Van Est differentiation and integration

Abstract

The classical Van Est theory relates the smooth cohomology of Lie groups with the cohomology of the associated Lie algebra, or its relative versions. Some aspects of this theory generalize to Lie groupoids and their Lie algebroids. In this paper, continuing an idea from Li-Bland and Meinrenken (Enseign Math 61(1–2):93–137, 2015), we revisit the van Est theory using the Perturbation Lemma from homological algebra. Using this technique, we obtain precise results for the van Est differentiation and integrations maps at the level of cochains. Specifically, we construct homotopy inverses to the van Est differentiation maps that are right inverses at the cochain level.

Appendix A. Proof of Theorem 4.4

Taking $Q=G$ in (25), we have $p+1$ commuting G-actions on $E_pG=B_pG\times _M G$; these commute with the principal action and descend to the actions on $B_pG$. The projection $\pi _p:E_pG\rightarrow M$ intertwines each of these actions with the trivial action on M; hence we obtain commuting G-actions on the vector bundles

$$\begin{aligned} \pi _p^*(V\otimes \wedge ^q A^*)=E_pG\times _M (V\otimes \wedge ^q A^*) \subseteq E_pG\times (V\otimes \wedge ^q A^*), \end{aligned}$$

(50)

using the trivial action on the $V\otimes \wedge ^q A^*$ factor. The infinitesimal action gives covariant derivatives $\nabla _\xi ^{(i)}$ on $\mathsf {D}^{p,q}(G,V)=\Gamma (\pi _p^*(\wedge ^q A^*\otimes V))$; the derivatives for different i’s commute. They ‘lift’ the operators $\nabla _\xi ^{(i)}$ on $\mathsf {C}^p(G,V)=\Gamma (B_pG\times _M V)$ introduced earlier.

Lemma A.1

• (a) The maps ${\mathsf {j}}:\mathsf {C}^p(G,V)\rightarrow \mathsf {D}^{p,0}(G,V)$ intertwine $\nabla _\xi ^{(i)}$ for $i=0,\ldots ,p$.

• (b) The operators $\nabla _\xi ^{(i)}$ on the double complex commute with the vertical differential $\mathsf {d}$, and also with contractions $\iota _\zeta ,\ \zeta \in \Gamma (A)$ and Lie derivatives $\mathcal {L}_\zeta$.

• (c) The maps $\mathsf {h}=(-1)^p h_{p-1}^*:\mathsf {D}^{p,q}(G,V)\rightarrow \mathsf {D}^{p-1,q}(G,V)$ intertwine $\nabla _\xi ^{(i)}$ for $i=0,\ldots ,p-1$, while

  $$\begin{aligned} \mathcal {L}_\xi \circ \mathsf {h}=\mathsf {h}\circ (\nabla _\xi ^{(p)}+\mathcal {L}_\xi ). \end{aligned}$$

  (51)

Proof

• (a) follows from the equivariance of the map $\kappa _p$ with respect to the i-th action.

• (b) Since $\iota _\zeta$ is equivariant for i-th action, it intertwines the operators $\nabla _\xi ^{(i)}$. Next, since $\kappa _p:E_pG\rightarrow B_pG$ is equivariant for the i-th action; the foliation $\mathcal {F}$ of $E_pG$ is preserved; i.e., the infinitesimal action of $\Gamma (A)$ is by infinitesimal automorphisms of the Lie algebroid $T_\mathcal {F}E_pG$. It follows that the action on $\mathsf {D}^{p,\bullet }(G,V)$ preserves the differential $\mathsf {d}_{CE}$ and hence also $\mathsf {d}=(-1)^p\mathsf {d}_{CE}$. Finally, since $\mathcal {L}_\zeta =[\mathsf {d}_{CE},\iota _\zeta ]$ it also intertwines the Lie derivatives (for the principal G-action); alternatively this follows directly because the i-th action commutes with the principal action.

• (c) The first part follows since the maps

  $$\begin{aligned} h_{p-1}:E_{p-1}G\rightarrow E_pG,\ (g_1,\ldots ,g_{p-1};g)\mapsto (g_1,\ldots ,g_{p-1},g;\mathsf {s}(g)) \end{aligned}$$

  (see (11)) are equivariant for the actions labeled by $i=0,\ldots ,p-1$. For (51), we need to consider both the generating vector fields $\xi ^{(p)}$ for the p-th G-action and the generators $\xi _{E_pG}\in \mathfrak {X}(E_pG)$ of the principal action. In terms of $E_pG=B_pG\times _M G$,

  $$\begin{aligned} \xi ^{(p)}=(\xi ^{L,p},-\xi ^R),\ \ \ \xi _{E_pG}=(0,\xi ^L),\ \ \end{aligned}$$

  where $\xi ^{L,p}$ is the left-invariant vector field sitting on the last G-factor of $B_pG$. Since $\xi ^L\sim _{u\circ \mathsf {s}} \xi ^L-\xi ^R$ (where $u:M\rightarrow G$ is the inclusion of units), we see that

  $$\begin{aligned} \xi _{E_{p-1}G}\sim _{h_{p-1}} \xi _{E_pG}+\xi ^{(p)},\end{aligned}$$

  which implies Eq. (51).

$\square$

We are now in position to give the proof of Theorem 4.4.

Proof of Theorem 4.4

On elements of $\mathsf {C}^p(G,V)=\Gamma (B_pG\times _M V)$, we have that

$$\begin{aligned} {\text {VE}}_G=(-1)^p \mathsf {p}\circ (\mathsf {d}\mathsf {h})^p\circ {\mathsf {j}}. \end{aligned}$$

Using

$$\begin{aligned} {\mathsf {j}}&=\kappa _p^*:\mathsf {C}^p(G,V)\rightarrow \mathsf {D}^{p,0}(G,V),\\ \mathsf {d}\mathsf {h}&=-\mathsf {d}_{CE}\circ h_{i-1}^*:\mathsf {D}^{i,p-i}(G,V)\rightarrow \mathsf {D}^{i-1,p-i+1}(G,V),\\ \mathsf {p}&=u^*:\mathsf {D}^{0,p}(G,V)\rightarrow \mathsf {C}^p(A,V),\\ \end{aligned}$$

this means that ${\text {VE}}_G= u^*\circ \mathsf {d}_{CE}\circ h_0^*\circ \mathsf {d}_{CE} \circ \cdots \circ h_{p-1}^*\circ \kappa _p^*$. Given $\xi _1,\ldots ,\xi _p\in \Gamma (A)$ and $\sigma \in \Gamma (B_pG\times _M V)$, we want to compute

$$\begin{aligned} {\text {VE}}_G(\sigma )(\xi _1,\ldots ,\xi _p)= \iota _{\xi _p}\cdots \iota _{\xi _1} u^* \mathsf {d}_{CE}\, h_0^*\, \mathsf {d}_{CE}\, h_1^*\, \cdots \mathsf {d}_{CE}\, h_{p-1}^*\, \kappa _p^*\, \sigma . \end{aligned}$$

Our strategy is to move the variables $\xi _p,\ldots ,\xi _1$ to the right, while retaining their ordering (keeping $\xi _i$ to the left of $\xi _j$ if $i>j$). The commutators of contractions $\iota _\xi$ with $\mathsf {d}_{CE}$ produces Lie derivatives $\mathcal {L}_\xi =[\iota _\xi ,\mathsf {d}_{CE}]$. Using Lemma A.1 and $\mathcal {L}_\xi \circ \kappa _p^*=0$, we find

$$\begin{aligned} \mathcal {L}_\xi \circ h_{i-1}^*\cdots \mathsf {d}_{CE} \circ h_{p-1}^*\circ \kappa _p^* =h_{i-1}^*\circ \cdots \mathsf {d}_{CE} \circ h_{p-1}^*\circ \kappa _p^* \circ \widehat{\nabla }_\xi ^{(i)} \end{aligned}$$

where we introduced the hat notation

$$\begin{aligned} \widehat{\nabla }_\xi ^{(i)}=\nabla _\xi ^{(i)}+\ldots +\nabla _\xi ^{(1)}, \end{aligned}$$

corresponding to the diagonal action for the actions labeled $1,\ldots ,i$. (Note that the 0-th action is not included.) We therefore obtain

$$\begin{aligned} {\text {VE}}_G(\sigma )(\xi _1,\ldots ,\xi _p)&=u^*h_0^*\cdots h_{p-1}^*\kappa _p^* \sum _{s\in \mathfrak {S}_p}{\text {sign}}(s) \widehat{\nabla }_{\xi _p}^{(s(p))} \cdots \widehat{\nabla }_{\xi _1}^{(s(1))}\sigma \\&=\left( \sum _{s\in \mathfrak {S}_p}{\text {sign}}(s) \widehat{\nabla }_{\xi _p}^{(s(p))} \cdots \widehat{\nabla }_{\xi _1}^{(s(1))}\sigma \right) \bigg |_M; \end{aligned}$$

here the second equality follows since the composition $\kappa _p\circ h_{p-1}\circ \cdots \circ h_0\circ u$ is just the inclusion $M\rightarrow B_pG$. To complete the proof, we argue that

$$\begin{aligned} \sum _{s\in \mathfrak {S}_p}{\text {sign}}(s) \widehat{\nabla }_{\xi _p}^{(s(p))} \cdots \widehat{\nabla }_{\xi _1}^{(s(1))} \end{aligned}$$

(52)

is equal to a similar sum with all hats removed. Given $s\in \mathfrak {S}_p$, let $i=s^{-1}(p)$. Since

$$\begin{aligned} \widehat{\nabla }_{\xi _i}^{(p)}=\nabla _{\xi _i}^{(p)}+\widehat{\nabla }_{\xi _i}^{(p-1)} \end{aligned}$$

we see that the product

$$\begin{aligned} \widehat{\nabla }_{\xi _p}^{(s(p))} \cdots (\widehat{\nabla }_{\xi _i}^{(p)}-\nabla _{\xi _i}^{(p)}) \cdots \widehat{\nabla }_{\xi _1}^{(s(1))} \end{aligned}$$

coincides with the corresponding expression for the permutation $s'$, given as the composition of s with the transposition of the indices $p,p-1$. Since $s,s'$ have opposite signs, it follows that (52) does not change when we remove the hats from all $\widehat{\nabla }_{\xi _i}^{(s(i))}$ for which $s(i)=p$.

Having done so, and assuming $p>2$, consider for a given $s\in \mathfrak {S}_p$ the indices i, j for which $s(i)=p,\ s(j)=p-1$. (If $p=2$, we may simply put $\widehat{\nabla }_\xi ^{(1)}={\nabla }_\xi ^{(1)}$, completing the proof.) An argument similar to the first step shows that the expression

$$\begin{aligned} \widehat{\nabla }_{\xi _p}^{(s(p))} \cdots \nabla _{\xi _i}^{(p)} \cdots (\widehat{\nabla }_{\xi _j}^{(p-1)}-\nabla _{\xi _i}^{(p-1)}) \cdots \widehat{\nabla }_{\xi _1}^{(s(1))} \end{aligned}$$

(53)

coincides with a similar expression for the composition of s with transposition of the indices $p-1,p-2$. (We wrote (53) for the case that $i>j$; of course, if $i<j$ the $\nabla _{\xi _i}^{(p)}$ would appear to the right of $\widehat{\nabla }_{\xi _j}^{(p-1)}-\nabla _{\xi _j}^{(p-1)}$.) Since those permutations have opposite signs, it shows that we may also remove the hat from the factors $\nabla _{\xi _j}^{(s(j)}$ with $s(j)=p-1$. Removing all the hats in this manner, we have proved the Weinstein-Xu formula

$$\begin{aligned} {\text {VE}}_G(\sigma )(\xi _1,\ldots ,\xi _p)=\left( \sum _{s\in \mathfrak {S}_p}{\text {sign}}(s) {\nabla }_{\xi _p}^{(s(p))} \cdots {\nabla }_{\xi _1}^{(s(1))}\sigma \right) \bigg |_M. \end{aligned}$$

$\square$