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This is about the notion of representations upto homotopy of Lie algebroids. I am following the reference Representations up to homotopy of Lie algebroids by Camilo Arias Abad and Marius Crainic.

Let $M$ be a smooth manifold and $A\rightarrow M$ a Lie algebroid over $M$. The proposition $3.2$ in above mentioned paper says that a representation upto homotopy is given by the following data:

  • a graded vector bundle $E\rightarrow M$ (I think they are considering $\mathbb{Z}$ grading $E=\oplus_{i \in \mathbb{Z}}E^i$),
  • a degree $1$ operator $\partial :E\rightarrow E$ making $(E,\partial)$ a complex (with $\partial:E^i \rightarrow E^{i+1}$),
  • an $A$-connection on $E\rightarrow M$ (a nice map $\nabla:\Gamma(M,A)\times \Gamma(M,E)\rightarrow \Gamma(M,E)$) that is compatible with the complex structure $\partial:E\rightarrow E$,
  • and $\rm{End}(E)$-valued $2$-form $\omega_2$ of total degree $1$, i.e., $$\omega_2\in \Omega^2(A, \underline{\rm{End}}^{-1}(E))$$ satisfying the condition $\partial(\omega_2)+R_\nabla=0$, where $R_\nabla$ is the curvature of $\nabla$,
  • for each $i>2$ and $\rm{End}(E)$-valued $i$-form $\omega_i$ of total degree $1$, i.e., $$\omega_i\in \Omega^i(A,\underline{\rm{End}}^{1-i}(E))$$ satisfying the condition $\partial(\omega_i)+d_\nabla(\omega_{i-1})+\omega_2\wedge\omega_{i-2}+\cdots+\omega_{i-2}\wedge\omega_2=0$.

The question is as I mentioned in the title: how should one think about representations upto homotopy?

Representation of a Lie algebroid $A\rightarrow M$ is a vector bundle $E\rightarrow M$, together with ''an action'' of $A\rightarrow M$ on $E\rightarrow M$. I used the term action to remember (and relate) the notion of representation of a Lie group/Lie algebra etc. Here, the "action" is the data of a map $$\nabla:\Gamma(M,A)\times \Gamma(M,E)\rightarrow \Gamma(M,E)$$ that is nice to some extent; precisely, we want $\nabla$ to be a flat $A$-connection. The curvature $R_\nabla$ would then be zero. This is what I understand about representations of Lie algebroids.

The punch line I heard multiple times is, representation upto homotopy of a Lie algebroid $A\rightarrow M$ is a graded vector bundle with an $A$-connection that is flat upto some correction.

So, the first three conditions are understandable. The third condition says, the curvature $R_\nabla$ may not be zero, but there is a $2$-form $\omega_2$ with $\partial(\omega_2)+R_\nabla=0$.

If the degrees of $\omega_2, \omega_{i-2}$ matches appropriately, then we would just have $\omega_2\wedge \omega_{i-2}+\omega_{i-2}\wedge\omega_2=0$. Assuming this is the case for each component in $4$th condition, we would have $\partial(\omega_i)+d_\nabla(\omega_{i-1})=0$. But, the degrees may not match that well to give us $\partial(\omega_i)+d_\nabla(\omega_{i-1})=0$. The condition $\partial(\omega_i)+d_\nabla(\omega_{i-1})=0$ is still fine. But, the condition $$\partial(\omega_i)+d_\nabla(\omega_{i-1})+\omega_2\wedge\omega_{i-2}+\cdots+\omega_{i-2}\wedge\omega_2=0$$ is too difficult to digest. How should one think about this condition?

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  • $\begingroup$ @მამუკაჯიბლაძე Yes. Fixed it. many thanks :) $\endgroup$ Commented Jan 28, 2022 at 13:09
  • $\begingroup$ Most likely this relates to the Maurer-Cartan forms: the equation for them is $d\omega+\omega\wedge\omega=0$; here it seems to be satisfied with $d=\partial+d_\nabla$... $\endgroup$ Commented Jan 28, 2022 at 19:30
  • $\begingroup$ @მამუკაჯიბლაძე that could possibly be true.. If you come up with some more explanation, please add it here... $\endgroup$ Commented Jan 29, 2022 at 4:04
  • $\begingroup$ One way to think about this is by considering the (algebroid) differential forms valued in $E$: $\Omega^{\bullet}(E)$. Recall that for a vector bundle $E$, a connection on $E$ is equivalent to a degree $1$ operator $d$ on $\Omega^{\bullet}(E)$. The curvature is given by $d^2$, so that the connection is flat precisely when $d^2 = 0$. In this case, $\Omega^{\bullet}(E)$ is a dg module over the dg algebra of Lie algebroid forms. $\endgroup$ Commented Feb 1, 2022 at 21:23
  • $\begingroup$ When $E$ is a graded vector bundle, we may still consider $\Omega^{\bullet}(E)$. It has a total grading that combines the form degree and the grading on $E$. We can still try to equip it with a differential $d$, turning it into a dg module. But now this no longer corresponds to a flat connection on $E$. Instead, it is a representation up to homotopy. So I believe that all the conditions involving $\partial$, $d_{\nabla}$, and $\omega_{i}$, all combine into the single equation $d^2 = 0$. $\endgroup$ Commented Feb 1, 2022 at 21:27

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