May 25, 2007

Derivation Lie 1-Algebras of Lie n-Algebras

Posted by Urs Schreiber Here is a fact which should be important, but whose true meaning is not entirely clear to me at the moment:

To every Lie $n$-algebra $g_{(n)}$ is canonically associated an ordinary Lie algebra $\mathrm{derivs}(g_{(n)})$.

This Lie 1-algebra is, in some sense, a Lie 1-algebra of derivations of the Lie $n$-algebra. But it is not the derivation Lie $(n+1)$-algebra $\mathrm{DER}(g_{(n)})$.

What is $\mathrm{derivs}(g_{(n)})$, conceptually?
Why does it exist, as an ordinary Lie algebra?
What is its analog at the level of Lie $n$-groups?

Below I give the detailed description and mention applications where this is relevant.

For more details see section 3.3.1 of

Structure of Lie $n$-Algebras

and section 5.3 of

Zoo of Lie $n$-Algebras

(which has grown out of the original zoo).

The Definition

This is how to canonically associate a Lie 1-algebra to any Lie $n$-algebra (it’s a perfectly standard construction in homological algebra, though not usually interpreted in terms of Lie $n$-algebras):

Conceive the Lie $n$-algebra as a free graded commutative algebra with a graded differential $d$ of degree 1 on it, such that $d^2 = 0$. Then on the vector space of graded derivations $\tau$ of degree -1 on the algebra we have the following Lie bracket $[\tau,\tau^'] := L_\tau(\tau') \,,$ where $L_\tau := [d,\tau]$ is the “Lie derivative” associated with $\tau$.

First Examples

I need to work out more examples. But it’s tedious. So far I have this:

1) Let $g_{(1)}$ be an ordinary Lie algebra. Then it coincides with its own Lie algebra of derivations as defined above.

2) Let $g_{(2)} = (t : h \to g)$ be the Lie 2-algebra corresponding to an infinitesimal crossed module. Then for the above Lie 1-algebra associated with it I find this funny result:

As a vector space, it is $g \oplus (g^* \otimes h) \,.$ This is equipped with a semidirect sum structure of Lie algebras:

The bracket on the first $g$-summand is that of $g$. The bracket on the summand $(g^* \oplus h)$ is given by $[(\alpha_1,b_1), (\alpha_2,b_2)] = \alpha_2(t(b_1))\, (\alpha_1, b_2) - \alpha_1(t(b_2))\, (\alpha_2, b_1)$ for all $\alpha_i \in g^*$ and all $b_i \in h$.

The action of $g$ on $(g^* \otimes h)$ is given by $x ((\alpha,b)) := ( (\mathrm{ad}_x)^* \alpha, x(b) ) \,,$ for all $x \in g$, $\alpha \in g^*$ and $b \in h$.

3) Let, for $g$ simple, $g_\mu$ be the Baez-Crans Lie 2-algebra coming from the canonical 3-cocycle. Then the above Lie algebra is the direct sum Lie algebra $g \oplus \mathbb{R}^{\mathrm{dim}(g)} \,.$

Why I think understanding this is important

My motivation for looking into this is Castellani’s observation that there is a Lie 1-algebra of derivations of the supergravity Lie 3-algebra, which coincides with the famous “M-algebra” polyvector extension of the super-Poincaré Lie algebra.

Now, what Castellani does is not precisely what I consider above. Rather, he looks at something like derivations acting on connections with values in the Lie 3-algebra. But since it is still not entirely clear to me what the operation he uses really is, conceptually (in particular how it extends from generators to the entire differential algebra), I thought it would be worthwhile first considering the related problem described above.

In any case, there seems to be something going on here relating Lie algebra cocycles and Lie algebra extensions in two ways: extensions as Lie 1-algebras and extensions as Lie $n$-algebras. Something important is going on, but it escapes me presently.

What does it correspond to at the level of Lie $n$-groups?

The problem is that the Lie 1-algebras discussed here are obtained from the Lie $n$-algebra in a non-conceptual manner: it so happens that it is convenient to compute with Lie $n$-algebras by instead using the equivalent differential graded algebras. But then one runs the risk of performing steps whose categorical interpretation is not transparent. This is what is happening here.

So I started making educated guesses for what is going on. Before I computed the derivation Lie 1-algebra for the $(t : h \to g)$ case I was sure I would simply get the Lie algebra of the Lie group of morphisms of the Lie 2-group coming from an integrating crossed module $(H \to G)$. That would have been the Lie algebra of the semidirect product group $H \ltimes G$.

Clearly, this is not unrelated to the true result mentioned above, but still different from that.

So what’s going on?

Posted at May 25, 2007 1:58 PM UTC

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Re: Derivation Lie 1-Algebras of Lie n-Algebras

If I understand correctly, a Lie n-algebra is just (the dual of) an FDA. An FDA is typically defined as a system of form equations w = 0, where w is some collection of forms, e.g. Castellani’s (3.1). The group of diffeomorphisms acts on w, and the diffeos which preserve the equation w = 0 is evidently a subgroup. Is not this what you have found?

One can quite generally describe Lie groups as subgroups of the diffeomorphism group which preserve some structure. E.g., the Poincare group preserves the length element ds^2, and the conformal group preserves the lightcone ds^2 = 0.

Posted by: Thomas Larsson on May 26, 2007 7:05 AM | Permalink | Reply to this

Re: Derivation Lie 1-Algebras of Lie n-Algebras

Larsson’s comments bring to mind a cultural difference, especially in the context of gravity. The mathematical study of Lie groups makes no reference to diffeomorphisms, though I guess the regular representation of a finite dim Lie group g could be regarded as acting by diffeomorphims of the underlying vector space with its standard smooth structure.

He also reminds us that the qFDA is often tought of as forms on some *unspecified* manifold, cf. the socalled soft group manifold.

Posted by: jim stasheff on May 27, 2007 1:29 PM | Permalink | Reply to this

Re: Derivation Lie 1-Algebras of Lie n-Algebras

Any finite-dimensional Lie algebra g can be embedded in the algebra vect(n) of vector fields in n dimensions, for suitable n. E.g., if g has a rep R with dim R = n, then g can be embedded in gl(n), which in turn can be embedded in vect(n). Take the gl(n) generators to be the vector fields Eij = xi d/dxj. But there are also other embeddings, e.g. the usual realization of so(d+1,1) as conformal transformations.

Posted by: Thomas Larsson on May 29, 2007 6:48 AM | Permalink | Reply to this

Re: Derivation Lie 1-Algebras of Lie n-Algebras

Any finite-dimensional Lie algebra $g$ can be embedded in the algebra $\mathrm{vect}(n)$ of vector fields

Okay. I am not sure yet if this helps answering the question, but it might be relevant.

But I should maybe emphasize the following: where I say “Lie derivative” in the above, I don’t necessarily mean the ordinary Lie derivative on vectors or differential forms.

Rather, I am using the obvious generalized version:

- for any differential graded commutative algebra, with the differential $d$ of degree +1 (which might be that of differential forms on some space, but need not),

- and for every degree -1 derivation $\tau$ on that (which might be the interior product of differential forms with vectors, but need not),

- we can form the degree 0 derivation $L_\tau := [d,\tau] \,.$

This I call the Lie derivative coming from $\tau$, since in the case where the algebra is that of differential forms, $d$ is the deRham differenmtial and $\tau = \iota_v$ is contraction with a vector field $v$, we obtain the ordinary Lie derivative $L_\tau = L_v$ along $v$, acting on differential forms.

These Lie derivatives always form a Lie algebra since their bracket closes: if $L_\tau$ and $L_{\tau'}$ are Lie derivatives, then their commutator $[L_\tau, L_{\tau'}] = L_\tau \circ L_{\tau'} - L_{\tau'} \circ L_{\tau}$ is again of the form $[d,\tau'']$, for some degree -1 derivation $\tau''$.

This is true whether or not we restrict to the special ordinary case of differential forms and contractions. It is a general construction which works for every differential graded commutative algebra.

And when that algebra is the dual of a Lie algebra, there is also no mystery what this construction amounts to.

But in the case where that differential graded algebra is of the form $(\wedge^\bullet V^*, d)$ with $V$ a graded vector space which is nontrivial in degrees higher than 1, the construction still works. But since such differential algebras can be interpreted as higher Lie algebras, I now find it a little surprising that I still obtain a Lie 1-algebra by the above construction.

I mean, technically it is clear that the above always yields a Lie algebra of (generalized) Lie derivatives. I would like to understand this conceptually: in terms of Lie $n$-algebras: what is this Lie 1-algebra of derivations, conceptually?

That’s what puzzles me.

Posted by: urs on May 29, 2007 3:08 PM | Permalink | Reply to this

Re: Derivation Lie 1-Algebras of Lie n-Algebras

As Jim indicates, there are at least two issues here:

the first is: mostly (but not always!) in literature such as that of the school which that Castellani paper originates in, what is actually considered is not the “FDA” $g_{(n)}^*$ itself (Jim rightly emphasizes that it must be called quasi-free differential graded commutative algebra “qfDGCA”, instead, where “quasi-free” means free as a graded commutative algebra but not necessarily as a differential algebra), but rather the space of (flat) connections with values in this, namely morphisms $f^* : (g_{(n)})^* \to \Omega^\bullet(X) \,.$ In this case it makes sense to ponder the action of diffeomorphisms on $X$.

But right at the moment, as I mentioned above, I would like to concentrate on just $g_{(n)}$ itself.

Secondly, to emphasize the main point, I should maybe have phrased my question more generally:

for any qfDGCA, the space of derivations on it of degree 0 form a Lie algebra.

(Above I concentrated on certain “inner” derivations, those that are of the form $[d,\tau]$, for $\tau$ a derivation of degree -1.)

What I find puzzling is that these always form a Lie algebra. A Lie 1-algebra, that is.

It is much easier to understand why derivations (of non-positive degree) of a qfDGCA which is concentrated in the first $n$-degree form a Lie $(n+1)$-algebra, and hence themselves as qfDGCA concentrated in the first $n+1$ degrees.

So the real point of the question is to understand from the conceptual Lie $n$-algebra level why we have a Lie 1-algebra of 0-degree derivations for every Lie $n$-algebra.

I am still not sure what is going on here.

Posted by: urs on May 28, 2007 4:51 PM | Permalink | Reply to this

Re: Derivation Lie 1-Algebras of Lie n-Algebras

I believe I found the answer.

Recall, the puzzle was the following:

to every Lie $n$-algebra is canonically associated an ordinary Lie algebra (“Lie 1-algebra”). Which role does this play? What is the relation of the corresponding Lie group to the corresponding Lie $n$-group?

The conjectured solution.

There are actually sufficiently many hints available to make the answer obvious.

Hint 1) How do we get something strict (an ordinary Lie algebra) from something arbitrarily weak (a general Lie $n$-algebra). Well, no matter what kind of weak $n$-categories one considers, what is always strict is composition of $n$-functors – bcause no matter what, this is just composition of maps.

So if we get a Lie algebra for every Lie $n$-algebra, chances are that it is the Lie algebra of a Lie group whose elements are $n$-functors on our Lie $n$-algebra and whose product is just composition of these.

Hint 2) In my second example discussed above, I notice that in the case the $n=2$ and we are talking about the Lie 2-algebra $g_{(2)}$ coming from a differential crossed module $(t : h \to g)$ of Lie algebras $g$ and $h$, the corresponding Lie 1-algebra we are after is defined on the vector space $g \oplus (g^* \otimes h) \,.$ Now, $g$ is the space of objects of $g_{(2)}$ while $h$ is the space of morphisms which start at the 0-object.

So this tells us that the Lie algebra is one generated from objects of $g_{(2)}$ and from maps $f : g \to h$ which assign a morphism starting at the 0-object to every object.

Pondering this for a moment, an obvious answer arises:

I believe the Lie algebra here is that of the Lie 1-group $Conj(G_{(2)})$ associated with the strict Lie 2-group $G_{(2)}$ whose objects are automorphisms $f : \mathrm{Obj}(G_{(2)}) \to \mathrm{Mor}(G_{(2)})$ which arise

- either from conjugating horizontally with a 1-morphism in $\Sigma(G_{(2)})$

$\left( \array{ & \nearrow\searrow^g \\ \bullet &\Downarrow^h& \bullet \\ & \searrow\nearrow_{g'} } \right) \mapsto \left( \array{ &&& \nearrow\searrow^g \\ \bullet &\stackrel{q}{\to}& \bullet &\Downarrow^h& \bullet &\stackrel{q^{-1}}{\to}& \bullet \\ &&& \searrow\nearrow_{g'} } \right)$

-or from conjugating vertically $\left( \array{ g \\ \downarrow^{h} \\ g' } \right) \mapsto \left( \array{ t(f(g)^{-1})g \\ \downarrow^{f(g)^{-1}} \\ g \\ \downarrow^{h} \\ g' \\ \downarrow^{f(g')} \\ t(f(g'))g' } \right)$

- or from an operation generated from these two basic operations.

Notice that this are the obvious two ways of “conjugating” in a 2-group. Also notice the crucial difference to the “inner automorphism 3-group$\mathrm{INN}(G_2) \,.$ This comes from all the horizontal conjugations only (!) but together with all transformations and modifications on these. This procedure produces an $(n+1)$-group $\mathrm{INN}(G_{(2)})$ of inner automorphisms from any $n$-group $G_{(n)}$.

Here, instead, for forming the (1-)group $\mathrm{Conj}(G_{(n)})$, we allow conjugations not just horizontally, but in all possible directions – but don’t care about the transformations and other higher morphisms between these (well, of course we could and should eventually, but this I won’t discuss right now).

It is easy to compute what the group $\mathrm{Conj}(G_{(2)})$ looks like for any strict 2-group $G_{(2)}$. I haven’t yet completely computed the Lie algebra obtained from that, being lazy, but one can clearly see that it will be of the kind I described in the second example of the above entry.

A possible application

Maybe I should mention why I am interested in this. This goes back to something that Jim Stasheff made me think about:

on a principal $G$-bundle $P$ with right $G$-action $R : P \times G \to P$ we have an induced action of the Lie algebra $g = \mathrm{Lie}(G)$, coming from a morphism of Lie algebras $R_* : g \to \Gamma(T P) \,.$ This action is what crucially enters the definition of a Cartan connection form on $P$.

This is nice, because it gives a way to think of connections on principle bundles entirely in the world of Lie algebras.

Maybe we want to category this description.

So given a principal 2-bundle $p : P \to X$ (and, yes, now I do need the total space of the 2-bundle, even though I recently ranted “against total spaces”) with a right principal action $R : P \times G_{(2)} \to P$ what would be a “Cartan 2-connection” on the total space $P$? What would be the analog of the differential action $g \to \Gamma(T P)$ from the 1-Cartan case?

In principle this is a straightforward matter, I think: pull the 2-bundle back to its own space of morphisms, there trivialize it canonically, obtain the differential cocycle description of a 2-connection with respect to this chosen trivialization, then differentiate everything with respect to the $G_{(2)}$-action.

While straightforward, this is immensely tedious (unless there is some shortcut which I am missing).

So I am currently trying to guess the right answer. I have reasons to believe that in the definition of a “Cartan 2-connection” the action $g \to \Gamma(T P)$ is replaced by an action $\mathrm{derivs}(g_{(2)}) \to \Gamma(T \mathrm{Mor}(P)) \,,$ where $\mathrm{derivs}(g_{(2)}) = (g \oplus (g ^* \otimes h))$ is the Lie 1-algebra associated with the structure Lie 2-algebra $g_{(2)} = \mathrm{Lie}(G_{(2)})$ which I keep talking about.

Posted by: urs on May 30, 2007 10:29 AM | Permalink | Reply to this

Conjugation (n+1)-groups

Here is a more readable depiction of the two different conjugation automorphisms of a 2-group: It is only the horizontal conjugation which strictly deserves to be called conjugation (hence “inner automorphism”) since this comes from the product functor on the 2-group.

In a double category setup the “vertical conjugation” would be entirely on the same footing as the horizontal conjugation. Here it plays a slightly different role.

One way to think of these vertical conjugations by maps $f : \mathrm{Obj}(G_{(2)}) \to \mathrm{Mor}(G_{(2)})$ is as transformations starting at the identity on $G_{(2)}$: $\array{ & \nearrow\searrow^{\mathrm{Id}} \\ G_{(2)} &\Downarrow f& G_{(2)} \\ & \searrow\nearrow } \,.$

Question: Can anyone tell me what the right canonical terminology would be for the sub 3-group of $\mathrm{Aut}_{2\mathrm{Cat}}(\Sigma G_{(2)})$ which contains only automorphisms generated from horizontal and vertical conjugations as above?

I am already using $\mathrm{INN}(G_{(2)}) \subset \mathrm{Aut}_{2\mathrm{Cat}}(\Sigma G_{(2)})$ for that 3-group coming only from the horizontal conjugations.

Should I instead better be using the “inner” terminology for the full thing? Above I argued that, strictly speaking, only horizontal composition is an “inner” automorphism for the 2-group.

What other terminology would suggest itself?

Posted by: urs on May 30, 2007 12:30 PM | Permalink | Reply to this

Re: Conjugation (n+1)-groups

I wrote:

What other terminology would suggest itself?

I’d still be interested if anyone has an opinion on this. But I should maybe point out that the issue is actually slightly more general, in the following sense.

For any given $n$-group $G_{(n)}$, we have an entire chain of inclusions of $(n+1)$-groups

$\mathrm{INN}_1(G_{(n)}) \subset \mathrm{INN}_2(G_{(n)}) \subset \cdots \subset \mathrm{INN}_n(G_{(n)}) \subset \mathrm{AUT}(G_{(n)}) \,.$

Here $INN_k(G_{(n)})$ is that sub $(n+1)$-group of $\mathrm{AUT}(G_{(n)})$ whose objects come from conjugation with $j$-morphisms as described above, for $j \leq k \,.$ So the “true” inner automorphism $(n+1)$-group is the first in this chain:

$\mathrm{INN}(G_{(n)}) = \mathrm{INN}_1(G_{(n)}) \,,$ where we conjugate only with 1-morphisms in $\Sigma G_{(n)}$.

Moreover, the Lie $(n+1)$-group which I claimed solves the puzzle I discussed in the above entry is $\mathrm{Conj}(G_{(n)}) = \mathrm{INN}_n(G_{(n)}) \,.$

Accordingly, one can define for any Lie $n$-algebra $g_{(n)}$ the chain of inclusions of Lie $(n+1)$-algebras

$\mathrm{inn}_1(g_{(n)}) \subset \mathrm{inn}_2(g_{(n)}) \subset \cdots \subset \mathrm{inn}_n(g_{(n)}) \subset \mathrm{DER}(g_{(n)}) \,.$

Here $\mathrm{inn}_k(g_{(n)})$ is defined to come from those degree 0 derivations $L$ on the corresponding dual differential algebra, which are “Lie derivatives” $L = [d,\iota]$ (compare the comment above concerning the notion of “Lie derivative” here) for $\iota$ a derivation (necessarily of degree -1) such that $\iota(\omega) = 0$ when $\omega$ is homogenous and of degree $|\omega| \gt k$.

This is now the terminology used in section 3.2 of Structure of Lie $n$-algebras.

Posted by: urs on May 30, 2007 3:17 PM | Permalink | Reply to this
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