3

I've been reading the algebraic topology book by Massey and I've been trying to do the next exercise, from chapter 4, about the van Kampen's theorem:

Construct for every integer $n>2$ a space such that it's fundamental group is cyclic and of order $n$

From what I have seen we can do this exercise fairly easy with covering maps. The problem is that the exercise appears before the chapter that covers covering maps. I asked my teacher and he told me that you can do this exercise considering a $n$-polygon where each edge is identified as the same and we consider the same orientation.

For example the projective plane is a $2$-polygon where each edge is identified as the same and we follow the same orientation. When we consider a $n$-polygon we are obviously not treating with a surface.

How can we use van Kampen's theorem, to prove that the group is (I assume) a group of 1 generator and relations that is isomorphic to $\mathbb{Z}/n\mathbb{Z}$?

Could you please elaborate on how to do it? I have only seen the fundamental group of surfaces and I don't know how to get it on this particular example.

  • 1
    Does this answer your question? – Lee Mosher Mar 23 '24 at 13:20
  • I'm afraid not, I am looking for the fundamental group of the n-polygon described. Also, in your link the edges don'tt appear in the orientation I am looking for. –  Mar 23 '24 at 13:26
  • 3
    Your space is not a polygon. Your space is a quotient space of a polygon, obtained by identifying all of the edges of the polygon (oriented in the same way around the boundary of the polygon) to a single loop. The link in my comment contains a more explicit description of that quotient space. – Lee Mosher Mar 23 '24 at 13:28
  • 1
    Lens space is another example. – Bob Dobbs Mar 23 '24 at 13:31
  • 1
    @BobDobbs: One might say that a lens space example would require covering space theory, although one can also show that a lens space has a cell decomposition whose 2-skeleton is homeomorphic to the object described in this post. – Lee Mosher Mar 23 '24 at 13:32
  • @LeeMosher It seems like the quotient space appears in the next chapter in Massey's book. In the link they end up talking about the p-fold Dunce cap, which is similar to the one my teacher suggested. But it isn't exactly the same, right? It seems in the dunce hat there is a edge that don't follow the orientation as the others. Or I am seeing something wrong? –  Mar 23 '24 at 13:46
  • @LeeMosher Also, in the link you have provided, when they consider p=2 it is a Möbius band, meanwhile in my example it's a projective plane. –  Mar 23 '24 at 14:08
  • No, when $p=2$ they do indeed get the projective plane. – Lee Mosher Mar 23 '24 at 14:53
  • Anyway, despite the fact that the book you are reading does not get to quotient spaces until the next chapter, in order to make any sense of your own post one needs concepts of quotient spaces; furthermore, application of Van Kampen's theorem can only take place in the quotient space. – Lee Mosher Mar 23 '24 at 14:54
  • This may be tangential, but do you (OP) see intuitively (geometrically) how the proposed "$n$-fold dunce cap" example works? – Andrew D. Hwang Mar 23 '24 at 16:22
  • do you know how to prove this for the $\mathbb RP^2$ case via van kampen? – Andres Mejia Mar 23 '24 at 20:33
  • @LeeMosher Here it is said that p=2 is a Möbius strip, is it incorrect? https://chat.stackexchange.com/rooms/16497/discussion-between-joseph-and-daniel-rust

    Anyway, I think I see it more clearly now, I'll try to do it tomorrow, thank so much for the help!

     –  Mar 23 '24 at 20:55
  • @AndrewD.Hwang I think I do understand the majority of it. I have the doubt if, for example, when given the 3-fold dunce cap, the one in the wikipedia article, can we consider it a surface? –  Mar 23 '24 at 21:08
  • I shuould have searched first, there is a proof that it isn't a surface here –  Mar 23 '24 at 21:10
  • @q19a: That conversation that you linked to is in the middle of a discussion about how to apply Van Kampen's Theorem in the quotient space. Van Kampen's Theorem uses a certain decomposition of the quotient space into a union of two open connected subsets. One of those subsets, denoted $U$, is a Möbius band, yes. But, like in any application of Van Kampen's Theorem, that subset $U$ is not the whole of the quotient space, it is a proper subset thereof. – Lee Mosher Mar 23 '24 at 23:56
  • It seems that you have some misunderstandings of the concepts of quotient spaces and how they are used to construct the spaces in your post. Let me suggest that you focus more directly on understanding those constructions first, rather than on the harder question of how to compute their fundamental groups. – Lee Mosher Mar 24 '24 at 00:02
  • @LeeMosher The construction of the space begins with the n-polygon, we consider the quotient space identifying every edge with each other in the same orientation. This yields a space that is contractible to a point and thus is not a surface. Am I missing something? –  Mar 24 '24 at 09:07
  • That last comment of yours would make an excellent question, if you also included your argument for why you think it is contractible to a point. – Lee Mosher Mar 24 '24 at 12:51
  • @LeeMosher When we consider the identification of the edges we get a 1-cell, which is homeomorphic to the 1-dimensional ball, which is essentially a line segment. For that reason it is contractible to a point. –  Mar 24 '24 at 15:13
  • Let me add to my previous comment: if you are arguing that the quotient space is contractible, then you should edit your post to insert this argument *and* to clarify what you are asking. That way, your argument can be seen by anyone on this site, and therefore it can be critiqued and answered by anyone on this site, rather than just by those who are willing to scroll down through the comments. – Lee Mosher Mar 24 '24 at 16:24

1 Answers1

0

Let's call $X$ the space, the quotiented polygon, so the trick to computing this homotopy group is the following, let $*$ be a point in the center of the polygon, let $U$ be a neighbourhood around it wich doesn't touch the boundary, and $V=X\\{*\}$. Now this two open sets satisfie the hypothesis of Seifert Van Kampen, and each open set involved has the following homotopy type:

U is contractible

$U \cap V = U\setminus \{*\} \cong \mathbb{S}^1$

and the most difficult seeing is $V$, but to figure it out, you can see that before the identifications, $V$ has the $\mathbb{S}^1$ as a deformation retract, and after the identifications this is also an $\mathbb{S}^1$, to see this check that you are partiotioning the circle into equal pieces and pasting one after the other, this is just another $\mathbb{S}^1$. However the important point is that in the inclusion of $U \cap V$ in $V$ after doing the different reformation retractions this is not an inclusion anymore but can be represented as winding the first $\mathbb{S}^1$ $n$ times around the other.

Now, Seifert-Van Kampen tells you that the fundamental group of the pasted polygon is $\mathbb{Z}$, free product with the trivial group, which doesn't change anything, and with the quotient by the image of the inclusion of the intersection of the two open sets into the second one (this is the quotient because it must equal the map induced through the other inclusion, but as that other fundamental group is trivial, this must be 0). Thus the fundamental group is $\mathbb{Z}/n\mathbb{Z}$ as we wanted to see.

Pedro Mayorga
  • 11