I am reading this notes of Hitchin to understand about gerbes. He defines gerbe by giving a collection of $2$ cocycles $g_{\alpha\beta\gamma}:U_\alpha\cap U_\beta\cap U_\gamma\rightarrow S^1$ with some conditions as here.

He then

define a connection on a gerbe given by cocycles $g_{\alpha\beta\gamma}:U_\alpha\cap U_\beta\cap U_\gamma\rightarrow S^1$ by forms which satisfy $$G|_{U_\alpha}=dF_\alpha$$ $$F_\beta-F_\alpha=dA_{\alpha\beta}$$ $$i A_{\alpha\beta}+iA_{\beta\gamma}+iA_{\gamma\alpha}=g_{\alpha\beta\gamma}^{-1}dg_{\alpha\beta\gamma}.$$ We call the closed $3$ form $G$ the curvature of the gerbe connection.

We say that a connection on a gerbe is flat if its curvature $G$ vanishes. Suppose a gerbe is flat, we have $G=0$ i.e., $dF_{\alpha}=0$. As we are considering contractible $U_\alpha$ (good open cover) we see that $F_\alpha$ being closed implies $F_\alpha$ is exact (by Poincare lemma) i.e., $F_\alpha=dB_{\alpha}$ on $U_\alpha$.

On $U_\alpha\cap U_\beta$ we have $F_\beta-F_\alpha=d(B_\beta-B_\alpha)$.

As $F_\beta-F_\alpha=dA_{\alpha\beta}$ we have $d(B_\beta-B_\alpha)=dA_{\alpha\beta}$ i.e., $A_{\alpha\beta}-B_\beta+B_\alpha=df_{\alpha\beta}$ (again by Poincare lemma) on $U_\alpha\cap U_\beta$.

He then says,

as $iA_{\alpha\beta}+iA_{\beta\gamma}+iA_{\gamma\alpha}=g_{\alpha\beta\gamma}^{-1}dg_{\alpha\beta\gamma}$ , we have $d(if_{\alpha\beta}+if_{\beta\gamma}+if_{\gamma\alpha}-\log g_{\alpha\beta\gamma})=0$

I am not able to see how this is true. I did write down what $A_{\alpha\beta}$ and $f_{\alpha\beta}$ are but I got some relation that does not look anyway close to this.

Assuming $d(if_{\alpha\beta}+if_{\beta\gamma}+if_{\gamma\alpha}-\log g_{\alpha\beta\gamma})=0$, it says the following:

Of course $\log(g)$ is defined only modulo $2\pi i \mathbb{Z}$ so what we have here is a collection of constants $c_{\alpha\beta\gamma}\in 2\pi \mathbb{R}/\mathbb{Z}$. The cocycle $c_{\alpha\beta\gamma}/2\pi$ represents a Cech class in $H^2(X,\mathbb{R}/\mathbb{Z})$ which we call the holonomy of the connection.

I do not understand completely what this means.

As $if_{\alpha\beta}+if_{\beta\gamma}+if_{\gamma\alpha}-\log g_{\alpha\beta\gamma}:U_\alpha\cap U_\beta\cap U_\gamma\rightarrow S^1$, $d(if_{\alpha\beta}+if_{\beta\gamma}+if_{\gamma\alpha}-\log g_{\alpha\beta\gamma})=0$ implies that $if_{\alpha\beta}+if_{\beta\gamma}+if_{\gamma\alpha}-\log g_{\alpha\beta\gamma}$ is a constant and I guess this is what they are calling $c_{\alpha\beta\gamma}$. Identifying $S^1$ with $\mathbb{R}/\mathbb{Z}$ he says $c_{\alpha\beta\gamma}\in 2\pi \mathbb{R}/\mathbb{Z}$. Seeing constants $c_{\alpha\beta\gamma}$ as constant functions, this defines maps $c_{\alpha\beta\gamma}:U_\alpha\cap U_\beta\cap U_\gamma\rightarrow \mathbb{R}/\mathbb{Z}$. This defines $2$ cocycle, thus an element of $H^2(X,\mathbb{R}/\mathbb{Z})$, this they are calling it as a holonomy.

I am not very sure if this is what it means. Any comments are welcome.

Any reference for concept of holonomy on gerbes would be useful.

EDIT : I thank user Tsemo for proving the equality that I said I was not able to prove. As I have not clearly stated what my question is, I would like to say it now. Any thoughts on motivation behind calling this holonomy is welcome. Is this collection $\{c_{\alpha\beta\gamma}\}$ restircted to some subset is holonomy of some (line) bundle?

  • $\begingroup$ As I said elsewhere, it would be very helpful for you to work out the analogous situation for a circle bundle, and see how the construction maps to holonomy there. $\endgroup$ May 2, 2018 at 17:26
  • $\begingroup$ @AaronBergman Can you say little more than what you said here, I do not completely understand what does it mean. $\endgroup$ May 2, 2018 at 17:53
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    $\begingroup$ Do everything for one degree less in Cech cohomology, and you’ve got a circle bundle. Everything is analogous to that. $\endgroup$ May 2, 2018 at 19:55
  • $\begingroup$ I think how you understand this "holonomy" bit depends largely on the definition you are comfortable with. I think of holonomy as a function from closed n-surfaces, mapped into X, into a Lie Group. Based on my way of thinking, I agree that this line is a bit confusing. But I would just have to answer the question: why could constants on triple intersections give me all I need to know for a function from [closed 2-surfaces into X] into S^1!? $\endgroup$
    – cheyne
    Aug 30, 2018 at 19:31
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    $\begingroup$ @cheyne your comment makes the question clearer... Do you have one or two lines to say why it is related to the notion of holonomy of a connection? $\endgroup$ Aug 31, 2018 at 2:07

2 Answers 2


$A_{\alpha\beta}=df_{\alpha\beta}+B_\beta-B_\alpha$ implies that



$=i(df_{\alpha\beta}+df_{\beta\gamma}+df_{\gamma\alpha})=dlog(g_{\alpha\beta\gamma})$ implies that $i(d(f_{\alpha\beta}+f_{\beta\gamma}+f_{\gamma\alpha}-log(g_{\alpha\beta\gamma}))=0$.

$c_{\alpha\beta\gamma}$ obtained here is a $2$-Cech cocycle which is by definition the holonomy of the gerbe.

  • $\begingroup$ Thanks for your answer. Question was more about paragraph above the line I dont completely understand what does this mean Its more about holonomy definition than about the equality that you have explained. I was able to see the equality and wanted to edit the question saying I got this point but as you have already gave answer for that It is not reasonable to do so. Please say something about my understanding of holonomy that I have said above.. $\endgroup$ Apr 29, 2018 at 14:30
  • $\begingroup$ It can be accepted as a definition of the holonomy of a gerbe. $\endgroup$ Apr 29, 2018 at 14:31
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    $\begingroup$ Can you say anything more than that, please. $\endgroup$ Apr 29, 2018 at 15:07

Take a closed surface, $\Sigma$, and map it into your manifold. If $\Sigma \hookrightarrow U_{i}$, then you can simply integrate $F_{i}$ over $\Sigma$ and obtain the holonomy in $S^1$, evaluated on $\Sigma$. Note that I'm skirting around the issues involved with thinking of holonomy as a functor/function, etc, but I think my comment will make you happy nonetheless.

OK, but what if $\Sigma \hookrightarrow U_i \cup U_j$? Well, then you could cut $\Sigma$ into two "2-cells": one which lays entirely in $U_i$, one which lays entirely in $U_j$ and the edge lies entirely in $U_i \cap U_j$. So now you integrate over the pieces with the corresponding $F_{\alpha}$, and then glue them together using $iA_{ij}$ integrated over the edge. Now, since closed surfaces have (in general) non-trivial degree two Cech Cohomology, this essentially means we have to worry about how this gluing remains compatible over triple intersections. So now for each triple $U_{ijk}$, you repeat my glueing process for each "2-cell" in a $U_{\alpha}$, glueing them together over "1-cells" using $iA_{\alpha \beta}$, and then glueing all of this together over vertices, using $g_{\alpha \beta \gamma}$.

I tried to keep things informal because of course the real question is "what do I mean by compatible", but then now you want to study this whole theory a bit more than a post like this one.

I wasn't really clear on where all of your forms/functions were taking their values, but it turns out that the process I outlined is well-defined for non-abelian situations to a certain extent as well so it remains useful.


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