For a strange reason the literature on gerbes is abound with a certain confusion of terms, which somewhat hides a rather beautiful simple picture. The worst problem in the literature is that “gerbe” in the sense of Giraud, Breen etc. is a different notion to the “bundle gerbes” of Murray, Stevenson, etc. and then there is the tendency to drop the “bundle” in “bundle gerbe”. Even though these notions are of course related, they are conceptually crucially different and *not equivalent*. This non-equivalence is effectively what your question aims at.

Here is the simple grand picture and general classification. Allow me to advertize for further details part I, section 4.4 “Gerbes” of our *Principal ∞-bundles – theory, presentations and applications (schreiber)* and for a bit more context sections 3.6.10 to 3.6.15 in *differential cohomology in a cohesive topos (schreiber)*.

So pick some ambient $\infty$-topos $\mathbf{H}$ which contains all the sheaves, stacks, etc. over your favorite site.

Then we have the following definitions and classifications, which should all at least sound entirely evident.

For any $V, X \in \mathbf{H}$ a $V$-

$\array{ V \times U &\to& E \\ \downarrow && \downarrow \\ U &\to& X }$*fiber ∞-bundle*over $X$ is a map $E \to X$ such that there exists a 1-epimorphism $U \to X$ and an $\infty$-pullback diagramIf $V \simeq \mathbf{B}G$ is pointed connected and hence equivalently the delooping/moduli $\infty$-stack of an ∞-group $G$, then this is equivalentlya $G$-∞-gerbe.

To repeat for emphasis: $G$-gerbes are $\mathbf{B}G$-fiber bundles.

A map $P \to X$ with a $G$-∞-action on $P$ over $X$ is a $G$-principal ∞-bundle if the map is equivalently the quotient map $P \to P//G \simeq X$.

Now the classification results:

$V$-fiber $\infty$-bundles are classified by $\mathbf{Aut}(V)$-cohomology, where $\mathbf{Aut}(V)$ is the automorphism ∞-group of $V$ as an internal group:

$V \mathbf{Bund}(X) \simeq \mathbf{H}(X, \mathbf{B}\mathbf{Aut}(G)) \,.$Moreover, the equivalence is established by sending a cocycle $g \colon X \to \mathbf{B}Aut(G)$ first to its homotopy fiber $P \to X$, which is the corresponding $\mathbf{Aut}(V)$-principal $\infty$-bundle, and then forming the associated ∞-bundle $E \coloneqq P \times_{\mathbf{Aut}(V)} V$. Equivalently, it is given by directly pulling back the *universal* $V$-fiber ∞-bundle, which is the homotopy fiber sequence

which itself is the pullback of the object classifier $\widehat Obj \to Obj$ of $\mathbf{H}$ along the inclusion of the 1-image of the name $* \stackrel{\vdash V}{\to} Obj$ of $V$.

Anyway, in particular therefore for $G \in Grp(\mathbf{H})$ an $\infty-group$ $G$-∞-gerbes $\simeq$ $\mathbf{B}G$-fiber ∞-bundles are classified by $\mathbf{Aut}(\mathbf{B}G)$-cohomology

$G Gerbes(X) \simeq \mathbf{H}(X, \mathbf{B}\mathbf{Aut}(\mathbf{B}G)) \,.$This is nonabelian cohomology and thus often regarded as something exotic. What is true is that you cannot use off-the-shelf homological algebra to compute it, but otherwise it’s an entirely mundane concept and already just slightly more sopisticated homological algebra in fact almost does the trick.

For $G$ an ordinary group object (a 0-truncated $\infty$-group), the $\mathbf{Aut}(\mathbf{B}G)$ is the “automorphism 2-group” of $G$ given by the crossed complex

$\mathbf{Aut}(\mathbf{B}G) \simeq [G \stackrel{Ad}{\to} \mathbf{Aut}(G)] \,.$Accordingly there is a canonical map

$\mathbf{B}G \to \mathbf{Aut}G$and if $G$ is at least a braided ∞-group (for instance an abelian ∞-group) then this is an $\infty$-group homomorphism and hence has a further delooping to a map

$\mathbf{B}^2 G \to \mathbf{B}\mathbf{Aut}(G) \,.$This hence induces a map from ordinary degree-2 $G$-cohomology to $\mathbf{Aut}(G)$-cohomology:

$\mathbf{H}(X, \mathbf{B}^2 G) \to \mathbf{H}(X, \mathbf{B}\mathbf{Aut}(G)) \,.$And this map is, finally, what injects $G$-principal 2-bundles which you may think of as $G$-bundle gerbes into actual $G$-gerbes.

]]>The short answer is: for $A$ an abelian group object, $A$-1-gerbes are classified by $\mathbf{Aut}(\mathbf{B}A)$-cohomology, where $\mathbf{Aut}(\mathbf{B}A)$ is the automorphism 2-group of $A$. This “contains” $\mathbf{B} A$-cohomology but is richer.

For instance if we are over the smooth site, then $U(1)$-gerbes are not equivalently $U(1)$-bundle gerbes (which are but a special case of the former), but are associated to $(U(1) \to \mathbb{Z}_2)$-principal 2-bundles, where $(U(1) \to \mathbb{Z}_2) = \mathbf{Aut}(\mathbf{B}U(1))$ is the 2-group coming from the crossed complex as indicated, with $\mathbb{Z}_2 \simeq \mathbf{Aut}(U(1))$ acting canonically.

The usual $U(1)$-bundle gerbes are instead $(U(1) \to 1) = \mathbf{B}U(1)$-principal 2-bundles (only).

I’ll say more in the next message…

]]>My question is: If I get rid of the assumption on n, but still have an abelian group object, etc., then what goes wrong with this classification? That is: In the special case that X corresponds to a space, and A corresponds to an abelian group, what is the difference between an (EM) 0-gerbe banded by A and an A-torsor? Similarly, what is the difference between an (EM) 1-gerbe banded by A and an A-banded gerbe (in the classical setting)?

I tried to parse the nonabelian classification given on the nLab involving a slightly different notion of gerbe in order to apply it to this question, but couldn't manage it.

I suppose another answer to my question would be a clear definition (for all n) of some objects that H^n+1(X, A) does classify. The end of the infty-gerbe article promises this in section 2.3 of something that Urs wrote, but I couldn't find a section 2.3 in the linked page... and clicking on "2." in the description of sections didn't seem to do anything... Maybe something's wrong with my browser. ]]>