IIP 2-cocycle for trivial group action

BEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]

For its use outside the wiki, please define the term when using it. If you are aware of an equivalent standard term, please leave a comment on the talk page
VIEW: Definitions built on this | Facts about this: (facts closely related to IIP 2-cocycle for trivial group action, all facts related to IIP 2-cocycle for trivial group action) |Survey articles about this | Survey articles about definitions built on this
VIEW RELATED: Analogues of this | Variations of this | Opposites of this |
Learn more about terminology local to the wiki | view a complete list of such terminology

Definition

Suppose $G$ is a group, $A$ is an abelian group. Consider the following three conditions for a function $f:G\times G\to A$:

Condition name	Expression for condition
2-cocycle for trivial group action	$f(g,hk)+f(h,k)=f(gh,k)+f(g,h)\ \forall \ g,h,k,\in G$
identity-preserving, also known as "normalized"	$f(g,1)=f(1,g)=0$ for all $g\in G$, and $1$ and $0$ are the identity elements of $G$ and $A$ respectively
inverse-preserving	$f(g,g^{-1})=0$ for all $g\in G$

$f:G\times G\to A$ is termed an IIP 2-cocycle for trivial group action if it satisfies the following equivalent conditions:

1. $f$ satisfies all three of the above conditions: it is a 2-cocycle, it is identity-preserving (normalized) and it is inverse-preserving.
2. $f$ is a 2-cocycle and it is inverse-preserving (the identity-preserving / normalized condition is not part of this definition).
3. $f$ corresponds to a system of coset representatives for a central extension (element of $H^{2}(G,A)$, the second cohomology group for trivial group action) where the representative for the identity element is the identity element and the representative of the inverse of an element is the inverse of its representative.

The group of IIP 2-cocycles is denoted $Z_{IIP}^{2}(G,A)$.

Note that the term IIP is not universally used. In some contexts, terms such as standard 2-cocycle may be used for IIP 2-cocycle.

Equivalence of definitions

To prove equivalence of definitions (1) and (2), we need to show that the condition of being normalized is redundant when the other two conditions are present. This can be shown using 2-cocycle for trivial group action is constant on axes, which says that if $f$ is a 2-cocycle, $f(g,1)=f(1,g)=f(1,1)\ \forall \ g\in G$.

Specifically, here's how we use the fact. From the inverse-preserving condition, setting $g=1$ in it, we get $f(1,1)=0$. Now we use the fact to show that in fact $f(g,1)=f(1,g)=f(1,1)$ also has to be $0$, establishing that $f$ is normalized.

The equivalence with definition (3) is discussed in more detail in the #Relation with central extensions section below.

Importance

Converting a 2-cocycle to an equivalent IIP 2-cocycle (differing by a 2-coboundary)

It's not always possible to convert a 2-cocycle to an equivalent IIP 2-cocycle. However, it is possible in either of these cases:

• $G$ has no element of order two.
• $A$ is 2-divisible, i.e., every element has a half. The half need not be unique.

In particular, it's always possible to find an IIP 2-cocycle if either $G$ or $A$ is an odd-order group.

Case that G has no element of order two

Start with $f:G\times G\to A$ that is normalized (we can start normalized without loss of generality, since any 2-cocycle can be converted to a normalized 2-cocycle by subtracting $f(1,1)$).

Define $\alpha :G\to A$ as follows:

• $\alpha (1)=0$.
• Divide the non-identity elements of $G$ into pairs of mutually inverse elements (that are always distinct from each other because there is no element of order two). We define $\alpha$ on these pairs one at a time. For any pair $(g,g^{-1})$. Pick $\alpha (g)$ and $\alpha (g^{-1})$ so that they sum up to $f(g,g^{-1})$.

Denote by $d\alpha$ the coboundary of $\alpha$. In other words:

$(d\alpha )(g,h)=\alpha (g)+\alpha (h)-\alpha (gh)$

$f-d\alpha$ therefore differs from $f$ by a 2-coboundary. We want to show that $f-d\alpha$ is IIP. It's already a 2-cocycle by construction, so we only need to show the inverse-preserving condition.

To see this, let's evaluate:

$(f-d\alpha )(g,g^{-1})=f(g,g^{-1})-(d\alpha )(g,g^{-1})=f(g,g^{-1})-(\alpha (g)+\alpha (g^{-1})-\alpha (1))$

We can make two cases:

• $g=1$: In this case, all terms on the right side simplify to 0, because $f$ is already normalized, and $\alpha (1)=0$ by construction.
• $g\neq 1$: In this case, by construction, $\alpha (g)+\alpha (g^{-1})=f(g,g^{-1})$ and $\alpha (1)=0$, so the right side simplifies to $0$.

We've thus shown that $(f-d\alpha )(g,g^{-1})=0$, demonstrating the inverse-preserving condition.

Case of 2-divisible base

Start with $f:G\times G\to A$ that is normalized (we can start normalized without loss of generality, since any 2-cocycle can be converted to a normalized 2-cocycle by subtracting $f(1,1)$).

Define $\alpha :G\to A$ as follows: $2\alpha (g)=f(g,g^{-1})$, with the following additional limitation: we always select $\alpha (g)$ and $\alpha (g^{-1})$ to be equal. We can do this because $f(g,g^{-1})=f(g^{-1},g)$ because 2-cocycle for trivial group action is symmetric between element and inverse.

Note that if $A$ is 2-divisible but not uniquely 2-divisible, $\alpha$ may not be uniquely defined, but there should still exist at least one $\alpha$ satisfying the above limitation.

Denote by $d\alpha$ the coboundary of $\alpha$. In other words:

$(d\alpha )(g,h)=\alpha (g)+\alpha (h)-\alpha (gh)$

$f-d\alpha$ therefore differs from $f$ by a 2-coboundary. We want to show that $f-d\alpha$ is IIP. It's already a 2-cocycle by construction, so we only need to show the inverse-preserving condition.

To see this, let's evaluate:

$(f-d\alpha )(g,g^{-1})=f(g,g^{-1})-(d\alpha )(g,g^{-1})=f(g,g^{-1})-(\alpha (g)+\alpha (g^{-1})-\alpha (1))$

Per our construction, $\alpha (g)=\alpha (g^{-1})$, so this simplifies to:

$f(g,g^{-1})-2\alpha (g)+\alpha (1)$

Since $f$ is normalized, $f(1,1)=0$, so $2\alpha (1)=0$ forcing $\alpha (1)=0$. This gives:

$f(g,g^{-1})-2\alpha (g)$

We now use that $2\alpha (g)=f(g,g^{-1})$ by definition, so this simplifies to zero.

We've thus shown that $(f-d\alpha )(g,g^{-1})=0$, demonstrating the inverse-preserving condition.

Relation with central extensions

This subsection looks at 2-cocycles in terms of their interpretation in terms of central extensions. It leads to a restatement / rediscovery / different perspective on some of the results of the preceding subsections.

Let $E$ be a group with a central subgroup isomorphic to (and explicitly identified with) $A$, and a quotient isomorphic to (and explicitly identified with) $G$, such that the induced action of the quotient on the subgroup (in the sense of action by conjugation, see quotient group acts on abelian normal subgroup). Let $S$ be a system of coset representatives for $G$ in $E$ with $s:G\to S$ being the representation map. Then, define $f:G\times G\to A$ such that

$s(gh)=f(g,h)s(g)s(h)$

In other words, $f$ measures the extent to which the collection of coset representatives fails to be closed under multiplication.

Such an $f$ is a 2-cocycle for trivial group action of $G$ on $A$. The different possible 2-cocycles for a given central extension form a single coset of the subgroup of 2-coboundaries, and hence a single element of the second cohomology group for trivial group action $H^{2}(G,A)$.

The key insight is as follows:

The 2-cocycle is an IIP 2-cocycle if and only if both these conditions are satisfied: (1) $s(1)$ (which needs to be an element of $A$ inside $E$ is the identity element of $E$ (and hence also of $A$ as a subgroup of $E$), and (2) $s(g^{-1})=s(g)^{-1}$ for all $g\in G$.

We now revisit the two cases discussed in the preceding section where we can guarantee an IIP 2-cocycle: $G$ has no element of order two, and $A$ is 2-divisible.

Case that G has no element of order two

If $G$ has no element of order two, then it should be possible to select a system of coset representatives satisfying this condition, because the cosets for $g$ and $g^{-1}$ for non-identity $g$ will always be different, so their representatives can be chosen to be inverses of each other. Essentially, we split all the non-identity elements into pairs of mutually inverse elements, and select the coset representatives for each pair together so that the representatives are also mutual inverses.

Case of 2-divisible base

Let's show that if $A$ is 2-divisible (i.e., every element has a half) then we can get a system of coset representatives that corresponds to an IIP 2-cocycle. We've already shown that elements of $G$ that don't have order two are no trouble, so we focus on elements in $G$ of order two. The goal is to find a coset representative that also has order two.

Let's say $g\in G$ has order two. Pick a random element $u\in E$ in the coset of $G$. We have $u^{2}\in A$. Let $v\in A$ be the half of $u^{2}$. Then, $uv^{-1}$ has order two (since $A$ is central), and we can pick $uv^{-1}$ as the coset representative for $g$.

Relation with other properties

Stronger properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Difference measure	Intermediate notions
IIP 2-coboundary for trivial group action	2-coboundary for trivial group action that is also an IIP 2-cocycle			IIP subgroup of second cohomology group for trivial group action	|FULL LIST, MORE INFO
cyclicity-preserving 2-cocycle for trivial group action	takes the value $0$ whenever both elements are in a common cyclic subgroup				|FULL LIST, MORE INFO

Weaker properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Difference measure	Intermediate notions
normalized 2-cocycle for trivial group action	2-cocycle with just the identity-preserving part				|FULL LIST, MORE INFO
2-cocycle for trivial group action					Normalized 2-cocycle for trivial group action|FULL LIST, MORE INFO