### Modules for Lie infinity-Algebras

#### Posted by Urs Schreiber

This here is mainly a question to Jim Stasheff – and possibly to his former student Lars Kjeseth in case he is reading this – concerning the general issue addressed in the article

Lars Kjeseth
*Homotopy Lie Rinehard cohomology of homotopy Lie-Rinehart pairs*

HHA 3, Number 1 (2001), 139-163.

which we were discussing in BV for Dummies.

The question is

What is the right $\infty$-categorification of a Lie-Rinehart pair?

A Lie-Rinehart pair is a pair $(B,L)$ consisting of an associative algebra $B$ and a Lie algebra $L$, such that $B$ acts on $L$ and $L$ acts on $B$ in a compatible way, where the two compatibility conditions are the obvious ones you find when looking at the archetypical example $(B = C^\infty(X), \; L = \Gamma(T X))$ of the Lie-Rinehart pair obtained from the smooth functions on a smooth manifold $X$ and the vector fields on $X$ acting on these.

This example clearly encodes the same information as the tangent Lie algebroid of $X$, and in fact it is rather manifest that whenever $B = C^\infty(X)$ for some space $X$, a Lie-Rinehart pair $(B,L)$ is precisely a Lie algebroid structure over $X$, and vice versa.

We have discussed that people are thinking that a *Lie $n$-algebroid*, whatever it is in direct terms, is dually encoded precisely in *non-negatively* graded dg-manifolds.

I found that disturbing. In light of the fact that non-negatively graded dg-algebras beautifully and neatly capture *everything* about semistrict Lie $\infty$-algebras, with the latter being a very natural categorical concept, I am not prepared to accept that there should be no equally nice categorical picture for arbitrarily graded dg-manifolds.

My conjecture therefore:

- non-negatively graded dg-manifolds appear when in a Lie-Rinehart pair you categorify only the $L$, not the $B$.

- as we categorify both $L$ and $B$, the categorified $L$ will give a dg structure in positive degree, whereas the $B$ gives a dg-structure in negative degree

- and *together* the $L$ and the $B$ fuse to form a single dg-structure, the differential on which is

$\;\;\;$ - restricted to the $L$-part just the categorified Lie bracket etc. on $L$

$\;\;\;$ - restricted to the $B$-part essentially just the differential on a Baez-Crans type $\infty$-vector space, which is essentially nothing but a chain complex.

$\;\;\;$ - on the intersection of both precisely the action of $L$ on $B$.

After mentioning this idea a couple of times on the Café, Jim Stasheff kindly pointed me to the work of Lars Kjeseth (possibly closely related to Marius Crainic’s work, but I can’t tell yet).

Now I have looked at Lars Kjeseth’s article, and have come back with the impression that it essentially supports this point of view.

Just to make sure, I’d like to ask a couple of questions about this, though. And would generally enjoy discussing this further.

So, instead of trying to review what Lars Kjeseth says in his article, I shall go the other way round: I’ll try to reproduce the main point in my own words, and am asking if we agree that what I am saying harmonizes with what Lars Kjeseth was saying. If not, I hope somebody will correct me.

First, recall the following well-known fact (more details about which you can find for instance in Lie $n$-algebra cohomology):

- A Lie $\infty$-algebra ($\infty$-category internal to vector spaces equipped with a skew-symmetric and coherently Jacobi bracket functor)

is the same thing as

- an $L_\infty$-algebra

which in turn is the same thing as

- a $\mathbb{N}_+$-graded vector space $V$ equipped with a degree -1 coderivation $D : S^\bullet(V) \to S^\bullet(V)$ on the free co-commutative coalgebra over $V$, such that $D^2 = 0 \,.$

If $V = V_1$, then this is an ordinary Lie algebra.

So the question is: what is a module over such an Lie $\infty$-algebra $L$?

The answer that Lars Kjeseth gives in his definition 4.4 (p. 15) is this (recall: in my words, please compare carefully):

- a possibly unbounded chain complex $B$, i.e. a $\mathbb{Z}$-graded vector space $M$ equipped with a degree -1 endomorphism $m_1 : M \to M$ such that $m_1^2 = 0$

- a coderivation $D_{(B,L)} : S^\bullet(V \oplus M) \to S^\bullet(V \oplus M)$ which extends both the $L_\infty$-structure as well as the complex structure on $M$ in that $D_{(B,L)}|_{S^\bullet(V)} = D_L$ and $D_{(B,L)}|_{S^\bullet(M)} = m_1$

Am I right about this?

If so, this would be what I was hoping to see – if we restrict $M$ to be *negatively* graded.

Let me remark on that:

once we dualize everything to quasi-free differential graded algebra by defining
$d : S^\bullet ((V \oplus M)^*) \to S^\bullet ((V \oplus M)^*)$
as
$d \omega := \omega(D_{(B,L)}(\cdot))$
for all $\omega \in (V \otimes M)^*$ we find that the positive degree parts of the complex induce an action of a Lie $\infty$-algebra on the complex in negative degree, which is now a cochain complex with a degree *plus one* differential
$d : M_{-|n|} \to M_{-|n-1|}
\,.$
But that really means that $(M^*,d)$ is naturally a *chain complex*, hence really an $\infty$-vector space.

I am trying to emphasize here a trivial point, which however deserves attention, I think, as long as we are still trying to figure out the right way to think about the question at hand:

it is *natural* for the module to be in negative degrees, since the module is really supposed to be a (Baez-Crans type) $\infty$-vector space, which is really a *chain* complex – which means that its differential seems to run in the “opposite” direction of the differential that encodes the Lie brackets.

For instance if we just have an ordinary Lie algebra $g$ in degree 1 together with a Lie action of $g$ on $M_{-|k|}^*$, i.e. $\rho : g \otimes M_{-|k|}^* \to M_{-|k|}^*$ then the differential $d$ on $M_{-|k|}^*$ would be $d \omega = \rho(\cdot)(\omega)$ with the right hand side regarded as being in $g^* \otimes M_{-|k|}^*$. So it’s $M^*$ rather than $M$ that $L$ is naturally acting on.

We can see this nicely exemplified in the BV complex (following the discussion in BV for dummies):

there we have a Lie algebra acting on the *two-vector* space that comes from the complex
$\Gamma(T X ) \stackrel{d S(\cdot)}{\to} C^\infty(X)
\,.$
This complex is to be read as:

the 2-vector space whose space of objects is the space of smooth functions on $X$. There is a morphism from the function $f$ to the function $g$ whenever $f$ and $g$ differ by a function that vanishes “on shell” (i.e. on the critical points of the function $S \in C^\infty(X)$).

Here $d S (\cdot)$ would be identified with the $m_1^*$ from above (but we have to be careful about the dualization now that the spaces are infinite dimensional, I am actually now taking the dg-picture with the lower case $d$ as the defining one).

Then the action of a Lie algebra (“of symmetries”) $g$ on $X$ by a Lie algebra homomorphism $\rho : g \to \Gamma(T X)$ defines the BV differential on $C^\infty(X)$ as $d f := \rho(\cdot)(f)$ and so on.

So from that point of view, I am thinking I can look at the entire BV complex generated from (I keep sticking to my toy example from BV for dummies)

$\array{ 0 &\stackrel{d|}{\to}& M_{-2}^* &\stackrel{d|}{\to}& M_{-1}^* &\stackrel{d|}{\to}& M_{0}^* &\stackrel{d|}{\to}& V_{1}^* &\stackrel{d|}{\to}& 0 \\ \\ 0 &\to& \mathrm{ker}(d S(\cdot)) &\hookrightarrow& \Gamma(T X ) &\stackrel{d S(\cdot)}{\to}& C^\infty(X) &\stackrel{\rho}{\to}& C^\infty(X) \otimes g^* &\to& 0 \\ \\ && -2 && -1 && 0 && 1 \\ && antighosts && antifields && fields && ghosts \\ \\ && (three- &-& -vector- &-& space) && (Lie algebra) \\ \\ && (--- &-& -3-module- &-& ---) && }$

(Here the $d|$ denotes the result of restricting domain and codomain of $d$ to be $V^* \otimes M^*$ (i.e. just the first tensor power). )

I understand (from talking to Zoran and Danny) that there are other definitions of modules for $L_\infty$-algebras, for instance in terms of $L_\infty$-morphisms from $L$ to the dg-Lie algebra $\mathrm{End}(M)$. I haven’t looked at that closely enough to say anything intelligent about it. Except that I am hoping that every reasonable definition of an $L_\infty$-module fits into the above picture.

## Re: Modules for Lie infinity-Algebras

Aha. Let us see if I got this right.

A 3-vector space is the same thing as the linear part of a Koszul-Tate resolution with antifields of degree -1 and -2. Arbitrary functions of the fields, but only linear functions of antifields.

It is a -3-module (why minus?) because g acts on this 3-vector space. IOW, g acts on each M

^{*}_{-n}and commutes with the differential.But g is still an ordinary 1-algebra, and the cohomology groups are ordinary g modules, right?