## November 13, 2007

### 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.

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 $D_{(B,L)}|_{S^\bullet(M)} = m_1$

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.

Posted at November 13, 2007 7:41 PM UTC

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### 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?

### Re: Modules for Lie infinity-Algebras

I’m linguistically challenged - linear part refers to the linear part of the differential or the space of generators?

The fields are special? elements of degree 0?

The rest seems ok.

will respond directly to Urs post as soon as I digest

Posted by: jim stasheff on November 14, 2007 1:11 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Yes, that’s what I meant. Arbitrary functions over elements of degree 0, linear functions over elements at negative degree.

### Re: Modules for Lie infinity-Algebras

It is a -3-module (why minus?)

Sorry, that is just an artefact of my imperfect MathML rendering. It is supposed to read just “3-module”. What looks like a minus sign here was supposed to be a big inter-column hyphen. I’ll see if I can improve the typesetting here.

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.

Yes!

More generally, a 3-vector space (of “Baez-Crans-type”, namely a module over the monoidal 2-category $\mathrm{Disc}(\mathrm{Disc}(\mathbb{C}))$) is (equivalent to) a 3-term chain complex.

And the Koszul-Tate part of the BV complex is indeed an example of this.

Even better, actually: I just checked with Zoran Škoda that the Koszul-Tate 3-vector space is indeed an associative 3-algebra in BC 3-vector spaces, which is equipped with an action of the Lie algebra of symmetries by derivations.

(For this to work as it should, one needs to be sure to use the right tensor product on chain complexes: not the naive one but the one induced from the equivalence with $BC3Vect$, namely the Loday-Pirashvili tensor product)

This proves one part of my conjecture: it shows that we can think of the BV complex as a categorified Lie-Rinehart pair $(B_3,L_1) \,.$

The Lie 3-algebra happens to be just an ordinary Lie (1-)algebra $L_1 = g$, but the associative 3-algebra $B_3 = (\mathrm{ker}(d S(\cdot)) \to \Gamma(T X) \to C^\infty(X))$ it acts on is an honest associative 3-algebra – albeit one which is equivalent to just an ordinary associative algebra (namely that of on-shell functions).

I am in the process of writing this up in more detail.

But $g$ is still an ordinary 1-algebra, and the cohomology groups are ordinary $g$ modules, right?

That’s right!

This is because we are looking at the “ordinary” BV formalism here, for quantization of ordinary (1-)gauge theory.

We find (I think) that the BV complex that we were talking about is

a 3-Lie-Rinehart pair which is a resolution (is equivalent to) the 1-Lie-Rinehart pair of on-shell functions acted on by the gauge symmetries.

That’s actually a fun little exercise to spell out. I need to do one or two other things today, then I try to write this up. (But I need to check with Zoran on how secret this will have to be…)

Posted by: Urs Schreiber on November 14, 2007 1:13 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

This is a reply just to the original post - I don’t see a reply option there!

First, the positive versus negative grading is an artifact/historical accident unless
you start with a topological space or an alg of C^\infty functions. Linguistically the tradition is: d of degree -1 for a chain complex, d of degree +1 for a cochain complex. Physicists are motivated cohomologically, cf. diff forms, so the literature for BV had degre +1 dominant.

As for dualizing, certainly graded duals should be used throughout, but if the pieces are also inf dim, are they still inf dim over the ground RING?

For classical Chevalley Eilenberg, they got it right to begin with; their cochains are alternating multilin fcns on the Lie alg, even if inf dim.

When talking about the L versus A part,
do we mean in L\oplus A or in S(L\oplus A)?

In Lie-Rinehart, each is a module over the other and L acts on A by derivations. Oh, lets start with A a comm assoc algebra!
So all of those structures should be up to strong homotopy when we categorify’.

The notions of sh derivations has been codified by ?Lada and?? and is reprised in my papers with Kajiura on the arXiv.

yes, M is an L_\infty module over L
is the same as an L_\infty morphism from L to End(M), but if M is an algebra,
the imagew is in the part of End that sees the alg structure

functions agreeing on shell (called in physspeak something like weakly equal?)
translates to f-g is in the image of Koszul differential - by definition

Posted by: jim stasheff on November 14, 2007 1:37 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

“Post a New Comment” at bottom of page.

### Re: Modules for Lie infinity-Algebras

Jim Stasheff wrote:

First, the positive versus negative grading is an artifact/historical accident

I am slightly afraid that I didn’t make the central point in my above entry clear enough:

the issue of positive and negative grading that I am going on about is not the fact that there is an arbitrary convention involved in whether we take the differential (either kind) to be of degree +1 or -1.

Rather, fix any one convention. Then I am talking about the question I first asked here and the reply I got to that from David Ben-Zvi here, here and here:

the issue is:

- in a dg structure like the BV complex, we have generators in both, negative as well as positive degree (whichever convention we use)

- whereas the quasi-generally accepted definition of a Lie $n$-algebroid (maybe going back to Ševera?) dually as an NQ-manifold says that the Lie $\infty$-algebroid inhabits just one half (either non-negative or non-positive, depending on your convention) of the gradings.

Back then you and David Ben-Zvi told me something like:

“sure, right, the degrees of one sign contain the symmetries, the ‘dg-directions’, while the degrees of the other sign contain the things quotiented out, the ‘stacky directions’”.

I found that very helpful back then. But now I want more!

I am saying: a dg-structure in just one half of the available degrees (a Lie $\infty$-algebroid) has a very natural $n$-categorical interpretation. I want an equally nice, satisfactory $n$-categorical interpretation for a dg-structure with no restrictions on the gradings.

One single thing. Nice and natural.

The kind of statements that I will accept as satisfactory are of the kind:

“A dg-structure without restriction on the grading is a Lie-Rinehart $\infty$-pair.”

Notice: I am not sure yet if this statement is actually right (thought it cannot be completely off, it seems). What I am saying is that this is the kind of statement that I am looking for.

If the statement is not quite right as stated: help me find the right version!

Posted by: Urs Schreiber on November 14, 2007 10:49 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I am slightly afraid that I didn’t make the central point in my above entry clear enough

I should maybe try to exhibit the main point mor vividly:

- we have an $\infty$-vector space $B$, equipped with an associative algebra structure. Let’s keep the convention that $n$-vector spaces come to us non-positively graded.

$B = \cdots B_{-3} \to B_{-2} \to B_{-1} \to B_0 \to 0$

- we have another $\infty$-vector space $L$, equipped with a Lie algebra structure

$L = \cdots L_{-3} \to L_{-2} \to L_{-1} \to 0 \,.$

So our Lie-Rinehart $\infty$-pair has only non-positive degrees

$\array{ ( & B &,& L &) \\ & -\infty \lt d \leq 0 && -\infty \lt d \lt 0 } \,.$

But the imporrtant thing happens when we form the Chevalley-Eielenberg algebra of this $\infty$-pair:

$L$ gets dualized, while $B$ does not!

The elements of the CE algebra look like

$L^* \otimes L^* \otimes \cdots L^* \otimes B \to B \,.$

For instance the binary bracket give $L^* \stackrel{[\cdot,\cdot]^*}{\to} L^* \otimes L^*$

and

$B \to L^* \otimes B \,.$

That’s crucial. Now, the dual of an $n$-vector space $V$ is $V^* := \mathrm{Hom}(V, b^n \mathbb{C}) \,.$

You have to use the Loday-Pirashvili closed structure on $n\mathrm{Term}$ to compute this. The result is, for the example of a 3-vector space

$(V_{-2} \stackrel{d_{-1}}{\to} V_{-1} \stackrel{d_0}{\to} V_0)^* = (V_0^* \stackrel{d_0^*}{\to} V_1^* \stackrel{d_{-1}^*}{\to} V_{-2}^*) \,.$

Hence: under dualization the order of the homogeneous degree vector spaces reverses.

Therefore, since our Chevalley-Eielenberg algebra structure on our Lie-Rinehart $\infty$-pair will mix the differentials in $B$ with those – not in $L$ but – in $L^*$, we should then think of $L^*$ being in positive degree

$\array{ ( & B &,& L^* &) \\ & -\infty \lt d \leq 0 && 1 \leq d \leq \infty }$

because that’s the way the differentials match.

So, this is my point:

a Lie-Rinehart $\infty$-pair should be a dg structure (its “Chevalley-Eileneberg $\infty$-algebra”) of unrestricted degree with

$B$ in the non-positive degrees

and

$L^*$ in the positive degrees.

Posted by: Urs Schreiber on November 15, 2007 10:41 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Except for the insistence on the L-P tensor product (which I don’t have access to yet),
looks correct

Posted by: jim stasheff on November 15, 2007 1:46 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

L-P tensor product

The LP tensor product (see my description of it here) is quite possibly well known under some other name to other people.

It is the tensor product induced on the category of $n$-term chain complexes by the equivalence of that with $n$-categories internal to vector spaces.

It amounts to the obvious thing:

the $k$th term $(V \otimes W)_k$ in the tensor product of two chain complexes $V = \cdots \to V_0 \to V_1 \to \cdots$ and $W = \cdots \to W_0 \to W_1 \to \cdots$ is the direct sum of all tensor products of elements of the two complexes that add up to the total degree $k$, hence

$(V \otimes W)_k := \oplus_{n+m = k} V_n \otimes W_m \,.$

As you can see, yet another way to regard this tensor product is this:

turn both $V$ and $W$ into qDGCAs, then take the ordinary tensor product of these qDGCAs and extract the chain complex that the result is freely generated from. That gives the same result.

As a special case of this one finds the tensor product of chain complexes that is mentioned in the Wikipedia article on the Koszul complex.

I am sure you know this under some name or other. I happen to have seen it stated explicitly first in that paper by Loday and Pirashvili.

Posted by: Urs Schreiber on November 15, 2007 3:07 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I don’t see any tensor product in that Wiki article. Is it anything other than the standard tensor prod of chain complexes?
perhaps involving a completion if they are both Z-graded?

Posted by: jim stasheff on November 16, 2007 1:56 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Jim Stasheff wrote:

Is it anything other than the standard tensor prod of chain complexes?

No, it is just that, the
standard tensor product on chain complexes.

For some reason I thought I should emphasize that it is not the naive tensor product, where you tensor the components degree-by-degree.

Posted by: Urs Schreiber on November 16, 2007 12:00 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Since an NQ-manifold is already defined
and does NOT suit your purposes, no problem - onward! Similarly for Lie_\infty algebroid. Something about putting new wine into old wineskins is not a good idea.

As to what the new wine shoudl be called….

Posted by: jim stasheff on November 15, 2007 1:49 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

The only other defn of L_\infty module is that obtained by using the adjoint
L –> End(M) is adjoint to L\otimes M –> M

oh, perhaps you mean an L_\infty algebra on L\oplus M such that???

Posted by: jim stasheff on November 16, 2007 2:00 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Huebschmann responds to Urs:

However I discovered U. Schreiber’s question

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

I actually have a paper on this question, published in the Alanfestschrift

Higher homotopies and Maurer-Cartan algebras: quasi-Lie-Rinehart,
Gerstenhaber-, and Batalin-Vilkovisky algebras.
The Breadth of Symplectic and Poisson Geometry,
Festschrift in Honor of Alan Weinstein, J. Marsden and T. Ratiu,
eds., Progress in Mathematics, 2004, pp. 237–302,
{\tt math.DG/0311294}.

Posted by: jim stasheff on November 14, 2007 2:25 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Huebschmann responds

Thanks a lot! I’ll have a look at that article.

But in case somebody can reply more quickly than I can read:

does that paper establish a correspondence between categorified Lie-Rinehart pairs and arbitrarily graded dg-algebras and/or arbitrarily graded dg-manifolds?

Posted by: Urs Schreiber on November 14, 2007 2:30 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I did have a look at

The crucial definition – as far as the discussion here is concerned – seems to be

equations (4.1)-(4.3) and then def. 4.9 on p. 33.

Do I understand correctly that this is (just) about what should maybe be called 2-Lie-Rinehart pairs: the Lie $\infty$-algebra part is essentially assumed to be strict (a dg Lie-algebra, i.e. only the unary and binary brackets exist (i.e (4.1) and (4.2))) and also the action is assumed to be weakened only up to the first homotopy (i.e (4.3) but nothing higher).

That’s all fine, but somehow I would like to see the following very simple statement to be discussed:

A Lie-Rinehart pair is an associative algebra $B$ and a Lie algebra $L$ such that…

Dually, whenever $B = C^\infty(X)$ for some $X$, this is a dg-manifold over $X$ with generators in degree 0 and 1.

This must have a correspondingly simple analog of the following kind:

A Lie-Rinehart $\infty$-pair is an $A_\infty$-algebra $B$ and a Lie $\infty$-algebra $L$ such that…

Dually, whenever $B = something of X$ for some $X$, this is a dg-manifold over $X$ such that …

How do we fill in these blanks?

I think I have an idea about it, but I am trying to get an impression what has been done on this in the literature and what not.

Posted by: Urs Schreiber on November 14, 2007 8:50 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Let R be a commutative ring, A a commutative R-algebra, L a projective A-module of finite type. Lie-Rinehart structures on (A,L) correspond bijectively to differentials d on the graded algebra Alt_A(L,A) of A-valued A-alternating forms on L that turn Alt_A(L,A) into a differential graded R-algebra, the Maurer-Cartan algebra of the Lie-Rinehart algebra. The terminology ”Maurer-Cartan” is due to van Est. The distinction between ‘A-valued A-multilinear alternating forms on L’ and ‘differential graded algebra merely over R’ is crucial. Thus the Lie-Rinehart structure amounts to

an A-module structure on L,

a bracket on L that turns L into an R-Lie algebra,

an action of L on A on L by derivations;

these pieces of structure satisfy certain compatibility requirements which correspond to the vanishing of the square of the differential on Alt_A(L,A).

Let A be an A_{\infty}-algebra and L an L_{\infty}-algebra. We suppose that A is augmented; let BA be the bar tilde construction, that is, the coaugmented
differential graded coalgebra defining the A_{\infty}-structure, and let \Omega BA be the corresponding loop algebra. Likewise, let C[L] be the Cartan-Chevalley-Eilenberg
tilde construction, that is, the coaugmented differential graded cocommutative coalgebra defining the L_{\infty}-structure, and let L C[L] be the corresponding loop Lie algebra.
The naive approach to extending the notion of Lie-Rinehart algebra to this situation
proceeds via the following pieces of structure:

An A_{\infty}-module structure on L,
that is, an ordinary \Omega BA-module structure \Omega BA \otimes L \to L
on L;

an L_{\infty}-action L \otimes A \to A of L on A, that is, an ordinary action
L C[L] \otimes A \to A of L C[L] on A, viewed at first merely as a graded module
(the ”algebra” structure being ignored).

Then the following questions arise

1) While A is commutative, \Omega BA is not. The Lie-Rinehart axioms do not carry over to non-commutative algebras. The reason is that, given a non-commutative algebra B, the Lie algebra of derivations of B does not acquire a B-module structure. Thus we cannot simply take a requirement of the kind that (\Omega BA, L C[L]) be a (differential graded) Lie-Rinehart algebra as the definition of a Lie-Rinehart {\infty} algebra.

2) What do we mean by an L_{\infty}-action of L on the A_{\infty} algebra A to be by derivations?

3) Do we wish to keep a complex of ”alternating” forms, the term ”alternating” being in general understood the graded sense.

Here is another way to describe some of the difficulties: A skew-symmetric bracket on a vector space g is a Lie-bracket (i.~e. satisfies the Jacobi identity) if and only if the coderivation \partial on the graded exterior coalgebra
\Lambda’[s\fra g] corresponding to the bracket on g has square zero, i.e. is a differential; this coderivation is then the ordinary Lie algebra homology operator, and \Lambda’[s\fra g] is the object dual to the Maurer-Cartan algebra
for g. This kind of characterization
immediately extends to _{\infty}-algebras, in fact, the extension leads to the definition of the notion of L_{\infty}-algebra. The extension is not available for a general Lie-Rinehart algebra, though: Given a commutative algebra A and an A-module L, a Lie-Rinehart structure on (A,L) cannot be characterized in terms of a coderivation on \Lambda_A[sL] with reference to a suitable coalgebra structure on \Lambda_A[sL] (unless the L-action on A is trivial); in fact, in the Lie-Rinehart context, a certain dichotomy between A-modules and chain complexes which are merely defined over R persists thoughout. However, as explained above, the characterization in terms of a Maurer-Cartan algebra is still available
for Lie-Rinehart algebras, and the Maurer-Cartan algebra structure handles
the dichotomy very well.

To come back to the problem of isolating Lie-Rinehart \infty algebras:

1) A possibly way out, which takes care of the problem with the action being by derivations, is to keep the commutative algebra structure and to generalize only
the Lie algebra variable to L_{\infty}-algebras. Thus, given the differential graded algebra A and the L_{\infty}-algebra L, a Lie-Rinehart \infty algebra structure on (A,L) in this sense is an ordinary differential graded Lie-Rinehart structure on the pair (A,LC[L]) .

2) If, instead, one wants to keep an algebra of forms, one arrives at a generalization in terms of a suitable Maurer-Cartan algebra, as explained in my paper

Higher homotopies and Maurer-Cartan algebras: quasi-Lie-Rinehart, Gerstenhaber, and Batalin-Vilkovisky algebras. In: The Breadth of Symplectic and Poisson Geometry, Festschrift in Honor of Alan Weinstein, J. Marsden and T. Ratiu, eds., Progress in Mathematics 232 (2004), 237–302, math.DG/0311294.

The situation worked out there arises in nature: A foliation gives rise to the kind of Lie-Rinehart \infty algebra (A,Q)
described there. The dga A includes ”the algebra of functions and their history” on the space of leaves, and the object Q, not an ordinary Lie algebra, includes ”the Lie algebra of vector fields and their history” on the space of leaves.
In particular, the homology of the object is an ordinary differential graded Lie-Rinehart algebra.

In the quoted paper, only that part of higher structure necessary for foliations has been worked out. The generalization to higher order operations is straightforward, but I do not know whether the more general situation arises in nature.

Posted by: Johannes Huebschmann on November 20, 2007 6:56 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Processing…

Posted by: Urs Schreiber on November 20, 2007 7:05 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Johannes Huebschmann wrote:

The generalization to higher order operations is straightforward, but I do not know whether the more general situation arises in nature.

I am to a large extent motivated by the conviction that the BV complex ought to have a natural interpretation as the dual (whatever we call that, CE or MC) of a Lie-Rinehart $\infty$-pair $(g,B)$, where $g$ is the Lie $n$-algebra of local symmetries and $B$ the $\omega$-vector space of fields.

Often $g$ here is just taken to be a Lie 1-algebra and $B$ just regarded as a puffed-up version of an ordinary vector space, but it is clear, for instance as described by AKSZ, that this setup wants to generalize to higher gauge theories that involve genuinely higher Lie algebras and higher vector spaces of fields. Like Chern-Simons theory.

Posted by: Urs Schreiber on November 20, 2007 8:33 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Johannes Huebschmann wrote

Let $A$ be an $A_{\infty}$-algebra

[…]

Then the following questions arise

1) While $A$ is commutative, $\Omega B A$ is not. The Lie-Rinehart axioms do not carry over to non-commutative algebras.

This is related to something I thought about a bit, lately. Currently I am holding the following opinion:

maybe we don’t want to categorify Lie-Rinehart by trying to extend the associative algebra appearing there to an $\infty$-algebra of sorts.

It seems to me that applications suggest that instead all we want to demand is that we have a complex $V$ of modules for an ordinary associative. algebra, which in degree 0 is that algebra itself.

In my conventions this $V$ would be, since it is supposed to be an $\omega$-vector space, a non-positively graded cochain complex.

Then together with the positively graded cochain complex $g^*$ underlying the $L_\infty$-algebra (or rather its dual) we can form the total complex $g^* \oplus V \,,$ now of arbitrary degree, form its graded-symmetric tensor algebra $\Lambda(g^* \oplus V)$ and put a differential on that, extending the one induced from $g^*$ and $V$ (hence extending the “co-unary” differential to a differential that potentially generates arbitrary word lengths).

This idea I try to describe in my notes. It seems to me that what I describe there is what actually happens in BV formalism.

Posted by: Urs Schreiber on November 20, 2007 8:45 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I don’t understand the last complex in your post. If you act with d on C(X) x g* you typically get a nonzero element in C(X) x Λ2 g*, no?

### Re: Modules for Lie infinity-Algebras

I don’t understand the last complex in your post.

Don’t worry, everything is as you expect it should be. Notice my remark, right below the table:

(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). )

Just as you would expect, once I actually pass from this complex of vector spaces to the differential graded-commutative algebra it induces (by taking graded-symmetric tensor powers of everything) the differential restricted to the non-negative degrees will be the Chevalley-Eilenberg differential that you expect.

It’s just that for the purpose of the exposition so far, I was mostly concerned with highlighting what the underlying complex of vector spaces itself is. So I just displayed $d|$ instead of $d$, which is supposed to be what you get when you apply $d$ only to generators and consider only the single powers of generators in the result.

Not before long, I’ll get to talking about the full qDGCA.

Sorry for being so long-winded and thanks for your all patience.

Posted by: Urs Schreiber on November 15, 2007 10:06 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I don’t understand the last complex in your post. If you act with $d$ on $C(X) \otimes g^*$ you typically get a nonzero element in $C(X) \otimes \wedge^2 g^*$, no?

To which I replied:

Not before long, I’ll get to talking about the full qDGCA.

So now I am, in the entry with the enchanting title Something like Lie-Rinehart infinity-pairs and the BV-complex (BV, part VII).

Because I didn’t do it well in the first version of the notes that I provided there (but now I have improved the discussion a little), let me emphasize here what I think the point is:

First an observation (currently prop. 1 on p. 15 of these notes):

Let $V$ be an element in $\mathbf{Ch}^\bullet(A)$, i.e. a cochain complex of $A$-modules, where for us $A = C(X)$ is the algebra of functions on our space $X$.

Then let $\wedge^n V := \mathrm{image}\left( \underbrace{V \otimes \cdots \otimes V}_{n factors} \stackrel{\mathrm{sym}}{\to} \underbrace{V \otimes \cdots \otimes V}_{n factors} \right)$

be the “symmetrized” $n$-fold tensor product of a complex with itself, where the symmetrization is that with respect to the natural symmetric braiding of $\mathbf{Ch}^\bullet(A)$ (which is what you expect it is: it introduces a sign whenever two odd graded modules change place).

Let $I = (\cdots \to 0 \to 0 \to A \to 0 \to 0 \to \cdots)$ be the tensor unit in $\mathbf{Ch}^\bullet(A)$ and let

$\Lambda V := \bigoplus_{n = 1}^\infty \wedge^n V$

be the usual notation for what should really be thought of as $\exp(V\wedge)I$.

Then, unless I am confused, it should be true that

$\exp(V\wedge \cdot)I \simeq \wedge^\infty (I \oplus V) \,.$

(All infies here are the obvious direct limits.)

The point is this:

if I have the complex $g^*$ underlying a dual $L_\infty$-algebra, then this is in positive degree (nothing in degree 0, in particula) and the differential on it is the (co)-unary part of the $L_\infty$ differential.

Forming

$\exp(g^*\wedge \cdot)I$

is then the same as forming the differential graded-commutative algebra over this complex, containing all the co-unary contributions of the differential. The full dg-structure defining the $L_\infty$ structure is then obtained by adding all the other co-$n$-ary parts of the differential, to obtain the full $d = d_1 + d_2 + \cdots \,.$

The resulting complex $(\exp(g^*\wedge \cdot)I , d)$ sits over the original one $\array{ (\exp(g^*\wedge \cdot)I , d) \\ \downarrow \\ g^* } \,,$ which says that the full differential indeed extends the co-unary one that existed before. Here the projection is the obvious one from $I \oplus g^* \oplus \wedge^2 g^* \oplus \cdots$ on the first $g^*$-summand.

But now in the Lie $n$-algebroid context, the BV-context, which I am trying to think of as an example for a Lie-Rinehart $\infty$-pair, this method needs modification:

the complex $V$ we are now starting with contains something in degree 0. Like in the example above, where

$V = (0 \to \mathrm{ker}(d S)\hookrightarrow \Gamma(T X) \stackrel{d S(\cdot)}{\to} C(X) \to C(X)\otimes g^* \to 0)$

(here $g^*$ now just an ordinary Lie algebra!, in this example)

$C(X)$ is in degree 0 and everything is a module over it. This is crucial. If this is the case, we can do Lie-Rinehart $\infty$-theory, I think.

So the differential above is the co-unary one of the full BV differential which we want and which Thomas was looking for in my discussion.

So let’s build it.

One point is: we should not form $\Lambda V = \exp(V \wedge \cdot)I$ to build the dg-algebra generated from $V$. Since $V$ contains the tensor unit in degree 0, doing that would produce not one, but infinitely many partial copies of the dg-algebra we are after.

$\wedge^\infty V \,.$

That’s the dg-algebra freely (graded-commutatively) generated from $V$. It’s just the standard construction in the physicss (and other) texts, phrased using symmetric tensor powers in $\mathbf{Ch}^\bullet(A)$.

So then, finding the full BV-differential means extending the differential on $\wedge^\infty V$ – which contains just the co-unary contributions – and adding all the higher co-$n$-ary contributions $d = d_1 + d_2 + d_3 + \cdots \,,$

where $d_n$, acting on an element in $V$, increases the “word-lenght” from 1 to $n$.

That we don’t change the co-unary component means that we obtain an extension $\array{ (\wedge^\infty V, \; d) \\ \downarrow \\ V \,. }$

In the given example we would in particular have the co-binary part from the Chevalley Eilenberg complex

$d_2 : g^* \to \wedge^2 g^*$

which is the one Thomas was looking for.

And $\wedge^2 g^*$ gets projected out as we map down along $\array{ (\wedge^\infty V, \; d) \\ \downarrow \\ V }$ which amounts to retaining just the first powers of $V$.

Posted by: Urs Schreiber on November 21, 2007 10:24 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I agree with the conclusion
but all that exp and I jazz just obscures for me what is going on. Suggested reading:
BFV, BV or HandT or my early expositions of their work

Posted by: jim stasheff on November 22, 2007 12:55 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I am in the process of absorbing standard stuff by metabolizing it thoroughly. Sorry if that comes across as if I were trying to reinvenbt wheels.

One of the points I tried to clarify here for myself was this:

given a complex $V$, what is the dg-algebra generated from it?

and there is an ever so sligfht subtlety, it seemed to me:

if the complex has nothing in degree 0, then one wants to form $\Lambda V = \oplus_{k=0}^\infty \wedge^n V \,.$

If, on the other hand, it has the underlying algebra in degree 0, then one should form $\wedge^\infty V \,.$

Sorry for this triviality. But it helped me to see more clearly.

Posted by: Urs Schreiber on November 22, 2007 3:01 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

If it has something in degree 0, then that is itself an algebra, hence treat the positive part as a module over V_0 and
take the tensor algebra over V_0

If V_0 is commutative, then you can take \Lambda

Posted by: jim stasheff on November 23, 2007 1:42 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Luckily, Danny points out that I am being ignorant, the definition of an $L_\infty$-module appearing as def. 5.1 in Lada-Markl, Strongly homotopy Lie algebras. It seems to me to pretty much agree with what Kjeseth has and what I wrote above. Not too surprising.

On the other hand, Lada and Markl in that old paper do not discuss am associative algebra structure on the module, hence no Chevalley-Eilenberg complex associated with it and no Lie-Rinehart $\infty$-structure on it. As far as I can see.

Posted by: Urs Schreiber on November 15, 2007 4:37 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Urs, I read BV for dummies but since I am not a physicist could you please say why one would consider manifolds as opposed to supermanifolds? More specifically, A. Schwarz showed that SP-manifolds are the important geometric objects in BV quantization.

To see what an SP-manifold is please take a look at section 3.4.2 on page 19-20 of this paper. Thanks.

Posted by: Charlie Stromeyer Jr on November 16, 2007 4:25 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

consider manifolds as opposed to supermanifolds?

The manifold I am considering here is the manifold underlying the supermanifold which is dually encoded in the full dg-algebra (“of functions on that supermanifold”) generated from the BV complex.

To see what an SP-manifold is please take a look at section 3.4.2 on page 19-20 of this paper.

I saw you reference this article before and I would like to know what it says, but haven’t found the time yet to look at it. Maybe you can summarize it a little, so that I get an idea of what its main point is?

Posted by: Urs Schreiber on November 16, 2007 5:44 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Thanks for the clarification. The main point of the article I refer to is that one does not usually encounter Maurer-Cartan equations within theoretical physics but instead master equations which involve a divergence operator. These equations cannot be understood via ordinary PROPs nor operads since they can’t encode the concept of trace.

Instead, the authors define wheeled PROPs and operads which are called “wheeled” because free objects in this new category are based on directed graphs which may contain wheels, i.e., directed closed paths of oriented edges.

I can’t draw what a PROP looks like here but you can see what it is like in the comparitive chart on page 34 of this paper. The authors of the first paper above then construct a dg free wheeled PROP whose representations are in one-to-one correspondence with formal germs of SP-manifolds (which involves what are called differential Gerstenhaber-Batalin-Vilkovisky algebras and you can see what these are on page 17 of the paper).

Urs, I think I have just found what I was looking for earlier but I want to ask you about this since you have expertise in both category theory and string theory physics:

Raphael Bousso once asked if string theory could be discretized in the following sense: Is there a sequence of theories with finite dimensional Hilbert spaces such that string theory emerges in the infinite dimensional limit?

I have just found this interesting paper about profinite categories by J. Almeida and P. Weil. On page 5, it says that the profinite category can be obtained as the projective limit (also called an inverse limit) of discrete topological spaces. Would this not be the correct mathematical framework for addressing Bousso’s physical question?

Similarly, would profinite math also be applicable to Edward Witten’s conjecture involving U(finite)-valued and U(infinity)-valued Chan-Paton factors on page 11 of his paper? Thanks.

Posted by: Charlie Stromeyer Jr on November 17, 2007 3:42 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

super - fine
but manifold!
only in a very special sense
99% of what’s going on is algebra or homological algebra

some people get some intuition from speaking in terms of manifolds but not me
until you invokd Poincare’ duality

Posted by: jim stasheff on November 22, 2007 1:00 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Jim Stasheff wrote:

super - fine – but manifold! only in a very special sense 99% of what’s going on is algebra or homological algebra

This is, by the way, a remarkable fact, it seems to me: to put it in a provocative manner:

There is surprisingly little difference between superification and categorification.

A dg-algebra is a monoid in complexes is a monoid in $\omega$-vector spaces, is a monoidal linear $\omega$-category.

The grading is the morphism label:

degree $n$-generators are tangents to $n$-morphisms starting at the zero $(n-1)$-morphism.

The “superpoint” is not far away from the category $2 = (\bullet \to \circ)$.

I find that striking. It has occupied my thinking quite a bit. But I wish I could end this comment here with a sence of the form

“This is telling us that…”

Which I can’t.

Posted by: Urs Schreiber on November 22, 2007 7:10 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Regarding super in the black dot, white dot sense?
where do you see that in cats?
and then superMANIFoLDS categorify how?

Posted by: jim stasheff on November 23, 2007 1:45 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

some people get some intuition from speaking in terms of manifolds but not me

Same for me. Most of the super we are talking about here is really $\infty$ (homotopy) Lie theory.

But the people who think of this business in terms of supermanifolds currently have the following advantage over the people who think about it in terms of Lie $\infty$-theory:

from the supermanifold perspective the Poisson structure on a Lie $\infty$-algebroid is naturally understood.

I keep asking: what does a Poisson structure on the supermanifold representing an Lie $\infty$-algebroid mean in terms of Lie theory??

(I keep standing by my conjecture that it’s about weak inverses in Lie $\infty$-groupoids. But I’d better come up with some more concrete evidence for this.)

Posted by: Urs Schreiber on November 22, 2007 7:15 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

from the supermanifold perspective the Poisson structure on a Lie ∞-algebroid is naturally understood.

it is???

Posted by: jim stasheff on November 23, 2007 1:46 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I wrote:

from the supermanifold perspective the Poisson structure on a Lie $\infty$-algebroid is naturally understood.

it is???

I mean: a Poisson structure is something we are used to putting on a manifold. Hence its a small step to put a Poisson structure also on a supermanifold.

Pretty much all differential geometric notions naturally carry over to supermanifolds.

But now it so happens that we are dealing with supermanifolds which have in fact to be interpreted as being the linearized version of certain $\omega$-groupoids. It’s not so clear anymore what a Poisson structure on an $\omega$-groupoid should mean. Since it was a natural concept in the world of supermanifolds, it must translate into a natural concept in the world of $\omega$-groupoids. But which?

Posted by: Urs Schreiber on November 26, 2007 7:27 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I really need to know the following, maybe somebody can help me:

Is the category of chain complexes star-autonomous?

I.e. is it true that for any two chain complexes $A$ and $B$ we have

$\mathrm{hom}(A,B) \simeq (A \otimes B^*)^*$

??

It’s in principle not hard to check, but I seemed to have had a sign error somewhere and now I am running out of time.

This is how it seems it should work:

The dual of the chain complex $V = \cdots \to V_{2} \to V_1 \to V_0 \to V_{-1} \to V_{-2} \to \cdots$

should be

$V^* = \cdots \to (V_{-2})^* \to (V_{-1})^* \to (V_0)^* \to (V_{1})^* \to (V_{2})^* \to \cdots$

with the obvious dual differentials everywhere.

And maybe with some signs sprinkled in.

Then just turn the crank.

Clearly we get

$(\mathrm{hom}(A,B))_n = ((A \otimes B^*)^*)_n$

term by term, but when writing out the differentials I had some sign difference on the left and on the right. But probably I was just making a mistake.

Have to run to catch my train now. If anyone happens to know the quickj answer, do me a favor and drop me a note.

Posted by: Urs Schreiber on November 16, 2007 5:55 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Okay, now it works. Every second differential of the dual complex needs a sign. Then chain complexes are star autonomous.

Posted by: Urs on November 16, 2007 8:45 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Urs, I’m not so sure this works, but let me think out loud a bit (this will also be a composite reply to some things you were asking in the earlier star-autonomy thread to which you referred).

We know chain complexes are symmetric monoidal closed: the tensor product is given by the formula

$(A \otimes B)_n = \sum_{p+q = n} A_p \otimes B_q$

with a well-known differential which I won’t bother writing down. From this one can work out the formula for the internal hom:

$hom(B, C)_p = \prod_{n-q = p} hom(B_q, C_n)$

and its differential (which again I won’t write down). Now suppose we consider just chain complexes with 0 differentials (i.e., $\mathbb{Z}$-graded modules), say where every component is the ground ring $k$. Then the tensor of two such involves a countable sum of copies of $k$ at every component, the hom of two such involves a countable product at every component, and there’s no way that we’re going to be able to get $hom(A, B) \cong (A \otimes B^\star)^\star$ from your proposed star-operator.

There are other ways of formulating the notion of star-autonomous category which I find helpful. Put $D = I^*$, where $I$ is the monoidal unit. Then

$hom(A, D) = (A \otimes I)^* \cong A^*$

so that the dualization functor is always obtained by homming into this special object $D$, the dualizing object. Also the natural double-dual map

$A \to hom(hom(A, D), D) = A^{**}$

is an isomorphism. In other words, a star-autonomous category is essentially the same thing as a closed symmetric monoidal category equipped with an object $D$ for which double $D$-dualization is isomorphic to the identity. But I’m pretty sure there’s no $\mathbb{Z}$-graded chain complex which can serve as a $D$ here. Sorry.

$hom(A, B) \cong (A \otimes B^{\star})^{\star}$

and why not the simpler formula for the right, $A^{\star} \otimes B$ (compact closure)? There are reasons coming from within categorical logic: one thinks of the dualization operator as a generalization of logical negation. Thus, star-autonomous categories obey a kind of algebra which generalizes Boolean algebra: we don’t have $a \Rightarrow b = \neg a \wedge b$, but we do have $a \Rightarrow b = \neg(a \wedge \neg b)$.

Posted by: Todd Trimble on November 17, 2007 2:21 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Todd: why not the simpler formula for the right, A ⋆⊗B (compact closure)?

Are you saying that’s not good enough categorically
although it’s fine for complexes of finite type??

what’s compact closure?
important when not of finite type?

Posted by: jim stasheff on November 17, 2007 2:58 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

It’s possible that I misinterpreted what Urs was saying; maybe he was considering only complexes of finite type (which I take to mean complexes which are 0 at almost every component, and finite-dimensional at every component). But in that case, we do have compact closure (a special case of star-autonomy), unless I’m making a terribly bad mistake.

“Compact closure” simply means we have a symmetric monoidal category such that for every object $A$, there’s an object $A^\star$ for which $A^\star \otimes -$ is right adjoint to $A \otimes -$. Or, in perhaps less scary language, that we have maps

$\eta: I \to A^* \otimes A$

$\varepsilon: A \otimes A^* \to I$

which are compatible in the sense of triangular equations, or yanking moves on string diagrams, what have you. And that’s what I think Urs has for complexes of finite type (it seems you and I agree about that). “Compact closure” is a perfectly lousy name for it, but I think it originates from the example of finite-dimensional representations of a compact group.

Compact closure is of course perfectly fine and dandy as a categorical condition on symmetric monoidal categories, but in certain contexts of categorical logic which need not concern us here, namely in the subculture of linear logic, categorical coherence theory, and their game semantics, compact closed categories are considered to be somewhat degenerate examples of star-autonomous categories. There are many significant examples of star-autonomous categories which are not compact closed; one would be the category of sup-lattices. [If pressed I think I can say more, and certainly people like Robin Houston would also have lots to say about this.]

Not sure this answer will be found satisfactory, but I’ll stop here for now.

Posted by: Todd Trimble on November 17, 2007 4:02 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

the formula for the internal hom:

$\mathrm{hom}(B,C)_p = \prod_{n-q = p} \mathrm{hom}(B_q, C_n)$

Ahh!. I was conidering a $\sum$ instead of a $\prod$ here. Grr…

All right, thanks a lot for correcting that.

maybe [Urs] was considering only complexes of finite type

Yes!

But in that case, we do have compact closure (a special case of star-autonomy),

Thanks for suggesting that, possibly that’s what I really need.

“Compact closure” simply means we have a symmetric monoidal category such that for every object $A$, there’s an object $A^*$ for which $A^*\otimes -$ is right adjoint to $A \otimes -$.

Okay, so in that case $\mathrm{Hom}(A,B) \simeq \mathrm{Hom}(I, A^* \otimes B) \,.$

Right?

That would indeed suffice for what I am trying to understand. Which is this:

given a Lie algebra $g$ and an associative algebra $B$ internal to chain complexes, I want to be able to pass from the corresponding CE-complex with things of the form $\mathrm{Hom}(g \wedge g \wedge \cdots g, B)$ to expressions involving $g^* \wedge g^* \wedge \cdots g^* \otimes B \,.$

Posted by: Urs Schreiber on November 18, 2007 12:20 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Okay, so in that case $Hom(A,B)\simeq Hom(I, A^* \otimes B)$. Right?

Right. By the way, Robin Houston pointed out to me privately that my saying

“Compact closure” simply means we have a symmetric monoidal category such that for every object $A$, there’s an object $A^\star$ for which $A^\star \otimes -$ is right adjoint to $A \otimes -$

contains a subtle error. The correct formulation is that in the suspension bicategory $\Sigma C$ of the monoidal category $C$, that every 1-cell $A$ has a right adjoint, i.e., in $C$ there is an object $A^\star$ together with a unit $I \to A^\star \otimes A$ and a counit $A \otimes A^\star \to I$ satisfying triangular equations (as I wound up saying in the end). The way I put it above appears to be ever so slightly weaker; to put it right, I should have added that the unit and counit of $(A \otimes -) \vdash (A^\star \otimes -)$ must be compatible with the canonical strengths of these functors.

Posted by: Todd Trimble on November 18, 2007 2:02 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Thanks again, Todd!

Okay, so let’s summarize that:

the category of chain complexes of finite type is

- symmetric monoidal

- with duals

- closed

- compact closed.

The dual of a complex $V = (V_n)$ is

$(V^*)_n = (V_{-n})^*$

with the obvious dual differentials (and no signs sprinkled in).

We have compact closure and the $A^*$ with respect to the compact closure happens to coincide with the $A^*$ obtained from duality (is there a special way to say this?)

$A^* := \mathrm{hom}(A,I)$

and the tensor unit $I = I^*$ is the obvious complex concentrated in degree 0.

the suspension bicategory $\Sigma C$ of the monoidal category $C$

By the way, as you have probably seen, I am being pressured not to say “suspension bicategory” anymore, and not to denote it by $\Sigma$ – because that might not harmonize well with the usage of the same terms and symbols in topology. I am feeling unsure about this, for various reasons. Do you have an opinion on that?

Posted by: Urs Schreiber on November 18, 2007 2:27 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

One usuful convention is to use lower indices for chain comolexes (d of degree -1)
and upper for cochain complexes (d of deg 1)
then the dual of V_n is V^{-n}
Note that if V^{-n}= Hom(V_n,k)
then its elements are indeed of degree -n
Denote the dual of d as \delta
then (\delta h)(v) = \pm h(dv)
so the sign does enter

Posted by: jim stasheff on November 18, 2007 8:01 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

All this looks good. My own use of the terminology ‘compact closed’ automatically means ‘symmetric monoidal with duals’ and all the rest, but I’m not certain that’s the universally accepted meaning. I seem to recall that’s how Max Kelly used this terminology (in his paper with LaPlaza on coherence theorems for these structures); there may be further information on this in David Yetter’s book Functorial Knot Theory (which however I don’t have with me). If that’s not the usually accepted meaning of ‘compact closed’, I hope someone will let us know.

Yes, in a compact closed category, it’s automatic that $A^\star \cong hom(A, I)$ and that $hom(A, B) \cong A^\star \otimes B$; often the hom notation is generally suppressed and this tensor formula used instead throughout. Everything you said goes through fine for (finite type) chain complexes of f.g. projective modules over a commutative ring $A$ (that’s just an example of course; I realize you may have other things in mind).

By the way, as you have probably seen, I am being pressured not to say “suspension bicategory” anymore, and not to denote it by $\Sigma$ – because that might not harmonize well with the usage of the same terms and symbols in topology. I am feeling unsure about this, for various reasons. Do you have an opinion on that?

That’s a tough one (actually, I hadn’t seen that pressure). Within the $n$-categorical community this notation is widely accepted, I believe. Would you direct me (either publicly or privately) to where that discussion took place? Despite something I once said at the Café, I don’t think I have a problem with that particular notation, and I’m not so sure it greatly conflicts with the use in topology.

(You realize it’s hard for me to offer any objective “words of wisdom” here, because I too have my own notations that I love and am even married to, which help me a lot in my own thinking but which might [or do] put off certain people. Obviously these matters have to be decided on a case by case basis [who is the target audience and so on]. I wonder what John Baez has to say about this?)

Posted by: Todd Trimble on November 18, 2007 8:40 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Hopefully it is clear that “pressure” here is to be read with a ;-). But Jim Stasheff expressed here concern about potential conflict with notion of suspension in topology.

Maybe you want to scroll up and down that thread and compare with the other translations

$n$-categories $\leftrightarrow$ spaces

discussed there, like

weak cokernel $\leftrightarrow$ homotopy quotient

and

weak kernel $\leftrightarrow$ homotopy fiber.

I think I understand under certain circumstances how these concepts match, but I still feel that there must be some big general abstract statement making a precise connection, which I am still missing.

In particular, I am quite aware that the categorical $\Sigma$ translates to a topological $B$ under taking nerves and realizing. But I really feel unsure now which notation is the most appropriate.

Posted by: Urs Schreiber on November 19, 2007 9:13 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Here is my take on these things.

1. Let’s talk first about degenerate higher categories and the symbol ‘$\Sigma$’. I distilled this viewpoint from the canon in HDA 0 and things Eugenia Cheng and Simon Willerton taught me.

A $k$-fold degenerate $(n+k$)-category is an $(n+k)$-category which is trivial up to dimension $k$, i.e. it has only one 0-cell, one 1-cell, …, one k-cell. We write the ‘totality’ of all these guys as $(n+k)Cat_{[k]}$.

Given a $k$-fold degenerate $(n+k)$-category $C$, we can reindex it (by thinking of the (k+1)-cells as objects, etc.) to form an $n$-category. We call $n$-categories of this form (in the image of the “reindex map”) $k$-tuply monoidal $n$-categories, and we write $nCat_k$ for the ‘totality’ of them. (We need to use philosophical words like ‘totality’ because this stuff is more subtle than it seems, see Eugenia and Nick Gurski’s papers I and II).

In this picture, the $k$-fold degenerate higher categories are the ‘fundamental’ structures whilst the $k$-tuply monoidal higher categories are derived from them via reindexing. In this way, the forgetful operation

(1)$F : nCat_{k+1} \rightarrow nCat_{k}$

and suspension operation

(2)$S : nCat_k \rightarrow nCat_{k+1}$

from HDA 0 are really the images, under reindexing, of the more fundamental operations of based loops ‘$\Omega$’ and suspension ‘$\Sigma$’ in the world of $k$-fold degenerate higher categories:

(3)\begin{aligned} (n+k)Cat_{[k]} & \overset{\quad \mathrm{reindex} \quad}{\to} && nCat_k \\ \Sigma \downarrow \uparrow \Omega &&& S \downarrow \uparrow F \\ (n+k+1)Cat_{[k+1]} & \overset{\quad \mathrm{reindex} \quad}{\to} && nCat_{k+1} \end{aligned}.

For example, $\Sigma$(a set $X$) is the free one-object category generated by $X$; $\Sigma$(a category $C$) is the free one-object 2-category generated by $C$, and so on.

This lines up with the topology use of the symbols $\Sigma$ and $\Omega$. But it clashes with the way Urs (and I) have been using $\Sigma$. For instance, if we regard a group $G$ as a set with structure, then $\Sigma (G)$ would be the one-object category freely generated by $G$, whereas we intended it to mean ‘the one object category with $G$ as its morphisms’.

So, yes, I am one of those that thinks the notation needs to be changed. I suggest a shifting notation, like $G[1]$ to mean $G$ sitting in the one-cells’.

2. What about the language, eg. “weak kernel” vs “homotopy fiber” and so on? I strongly advocate for the geometric language of homotopy fibers, etc.! That’s just because I like to think geometrically, in terms of pictures. Perhaps Urs misunderstands the proposal. I am not saying we should blur the distinction between topological spaces and higher categories. I’m proposing we define the ‘homotopy fiber’ in the world of higher categories as the such and such bla bla (weak kernel). In other words, it means a precise higher categorical thing, and when we translate everything into topology, the “homotopy fiber” from higher categories becomes the “homotopy fiber” of topology/homotopy theory.

By the way, I realized recently that homotopy fibers are really cool for describing induced representations coming from maps $f : \mathcal{H} \rightarrow \mathcal{G}$ between groupoids! The induced representation $f_*(\rho)$ assigns to $g \in \mathcal{G}$ the space of flat sections of the homotopy fiber $f^{-1}(g)$. Cool!

Posted by: Bruce Bartlett on November 19, 2007 11:02 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

I agree totally!
while we are at it, how about k-fold degenerate
as k-connected
or is it k-1?
e.g. no (non-trivial) 1-cells would be 1-connected (aka simply connected)

Posted by: jim stasheff on November 19, 2007 1:14 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

how about k-fold degenerate as k-connected?

Heh, we’ll have to ask John :-) I’m up for that in spirit, but I’m worried that it might perhaps cause some confusion because a 1-connected space doesn’t mean the space has only one path; it means it has only one path up to homotopy.

So in the Grothendieck approach, where we think of a space $X$ via its fundamental $n$-groupoid $\Pi_n (X)$ of points, paths, homotopies between paths, etc. up to homotopy classes of homotopies between …. between paths, we might cause confusion. Because in the current “strict” paradigm, if $X$ is $k$-connected we wouldn’t say that $\Pi_n (X)$ is $k$-fold degenerate because is has $|X|$ worth of zero-cells, plenty of 1-cells, etc.

Of course, perhaps we should be saying that $\Pi_n (X)$ is $k$-fold degenerate… but that’s the age-old conundrum of higher categories ‘not yet’ reaching the Mount Everest of homotopy theory.

Posted by: Bruce Bartlett on November 19, 2007 2:36 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Todd wrote:

Obviously these matters have to be decided on a case by case basis [who is the target audience and so on]. I wonder what John Baez has to say about this?

I strongly agree with your first sentence here. When we’re talking to ourselves, we can use whatever notation we like. When we’re talking to other people, it’s crucial to start by considering who we’re talking to, and what we’re trying to say. Only after this can we decide how much notation we need — the less the better! — and choose notations that our audience will feel comfortable with.

Occasionally it’s good to use a strange notation that people feel uncomfortable with, but only when we’re trying to make a specific point.

For example, writing the permutation group $S_n$ as $n!$ is great if you’re trying to convince kids that categorification is just a matter of seeing familiar things in new and deeper ways. It’s probably not so good if you’re addressing a crowd of hardened finite group theorists — they’ll just think you’re an idiot.

Topologists often use $\Sigma$ or $S$ to denote the suspension of a space. I’m fond of $S$, because then $S^n$ becomes a wonderful kind of pun: again, a matter of looking at something familiar in a new way!

Algebraic topologists also use $\Sigma$ to denote the suspension of a chain complex, or a spectrum, or an object in the homotopy category of any pointed model category. This lets us talk very generally about stable model categories.

I’m happy with all this. I’m less happy about using $\Sigma$ to stand for the obvious map from $(k+1)$-tuply monoidal $n$-categories to $k$-tuply monoidal $(n+1)$-categories — e.g. the one we use to reinterpret a monoid as a special sort of category.

While writing the paper Categorification I got very confused about this, but finally decided it’s better to think of this process as ‘delooping’. So, I called it $B$ rather than $\Sigma$ or $S$ — see page 13.

I used $S$ to stand for the obvious map from $k$-tuply monoidal $n$-categories to $(k+1)$-tuply monoidal $n$-categories – the one that’s ‘left adjoint’ in some weak sense to the forgetful functor going the other way. For example, this sends the 1-element set to the monoid $\mathbb{N}$.

The point is, all these processes should reduce to the topological processes we know and love when we work with $n$-groupoids instead of $n$-categories. The left adjoint to the forgetful functor from groups to sets sends the 1-element set to the group $\mathbb{Z}$. Thought of as a space, this group is secretly the circle. And that’s how suspension works, topologically.

Posted by: John Baez on November 27, 2007 1:06 AM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

And while we are at it, maybe you can also help me with the other question lurking in the background.

Somehow I am trying to understand something related to Dold-Kan here.

First of all, chain complexes concentrated in non-negative degree are supposed to be equivalent to $\omega$-categories internal to vector spaces.

In fact, it seems to me that the fact that the tensore product of chain complexes of maximal degree $d_1$ and $d_2$, respectively leads to a chain complex of maximal degree $d_1 + d_2$ is directly related to Sjoerd Crans’s Gray tensor product of $\omega$-categories. Is that right?

I wanted to say more, but have to run now. To be continued.

Posted by: Urs Schreiber on November 18, 2007 2:33 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

That’s a very interesting idea, Urs, and it certainly feels right to me, although I haven’t been through the calculation myself. Someone who may have thought about this is Jim Dolan; I’m supposed to be meeting him tonight. Maybe he’ll see this message and we can talk about it.

Off-hand, it occurs to me that there is a familial resemblance between these pasting schemes that Sjoerd uses (to construct the tensor product of $\omega$-categories) and Ross Street’s parity complexes, and I know for a fact that “abelianization” maps tensor products of parity complexes to tensor products of chain complexes. So it seems exceedingly likely to me that there is a precise comparison along the lines you suggest. By the way, is there a Dold-Kan correspondence for cubical complexes of abelian groups?

Posted by: Todd Trimble on November 18, 2007 9:28 PM | Permalink | Reply to this

### Re: Modules for Lie infinity-Algebras

Todd wrote:

That’s a very interesting idea, Urs, and it certainly feels right to me, although I haven’t been through the calculation myself.

Thanks.

Actually, I realize I might need a reminder on a more elementary statement even.

I understand that the Dold-Kan correspondence states an equivalence of categories

$\mathrm{Ch}(C) \simeq C^{\Delta^{\mathrm{op}}}$

between (non-negatively graded) chain complexes in $C$ and simplicial objects in $C$, for $C$ some suitable category like (maybe help me here: what is the most general $C$ we can use?) $C = \mathrm{FinVect}$.

Now, I keep forgetting what the precise relation of this to $n$-categories internal to $C$.

Certainly we have a functor from $\omega$-catgories internal to $C$ to simplicial $C$-objects, the nerve. But is that an equivalence? And how much does it depend on which notion of “infinity-category” one unses?

I’d be quite grateful for a brief reply.

(Preferably one that pertains also to $C = A\mathrm{Mod}$ for $A$ an abelian algebra.)

Posted by: Urs Schreiber on November 21, 2007 6:10 PM | Permalink | Reply to this
Read the post Something like Lie-Rinehart infinity-pairs and the BV-complex (BV, part VII)
Weblog: The n-Category Café
Excerpt: Notes on something like Lie infty-algebroids in the light of the BV complex.
Tracked: November 20, 2007 8:10 PM
Read the post On BV Quantization, Part VIII
Weblog: The n-Category Café
Excerpt: Towards understading BV by computing the charged n-particle internal to Z-categories, secretly following AKSZ.
Tracked: November 29, 2007 10:29 PM

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