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I am reading Nigel Hitchin's notes to understand about gerbes.

It starts the article by saying the following :

Before giving a definition, it’s worthwhile to recognize when we, as mathematicians, might be in a situation where the language of gerbes could be relevant. We are basically in gerbe territory (for smooth manifolds) if a ny one of the following is being considered:

  • a cohomology class in $H^3(X,\mathbb{Z})$
  • a Cech cocycle $[g_{\alpha\beta\gamma}]\in H^2(X,C^\infty(S^1))$

In the last case, this means a 2-cocycle for the sheaf of germs of $C^\infty$ functions with values in the circle.

Here, sheaf $C^\infty(S^1)$ on $X$ defined as $C^\infty(S^1)(U)=\{\text{smooth }f:U\rightarrow S^1 \}$ for open $U\subseteq X$.

To understand gerbes, we need to consider the other creatures in a hierarchy to which gerbes belong, and here the lowest form of life consists of circle valued functions $f:X\rightarrow S^1$. Consider the following features of such a function:

  • a cohomology class in $H^1(X,\mathbb{Z})$
  • a Cech cocycle $[g_\alpha]\in H^0(X,C^\infty(S^1))$

The next step in hierarchy consists of unitary line bundle $L$, or its principal $S^1$ bundle of unitary frames. Here we have

  • a cohomology class in $H^2(X,\mathbb{Z})$.
  • a Cech cocycle $[g_{\alpha\beta}]\in H^1(X,C^{\infty}(S^1))$

I dont know what a unitary line bundle is. I only know what is a line bundle. I am guessing unitary line bundle is the one whose transition functions takes values in $S^1$ and not just in $\mathbb{C}^*$.

Given a line bundle $\pi:L\rightarrow X$, we have trivialization cover $\{U_\alpha\}=\mathcal{U}$ with trivializations $pi^{-1}(U_\alpha)\times U_\alpha\times \mathbb{C}$ and corresponding transition funcions $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow S^1$. So, we have $(g_{\alpha\beta})\in \mathcal{C}^1(\mathcal{U},C^{\infty}(S^1))$. We also have $g_{\alpha\beta}g_{\beta\gamma}=g_{\alpha\gamma}$ on $U_\alpha\cap U_\beta\cap U_\gamma$. This says that $\underline{g}=(g_{\alpha\beta})\in \mathcal{C}^1(\mathcal{U},C^{\infty}(S^1))$ is actaully a $1$ cocycle in the sense of Cech cocycles. This gives an element in first Cech cohomology $H^1(X,C^{\infty}(S^1))$.

It says further that

Now take an open covering of $X$ and a map $g_{\alpha\beta\gamma}:U_\alpha\cap U_\beta\cap U_\gamma\rightarrow S^1$ on each threefold intersection with $g_{\alpha\beta\gamma}=g_{\beta\alpha\gamma}^{-1}=g_{\alpha\gamma\beta}^{-1}=g_{\gamma\beta\alpha}^{-1}$ and satisfying the cocycle condition $$g_{\beta\gamma\delta}g_{\alpha\gamma\delta}^{-1}g_{\alpha\beta\delta}g_{\alpha\beta\gamma}^{-1}=1 \text{ on } U_\alpha\cap U_\beta\cap U_\gamma\cap U_\delta.$$ We shall say that this data defines a gerbe. By this we mean that it suffices to define a gerbe in the same way that a collection of transition functions defines a line bundle or a collection of coordinate charts defines a manifold.

I understand that we are trying to go a step above to second Cech cohomology group to define a gerbe. So, we need to give an element of $\mathcal{C}^2(\mathcal{U},C^{\infty}(S^1))$ that is a $2$ cocycle. This only means that we are asked to give $\overline{g}=(g_{\alpha\beta\gamma})$ with $g_{\alpha\beta\gamma}\in C^{\infty}(S^1)(U_\alpha\cap U_\beta\cap U_\gamma)$ such that $$g_{\beta\gamma\delta}g_{\alpha\gamma\delta}^{-1}g_{\alpha\beta\delta}g_{\alpha\beta\gamma}^{-1}=1 \text{ on } U_\alpha\cap U_\beta\cap U_\gamma\cap U_\delta$$ which is just the multiplicative version of coycle condition.

Can some one help me to realize why to impose the condition $$g_{\alpha\beta\gamma}=g_{\beta\alpha\gamma}^{-1}=g_{\alpha\gamma\beta}^{-1}=g_{\gamma\beta\alpha}^{-1}$$ on threefold intersection. I understand that if we allow $\underline{g}$ to be a $2$ couboundary then this condition holds true but I am sure we are not allowing $\overline{g}$ to be a $2$ coboundary, otherwise cohomology class of $\overline{g}$ would just be $0$.

So, what is happening here when they impose condition $g_{\alpha\beta\gamma}=g_{\beta\alpha\gamma}^{-1}=g_{\alpha\gamma\beta}^{-1}=g_{\gamma\beta\alpha}^{-1}$?

Is this some generalization of a condition in case of $1$ coycles seen above? If so, can some one suggest me the way to see this?

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  • $\begingroup$ You need to read about Cech cohomology. The alternating condition is not needed, but when imposed it leads to the same Cech cohomology groups; see Section IV.5 of these notes by J.P. Demailly. www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf $\endgroup$ – Liviu Nicolaescu Apr 22 '18 at 9:32
  • $\begingroup$ @LiviuNicolaescu I started reading Cech cohomology.. I will see that book and come back if I have anything to ask. Thanks :) $\endgroup$ – Praphulla Koushik Apr 22 '18 at 9:39
  • $\begingroup$ I believe you are saying about equation $5.22$ in that notes.. Am I correct? Can you make it as an answer? @LiviuNicolaescu $\endgroup$ – Praphulla Koushik Apr 22 '18 at 10:56
  • $\begingroup$ Yes. It is the result in Section IV.5.D. $\endgroup$ – Liviu Nicolaescu Apr 22 '18 at 12:30
  • $\begingroup$ @LiviuNicolaescu Thank you :) So, it is just for convenience we are considering subcomplex which is still giving same cohomology group. Thanks. $\endgroup$ – Praphulla Koushik Apr 22 '18 at 12:51
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Isn't the condition you mention just saying that the $g_{\alpha\beta\gamma}$ depends only on the corresponding intersection and not on the order in which the three sets are listed in writing down that intersection, mod transposition of any two. It is thus a normalisation condition, like saying $g_{\alpha \beta}=g_{\beta\alpha}^{-1}$ or $g_{\alpha\alpha}=1$. Are you thinking it is more than that? If you know about Cech complexes and simplicial objects this is like a `compatibility with degeneracy' condition together with a symmetry condition. (I personally like the clarity that comes, for me, with using simplicial objects, although some people find the notation can get in the way.)

It may help to compare Hitchin's treatment with that of Larry Breen in https://arxiv.org/abs/math/0611317

or

https://link.springer.com/chapter/10.1007/978-1-4419-1524-5_5

Looking at several treatments will I am sure help resolve the difficulty.

As an aside it is interesting / fun to look at case in which the order of intersection does influence the assignment, i.e. in which $g_{\alpha\beta}$ need not be $g_{\beta\alpha}^{-1}$, as that to a limited extent reflects a more 'quantum theoretic' viewpoint in that an `observation' corresponding to $U_{\alpha}$ followed by one corresponding to $U_\beta$ need not be the same as when the observations are done in the opposite order.

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  • $\begingroup$ So, you are saying it is just a normalisation condition.. I am not very sure what more to ask.. I am planning to read Breen’s article also.. but I think it says about non abelian gerbes $\endgroup$ – Praphulla Koushik Apr 22 '18 at 7:23
  • $\begingroup$ I would prefer to say that I am guessing it is really about normalising the assignment. As to non-abelian gerbes, I believe that it is better to understand them before looking at the abelian case as then you see a general situation in which the combinatorics / geometry of the situation can be examined. Think of a group theory course that only looked at abelian groups. The vision that you would have would be very restrictive and the methods used would be likewise restrictive. What is more, if you understand non-abelian gerbes, the simplification of the abelian case is a relief,;-) $\endgroup$ – Tim Porter Apr 22 '18 at 8:25
  • $\begingroup$ I don’t understand your first line completely... in case of two indices there are only two choices $g_{\alpha\beta}$ and $g_{\beta\alpha}$. So it is alright to impose a condition that says $g_{\alpha\beta}=g_{\beta\alpha}^{-1}$ and I this actually happens in case of transition functions of Principal bundles.. but in case of three indices $\alpha,\beta,\gamma$ there are exactly $6$ choices $g_{\alpha\beta\gamma},g_{\alpha\gamma\beta},g_{\beta\alpha\gamma},g_{\beta\gamma\alpha},g_{\gamma\alpha\beta},g_{\gamma\beta\alpha}$.. so, why impose relation between only four of them? What am I missing Sir $\endgroup$ – Praphulla Koushik Apr 22 '18 at 8:30
  • $\begingroup$ Sir, I really like your approach.. I thought I would first read abelian gerbes and then go for non abelian gerbes.. but what you said is correct.. it would be too restrictive... thanks for the comment :) I will start reading Breen’s article now,. $\endgroup$ – Praphulla Koushik Apr 22 '18 at 8:33
  • $\begingroup$ The identity of the three with $-1$s on them together with the first identiyt tell you the other two (of the 6 ) as also equal to that value. $\endgroup$ – Tim Porter Apr 22 '18 at 8:34

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