example 2.9) yielding

Da1Ψ(α)(g2,,gp)=Ip-1(θ*α)(1γa1(τ(g2)),g2,,gp(0,t2,,tp),
1γg2,,gp(t2,,tp),,p-1γg2,,gp(t2,,tp))dt2dtp,

where I𝑑t1=1 has been factored out. Above, the first factor inside α is of a different nature from the rest: it is the only one depending on a1 while the others only depend on the fixed string g2,,gp. We introduce the following general notation for such terms: given aΓ(A), small (k1,,kn)G(n) and any (s1,,sn)In,

(A.3)vk1,,kna(s1,,sn):=1γa(τ(k1)),k1,,kn(0,s1,,sn);

thus, we have

vk1,,kna(s1,,sn)TσGγk1,,kn(s1,,sn).

Before stating the general inductive result, let us compute one more derivative fixing elements g3,,gpG(p-2) and letting h2(ϵ)=ϵa2(τ(g3)),

Da2Da1Ψ(α)(g3,,gp)=ddϵ|ϵ=0Da1Ψ(α)(h2(ϵ),g3,,gp)
=ddϵ|ϵ=0Ip-1(θ*α)(vh2(ϵ),g3,,gpa1,1γh2(ϵ),g3,,gp,,
p-1γh2(ϵ),g3,,gp)(t1,,tp-1)dt1dtp-1,

where we have relabeled the integration variables. Using (A.2) we compute ddϵ|ϵ=0, yielding the integral over dt1dtp-1 of

(θ*α)(vh2(0),g3,,gpa1(t1,,tp-1),vg3,,gpa2(t2,,tp-1),
1γg3,,gp(t2,,tp-1),,p-2γg3,,gp(t2,,tp-1)).

The key step for recognizing an inductive structure in our computation is to note that (recall that M is identified with the unit section)

(A.4)γg1,τ(g3),g3,,gn(t1,t2,,tn)=γg1,g3,,gn(t1t2,t3,,tn)

for any small composable string (g1,g3,g4,,gn)G(n-1) and any (t1,t2,,tn)In. Using (A.4) and the notation (A.3), we get

(A.5)vτ(g3),g3,,gna(t1,,tn-1)=t1vg3,,gna(t2,,tn-1)

for any aΓ(A), small (g3,,gn)G(n-2) and (t1,,tn-1)In-1. Thus, since

h2(0)=τ(g3),

we have

Da2Da1Ψ(α)(g3,,gp)=Ip-1t1(θ*α)(vg3,,gpa1,vg3,,gpa2,
1γg3,,gp,,p-2γg3,,gp)(t2,,tp-1)dt1dtp-1

from which It1𝑑t1=12 factors out yielding

Da2Da1Ψ(α)(g3,,gp)
  =12Ip-2(θ*α)(vg3,,gpa1,vg3,,gpa2,1γg3,,gp,,p-2γg3,,gp).

Continuing by induction, we obtain the following:

Lemma A.1.

For a1,,akΓ(A) and (gk+1,,gp)G(p-k) small enough, we have

(A.6)DakDa1Ψ(α)(gk+1,,gp)
=1k!Ip-k(θ*α)(vgk+1,,gpa1,,vgk+1,,gpak,
1γgk+1,,gp,,p-kγgk+1,,gp),

where vgk+1,,gpa:Ip-kTσG is defined in (A.3).

Proof.

We will consider an induction over k1 recalling that the case k=1 was already worked out above. Assume now (A.6) and compute

Dak+1Da1Ψ(α)(gk+2,,gp)
=ddϵ|ϵ=0DakDa1Ψ(α)(ϵak+1(τ(gk+2)),gk+2,,gp).

To use (A.6) on the right-hand side above, let us introduce the following notation:

wgk+1,,gpa1,,ak:=vgk+1,,gpa1vgk+1,,gpak

and b:=ak+1(τ(gk+2)), so that

(A.7)k!Dak+1Da1Ψ(α)(gk+2,,gp)
=ddϵ|ϵ=0Ip-k(θ*α)(wϵb,gk+2,,gpa1,,akdγϵb,gk+2,,gp).

By using (A.1), it follows that

dγϵb,gk+2,,gp(t1,,tp-k)=ϵdγb,gk+2,,gp(ϵt1,,tp-k),

and that

γb,gk+2,,gp(0,t2,,tp-k)=γgk+2,,gp(t2,,tp-k).

Using these to compute (A.7) gives

k!Dak+1Da1Ψ(α)(gk+2,,gp)
=Ip-k(θ*α)(wτ(gk+2),gk+2,,gpa1,,ak(t1,,tp-k)dγgk+2,,gp(t2,,tp-k))𝑑t1𝑑tp-k.

Applying (A.5), we get

wτ(gk+2),gk+2,,gpa1,,ak(t1,,tp-k)=t1kwgk+2,,gpa1,,ak(t2,t3,,tp-k).

The lemma thus follows by noticing that 01t1k𝑑t1=1k+1 factors out. ∎

Acknowledgements

The authors would like to thank Marius Crainic, Pedro Frejlich, Rui Loja Fernandes, Marco Gualtieri, Eckhard Meinrenken and Daniele Sepe for useful discussions, and the anonymous referee for their careful reading and suggestions.

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Received: 2017-04-06
Revised: 2018-04-05
Published Online: 2018-05-09
Published in Print: 2020-03-01

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