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I have two functions acting on $\mathbb{R}^2$, say $f(x,y)=(x+\frac{1}{2},-y)$ and $g(x,y)=(x,y+1)$. I want to find the subgroup of the group of all self-homeomorphisms of $\mathbb{R}^2$ that is generated by $f$ and $g$.

At first I thought this would just be the free group on two generators (because of the context, not relevant here), but it turned out that $fgf^{-1}g$ is the identity.

How can I know that I have found "all" necessary relations? Are there any general tips and tricks that can be used in situations like these?

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When you say "I thought this would just be", what is "this"? The subgroup of functions on $\mathbb{R}^2$ with composition generated by $f$ and $g$? –  Alex Becker Apr 26 '11 at 2:53
    
Ops, I forgot to be precise. Yep, you're right. I've added an explanation now. –  Fredrik Meyer Apr 26 '11 at 3:04
    
@Fredrik: And by "linear functions" you don't mean "linear transformations". Do you just mean that each coordinate is given by a polynomial of degree at most 1 in $x$ and $y$? –  Arturo Magidin Apr 26 '11 at 3:07
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The problem is hard enough when you "know" the target group (e.g., proving that a given presentation is the trivial group cannot be done purely algorithmically). Here, you don't even know what the group you are trying to describe is, which makes things more difficult... –  Arturo Magidin Apr 26 '11 at 3:16
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This is a really bad question. I thought you were asking how to prove this specific case but you're actually asking for a decision procedure. –  quanta Apr 26 '11 at 11:28

2 Answers 2

up vote 1 down vote accepted
  • The reflection $M:(x,y) \mapsto (x,-y)$ generates $\mathbb Z/2 \mathbb Z$ because $M^2 = I$ (note also that $M^{-1} = M$).
  • The translation $T(a,b):(x,y)\mapsto(x+a,y+b)$ is free (unless it's the identity) so it generates $\mathbb Z$ (and note that $T(a,b)^{-1} = T(-a,-b)$).

Furthermore we have $$M\circ T(a,b) = T(a,-b) \circ M.$$

Now $f = M T(1/2,0)$ and $g = T(0,1)$ so we already completely understand these elements on their own and how they commute. The result is that we can create a notion of "normalized form" which is syntactically unique for equal elements. Before defining it formally I will show an example:

$$\begin{align*} \bullet\ fgf^{-1}g &= M\circ T(1/2,0)\circ T(0,1)\circ M\circ T(-1/2,0)T(0,1)\\ &= M\circ T(1/2,0)\circ M\circ T(0,-1)\circ T(-1/2,0)\circ T(0,1)\\ &= M\circ M\circ T(1/2,0)\circ T(-1/2,0)\circ T(0,-1)\circ T(0,1)\\ &= I.\end{align*}$$

If you shuffle all the mirrors to the very left and all the y-axis translations to the very right we are left with something of the form $M^i\circ T(r\cdot 1/2)\circ T(0,s)$ where $i = 0,1$ and $r,s \in \mathbb Z$ this is a normal form in the sense that every element can be reduced to it - and if two elements are equal they have syntactically equal normal form. First of all it's obvious that we can reduce every element to this form since we know how to commute elements, secondly we can see that if two elements are equal they have equal normal form because were any $i,r$ or $s$ different we would have different elements!

The group structure is completely understood now, you could even write it as a "presentation" $\langle m,s,r\mid m^2 = I, sm = ms'\rangle$ (I'm not completely sure if I used this syntax right so don't take my word for anything). Of course what I said is quite general and you can use the idea of a normal form to deal with lots of groups in the $\mathbb R^n$.

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@Arturo, thanks for fixing the formatting! I will use that align if I want to write series of equations in future. –  quanta Apr 26 '11 at 15:20
    
No problem. We learn by watching others do better than we do. (-: –  Arturo Magidin Apr 26 '11 at 15:51

$f$ and $g$ are much more than homeomorphisms - they preserve distances, they map horizontal and vertical lines to themselves (meaning a horizontal line is mapped to a horizontal line, and the same goes for vertical, for both $f$ and $g$). You should try to understand what they do geometrically and come up with a conjecture around what that group is. For starters, what is $f^2$?

Once you guess the answer, it should be easier to prove it than to go through a search for relations. You may find lots of relations and not gain much insight into what the group is. Generally speaking, the Todd-Coxeter algorithm would be relevant to your question, but I wouldn't recommend trying it here.

If your textbook discussed "fundamental domains", this may be a useful viewpoint as well.

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I know what the group is, because I know that $\mathbb{R}^2$ quotiented by the action of this group is the Klein bottle (hence the group is $\mathbb{Z}*\mathbb{Z}/(fgf^{-1}g)$). What I am asking for, is if there is some method for deciding in general (not just with this particular example), how to know if all relations are found. –  Fredrik Meyer Apr 26 '11 at 5:21
    
@Fredrik: No, there is no algorithm. In general, if you know what the "target group" is, you can try to use van Dyck's Theorem to show the given presentation has the group you want as a quotient, and then try to establish this map is either one-to-one or has an inverse (provided the group you have is not Hopfian, that will do it). –  Arturo Magidin Apr 26 '11 at 13:53

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