2
$\begingroup$

Let $A$ be a group. Then we choose that $B$ is a subgroup of $A$.

Let us write the cohomology group cocycle of $A$, as $\alpha_{d}(\{a\}) \in H^d(A,U(1))$ where $\{a\}$ is a shorthand for a set of $a_j \in A$ for $j=1,2, \dots$.

(1) For any given subgroup $B$ (here $B$ may not be a trivial group), what are the conditions on the subgroup $B \subset A$ that the injective group homomorphism $$B \overset{i}{\to} A$$ such that $$ \alpha_{d}(\{b\})$$ is a coboundary (namely, a trivial element) in $H^d(B,U(1))$?

-

(2) What are the conditions that there exists a group homomorphism $$A \overset{f}{\to} B$$ where we name $f(a)=b$ for any $a \in A$ and some $b \in B$, such that $$\alpha_{d}(\{f(a)\})= \alpha_{d}(\{b\})$$ is a coboundary (namely, a trivial element) in $H^d(B,U(1))$?

We can consider $A$ and $B$ are finite groups.

p.s. Bonus questions: What if $A$ and $B$ are Lie groups?

Improve this question
$\endgroup$
10
  • 4
    $\begingroup$ Are you talking about the map in cohomology induced by the inclusion $B\subseteq A$? I'm a little confused by the homomorphism $f:A\to B$, which induces a map in the other direction. $\endgroup$
    – Mark Grant
    Commented Feb 17, 2017 at 7:27
  • 1
    $\begingroup$ Right. Usually if one has a subgroup $B\leq A$ and a homomorphism $f:A\to B$, one insists that $f$ is a retraction, meaning that $f|_B=\mathrm{id}_B$. Otherwise, what's the point of thinking of $B$ as a subgroup? In this case, writing $i$ for the inclusion $B\to A$, we have $i^*\circ f^*=\mathrm{id}_B^*$, so $f^*$ is necessarily injective. $\endgroup$
    – HJRW
    Commented Feb 17, 2017 at 11:32
  • $\begingroup$ Thank you for the nice comments. I am not 100% sure which way works, thus I take the suggestion and add a new possibility (1). $\endgroup$
    – miss-tery
    Commented Feb 17, 2017 at 16:35
  • $\begingroup$ For (1) you can always take $B$ to be the trivial group, provided $d>0$. On the other hand, (2) does not make sense, as cohomology is contravariant on the group. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 17, 2017 at 16:43
  • $\begingroup$ For (1), there are other subgroups $B$ of $A$, such that $B$ is not a trivial group, but (1) still holds. $\endgroup$
    – miss-tery
    Commented Feb 17, 2017 at 16:50

1 Answer 1

Reset to default
8
$\begingroup$

Here is one possible sufficient condition for (1). Suppose $B$ is normal in $A$, then you have an extension of groups $$ 1 \to B\stackrel{i}{\longrightarrow} A \stackrel{q}{\longrightarrow} A/B\to 1, $$ where $q:A\to A/B$ is the map to the quotient group. Then $\alpha\in H^d(A;M)$ will have $$0=i^*\alpha\in H^d(B;M)$$ if there exists some $$\gamma\in H^d(A/B;M)$$ such that $\alpha=q^*\gamma$. (Here I am assuming that the coefficient module $M$ comes from a module over $A/B$, or equivalently, it becomes trivial when regarded as a $B$-module.)

When $d=1$ this sufficient condition is also necessary, as can be seen by looking at the 5-term exact sequence of low-dimensional cohomology groups coming from the Lyndon-Hochschild-Serre spectral sequence of the extension. The same spectral sequence also hints that necessary conditions may be hard to come by in higher dimensions.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks, +1, this is a good type of answer that I am happy to receive. $\endgroup$
    – miss-tery
    Commented Feb 18, 2017 at 20:16
  • 2
    $\begingroup$ @Mark Grant, Actually, why do you wrote $0=i^*\alpha\in H^d(A;M)$ instead of $0=i^*\alpha\in H^d(B;M)$? Should it be trivial in the cohomology group of $B$? (Is that the same as trivial in the cohomology group of $A$?) $\endgroup$
    – wonderich
    Commented Feb 18, 2017 at 20:34
  • $\begingroup$ @wonderich: good catch. Now corrected. $\endgroup$
    – Mark Grant
    Commented Feb 18, 2017 at 22:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .