# Algebraic 1-cocycles and Galois gerbs

We have the following set up: $$K/F$$ is Galois, $$D$$ is an algebraic group of mult. type and $$E$$ is an extension of groups: $$1\to D(K)\to E\to Gal(K/F)\to 1$$ Now take a linear algebraic group G over F. An algebraic 1-cocycle is a map $$w \mapsto x_w$$ from $$E$$ to $$G(K)$$ s.t. $$x_{w_1 w_2}= x_{w_1} w_1(x_{w_2})$$, where $$E$$ acts on $$G(K)$$ via $$E\to Gal(K/F)$$.

Furthermore we have a Galois action on morphisms $$v:D(K)\to G(K)$$ via $$\sigma(v)(d)=\sigma(v(\sigma^{-1}(d))$$ (cf here section 2.2/2.3 for more details).

In the above paper it is claimed then that for $$w\in E$$ which maps to $$\sigma$$ $$x_w\cdot\sigma(v)\cdot (x_w)^{-1}=v$$ as a cosequence of the cocycle condition. Here $$v$$ is the restriction of the cocycle to $$D$$ (i.e. $$v(d)=x_d$$).

But my calculation only gives me: $$(x_w\cdot\sigma(v)\cdot (x_w)^{-1})(d)=x_w\cdot\sigma(x_{\sigma^{-1}(d)})\cdot (x_w)^{-1}=x_{w\sigma^{-1}(d)}\cdot \sigma( x_{w^{-1}}) =x_{w\sigma^{-1}(d)w^{-1}}$$ and I am not able to conclude the desired result.

## 1 Answer

It seems to me that Kottwitz's review of the notion of Galois gerb used by Langlands and Rapoport was not meant to be competely self-contained, and that it was meant only to highlight differences of a technical nature. In particular---even though he does not review this point---for any Galois gerb $$E$$, the induced conjugation action of $$G(K/F)$$ on $$D(K)$$ is required to coincide with the standard Galois action. In other words, given an element $$w$$ in $$E$$ that maps to $$\sigma$$ in $$G(K/F)$$, we require that$$wdw^{-1} = \sigma(d)$$for all $$d$$ in $$D(K)$$. Here the left-hand side is conjugation in the group $$E$$, while the right-hand side is the standard Galois action of $$\sigma$$ on $$D(K)$$. The requirement that this equality hold is part of the definition of Galois gerb.

For this reason the subscript$$w\sigma^{-1}(d)w^{-1}$$occurring at the far right end of your calculation simplifies to $$d$$.

• Yes, I already assumed these two actions to coincide, but I was not sure if this was just per definition. – Notone Oct 31 '18 at 19:49