7
$\begingroup$

Let $F$ be the free group of rank $2$ (or any finite rank if this does not matter). The set of injective group endomorphisms $F\to F$ forms a monoid $M$ by compositions. Is there a simple looking set of generators of $M$?

It is natural to ask this. For example if we would like to show some property $P$ holds for all injections in $M$ and is preserved under compositions, it is natural to prove $P$ for a set of generators. So by simple looking, I mean it would really save us some work by only proving $P$ for generators, for example, the generators all look similar to each other (like generating the mapping class groups by Dehn twists), or fall into finitely many (infinite) classes.

Obviously, $M$ contains $\mathrm{Aut}(F)$, which has a nice finite generating set: the elementary Nielsen transformations.

Also note that, $M$ is not finitely generated, since each homomorphism $F\to F$ induces a homomorphism on the abelianization $\mathbb{Z}^2\to \mathbb{Z}^2$, whose determinant could be any integer. Thus the determinants of any generating set of $M$ should be a generating set of the monoid $\mathbb{Z}$, where the operation is multiplication.

Any answer to the related question of generating the monoid of all (not necessarily injective) endomorphisms would be helpful as well. In the rank $2$ case, the non-injective ones are easily understood.

I don't have a good idea to think about this. Are there any related reference or theory? Thanks!

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ The set of monomorphisms $F_2 \hookrightarrow F_2$ can be parameterized by all pairs of noncommuting elements of $F_2$ (if $n\neq 2$ then the description is more complicated.) You could even put a monoid structure on this set by defining multiplications to be composition. Your observation about determinants is interesting and seems to indicate that there is some notion of "prime" or minimal monomorphisms. I'm not aware of any simplifying characterization for such elements, in fact, at some point, I was thinking about questions related to the density of such elements. $\endgroup$
    – NWMT
    Sep 23, 2019 at 15:04
  • $\begingroup$ A remark: iif $S$ is a generating subset then $S\cap\mathrm{Aut}(F)$ is a generating subset (because the set of non-surjective injections is an ideal). This also true for the whole monoid of endomorphisms (using that surjective endomorphisms are injective). So, the study of generating subsets of the automorphism group being granted, one can focus on generating subsets of the form $\mathrm{Aut}(F)\cup T$. $\endgroup$
    – YCor
    Nov 3, 2019 at 19:02
  • $\begingroup$ Another remark: Another submonoid, containing $\mathrm{Aut}(F)$, whose complement is an ideal is the submonoid $M^*$ of endomorphisms whose abelianization has nonzero determinant. Hence for every generating subset $S$ of $\mathrm{End}(F)$ (or of $\mathrm{End}_{\mathrm{inj}}(F)$, the intersection $S\cap M^*$ generates $M^*$. Hence the question of finding a generating subset for $M=\mathrm{End}_{\mathrm{inj}}(F)$ is (in principle) strictly harder than the one for $M^*$. (BTW I don't remember whether an endomorphism whose determinant is $\pm 1$ is an automorphism.) $\endgroup$
    – YCor
    Nov 3, 2019 at 19:06
  • $\begingroup$ @YCor, thank you for your comments! Regarding the first one, yes, it is really about finding generators outside $\mathrm{Aut}(F)$. For the second one, this is a good point. There are endomorphisms with determinant $1$ that are not automorphisms, for example $x\mapsto x^2 y, y\mapsto yx$. Right now, I don't see an obvious way to generate $M^*$. $\endgroup$
    – Lvzhou Chen
    Nov 5, 2019 at 2:00

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy