# substitute for Serre's twisting when the "twisting" is outer

Does anyone know if there is something that can be said (ideally under at most very mild hypotheses) in group cohomology (let's even restrict to degree 1) that is similar to Serre's twisting, but in the case where the thing you want to twist by is not inner? I'm not expecting the statement to hold word-for-word, just that there is some general and reasonably sharp relationship between the original cohomology set and the twisted one. Thanks!

Edit in response to Chris Schommer-Pries:

A complete discussion of Serre's twisting can be found in section 5.3 of his book "Galois Cohomology". It's both a "process" and a proposition I guess. It goes as follows. Start with a group $G$ and a (not necessarily abelian) group $A$ on which $G$ acts, and form the pointed set $H^1(G,A)$. Now pick a $1$-cocycle $c \in H^1(G,A)$. Let $A_c$ denote the same abstract group $A$, but now it has the "$c$-twisted" action defined by $g*a = c(g) \cdot (g \cdot a) \cdot c(g)^{-1}$ (the 1st and 3rd $\cdot$ are operations in $A$ and the 2nd $\cdot$ is the original action of $G$). That's the "process". The proposition (cf. Prop 35 bis in the book) is that $d \mapsto d \cdot c$ (product of functions) induces a bijection of pointed sets $H^1(G,A_c) \cong H^1(G,A)$ (note that the first set has a different basepoint than the second). It is frequently used to show that some "monomorphism of pointed sets" is actually injective as a function.

Edit in response to Mariano Suárez-Alvarez:

By "inner" above I am referring to the fact that, in Serre twisting, the action of $G$ on $A$ was modified by conjugating by an element of $A$. By "outer", I mean the case where you instead replaced the action by $g*a = f(g)( g \cdot a )$ where $f : G \rightarrow Aut( A )$ is a prescribed cocycle ($Aut( A )$ has the conjgation action using the original action of $G$ on $A$).

• Could you provide some background or reference for "Serre's twisting"? Dec 29, 2009 at 21:00
• What do you mean by "inner"? Dec 29, 2009 at 22:21
• Adding... XXXXX Dec 29, 2009 at 23:01

The short answer is that there is not much to say about the relationship between $H^1(G, B)$ and a twist $H^1(G, B_c)$ where $c$ is a cocycle taking values in $Aut(B)$. (I am going to write $B_c$ for the twist instead of Serre's notation $_cB$ for the sake of easy typesetting.) You can get a good feel for what is possible by soaking in sections I.5.7 and III.1.4 of Serre's Galois Cohomology.

## Section I.5.7

One thing you can do -- as exhibited in I.5.7 -- is twist all three terms in a short exact sequence of $G$-modules and get a new short exact sequence, assuming the obvious compatibility conditions hold. Serre starts with an exact sequence

$1 \to A \to B \to C \to 1$

where $A$ is assumed central in $B$. Then he fixes a 1-cocycle $c$ with values in $C$ and twists to get an exact sequence

$1 \to A \to B_c \to C_c \to 1$.

Note that this twist $B_c$ is not an inner twist of $B$, because $c$ need not be in the image of $H^1(G, B) \to H^1(G, C)$.

This may look like a lame example, in that the twist of $B$ is "pretty close" to being inner. But already here you don't have any results regarding a connection between $H^1(G, B)$ and $H^1(G, B_c)$. That's a pretty fuzzy statement; Serre says as much as you can say with precision in Remark 1: "it is, in general, false that $H^1(G, B_c)$ is in bijective correspondence with $H^1(G, B)$."

## Section III.1.4

This section discusses your question for the specific case where $G$ is the absolute Galois group of a field $k$ and $B$ is the group of $n$-by-$n$ matrices of determinant 1 with entries in a separable closure of $k$. Serre explains what you get as $B_c$ when you twist $B$ by a cocycle with values in $Aut(B)$. You can get, for example, a special unitary group.

You can find explicit descriptions of $H^1(G, B_c)$ for some $B_c$'s in The Book of Involutions, pages 393 (Cor. 29.4) and 404 (box in middle of page). Note that for $B_c$ as in Section I.5.7, $H^1(G, B_c)$ is a group (a nice coincidence) but in the case where you get a true special unitary group, $H^1(G, B_c)$ does not have a reasonable group structure--it is just a pointed set like you expect.

• Ah, thanks very much. I probably violated some basic etiquette of MathOverflow, but I somewhat answered my own question some time ago without updating this question. My conclusion was the same as yours "not much" and my reason was related to what you mentioned later about the unitary groups. Over a p-adic field, there are at most 2 unitary groups U(n) (outer forms of GL(n)), but there are quite a few inner forms of GL(n) for large n. But also things are different over other fields. So it would seem that is a very case-dependent phenomenon. Thanks again! Jun 3, 2010 at 22:46