### n-Curvature

#### Posted by Urs Schreiber

We have learned that parallel $n$-transport in an $n$-bundle with connection over a base space $X$ is an $n$-functor $\mathrm{tra} : \mathcal{P}_n(X) \to T$ from the $n$-path $n$-groupoid of $X$ to some $n$-category of fibers.

With every notion of connection we expect to obtain notions of

1) curvature;

2) Bianchi identity;

3) parallel sections;

4) covariant derivative.

Here we describe the functorial incarnation of these concepts. We find

1) To every transport $n$-functor $\mathrm{tra}$ is canonically
associated a curvature $(n+1)$-functor
$\mathrm{curv}_{\mathrm{tra}} : \Pi_{n+1}(X) \to T_{n+1}\,.$
The functor $\mathrm{tra}$ is *flat* precisely if
$\mathrm{curv}_{\mathrm{tra}}$ is trivial on all $(n+1)$-morphisms.

2)
The curvature $(n+1)$-functor, regarded as an $(n+1)$-transport itself,
is always flat.

3) Parallel sections $e$ of the $n$-bundle with connection associated with $\mathrm{tra}$ are equivalent to morphisms from the trivial $n$-transport into $\mathrm{tra}$: $e : \mathrm{tra}_0 \to \mathrm{tra} \,.$

4) General sections $e$ together with their covariant derivative $\nabla e$ are equivalent to morphisms from the trivial curvature $(n+1)$-transport into the curvature $(n+1)$-transport $(e,\nabla e) : \mathrm{curv}_0 \to \mathrm{curv}_{\mathrm{tra}} \,.$

What is the **curvature** associated with a
transport, *really*?

Whatever it is, it should vanish if our transport factors through
the *fundamental* $n$-groupoid $\Pi_n(X)$ of $X$. As opposed
to $\mathcal{P}_n(X)$, whose $n$-morphisms are *thin* homotopy
classes of $n$-paths, the $n$-morphisms of $\Pi_n$ are
ordinary homotopy classes of $n$-paths.

Hence we have a canonical projection $\pi : \mathcal{P}_n(X) \to \Pi_n(X)$ that sends any $n$-path to its homotopy class.

**Definition.**
*
We say that $\mathrm{tra}$ is flat precisely
if there is an $n$-functor
*
$f : \Pi_n(X) \to T$

*such that*$\array{ \mathcal{P}_n(X) &\stackrel{\pi}{\to}& \Pi_n(X) \\ {}^{\mathrm{tra}}\downarrow\;\;\, &\Downarrow^\sim& \;\;\;\downarrow^f \\ T &=& T } \,.$

Whatever the curvature of an $n$-functor is, it should be an obstruction for this construction.

**Proposition.**
*
Given any $n$-transport $\mathrm{tra} : \mathcal{P}_n(X) \to T$,
we canonically obtain an
$(n+1)$-category
*
$T_{n+1}$
*
and an $(n+1)$-functor
*
$\mathrm{curv}_\mathrm{tra}
:
\Pi_{n+1}(X) \to T_{n+1}$
*
such that $\mathrm{curv}_{\mathrm{tra}}$ is trivial
on $(n+1)$-morphisms (sends every $(n+1)$-morphism to an
identity $(n+1)$-morphism) if and only if $\mathrm{tra}$ is
flat.
*

When I wrote my first entry on this topic I think I had the right idea. But I did it by hand. This time I want to let the Dao do it by its own means:

## Lifting gerbes?

Here is a problem whose solution I do not understand yet.

The way I have discussed curvature above, it is the obstruction to a descent problem:

we have some $n$-thing $\array{ P_n(X) \\ \downarrow^{\mathrm{tra}} \\ T }$ on $P_n(X)$ and want to push it down to $\Pi_n(X)$ by completing a square to the

right$\array{ P_n(X) &\to& \Pi_n(X) \\ {}^{\mathrm{tra}}\downarrow \;\; &\Downarrow^\sim& \;\; \downarrow^f \\ T &=& T } \,.$The obstruction is an $(n+1)$-thing.

This curiously smells like it should be one aspect of a general mechanism of which lifting gerbes are another.

For lifting gerbes, the problem is essentially “the same but opposite”:

given an extension $U(1) \to H \to G$ of groups, and given a principal $G$-transport $\array{ \P_1(X) \\ \;\;\downarrow^{\mathrm{tra}} \\ G\mathrm{Tor} }$ we want to know if we can

liftby completing a square to theleft$\array{ \P_1(X) &=& \P_1(X) \\ {}^f \downarrow\;\; &\Downarrow^\sim& \;\;\downarrow^{\mathrm{tra}} \\ H\mathrm{Tor} &\to& G\mathrm{Tor} \,. }$We know from pedestrian reasoning that this lift of 1-things is obstructed by a 2-thing: the lifting gerbe.

Therefore I expect that there is, as there was for curvature, a canonical arrow-theoretic construction that reads in the above extension problem and canonically spits out the parallel transport 2-functor of the lifting gerbe.

The entire problem here looks like that for curvature, with arrows reversed. Therefore I thought it would be simple to see how it works. But I still don’t see it.