1
$\begingroup$

I have a set of matrices $A_1,\ldots,A_n$. Let $\mathcal{A} = \{A_i\}$.

What are some simple conditions under which for any sequence of indices between $1$ and $n$, $a_1,\ldots,a_m$, the product

$$\prod_{k=1}^m A_{a_k}$$

uniquely determines the sequence of indices?

This is clearly related to the idea of free product in group theory. Following my question here: https://math.stackexchange.com/questions/823194/under-what-conditions-a-product-of-matrices-is-the-identity-matrix-more-complic I took a look at de la Harpe's book (Topics in Geometric Group Theory), but did not find a satisfying answer (reading the chapter about free product, chapter 2).

The main issue is that I am not sure how to reduce this set of matrices into a free product. According to the book, the free product is created by taking disjoint union of elements of a family of groups. The construction is not clear for my case.

I was also advised to look at the Ping-Pong lemma (or the Table-Tennis lemma), and that one seems not very related either, especially since it talks about a free product generated by only two disjoint groups. If we take each $A_i$ to be the generating element of a group, and take the disjoint union of all of these groups, then the Table-Tennis lemma is way too simple.

I get the sense that if we think of a product by a matrix as taking us a step in some space, then the requirement is that there can't be a case for which $A_i = \prod_{j=1}^k A_{b_j}$ for some sequence of indices $b_1,\ldots,b_k$ which is not just $i$. But that's only a necessary condition? (and it is too close to the condition that no sequence of products equals to another sequence of products, which is both necessary, sufficient and makes the question trivial.)

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Wikipedia has a statement of the Ping-Pong lemma for several subgroups. $\endgroup$
    – Jeremy Rouse
    Commented Jul 1, 2014 at 14:54
  • $\begingroup$ I do not understand the question: It seems you find the statement of ping-pong lemma "too simple" and at the same time is asking for help using it. $\endgroup$
    – Misha
    Commented Jul 1, 2014 at 16:22
  • $\begingroup$ @Misha I looked at the Ping-Pong theorem on Wikipedia, and am I right in thinking that it is presented more generally there than in de la Harpe's book? It might be the case that Wikipedia's version could be used somehow for my cause. I will get back with other questions if I have any. thanks for the pointer. $\endgroup$
    – kloop
    Commented Jul 1, 2014 at 17:42
  • 1
    $\begingroup$ You just need ping pong for free product of several (n) infinite cyclic groups. Wikipedia is a good place to start learning about it. $\endgroup$
    – Misha
    Commented Jul 1, 2014 at 19:32
  • 2
    $\begingroup$ The answer strongly depends on whether your matrices are invertible or not. If they are, then the question is equivalent to asking whether the group they generate is free, and techniques like the ping-pong lemma should apply. If they are not all invertible, then you are asking whether the monoid that they generate is free, which is a more difficult problem and undecidable even for 3x3-matrices with integer coefficients, see this paper and references therein: arxiv.org/abs/0808.3112 This implies that there cannot exist any simple necessary and sufficient conditions. $\endgroup$
    – Tobias Fritz
    Commented Oct 29, 2014 at 20:38

1 Answer 1

Reset to default
-1
$\begingroup$

For $Gl_n$, I like to use the group presentation found in Springer's Linear Algebraic Groups chapters 8-9. Basically the generators are digonal matrices and unipotent matrices (triangular with all 1's on diagonal). If you choose your $A_1, ..., A_n$ to be a nice subset of the generators, then often elements will have a unique expression as a product of those generators. Decomposition theorems (e.g. Bruhat) are also useful. So here you may want to restrict your $m$ to be less than $n$ say.

Am I understanding the question right - you have a fixed set of $n$ matrices and want the (infinitely many) possible products to be distinct?

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ This does not answer the question. $\endgroup$
    – Misha
    Commented Jul 1, 2014 at 19:33

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