2
$\begingroup$

I'm working on a problem about Artin groups, and to simplify this problem I want to take a quotient that allow us to go to an easier Artin group, but I'm not sure if the quotient is well defined. This is the statement of my problem:

Let $A_\Gamma$ be an Artin group, where $\Gamma$ is a complete graph, $f:A_\Gamma\to\mathbb{Z}$ a group homomorphism and $a\in V(\Gamma)$ such that $f(a)=1$. The artin group is generated by the vertices of the graph, and there are two types of generators: the $v\in V(\Gamma)$ such that $f(v)=0$ and the $v\in V(\Gamma)$ such that $f(v)\neq0$. My objective is to define a normal subgroup $N\trianglelefteq A_\Gamma$ such that $A_\Gamma/N\cong A_{\Gamma'}$ where $\Gamma'\subset\Gamma$ is the induced subgraph by the $v\in V(\Gamma)$ such that $f(v)=0$ and $a$, i.e. we want a new Artin group given by colapsing all vertices with non-zero image to $a$. To do so, I wanted to define $N$ as the normal subgroup generated by the set: $$\lbrace v^{-1}a^{f(v)} \mid v\in V(\Gamma),f(v)\neq0\rbrace$$ (Remark: I $N\leq\ker(f)$ if possible, that is why I need to define those ''weird'' generators instead of $v^{-1}a$) It is obvious that in the quotient $A_\Gamma/N$ all the $v\in V(\Gamma)$ with $f(v)\neq 0$ are identified with some power of $a$, which allow us to get rid of those generators. However, I can't see if this quotient is indeed the Artin group I want or sth different.

Any hint will be thanked.

Edit: Inspired by @MoisheKohan I was able to see that I can reduce the problem to the right-angled case by first taking a quotient with the normal subgroup generated by the commutators of the generators. In this situation, since $\Gamma$ is a complete graph $A_\Gamma\cong\mathbb{Z}^n$, where $n$ is the number of generators of $\mathbb{Z}$, so the problem should be easier.

$\endgroup$
4
  • 1
    $\begingroup$ You did not specify what type of Artin groups you are working with: do you assume right-angled? $\endgroup$ Apr 17, 2023 at 22:22
  • $\begingroup$ @MoisheKohan I wasn't working with right-angled groups, but I managed to simplify the problem to this case. Hope it helps to solve the problem. $\endgroup$
    – Marcos
    Apr 18, 2023 at 9:47
  • $\begingroup$ What's with the $a^{f(v)}$ thing? You say you want to identify all the "living" $v$ with $a$, do you mean with a power of $a$? Or should your set of normal generators for $N$ consist of $v^{-1}a$ rather than $v^{-1}a^{f(v)}$? $\endgroup$ Apr 18, 2023 at 10:50
  • $\begingroup$ @MattZaremsky I wanted to do this because I want $N\leq \ker(f)$. And yes, then I have identified $v$ with a power of $a$, but that is not a problem for my purposes, since this is enough to get rid of the generator $v$. I'll edit the question. $\endgroup$
    – Marcos
    Apr 18, 2023 at 11:02

1 Answer 1

3
$\begingroup$

(Edit: Missed that the graph should be complete, now it is.)

If I'm understanding all this correctly, the answer is "no". Take three vertices, $a$, $b$, and $c$. Take an edge labeled 4 from $a$ to $b$, an edge labeled 4 from $a$ to $c$, and an edge labeled 2 from $b$ to $c$. Take $f$ to be $f(a)=f(b)=1$, $f(c)=0$. So now you're modding out that $a$ and $b$ get identified. In this quotient, $a$ and $c$ suddenly commute, which is not what the induced subgraph $\Gamma'$ wants (it's $a$ and $c$ with an edge labeled 4). If you want to phrase it in terms of quotient maps instead of normal subgroups, the problem is that the quotient map isn't well defined: you have that $b$ and $c$ commute, but their images are $a$ and $c$, which don't commute.

$\endgroup$
6
  • $\begingroup$ Thanks for your time! However, one of the hypothesis I haveis that $\Gamma$ is a complete graph, so when passing to the RAAG case this is equivalent to say that $A_\Gamma=\mathbb{Z}^n$. Since I've added the RAAG hypothesis later I forgot to mention that, sorry. $\endgroup$
    – Marcos
    Apr 18, 2023 at 11:13
  • $\begingroup$ Well, now I think about this it seems trivial that we can take a quotient of $\mathbb{Z}^n$ to get $\mathbb{Z}^m$ getting rid of the generators we want, am I right? $\endgroup$
    – Marcos
    Apr 18, 2023 at 11:23
  • $\begingroup$ Oh! Complete graph, yeah I missed that. But then it's just a question about free abelian groups, and, sure, there it's trivial. (But can you really reduce to the RAAG case while maintaining the graph? If there are odd labels on edges then things collapse when modding out commutators of generators. Like, aba=bab plus ab=ba equals a=b.) $\endgroup$ Apr 18, 2023 at 11:26
  • $\begingroup$ Actually I've written my question poorly. My idea was to make succesive quotients until I got to a group of this type, since I know how to work there and the propery I want passes to quotients, so it is not a problem if I lose generators in the process (and, of course I loose generators for the odd edges). Thanks for your help!! $\endgroup$
    – Marcos
    Apr 18, 2023 at 11:33
  • 1
    $\begingroup$ Thanks! I'm partially ok with that. Its nice to know that it is not possible in general, so I have to be careful about what I want to do. thanks! $\endgroup$
    – Marcos
    Apr 18, 2023 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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