The n-Category Café

So here is what I originally wrote on the Coffee Table, slighly abridged, followed by some comments that were exchanged between me and John, and, at the end, John’s comments on how Koszul duality plays a major role in this business.

Urs said:

In

Leonardo Castellani
Lie derivatives along antisymmetric tensors, and the M-theory superalgebra
hep-th/0508213

the author implicitly shows that

1)

the central extension by membrane charges $$\{Q_{\alpha },Q_{\beta }\}=i(C{\Gamma }^a{)}_{\alpha \beta }{P}_a+(C{\Gamma }_{\mathrm{ab}}{)}_{\alpha \beta }{Z}^{\mathrm{ab}}$$ of the super-Poincaré algebra in eleven dimensions defines a semistrict Lie 3-algebra;

2)

the local field content of 11D supergravity defines the local data for a connection on a 3-bundle with this gauge 3-group.

Recall ($\to$) that we expect on general grounds ($\to$) M-branes to couple to a 3-bundle (2-gerbe) with some gauge 3-group ($\to$).
1) survey of FDAs, Lie $n$-algebras and $n$-connections and $n$-bundles

Free differential graded algebra (we should really say “free differential graded commutative algebras”, as Ezra Getzler kindly emphasized to me today), FDA for short, are essentially the same ($\to$) as

$•$ semistrict Lie $n$-algebras and Lie $n$-algebroids ($\to$, $\to$)

$•$ $n$-term $L_{\mathrm{\infty }}$ algebras .

More precisely, from $L_{\mathrm{\infty }}$-algebras and Lie $n$-algebras, which come with complexes of vector spaces with lots of graded brackets on them, we obtain free differential graded algebras simply by passing to the dual vector spaces and equipping them with a differential whose nilpotency is equivalent to the (intricate) system of higher Jacobi almost-identities defining the original structure.

This is nice, for two reasons:

1) FDAs are much easier to compute with than Lie $n$-algebras.

2) Lie $n$-algebras have a much clearer conceptual interpretation in higher gauge theory ($\to$) than their mere FDA structure suggests.

So we can pass between the two pictures as convenient. In particular, if we encounter considerations in just one picture, we know how to translate it to the other.

The conceptual understanding of Lie $n$-algebras allows us to easily understand their role in higher gauge theory.

An (integrable) connection on an $n$-bundle ($\to$) is, by definition, something that allows us, consistently, to perform parallel transport ($\to$) over $n$-dimensional volumes.

In other words, it is a morphism $$\mathrm{tra}:P_n(X)\to {\mathrm{Trans}}_n(T)$$ from the $n$-groupoid of $n$-paths in the base space $X$ to the transport $n$-groupoid of the $n$-bundle $T\to X$ with connection ($\to$).

But this setup is easily differentiated. Passing to infinitesimals, $n$-groupoids become $n$-algebroids. Hence, infinitesimally, an (integrable) $n$-connection on an $n$ bundle is a morphism

$$d\mathrm{tra}:p_n(X)\to {\mathrm{trans}}_n(T)$$

of the corresponding algebroids ($\to$).

Knowing this, we may pass to the dual FDA description of this situation, and study connections on $n$-bundles in terms of morphisms of FDA algebras (differential graded algebras).

Motivated by the Poisson $\sigma$-model, Thomas Strobl and collaborators have looked at such morphisms ($\to$) from the point of view of gauge theory.

One finds a couple of nice, unifying structures in this context.

i) First of all, one should note that a morphism $d\mathrm{tra}:A\to B$ of $n$-algebroids corresponds to a chain map $(d_B,B^{•})\to (d_A,A^{•})$ of the corresponding dg-algebras.

ii) Naturally, then, 1-morphisms of $n$-algebroid morphisms correspond to chain homotopies, 2-morphisms to homotopies of homotopies, and so on.

iii) If we look at the double complex $(Q:=d_A\pm d_B,A^{•}\oplus B^{•})$, these conditions read as follows:

- a map $\Phi$ of dg-algebras (=FDAs) has to be $Q$-closed $[Q,\Phi ]=0$.

- a map of complexes $ϵ:B^{•}\to A^{•-1}$ is a 1-morphisms of maps of dg-algebras, with $\varphi \prime =\varphi +[Q,ϵ]$ (where the bracket is graded, hence now an anticommutator).

- similarly, a map ${ϵ}_p:B^{•}\to A^{•-p}$ is a $p$-morphisms of (the underlying) Lie $n$-algebras, relating two $(p-1)$-morphisms that differ by $[Q,{ϵ}_p]$ ($\to$).

iv) the failure of a morphism $\Phi$ to be a chain map in degree $p$ is, when this map is interpreted as a connection on an $n$-bundle, precisely the $p$-form curvature (for $p<n$ also known as “fake curvatures”)

v) Bianchi identites in the gauge theory sense are nothing but $Q^2\Phi =0$.

vi) Infinitesimal gauge transformations in the gauge theory sense are nothing but exact morphisms $[Q,ϵ]$.

vii) more generally, infinitesimal transformations are generated by generalized Lie derivatives $\{Q,i_t\}$. These are symmetries of $\Phi$ iff $[L_t,\Phi ]:=[\{Q,i_t\},\Phi ]=0$.

2) translating Castellani’s paper into Lie $n$-algebra language

We can now, step by step, go through Castellani’s paper hep-th/0508213 and interpret the FDA constructions there in the context of $n$-connections on $n$-bundles.

$•$ equations (2.1) and (2.2) are the dual formulation of a certain semistrict Lie $n$-algebra, which plays the role of the $n$-algebra of the gauge $n$-group. The number $n$ is detrmined, in this paper, by the highest $p$-form degree appearing, as $n=p$.

For $p=1$ we get only 1-forms and the formalism described charged points (section 13.5).

For $p=2$ we get 1- and 2-forms. If the 1-forms are trivial and the 2-form is abelian this described the Kalb-Ramond gerbe connection that the fundamental string couples to (section 13.6)

For $p=3$ we get 1-, 2, and 3-forms. The 3-form of 11D supergravity should be a realization of this (compare section 13.8).

Indeed, that’s the case the Castellani studies in section 3 of his paper.

$•$ We may interpret all the constants appearing there intrinsically. In particular, the coefficients $C^i{}_{A_1{A}_2{A}_2}$, which relate the $p$-forms to the connection 1-form encode nonvanishing Jacobiators (measuring the failure of the Jacobi identity to hold).

$•$ The concept referred to as soft group manifolds in the last paragraph of page 2 is secretly precisely the concept of a map from the dg-algebra characterizing a group to that of the ordinary deRham complex, i.e. a morphisms from $n$-paths to an $n$-group characterizing an $n$-connection.

$•$ Equations (2.6) and (2.7) give the curvatures, which encode the failure of this map to be a chain map.

$•$ Equation (2.16) is a realization of the statement that Lie derivates split into a pure gauge part and a contraction of the curvature $$[[Q,i_v],\Phi ]=\pm [[Q,\varphi ],i_v]\pm [Q,[i_v,\Phi ]].$$

$•$ The closure of the algebra of generalized Lie derivatives, given in the particular example in equations (2.28)-(2.30), is guaranteed by the general structure $$[[Q,i_v],[Q,i_w]]\propto [Q,[i_v,[Q,i_w]]].$$

3) Castellani’s result

The crucial new result of the paper is given in section 3. With hindsight, given the above considerations, I think can rephrase this main result as follows.

The author notes that there is semistrict Lie $3$-algebra ($\simeq$ 3-term $L_{\mathrm{\infty }}$-algebra $\simeq$ a certain dg-algebra) whose Lie algebra generated by the $[Q,i_v]$ is precisely the super-Poincaré Lie algebra in eleven dimensions, centrally extended by the central charge $Z^{\mathrm{ab}}$ corresponding to membrane (“M-branes”).

Moreover, a 3-connection with values in that Lie 3-algebra encodes, locally, precisely the field content of 11D supergravity.

There is one more nice fact, which builds on an older, well known, result, as discussed for instance in

L. Castellani, R. D’Auria & P. Fré
Supergravity and superstrings: a geometric perspective
World Scientific, Singapore (1991),

namely that imposing the condition that the curvature $[Q,\Phi ]$ of this 3-connection is horizontal, meaning that it takes values only in the algebra of objects of the gauge $3$-algebra, is equilvalent to the equations of motion of the graviton, the gravitino and the vielbein.

John said:

The stuff about free differential graded algebras and $L_{\mathrm{\infty }}$ algebras (= chain complexes that are Lie algebras up to coherent homotopy) is an example of a wonderful general pattern called “Koszul duality”.

One can also use Koszul duality to give efficient descriptions of $A_{\mathrm{\infty }}$ algebras (= chain complexes that are associative algebras up to coherent homotopy) and $C_{\mathrm{\infty }}$ algebras (= chain complexes that are commutative algebras up to coherent homotopy). There’s a nice treatment of it in Markl, Schnider and Stasheff’s book on operads and physics.

But, Koszul duality has many other aspects as well! I understand some of these but not others. I feel I still need to dig down into its essence. It’s one of those grand patterns that manifests in many different contexts.

Anyway, I’m digressing. What I want to understand now is this Lie 3-superalgebra associated to 11d supergravity. You write:

The author notes that there is semistrict Lie 3-algebra … whose Lie algebra generated by the $[Q,i_v]$ is precisely the super-Poincaré Lie algebra in eleven dimensions, centrally extended by the central charge $Z_{\mathrm{ab}}$ corresponding to membranes (“M-branes”).

Could you please help me understand this Lie 3-superalgebra? I’d eventually like to know the objects, morphisms, and 2-morphism… or if you prefer, the 0-chains, 1-chains and 2-chains… and the bracket, Jacobiator, and Jacobiatorator. But, any step in this direction would be great!

What I really want to understand is why this structure can only be built in 11 dimensions. A purely algebraic explanation of “what’s so great about 11 dimensions” - that’s been a dream of mine for some time.

But first, I need to understand what this structure is!

It sounds like the 0-chains include the super-Poincare algebra in 11 dimensions.

What else? I know you said somewhere that the loop Lie algbra of $E_8$ makes its appearance….

Could you please help out? At some point, I’m hoping that my understanding of exceptional groups will kick in and I’ll be able to see what’s special to 11 dimensions here. I understand how $E_8$ is built using the rotation groups in 8 and 16 dimensions. I understand how E₆ is related to 10-dimensional spacetime. But, I don’t know relations between these groups and 11-dimensional spacetime! It’s possible that Castellani’s construction will explain that.

Urs said:

Could you please help me understand this Lie 3-superalgebra?

Sure. It’s defined in equation (3.1) of hep-th/0508213.

In order to see how to read off the data in the form you are looking for, open for instance hep-th/0509163, page 342, example 13.1. and compare term-by-term.

Actually, that example applies to 2-groups, but the generalization of the pattern should be obvious.

So, the 3 Lie algebra in question

- has as objects the Lie algebra of the super Poincaré group

- has only trivial 1-morphisms

- has a 1-dimensional vector space of 2-morphisms on every 1-morphism.

Apart from the ordinary 2-ary bracket on objects, the only nontrivial bracket is the 4-ary one.

(I don’t know the category-theoretic name for that. Maybe “2-associator”? )

This is implicitly defined in the last line of equation (3.1) in Castellani’s paper. The bracket is nonvanishing precisely if two of its entries are generators of translations, and the other two are spinors. In components it is simply given by the commutator of “Gamma-matrices” $$C_{\overline{\alpha }\beta }^{\mathrm{ab}}:=[{\Gamma }^a,{\Gamma }^b{]}_{\overline{\alpha }\beta }.$$

You ask:

why this structure can only be built in 11 dimensions

Because (as Castellani recalls right below equation (3.1)) the Fierz identity $$\overline{\psi }{\Gamma }^{\mathrm{ab}}\psi \overline{\psi }{\Gamma }^a\psi =0,$$ which holds in $D=11$, is what ensures that the above “2-associator” (I’ll be glad to use a better term if you suggest one) does satisfy its coherence law - in other words, that the differential of the dg-algebra defined in (3.1) does indeed square to zero.

The existence of this extension of the super-Poincaré dg-algebra is implied by an old result by Chevalley on extensions of dg-algebras, reviewed somewhere in the FDA review papers cited in Castellani’s papers.

The new result of Castellani’s papers is that one can re-extract the centrally extended super-Poincaré algebra from this 3-algebra, as he describes in section 3.1 and 3.2.

$E_8$ makes its appearance

Yes, according to the Jurčo-Aschieri argument ($\to$) we expect the supergravity 3-form to contain a component of the lifting 2-gerbe of a twisted Chern-Simons 1-gerbe for the lift of crossed modules $$(\Omega E_8\to P{E}_8)\to (\hat{\Omega }{E}_8\to P{E}_8),$$ i.e. from $E_8$ to ${\mathrm{String}}_{E_8}$.

As Danny explains in his notes somewhere (possibly here), this 2-gerbe has as connection 3-form the Chern-Simons form for an $E_8$-bundle (he explains it for $\mathrm{Spin}(n)$ instead of $E_8$).

So, it looks like we should take something like the direct sum of the super Poincaré 3-algebra and the ${\mathrm{String}}_{E_8}$-3-algebra for the description of 11D SUGRA.

As a consistency check, we note that this predicts that the SUGRA 3-form has a component which is an $E_8$ CS form and a component coming from a Lorentz connection (aka spin connection). This is in accordance with what Diaconescu-Freed-Witten anomaly cancellation demands.

I was beginning to work out more details of this. But no chance - too much other things to do.

Here is a vague observation, though:

I think we can construct a 3-Lie algebra which in lowest two degrees is that of a trivial differential crossed module $$(\mathrm{Lie}G\to \mathrm{Lie}G),$$ thus giving rise to 2-connections with $\mathrm{Lie}G$-valued 1- and 2-forms.

In addition, let there be the Lie algebra of $U(1)$ in the next degree, with a 2-associator that leads to the 4-curvature $$d{C}_3+\mathrm{tr}(B_2\wedge B_2).$$ Now demand all curvatures to vanish (fake curvature and everything).

Vanishing of fake curvature says that $$B_2=F_A$$ is the curvature of an ordinary $G$-connection $A$. Vanishing of the top-level curvature then says that ${\mathrm{dC}}_3$ is the Pontryagin 4-form of the corresponding $G$-bundle $${\mathrm{dC}}_3\propto \mathrm{tr}(F_A\wedge F_A).$$ This is indeed the case for the Chern-Simons 2-gerbes that we are after. It implies that the 3-form is the CS 3-form of that $G$-bundle.

Now use $G=E_8$.

John said:

In your email to me you observed that detailed computations in this subject tend to bog down in a mess of Fierz identities. I’m glad you said this, because I’d been sort of embarrassed to admit that for me to understand this subject, the first thing I need to understand is where the $d=11$ Fierz identity comes from, and why it manages to make the Jacobiatorator satisfy the Jacobiatorator identity.

A note on terminology is probably warranted here:

Jacobi identity = identity satisfied by the bracket in a Lie algebra.

Jacobiator = 1-chain which replaces the Jacobi identity when we go from Lie algebras to Lie 2-algebras.

Jacobiator identity = identity satisfied by the Jacobiator in a Lie 2-algebra.

Jacobiatorator = 2-chain which replaces the Jacobiator identity when we go from Lie 2-algebras to Lie 3-algebras.

Jacobiatorator identity = identity satisfied by the Jacobiatorator in a Lie 3-algebra.

etcetera.

If this terminology seems too silly, which it probably is after “Jacobiator identity”, feel free to say l₂ for bracket, l₃ for Jacobiator, l₄ for Jacobiatorator, etc.

Anyway, I don’t understand the $d=11$ Fierz identity and why it just luckily happens to be the Jacobiatorator identity in disguise.

But, I’ve confronted this sort of issue before in my work on the octonions. Usually people say the Lagrangian in super-Yang-Mills theory gets its supersymmetry in $d=3$, $4$, $6$ and $10$ because of certain special Fierz identities that hold in these dimensions. However, a more illuminating explanation involves the reals, complexes, quaternions and octonions - which “just happen” to have dimensions 2 less than the above listed numbers.

The reals, complexes, quaternions and octonions are all alternative algebras - not in the counterculture sense of “alternative”, but in the technical sense: the associator

$$[a,b,c]=(\mathrm{ab})c-a(\mathrm{bc})$$

is completely antisymmetric. And, this fact is secretly the same as the relevant Fierz identities!

This is explained somewhat in Robert Helling’s Addendum to my week104.

Similar facts underlie the existence of the exceptional Lie algebras F₄, E₆, E₇ and E₈, which are closely related to the Lie algebras of rotations in 9, 10, 12 and 16 dimensions - which “just happen” to be 8 more than the dimensions of the reals, complexes, quaternions and octonions. I understand this pretty well.

So, I want to think of $d=11$ spinors in terms of the octonions, and see what the Fierz identity you mention is “really saying”.

Urs said:

see what the Fierz identity you mention is really saying

I see what you are after. While I don’t know the full answer, I could point out that a useful representation-theoretic explanation and list of the $D=11$ Fierz identities is given in section 3 of

R. D’Auria & P. Fré
Geometric supergravity in $D=11$ and its hidden supergroup
NPB 201 (1982) 101-140
(pdf) .

On p. 115 they write down all the $p$-form terms that one might naively expect and then use an irrep decomposition given on p. 112 to show that only the 2-form and the 11-form satisfy the required identity.

So they start by observing that for $A^{(p)}$ some $p$-form, the corresponding $l_{p+1}$ bracket (the Jacobiatoratoratoratorator…) must be of the form

$$\overline{\psi }\wedge {\Gamma }^{a_1\cdots a_{p-1}}\psi \wedge V_{a_1}\wedge \cdots V_{a_{p-1}},$$ where $\psi$ are spinor-valued 1-forms and $V$ vector-valued 1-forms.

This is, first of all, non-vanishing only for $p=2$, 3, 6, 7, 10 and 11.

The identity to be satisfied by this guy is $$(\overline{\psi }\wedge {\Gamma }^{a_1\cdots a_{p-1}}\psi )\wedge (\overline{\psi }\wedge {\Gamma }_{a_i}\psi )\wedge V_{a_2}\wedge \cdots V_{a_{p-1}}=0,$$ where I have put some brackets just to highlight the structure of this expression.

In other words, this says that $$(\overline{\psi }\wedge {\Gamma }^{a_1\cdots a_{p-1}}\psi )\wedge (\overline{\psi }\wedge {\Gamma }_{a_i}\psi )$$ must be a vanishing antisymmetric rank $p-2$-tensor.

So it boils down to checking if this term may contain any $(p-2)$-form contributions at all. We have four gravitinos, hence the representation $((\frac{1}{2}{)}^5{)}^{\otimes 4}$.

This can be decomposed in bosonic reps as indicated below equation 3.2 on p. 112.

More concretely, table 2 on p. 113 shows how to realize this decomposition by contracting the $\psi$ with gamma matrices.

Either way, the result is that (up to Hodge duality) precisely the 3-form, and 4-form reps do not appear. We cannot use 4-forms, since for them, as noted above, the Jacobiator vanishes in the first place. Hence we are left with the 3-form.

Maybe you can see a deeper truth by staring at that for a while and using some facts about triality and octonions.

John said:

Next I want to say a tiny bit about Koszul duality for Lie algebras, which plays a big role in the work of Castellani on the M-theory Lie 3-algebra, which I discussed in “week237”.

Let’s start with the Maurer-Cartan form. This is a gadget that shows up in the study of Lie groups. It works like this. Suppose you have a Lie group $G$ with Lie algebra $\mathrm{Lie}(G)$. Suppose you have a tangent vector at any point of the group $G$. Then you can translate it to the identity element of $G$ and get a tangent vector at the identity of $G$. But, this is nothing but an element of $\mathrm{Lie}(G)$!

So, we have a god-given linear map from tangent vectors on $G$ to the Lie algebra $\mathrm{Lie}(G)$. This is called a “$\mathrm{Lie}(G)-$valued 1-form” on $G$, since an ordinary 1-form eats tangent vectors and spits out numbers, while this spits out elements of $\mathrm{Lie}(G)$. This particular god-given $\mathrm{Lie}(G)$-valued 1-form on $G$ is called the “Maurer-Cartan form”, and denoted $\omega$.

Now, we can define exterior derivatives of $\mathrm{Lie}(G)$-valued differential forms just as we can for ordinary differential forms. So, it’s interesting to calculate $d\omega$ and see what it’s like.

The answer is very simple. It’s called the Maurer-Cartan equation:

$$d\omega =-\omega \wedge \omega$$ On the right here I’m using the wedge product of $\mathrm{Lie}(G)$-valued differential forms. This is defined just like the wedge product of ordinary differential forms, except instead of multiplication of numbers we use the bracket in our Lie algebra.

I won’t prove the Maurer-Cartan equation; the proof is so easy you can even find it on the Wikipedia:

14) Wikipedia, Maurer-Cartan form,

An interesting thing about this equation is that it shows everything about the Lie algebra $\mathrm{Lie}(G)$ is packed into the Maurer-Cartan form. The reason is that everything about the bracket operation is packed into the definition of $\omega \wedge \omega$.

If you have trouble seeing this, note that we can feed $\omega \wedge \omega$ a pair of tangent vectors at any point of $G$, and it will spit out an element of $\mathrm{Lie}(G)$. How will it do this? The two copies of $\omega$ will eat the two tangent vectors and spit out elements of $\mathrm{Lie}(G)$. Then we take the bracket of those, and that’s the final answer.

Since we can get the bracket of any two elements of $\mathrm{Lie}(G)$ using this trick, $\omega \wedge \omega$ knows everything about the bracket in $\mathrm{Lie}(G)$. You could even say it’s the bracket viewed as a geometrical entity - a kind of “field” on the group $G$!

Now, since

$$d\omega =-\omega \wedge \omega$$

and the usual rules for exterior derivatives imply that

$$d^2\omega =0$$

we must have

$$d(\omega \wedge \omega )=0.$$

If we work this concretely what this says, we must get some identity involving the bracket in our Lie algebra, since $\omega \wedge \omega$ is just the bracket in disguise. What identity could this be?

THE JACOBI IDENTITY!

It has to be, since the Jacobi identity says there’s a way to take 3 Lie algebra elements, bracket them in a clever way, and get zero:

$$[u,[v,w]]+[v,[w,u]]+[w,[u,v]]=0$$

while $d(\omega \wedge \omega )$ is a $\mathrm{Lie}(G)$-valued 3-form that happens to vanish, built using the bracket.

It also has to be since the equation $d^2=0$ is just another way of saying the Jacobi identity. For example, if you write out the explicit grungy formula for d of a differential form applied to a list of vector fields, and then use this to compute $d^2$ of that differential form, you’ll see that to get zero you need the Jacobi identity for the Lie bracket of vector fields. Here we’re just using a special case of that.

The relationship between the Jacobi identity and $d^2=0$ is actually very beautiful and deep. The Jacobi identity says the bracket is a derivation of itself, which is an infinitesimal way of saying that the flow generated by a vector field, acting as an operation on vector fields, preserves the Lie bracket! And this, in turn, follows from the fact that the Lie bracket is preserved by diffeomorphisms - in other words, it’s a “canonically defined” operation on vector fields.

Similarly, $d^2=0$ is related to the fact that d is a natural operation on differential forms - in other words, that it commutes with diffeomorphisms. I’ll leave this cryptic; I don’t feel like trying to work out the details now.

Instead, let me say how to translate this fact:

$d^2\omega =0$ IS SECRETLY THE JACOBI IDENTITY

into pure algebra. We’ll get something called “Kozsul duality”. I always found Koszul duality mysterious, until I realized it’s just a generalzation of the above fact.

How can we state the above fact purely algebraically, only using the Lie algebra $\mathrm{Lie}(G)$, not the group $G$? To get ourselves in the mood, let’s call our Lie algebra simply $L$.

By the way we constructed it, the Maurer-Cartan form is “left-invariant”, meaning it doesn’t change when you translate it using maps like this:

$$\begin{array}{rl}{L}_g:& G\to G\\ & x\mapsto x\end{array}$$

that is, left multiplication by any element $g$ of $G$. So, how can we describe the left-invariant differential forms on $G$ in a purely algebraic way? Let’s do this for ordinary differential forms; to get $\mathrm{Lie}(G)$-valued ones we can just tensor with $L=\mathrm{Lie}(G)$.

Well, here’s how we do it. The left-invariant vector fields on $G$ are just

$L$

so the left-invariant 1-forms are

$L^{*}$

So, the algebra of all left-invariant diferential forms on $G$ is just the exterior algebra on $L^{*}$. And, defining the exterior derivative of such a form is precisely the same as giving the bracket in the Lie algebra $L$! And, the equation $d^2=0$ is just the Jacobi identity in disguise.

To be a bit more formal about this, let’s think of $L$ as a graded vector space where everything is of degree zero. Then $L^{*}$ is the same sort of thing, but we should add one to the degree to think of guys in here as 1-forms. Let’s use $S$ for the operation of “suspending” a graded vector space - that is, adding one to the degree. Then the exterior algebra on $L^{*}$ is the “free graded-commutative algebra on ${\mathrm{SL}}^{*}$”.

So far, just new jargon. But this lets us state the observation of the penultimate paragraph in a very sophisticated-sounding way. Take a vector space $L$ and think of it as a graded vector space where everything is of degree zero. Then:

Making the free graded-commutative algebra on ${\mathrm{SL}}^{*}$ into a differential graded-commutative algebra is the same as making $L$ into a Lie algebra.

This is a basic example of “Koszul duality”. Why do we call it “duality”? Because it’s still true if we switch the words “commutative” and “Lie” in the above sentence!

Making the free graded Lie algebra on ${\mathrm{SL}}^{*}$ into a differential graded Lie algebra is the same as making $L$ into a commutative algebra.

That’s sort of mind-blowing. Now the equation $d^2=0$ secretly encodes the commutative law.

So, we say the concepts “Lie algebra” and “commutative algebra” are Koszul dual. Interestingly, the concept “associative algebra” is its own dual:

Making the free graded associative algebra on ${\mathrm{SL}}^{*}$ into a differential graded associative algebra is the same as making L into an associative algebra.

This is the beginning of a big story, and I’ll try to say more later. If you get impatient, try the book on operads mentioned in “week191”, or else these:

15) Victor Ginzburg and Mikhail Kapranov, Koszul duality for quadratic operads, Duke Math. J. 76 (1994), 203-272. Also Erratum, Duke Math. J. 80 (1995), 293.

16) Benoit Fresse, Koszul duality of operads and homology of partition posets, Homotopy theory and its applications (Evanston, 2002), Contemp. Math. 346 (2004), 115-215. Also available at http://math.univ-lille1.fr/~fresse/PartitionHomology.html

The point is that Lie, commutative and associative algebras are all defined by “quadratic operads”, and one can define for any such operad $O$ a “dual” operad $O^{*}$ such that:

Making the free graded $O$-algebra on ${\mathrm{SL}}^{*}$ into a differential graded $O$-algebra is the same as making $L$ into an $O^{*}$-algebra.

And, we have $O^{**}=O$, hence the term “duality”.

This has always seemed incredibly cool and mysterious to me. There are other meanings of the term “Koszul duality”, and if really understood them I might better understand what’s going on here. But, I’m feeling happy now because I see this special case:

Making the free graded-commutative algebra on ${\mathrm{SL}}^{*}$ into a differential graded-commutative algebra is the same as making $L$ into a Lie algebra.

is really just saying that the exterior derivative of left-invariant differential forms on a Lie group encodes the bracket in the Lie algebra. That’s something I have a feeling for. And, it’s related to the Maurer-Cartan equation… though notice, I never completely spelled out how.

Addenda: Let me say some more about how $d^2=0$ is related to the fact that $d$ is a canonically defined operation on differential forms. Being “canonically defined” means that $d$ commutes with the action of diffeomorphisms. Saying that $d$ commutes with “small” diffeomorphisms - those connected by a path to the identity - is the same as saying

$$d{L}_v=L_{v}d$$

where $v$ is any vector field and $L_v$ is the corresponding “Lie derivative” operation on differential forms. But, Weil’s formula says that

$$L_v=i_{v}d+d{i}_v,$$

where $i_v$ is the “interior product with $v$”, which sends $p$-forms to $(p-1)$-forms. If we plug Weil’s formula into the equation we’re pondering, we get

$$d(i_{v}d+d{i}_v)=(i_{v}d+d{i}_v)d$$

which simplifies to give

$$d2i_v=i_v{d}^2.$$

So, as soon as we know $d^2=0$, we know $d$ commutes with small diffeomorphisms. Alas, I don’t see how to reverse the argument.

Similarly, as soon as we know the Jacobi identity, we know the Lie bracket operation on vector fields is preserved by small diffeomorphisms, by the argument outlined in the body of this Week. This argument is reversable.

So, maybe it’s an exaggeration to say that $d^2=0$ and the Jacobi identity say that $d$ and the Lie bracket are preserved by diffeomorphisms - but at least they imply these operations are preserved by small diffeomorphisms.

Urs said:

We know that the supergravity 3-form $C$ should really come from Chern-Simons 3-forms of a Lorentz and an $E_8$ connection.

This information is not present in the classical FDA formulation of supergravity. Here is a general observation on connections on Chern-Simons 2-gerbes, which might be relevant.

Consider a Lie group $G$. Its Lie algebra is encoded in the fda defined by $$d{a}^a+\frac{1}{2}{C}^a{}_{\mathrm{bc}}{a}^b{a}^c=0,$$ where $\{C^a{}_{\mathrm{bc}}\}$ are the structure constant in some chosen basis. Nilpotency of $d$ is equivalent to the Jacobi identity.

Next, consider the crossed module $G\to G$. The FDA corresponding to its Lie 2-algebra is given by $$\begin{array}{rl}& d{a}^a+\frac{1}{2}{C}^a{}_{\mathrm{bc}}{a}^b{a}^c+b^a=0\\ & d{b}^a+C^a{}_{\mathrm{bc}}{a}^b{b}^c=0\end{array}.$$ A 2-connection with values in this Lie 2-algebra is given by a 1-form $A^a$ and a 2-form $B^a$ and has curvatures $$\begin{array}{rl}{F}_1& =F_A+B\\ F_2& =d_{A}B\end{array}.$$ Without mentioning anything like 2-connections, but implicitly considering precisely this, such 2-connections have been studied for instance in hep-th/0204059. The 2-group aspect is discussed very nicely in hep-th/0206130. Suppose we demand this 2-curvature to vanish. This is equivalent to $$B=-F_A.$$ Hence a flat $(G\to G)$-2-connection is the same as an ordinary $G$-connection. In fact, a trivial flat $(G\to G)$-2-bundle is the same as an ordinary $G$-bundle. Next, let’s also add a generator in degree 3 to our fda, to get the Lie 3-algebra encoded by $$\begin{array}{rl}& d{a}^a+\frac{1}{2}{C}^a{}_{\mathrm{bc}}{a}^b{a}^c+b^a=0\\ & d{b}^a+C^a{}_{\mathrm{bc}}{a}^b{b}^c=0\\ & dc+k_{\mathrm{ab}}{b}^a{b}^b=0\end{array},$$ where $k_{\mathrm{ab}}$ is proportional to the Killing form on the Lie algebra of $G$. This ensures that $d$ is still nilpotent. A 3-connection with values in this Lie 3-algebra is a 1-form $A^a$, a 2-form $B^a$ and a 3-form $C$. Its curvature is $$\begin{array}{rl}{F}_1& =F_A+B\\ F_2& =d_{A}B\\ F_3& =dC+\mathrm{tr}(B\wedge B)\end{array}.$$ Assume again that the curvature vanishes. In the first two degrees this is, as before, equivalent to $B=-F_A$. In third degree it now says that $$dC\propto \mathrm{tr}(F_A\wedge F_A).$$ But this means that $C$ itself must be, up to a closed part, the Chern-Simons 3-form of $A$ $$C\propto \mathrm{CS}(A).$$ Therefore a 3-connection of the above sort is the local connection of a Chern-Simons gerbe corresponding to some $G$-bundle with connection. Now take $G$ to be the product of $E_8$ with the Lorentz group in $D=11$ and there we go.