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Question: What surface is represented by $a_1 a_2 \cdots a_n a_1^{-1} a_2^{-1} \cdots a_n^{-1}$?

Attempt: I'm not sure where to even begin except I can make a few of the following observations:

EDIT: I think these two observations are non-sensical.

(1) If $G$ is an abelian group, the relation above trivially holds.

(2) With $n=2$ we have that the relation above holds if and only if $a_1$ and $a_2$ commute.

How should I proceed from here?

Michael Hardy
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user125103
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  • What does "represented by" mean? – Hugh Thomas Mar 25 '14 at 02:29
  • Here I mean $G$ is represented by $a_1 a_2 \cdot \ldots \cdot a_n a_1^{-1} a_2^{-1} \cdot \ldots \cdot a_n^{-1}$ if and only if $G = \left\langle a_1, a_2, \ldots, a_n \mid a_1 a_2 \cdot \ldots \cdot a_n a_1^{-1} a_2^{-1} \cdot \ldots \cdot a_n^{-1} \right\rangle$ – user125103 Mar 25 '14 at 02:35
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    That doesn't quite seem to mesh with what you wrote in your question: in your question, you are talking about a surface, while what you have just written defines a group. – Hugh Thomas Mar 25 '14 at 02:38
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    The question which someone asked you in a comment about your question on the Klein bottle (http://math.stackexchange.com/questions/725555/showing-that-left-langle-a-b-mid-abab-1-right-rangle-cong-pi-1k-cong) is also relevant: I think what you are writing here as "represented by" is exactly the sense in which you get $abab^{-1}$ from the standard picture of the Klein bottle. An introductory text on algebraic topology would probably be more helpful than having someone try to explain this specific question. (Also, my memory of this is foggy.) – Hugh Thomas Mar 25 '14 at 02:47
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    To try to say something a bit helpful: a word such as $a_1a_2\dots a_n a_1^{-1}\dots a_n^{-1}$ is telling you how to glue the edges of a 2n-gon together to form a surface without boundary. In this case, you are going to glue together the 1st and n+1-st edges, the 2nd and n+2-th, etc, because you see the same letter in the first and n+1-st position, etc. The inverses tell you that the orientation of the edges is clockwise (uninverted) the first n times, and counter-clockwise (inverted) the other n times. You glue the edges so the orientations match. The question is: what surface results? – Hugh Thomas Mar 25 '14 at 02:58
  • So in case $n = 2$, we have a square whose top side is marked $a_1$, right side is marked $a_2$, bottom side is marked $a_1^{-1}$, and left side is marked $a_2^{-1}$. We then take $a_1$ and match it with $a_1^{-1}$ s.t. $a_1^{-1}$'s orientation is inverted. We then do the same with $a_2$ and $a_2^{-1}$. We then think in the general case where $n$ is arbitrary and ask what surface is represented. Then is this what the question is asking? I'm having trouble visualizing the case even when $n=2$, but hopefully this will put me on the right track for what the question is asking. – user125103 Mar 25 '14 at 20:49
  • In the case n=2, you get a torus. It's just like you're taping the opposite sides of the square together to make a cylinder, and then taping the ends of the cylinder together to make a torus. – Hugh Thomas Mar 26 '14 at 00:45
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    What about the inversions? Wouldn't a torus be $a_1a_2a_1a_2$ as opposed to $a_1 a_2 a_1^{-1} a_2^{-1}$? – user125103 Mar 26 '14 at 02:22

2 Answers2

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The easiest way to determine the surface is the Euler characteristic. The surfaces are all orientable, since no the edge $a_i$ is not glued with a twist. So one can use the formula $2-2g=V-E+F$, where $g$ is the genus (number of holes) and $V,E,F$ count the number of vertices, edges and faces respectively. The number of edges and faces is easy to see from a picture $E=n$ and $F=1$. The number of vertices is trickier. There are $2n$ apparent vertices in the picture, but these are identified. If you follow around the identifications, you can see that if $n$ is odd, there are $2$ vertices (represented by alternate vertices as you go around the $2n$-gon) and if $n$ is even, there is just $1$ vertex. So, solving for $g$, we see that $g=\frac{n-1}{2}$ is $n$ is odd, and $g=\frac{n}{2}$ if $n$ is even.

Cheerful Parsnip
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For $n=2$, the square looks like this

. ->- .
|     |
v     v
. ->- .

If you start in the top left corner and you go around the square, first you see two arrows that go clockwise, then you see two that go counter-clockwise. When you glue them, you match the arrows. So the result is a torus.

Hugh Thomas
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  • I see -- thanks for clarifying. Roughly -- it seems that if $n=4$ you get an object that can be stretched into a taurus as well. If this is true my guess is that for even choices of $n$ you get the taurus. Am I on the right track? I'm not sure about odd choices of $n$. – user125103 Mar 26 '14 at 02:36
  • I don't think your statement for even $n$ is true. I believe the answer is that you always get a handlebody with $n-1$ holes, but it's not so easy to visualize (at least for me). Perhaps it's easier to start with a handlebody with two holes and try to cut it apart into a hexagon with opposite sides identified. An easier general strategy is to use homology and the classification of closed surfaces. But really, this is a very standard topic in elementary algebraic topology, so I strongly recommend that you look in a text like Massey. – Hugh Thomas Mar 26 '14 at 02:51