Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a module $A$ for a group $G$, and a subgroup $P\leq G$ with unipotent radical $U$, I have encountered the notation $[A,U]$ in a paper. Is this a standard module-theoretic notation, and if so, what does it mean.

In the specific case I am looking at, it works out that $[A,U]$ is equal to the submodule of the restriction of $A$ to $P$ generated by the fixed-point space of $A$ with respect to $U$, but whether this is the case in general I do not know. If anyone could enlighten me on this notation, it would be greatly appreaciated.

share|improve this question
1  
If U normalizes A, then [A,U] is the subgroup of A generated by the commutators [a,u] = a^-1 a^u, or in additive notation, -a + a*u = a*(-1+u). This not typically equal to the fixed point space C_U(A) = { a in A : [a,u]=1 }, but if U is unipotent then it is not impossible (though still a little unlikely). In other words, look at the semi-direct product of U and A and calculate as usual. –  Jack Schmidt Apr 12 '12 at 11:19
1  
every module is a group, and in particular, a module for a group algebra is a homomorphism from the group into the automorphism group of the underlying group of the module. this is exactly what is needed for the semi-direct product. –  Jack Schmidt Apr 12 '12 at 12:23
    
Sorry, I've just realised how silly my previous (now deleted) comment was, and appreciate now what you are saying. Thank you for your response. –  David Ward Apr 12 '12 at 13:03
    
Having checked the calculations involving the specific modules and groups involved, and the semi-direct product produces the correct result. Thank you once again for the help. –  David Ward Apr 12 '12 at 13:31
    
No problem :-) It is not too uncommon in finite group theory for a G-module to be an abelian quotient of a subgroup of some group containing G. In other areas, especially module theory or algebraic groups, one might be more interested in less "internal" modules, and so it could seem pretty strange to internalize them like this. –  Jack Schmidt Apr 12 '12 at 14:08

1 Answer 1

up vote 2 down vote accepted

In group theory it is standard to view $G$-modules $A$ as embedded in the semi-direct product $G \ltimes A$. Inside the semidirect product, the commutator subgroup $[A,U]$ makes sense for any subgroup $U \leq G$ and since $A$ is normal in $G \ltimes A$, we get $[A,U] \leq A$ and by the end we need make no reference to the semi-direct product. If we let $A$ be a right $G$-module written multiplicatively so that the $G$ action is written as exponentiation, then $$[A,U] = \langle [a,u] : a \in A, u \in U \rangle = \langle a^{-1} a^u : a \in A, u \in U \rangle$$

If you have a left $A$-module written additively and $G$-action written as multiplication, then we get $$[A,U] = \langle a - u\cdot a : a \in A, u \in U \rangle = \sum_{u \in U} \operatorname{im}(1-u)$$ which is just the sum of the images of $A$ under the nilpotent operators $1-u$, which is probably a fairly interesting thing to consider.

In some sense this is the dual of the centralizer: $A/[A,U]$ is the largest quotient of $A$ centralized by $U$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.