# Cocycle description of gerbes

I am trying to understand cocycle description of gerbes as in https://arxiv.org/pdf/math/0611317.pdf.

Let $\mathcal{P}$ be a gerbe on a topological space $X$ i.e., $\mathcal{P}$ is a stack over groupoids that is both locally non empty and locally connected.

As $\mathcal{P}$ is locally nonempty, we can choose an open cover $\mathcal{U}=\{U_i\}$ such that $\mathcal{P}(U_i)\neq \emptyset$.

So, we can choose $x_i\in \mathcal{P}(U_i)$. As $\mathcal{P}$ is a stack, we get a sheaf over $U_i$, $G_i=\underline{\text{Aut}}(x_i)$ over $U_i$.

Given any sheaf on a space $X$, one can associate a group bundle (Etale space) over $X$. I guess after identifying sheaf with etale space, they are calling $G_i$ to be bundle of groups on $U_i$ (pg no $10$).

Once we chose $x_i$ as above, we also choose arrows $\phi_{ij}:x_j|_{U_{ij}}\rightarrow x_i|_{U_{ij}}$ in $\mathcal{P}(U_{ij})$.

What I don’t understand is the map $\lambda_{ij}:G_j|_{U_{ij}}\rightarrow G_i|_{U_{ij}}$. I fail to see how they are identifying elements of $G_j|_{U_{ij}}$ as just maps $x_j\rightarrow x_j$. They are using notion of sheaf and that of etale space simultaneously and it is getting confused.

Any suggestion is welcome.

• I realized what I was missing and made it as an answer. Let me know if there is anything wrong. Thanks :) – Praphulla Koushik May 11 '18 at 18:09

It might help to divide this process into finer steps. We choose our open cover so that the groupoids $\mathcal{P}(U_i)$ and $\mathcal{P}(U_i \cap U_j)$ are connected and nonempty. Choose objects $x_i \in \mathcal{P}(U_i)$ and $x_{i,j} \in \mathcal{P}(U_i \cap U_j)$, and let $G_i = \operatorname{Aut}(x_i)$ and $G_{i,j} = \operatorname{Aut}(x_{i,j})$. By the fibered category property, the inclusions $\iota_{i,j}: U_i \cap U_j \to U_i$ yield restriction functors $\iota_{i,j}^*: \mathcal{P}(U_i) \to \mathcal{P}(U_i \cap U_j)$. By the connected property, there exist isomorphisms $\iota_{i,j}^\sharp: \iota_{i,j}^*(x_i) \to x_{i,j}$ in $\mathcal{P}(U_i \cap U_j)$, and any such choice yields a corresponding isomorphism of groups $\iota_{i,j}^\flat: \iota_{i,j}^*(G_i) \to G_{i,j}$. If you choose your isomorphisms $\iota_{i,j}^\sharp$, giving a choice of $$\phi_{ij} = (\iota_{i,j}^\sharp)^{-1} \circ \iota_{j,i}^\sharp$$ then you automatically get $$\lambda_{ij} = (\iota_{i,j}^\flat)^{-1} \circ \iota_{j,i}^\flat$$.

• Thank you for the answer. Your $G_i$ is a sheaf of groups? I used that notation for sheaf of groups. – Praphulla Koushik May 11 '18 at 0:51
• @PraphullaKoushik No, I wrote the answer just considering the group of automorphisms of the object $x_i$. However, you can use the automorphism group sheaf of $x_i$, and the answer is the same. – S. Carnahan May 11 '18 at 3:40
• Oh, sorry to say this but this is as confusing as the discussion in the article i have posted... Are you choosing $x_{i,j}\in \mathcal{P}(U_i\cap U_j)$ separately? Nothing to do with already chosen $x_i\in \mathcal{P}(U_i)$? Inclusion $U_i\cap U_j\rightarrow U_i$ gives restriction functor $\rho_{i,j}:\mathcal{P}(U_i)\rightarrow \mathcal{P}(U_i\cap U_j)$.. As $x_i\in \mathcal{P}(U_i)$, we have $\rho_{i,j}(x_i)\in \mathcal{P}(U_i\cap U_j)$. – Praphulla Koushik May 11 '18 at 4:37
• As $\rho_{i,j}(x_i),x_{i,j}\in \mathcal{P}(U_i\cap U_j)$ and as $\mathcal{P}(U_i\cap U_j)$ is connected, there exists an arrow $\rho_{i,j}(x_i)\rightarrow x_{i,j}$ in $\mathcal{P}(U_i\cap U_j)$.. As $\mathcal{P}$ is stack of groupoids, every arrow in $\mathcal{P}(U_i\cap U_j)$ is an isomorphism, in particular, $\rho_{i,j}(x_i)\rightarrow x_{i,j}$ that we have chosen above is an isomorphism which we are denoting by $\rho^{\#}_{i,j}$.. Now, let $\phi:x_i\rightarrow x_i$ be an element of $G_i$, in particular an arrow in $\mathcal{P}(U_i)$. – Praphulla Koushik May 11 '18 at 4:43
• This gives (an isomorphism) arrow $\rho_{i,j}(x_i)\rightarrow \rho_{i,j}(x_i)$ in $\mathcal{P}(U_i\cap U_j)$. As $\rho_{i,j}^{\#}:\rho_{i,j}(x_i)\rightarrow x_{i,j}$ is an isomorphism, this gives an isomorphism $x_{i,j}\rightarrow x_{i,j}$ giving an element of $G_{i,j}$.. Giving an isomorphism $\rho_{i,j}(G_i)\rightarrow G_{i,j}$ and if we interchange $i,j$ we get another map and this gives isomorphism $\rho_{i,j}(G_i)\rightarrow \rho_{j,i}(G_j)$ and this is what $\lambda_{i,j}$ is.. This is what you have said, correct me if I am wrong. – Praphulla Koushik May 11 '18 at 4:48

It says the following.

Let us keep the notations of section $3.1$. In addition to choosing local objects $x_i\in \mathcal{P}(U_i)$, in a gerbe $\mathcal{P}$ on $X$, we now choose arrows $x_j\rightarrow x_i$ in $\mathcal{P}(U_{ij})$. Since $G_i=\underline{\text{Aut}}(x_i)$, a chosen arrow $\phi_{ij}$ induces by conjugation a homomorphism of group bundles $$G_j|_{U_{ij}}\xrightarrow{\lambda_{ij}}G_i|_{U_{ij}}$$ with $\lambda\mapsto \phi_{ij}\gamma\phi_{ij}^{-1}$.

All that I know is, as $\mathcal{P}$ is a stack, $\underline{\text{Aut}}(x_i)$ is a sheaf of groups. There is a bijective correspondence between sheaf of groups on $X$ and group bundles(Etale spaces) on $X$. So, I was thinking I have to see $G_i$ as a group bundle on $U_i$ i.e., as a map $G_i\rightarrow U_i$ and then try to understand the homomorphism of group bundles. This is only because they have said homomorphism of group bundles. But, I realized it is easy and equivalent to see $\lambda_{ij}$ as a morphism of sheaves itself. It seems unnecessary and confusing as well.

So, we try to understand $\lambda_{ij}$ as a morphism of sheaves. sheaves are in bijective correspondence with group bundles so morphism of sheaves are in bijective correspondence with morphism of sheaves. So, it is just the same.

Now the question is,

Does it make sense to write $\gamma\mapsto \phi_{ij}\gamma\phi_{ij}^{-1}$ for morphism of sheaves $\lambda_{ij}:G_i|_{U_{ij}}\rightarrow G_j|_{U_{ij}}$?

Answer is, by abuse of notation, Yes.

Let $\mathcal{F},\mathcal{G}$ be sheaves on $X$. By a morphism of sheaf $\eta:\mathcal{F}\rightarrow \mathcal{G}$, we mean collection of maps $\eta(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)$ that are compatible with restriction maps on respective sheaves.

So, considering shevaes $G_i|_{U_{ij}}, G_j|_{U_{ij}}$ on $U_{ij}$, we define a morphsim $\lambda_{ij}:G_j|_{U_{ij}}\rightarrow G_i|_{U_{ij}}$ of sheaves . For that, we need to give a map $\lambda_{ij}(V):G_j|_{U_{ij}}(V)\rightarrow G_i|_{U_{ij}}(V)$ for each $V\subseteq U_{ij}$, this is just the same as $\lambda_{ij}(V):G_j(V)\rightarrow G_i(V)$.

To make sense of $G_i(V)$ and $G_j(V)$ we fix up some notation.

We have inclusions $\rho:V\rightarrow U_{ij},\alpha_i:U_{ij}\rightarrow U_i, \alpha_j:U_{ij}\rightarrow U_j$.

Then, by definition, $G_j(V)=\text{Hom}((\alpha_j\circ \rho)^*(x_j),(\alpha_j\circ \rho)^*(x_j))$.

Let $\eta:(\alpha_j\circ \rho)^*(x_j)\rightarrow (\alpha_j\circ \rho)^*(x_j)$ be an arrow in $\mathcal{P}(V)$. This is same thing as $$\eta:\rho^*(\alpha_j^*(x_j))\rightarrow \rho^*(\alpha_j^*(x_j))$$ in $\mathcal{P}(V)$. As $\alpha_j^*(x_j)$ is just $x_j|_{U_{ij}}$, we have $$\eta:\rho^*(x_j|_{U_{ij}})\rightarrow \rho^*(x_j|_{U_{ij}})$$ in $\mathcal{P}(V)$.

The map $\phi_{ij}:x_j|_{U_{ij}}\rightarrow x_i|_{U_{ij}}$ in $\mathcal{P}(U_{ij})$ gives an arrow $\rho^*(\phi_{ij}):\rho^*(x_j|_{U_{ij}})\rightarrow \rho^*(x_i|_{U_{ij}})$ in $\mathcal{P}(V)$. The composition $$\rho^*(x_i|_{U_{ij}})\xrightarrow{\rho^*(\phi_{ij})}\rho^*(x_j|_{U_{ij}})\xrightarrow{\eta} \rho^*(x_j|_{U_{ij}})\xrightarrow{\rho^*(\phi_{ij})^{-1}}\rho^*(x_i|_{U_{ij}})$$ gives an arrow $\rho^*(x_i|_{U_{ij}})\rightarrow \rho^*(x_i|_{U_{ij}})$.

As $x_i|_{U_{ij}}=\alpha_i^*(x_i)$, this gives an arrow $\rho^*(\alpha_i^*(x_i))\rightarrow \rho^*(\alpha_i^*(x_i))$, which is same as $(\alpha\circ \rho)^*(x_i)\rightarrow (\alpha\circ \rho)^*(x_i)$ an element in $G_i(V)=\text{Hom}((\alpha_i\circ \rho)^*(x_i),(\alpha_i\circ \rho)^*(x_i))$.

This gives map $G_j(V)\rightarrow G_i(V)$ given by $\eta\mapsto \rho^*(\phi_{ij})^{-1}\circ \eta\circ \rho^*(\phi_{ij})$. This by abuse of notation, can be denoted by $\eta\mapsto \phi_{ij}^{-1}\eta\phi_{ij}$.

• I was trying to see $G_i$ as a group bundle and was trying to see $\lambda_{ij}$ as morphism of group bundles just because it was given like that in the article. They could have simply given $\lambda_{ij}$ as morphism of sheaves. I do not see the point of creating little confusion for some one reading for the first time. – Praphulla Koushik May 11 '18 at 18:07
• This seems to be essentially correct, but your composition has $x_i$ in the third place, where it should be $x_j$. – S. Carnahan May 12 '18 at 10:52
• Thank you for your reading and pointing out typo. :) – Praphulla Koushik May 12 '18 at 11:18