### 10D SuGra 2-Connection

#### Posted by Urs Schreiber

We have seen ($\to$) that 11-dimensional supergravity is a gauge theory of a 3-connection, taking values in a certain Lie 3-algebra, $\mathrm{sugra}_{11}$, which is an extension of the super-Poincaré-1-algebra by a 4-cocycle.

I claimed ($\to$) that non-fake flat $n$-connections with values in an $n$-algebra $g$ are to be interpreted in terms of their curvatures, which are *flat* $(n+1)$-connections with values in

the $(n+1)$-algebra of inner derivations of $g$, where flatness encodes the $n$-Bianchi identity.

The task is hence to compute $\mathrm{inn}(\mathrm{sugra}_11)$.

That’s straightforward, but pretty hard. I am still hoping to figure out a shortcut, computing $\mathrm{inn}(g)$ directly at the level of FDAs. But I don’t see the pattern yet.

Meanwhile, it might be a good idea to study related but simpler examples. As John has already mentioned ($\to$) it might be easier to look at 10-dimensional supergravity first.

Here I present a discussion of what should be the bosonic part of the 2-connection governing 10-dimensional supergravity. The main point is to understand, from a categorical point of view, the relation

between the curvature 3-form $H$, the Kalb-Ramond 2-form $B$ and the $\mathrm{SO}(32) \oplus \mathrm{SO}(9,1)$-connection $A$, which governs the Green-Schwarz anomaly cancellation ($\to$) (Phys.Lett.B149:117-122,1984, ($\to$)).

Following Killingback ($\to$, $\to$), we expect 10-dimensional supergravity to be governed by the 2-group $\mathrm{String}_G$ ($\to$, $\to$), where $G$ is the the 10-dimensional Lorentz group times an internal factor $\mathrm{SO}(32)$ or $E_8\times E_8$.

The Lie-2-algebra of this 2-group has a weak skeletal incarnation which is, as noticed first by André Henriques ($\to$) nothing but the Baez-Crans Lie 2-algebra $g_\hbar$ (see example 50 of Alissa Crans’ thesis). The Koszul-dual FDA of this guy is particularly simple:

Let $g$ be some Lie algebra, and let $h = \mathrm{Lie}(\mathbb{R})$. On the free graded-commutative algebra

with $g^*$ in degree 1 and $h^*$ in degree 2, we define a differential of grade 1 in terms of a basis $\{a^a\}$ of $g^*$ and $\{b\}$ of $h^*$ by setting

Here $C^a{}_{bc}$ are the structure constants of $g$ in the chosen basis and $C_{abc} = k_{aa'}C^a{}_{bc}$, where $k$ is the Killing form of $g$.

This $d$ is nilpotent due to the Jacobi identity in $g$.

We would like to find the Lie 3-algebra $\mathrm{inn}(g_\hbar)$ of inner derivations of this Lie 2-algebra.

Instead of trying to strictly derive this, I’ll notice that due to various constraints there is not much of a choice and an obvious ansatz will do the job.

From the study of $\mathrm{inn}(h\to g)$ for $(h\to g)$ an arbitrary strict Lie-2-algebra (example 12 of the FDA Lab ($\to$)), we expect the algebra of $\mathrm{inn}(g_\hbar)$ to be based on a vector space consisting of two copies of $g_\hbar$, one of which shifted by one in degree. So consider

with the first $g^*$ in degree 1, the expression in brackets in degree 2 and the last $h^*$ in degree 3.

Let $\{a^a\}$ be a basis of $g^*$ in degree 1, $\{b\}$ a basis of $h^*$ in degree 2, $\{r^a\}$ a basis of $g^*$ in degree 2 and $\{c\}$ a basis of $h^*$ in degree 3.

We know (example 12 of the FDA Lab ($\to$)) that the 2-algebra of inner derivations of $g$ itself is encoded in the differential defined by

All available modifications of the original $d b + C_{abc}a^a a^b a^c = 0$ are hence given by

for $u \in \mathbb{R}$ some real parameter. For the particular choice $u = 1$ we find that $d$ is nilpotent precisely if

In summary, we have the FDA on $\bigwedge^\bullet (g^* \otimes (g^* \otimes h^*) \otimes h^*)$ determined by

I dare to call this FDA $\mathrm{inn}(g_\hbar)$, though I have only shown that it is one particular choice from a 1-parameter collection of admissable ansätze.

Given this, it is easy to derive the degrees of freedom of a flat (curvature-)$3$-connection with values in this Lie 3-algebra.

First of all, this is a degree-preserving map $\Phi$ from $\bigwedge^\bullet (g^* \otimes (g^* \otimes h^*) \otimes h^*)$ to the deRham complex of our base manifold, compatible with the algebra structure. In other words, we have a $g$-valued 1-form

a $g$-valued 2-form

and an ordinary 2-form

and finally a 3-form

In order for this to be a chain map, we need to have $[Q,\Phi] = 0$, where $Q = d_{\mathrm{inn}(g_\hbar)} \oplus d_{\text{deRham}}$. This condition is equivalent to requiring that

is the curvature of $A$ and that

is the 3-form curvature, where $\mathrm{CS}(A)$ denotes the Chern-Simons 3-form ($\to$) of $A$.

The last condition - the Bianchi-identity on $H$ - is then automatic

This is exactly the bosonic field content of the gauge sector that we expect.

In closing, I note that the above should essentially be the local infinitesimal version of a connection on the String-gerbe which Danny Stevenson describes in section 6 of his notes. Notice how in these notes, too, the String gerbe is governed by a flat CS-2-gerbe.

You can find notes on the computations involved here in the FDA Laboratory.

## Re: 10D SuGra 2-Connection

The progress you’re making is really impressive!

Let’s try to figure out the Lie 3-algebra $inn(g)$ for a Lie 2-algebra g without resorting to any guesses. I’m sure you’re on the right track, but it would be nice to have a systematic approach, especially for when we categorify.

The only way I can imagine is fairly strenuous. You may have already tried it. If $g$ is the Lie 2-algebra of a Lie 2-group $G$, $inn(g)$ should be the Lie 3-algebra of the Lie 3-group $Inn(G)$ - the Lie 3-group of inner automorphisms of $G$. I think we can, at least with sufficient energy, figure out the definition of $Inn(G)$ and then work out its Lie 3-algebra $inn(g)$. In fact, $inn(g)$ should only depend on $g$; we should be able to see how. Then, we can use this as a definition of $inn(g)$ for any Lie 2-algebra $g$, regardless of whether it has a corresponding Lie 2-group.

(In fact any Lie n-algebra $g$ will come from a Lie n-group in the sense of Henriques . For the present purposes, this simply amounts to saying we can

locallyintegrate the Lie n-algebra to a Lie n-group. In fact Henriques goes further, but we don’t need to worry about that here.)Before we dive into this, a question: why are you sure we need $inn(g)$ instead of the potentially larger $der(g)$?

There’s something funny about $inn(g)$, after all. It should be analogous to $Inn(G)$. When $G$ is a group, the 2-group $Inn(G)$ is equivalent to the trivial 2-group, right? I assume you mean to define this 2-group using a crossed module $t: G \to G$ with $t = 1$. So, all objects in this 2-group are uniquely isomorphic. So, it’s equivalent to the trivial 2-group. By analogy, I expect that for any Lie algebra $g$, the Lie 2-algebra $inn(g)$ should be equivalent to the trivial Lie 2-algebra. And, categorifying this idea rather blindly, I’d guess that for any Lie n-algebra $g$, the Lie (n+1)-algebra $inn(g)$ is equivalent to the trivial one!

Can we really get something interesting with a $inn(g)$-connection when $inn(g)$ is equivalent to something trivial?

I guess it’s easiest to check this for n = 1, and I assume you already have. If you could explain what’s really going on here, that would be great. You may recall a similar discussion over on David’s blog, where I was attacking the idea of Klein 2-geometry being interesting when the relevant 2-group is equivalent to a trivial one.

(Could there be a difference between 2-groups and

Lie2-groups here??? A smooth category with all objects uniquely isomorphic may not besmoothlyequivalent to the trivial category. But in fact, I don’t think that’s a way out - I think $Inn(G)$ is evensmoothlyequivalent to the trivial 2-group.)Anyway, if we wanted to compute $der(g)$, I’d first ponder $Aut(G)$ and then differentiate. If $G$ is a 2-group, $Aut(G)$ is the 3-group with

Sitting inside here should be $Inn(G)$, but I’m not quite sure how to define it.