This question is geared towards the experts, so I will only briefly gloss the definitions. Everything I say is in the category of finite-dimensional smooth manifolds, and whenever I say "$\mathbb Z$-graded" I'm implicitly using the Koszul or "super" rule for signs. Recall that a vector bundle $A\to X$ determines a $\mathbb Z$-graded commutative algebra whose degree-$k$ part is $\Gamma(\wedge^k A^*)$. A Lie algebroid is a vector bundle $A\to X$ along with a degree-one square-zero derivation (Koszul rule!) on $\Gamma(\wedge^k A^*)$, i.e. it makes $\Gamma(\wedge^k A^*)$ into a complex and in fact a differential graded algebra. Lie algebroid cohomology is the homology of this complex (closed sections mod exact sections). (The "co", as far as I can tell, comes from the typical example: for any $X$, the tangent bundle ${\rm T}X \to X$ is a Lie algebroid via the de Rham $d$, and the homology of the complex $\Gamma(\wedge^k {\rm T}^* X)$ is de Rham cohomology of $X$. Or maybe the etymology is different. The functor from Lie algebroids to their cohomology is contravarian, so maybe that's it.) A morphism of Lie algebroids is a morphism of vector bundles that induces a morphism of the corresponding complexes.

A VBLA is a "vector bundle in the category of Lie algebroids". I.e. it is a pair of Lie algebroids $D \to B$ and $A\to X$, a morphism of algebroids $\{D\to B\} \to \{A\to X\}$, and maps $0: \{A\to X\} \to \{D\to B\}$ and $+ : \{D\to B\} \times_{\{A\to X\}} \{D\to B\} \to \{D\to B\}$ satisfying the conditions you would think of if you write down the words "vector bundle" in categorical language. (Equivalently, the induced maps on manifolds $B \to X$ and $D\to A$ are vector bundles, and there's some compatibility conditions.) For details, peruse the papers by Mehta and Garcia-Saz, or by Mackenzie.

In particular, a VBLA is a morphism of algebroids, and so induces a map on cohomology. My question is:

What "geometric" conditions on the VBLA are equivalent to the map on cohomology being an isomorphism? For example, does every VBLA induce an isomorphism on cohomology? If not, is there a way to check whether the cohomologies are isomorphic without computing the full cohomologies of both?

I'm realizing that I don't know enough examples (that I can actually compute) to gather up intuition. And the answer must be known by the experts (I'm guessing that the answer is known much more generally than just for VBLAs?). As with any mathoverflow question, I'm perfectly happy with an answer consisting entirely of a good reference.


1 Answer 1


The short answer is that a VBLA always induces an injection on cohomology, but in general it isn't an isomorphism.

The less short answer is that the complex for $D \to B$ decomposes as a direct sum of a bunch of subcomplexes, the first of which is the complex for $A \to M$. Thus, you have an isomorphism if and only if all of the higher subcomplexes are acyclic.

I'll give more detail. First of all, it helps to take the point of view of graded geometry. Here, a vector bundle $A \to X$ is associated to a graded manifold $A[1]$, whose algebra of "functions" is $\Gamma(\wedge^k A^*)$. Then the differential (let's call it $d_A$) can be thought of as a degree 1 vector field satisfying the equation $[d_A,d_A] = 0$. From this point of view, a VBLA corresponds to a vector bundle in the category of graded manifolds $D[1] \to A[1]$, where the differential $d_D$ is a linear vector field over the base vector field $d_A$ (so as a derivation, it preserves linear functions). This means you can impose a grading on the "functions" on $D[1]$ (different from the homological grading), which is preserved by $d_D$. The decomposition via this grading is the one that I alluded to above.

You can go even further by using the product structure. As an algebra, the "functions" on $D[1]$ is, up to some closure, generated by the functions coming from $A[1]$ and the linear functions. I think you can use this fact to conclude that the cohomology of $D$ is isomorphic to the cohomology of $A$ if and only if the subcomplex of linear functions is acyclic. In our paper on VB-algebroids, Gracia-Saz and I studied the linear subcomplex in terms of Lie algebroid superrepresentations, and we gave a classification of those satisfying a regularity assumption. So take a look at that---we didn't directly address the question, but it should be fairly easy to use the classification data to describe which ones are acyclic.


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