# 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.

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 $$d$$.