The String Coffee Table

There are in particular two questions which I would like to comment on.

1) What about the 3-form really being a Chern-Simons 3-form?

2) What happens in degree 2?

1) What about the 3-form really being a Chern-Simons 3-form? We know that the 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

where $\{C^a{}_{bc}\}$ are the structure constant in some chosen basis. Nilpotency of $\mathbf{d}$ 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

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

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

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

where $k_{ab}$ is proportional to the Killing form on the Lie algebra of $G$. This ensures that $\mathbf{d}$ 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

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

But this means that $C$ itself must be, up to a closed part, the Chern-Simons 3-form of $A$

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. There is one problem. The known FDA description of SUGRA uses not the Lorentz group in lowest degree, but the Poincaré group. Plus, there is already something going on in degree 3. Both facts imply that applying the above construction to this case is not enirely straightforward. Partly motivated by this, in the next subsection I make some comments on the Poincaré versus Lorentz issue.

2) What happens in degree 2?

The Poincaré-group happens to be a semidirect product of the Lorentz and the translation group, with one factor being abelian. Every such semidirect product group is already naturally a 2-group. It’s Lie 2-algebra corresponds to the fda defined by

where $\{a^a\}$ are a basis of the dual of the Lorentz Lie algebra, $\{t^i\}$ a basis of the dual of the translation algebra, and $\alpha$ encodes the action of one on the other. Motivated by this observation, there have been attempts to formulate gravity in terms of the Poincaré 2-group ($\to$). But notice this: In the above fda, the $t^i$ live in degree 2. Hence a 2-connection with values in this guy contains a 2-form valued in the translation Lie algebra. What’s that? On the other hand, in the known fda formulation of sugra, one regards the $t^i$ as living in degree 1. Accordingly, the connection with values in this fda has a 1-form with values in the translation Lie algebra. This 1-form gets interpreted as the vielbein, which makes the fda SUGRA formulation something like an example for a Palatini-formulation of gravity ($\to$). Still, let’s assume we agree that it is unnatural to have the $t^i$ not in degree 2. But let’s also assume that we agree that we want a vielbein. There is one way out: Just like $so(n,1)$ acts on $\mathbb{R}^{n,1}$, it also acts on the abelian group $\bigwedge^2 \mathbb{R}^{n,1}$ of bivectors, surely. Hence we can just as well consider the 2-group coming from the crossed module

with trivial $t$. The corresponding fda looks precisley like the one above, just with $t^i$ and $\alpha$ suitably reinterpreted. A 2-connection with values in this Lie 2-algebra now consists of an $so(n,1)$-valued 1-form, together with a $\bigwedge^2 \mathbb{R}^{n,-1}$-valued 2-form. This now has a chance to encode a vielbein, since by wedging any vielbein with itself we do obtain such a 2-form. There is apparently an even better solution, using spin groups instead. This idea can be found discussed in section IV and V of hep-th/0204059 (see also gr-qc/9502037), whose authors consider precisely this, albeit without identifying $n$-connections explicitly (they call what they do “generalized differential calculus”). So consider $D = n + 1 = 3+ 1$ for a moment and pick the double cover of the Lorentz group group $G = \mathrm{SL}(2,\mathbb{C})$ and the translation group $T \simeq \mathbb{R}^{3,1}$. Using the 2-component spinorial language (and the Pauli matrices $\sigma_a^{AB'}$), we can write a vielbein 1-form $e^a$ as $e^{AB'}$ and wedge it with itself to form the $\mathrm{SL}(2,\mathbb{C})$-valued 2-form

So we get a 2-connection for our Lie 2-algebra given by an $\mathrm{SL}(2,\mathbb{C})$-valued connection 1-form $\omega^{AB}$ and that “surface-vielbein” 2-form $B^{AB}$. In terms of these the Einstein-Hilbert action (in $D=4$) is of BF-theory form ($\to$)

Consider again the situation where the curvature of our 2-connection vanishes in degree 2. This now says that our 1-form connection is torsion free with respect to the metric encoded by the vielbein. The vielbein-2-form $B$ should be precisely the “area vielbein” corresponding to the area metrics which have been argued to encode the geometry of string backgrounds ($\to$).

Update. Something closely related is also discussed in gr-qc/9804061.