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Let $k$ be a field and let $G$ be an algebraic group over $k$.

I encountered the following notion in an article:

"Let $\psi: \text{Gal}(\overline{k}/k) \to \text{Int}(\overline{G})$ be a cocycle. The twisted group $_\psi G$..."

How is this "twisted group" defined?

flag	asked Apr 19 2012 at 22:43 Wanderer 2,977●12●28

	... dare we presume there were more words of context? – some guy on the street Apr 20 2012 at 2:17
	Which article had this sentence? What is $\operatorname{Int}(\overline{G})$? – S. Carnahan♦ Apr 20 2012 at 3:11
	presumably int(g bar) is the group of inner automorphisms of the base change of G to a separable closure of k (and psi is a 1-cocylce etc...) – Peter McNamara Apr 20 2012 at 4:59
	Yes indeed, I thought this would be clear. The article is Borovoi's 1993 Duke paper, "Abelianization of the second nonabelian Galois cohomology". – Wanderer Apr 20 2012 at 8:50

Gjergji Zaimi · Accepted Answer · 2012-04-21 00:52:36Z UTC

You can find the definitions in Serre's book "Galois cohomology". The relevant sections are I.5.3 - I.5.7. The idea is that if you have a group $G$ acting on a group $A$ and a cocycle $\psi\in H^1(G,A)$. The twisted group $_\psi A$ has the same group structure as $A$ but it has a twisted action by $G$, which is given by $$g*a = \psi(g) \cdot (g \cdot a) \cdot \psi(g)^{-1}.$$ There is also a previous question which asked if there is an analog of this for outer automorphisms: Substitute for Serre’s twisting when the “twisting” is outer

Definition of “twisted group”

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