The n-Category Café

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.

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.