What is the fundamental group of X?

[quasar987]

Compute the fundamental group of the space

[tex]X:=((S^1\times S^1) \sqcup (S^1\times S^1))/\sim[/tex]

where ~ is the equivalence relation

[tex] (e^{it},e^{it}) \sim (1,e^{2it})[/tex]

meaning the diagonal of the first torus is identified and wrapped around twice the second generating circles.

Call T_A the first torus and T_B the second one. I tried using Van Kampen with U=[T_A u (nbhd of the 2nd generating circle of T_B)]/~ and V = [(nbhd of the diagonal in T_A) u T_B]/~.

We have that U n V had the homotopy type of a cicle and V has the homotopy type of a torus, but U has the homotopy type of T_A/~ i.e. a torus with the first half of its diagonal identified with the second half. What is the pi_1 of that? :(

Anyone sees a way to compute pi_1(X) by Van Kampen or otherwise?

 

[mathwonk]

well i'll just guess. it seems like the free product of two copies of ZxZ, with (1,1) in one copy identified with (0,2) in the other.?

 

[quasar987]

This would be my guess also...

More precisely, the group with presentation

< Z² x Z² | (1,1) = (0,2) >

But how to proof it?

 

[mathwonk]

except free product is not cross product. it is much bigger.

 

[mathwonk]

well that seems to be exactly what van kampen's theorem says.

 

[quasar987]

Well, there seems to be a missing piece in order to apply V-K.. that is to prove that the fundamental group of T² with the first half of its diagonal identified with its second half is indeed still just Z².

 

[mathwonk]

yes apparently i was too hasty.

 

[mathwonk]

ok, I am assuming that the torus with those identifications is essentially the same as a rectangle with all four edges identified. I.e. I am deforming the diagonal into the union of two contiguous edges. Then it seems to me that Van Kampen implies that space has fundamental group Z.

So then one can proceed with the Van Kampen result on the full space. I.e. then one has the free product Z*(Z^2). modded out by something.

 

[mathwonk]

hmmm.. i guess I'm not too good at this. now it looks to me as iof the fundamentl group of the space you ask about is generated by two generators a,b with one relation a.b^2 = b^2.a ?

My idea was that you can represent a torus as a rectangle with opposite sides identified. And then if you cut along the diagonal and glue two opposite sides, you can represent it as a parallelogram with opposite side identified, where now two of the opposite sides are the original diagonal.So now it seems to me that instead of the diagonal we can think of gluing one loop of the torus to twice a loop of the other torus. so now i seem to get the group with 4 generators a,b,c,d, with relations, cd = dc, ab^2 = b^2.a, and b = c.

so that seems to be the same as 3 generators a,b,d, and relations, bd = db, and a.b^2 = b^2.a.

i am not at all sure of this. kind of a fun problem.

 

[mathwonk]

In the second attempt i was assuming the torus with identifications was obtained by attaching a 2-cell (rectangle) to a wedge of 2 circles by attaching maps which had form ab^2.a^-1.b^-2, as you go around the edges of the rectangle, if that makes sense. Thus we killed that element of the fundamental group of the wedge of 2 circles, instead of the usual; attaching map aba-1b-1, for a usual torus. I have not reviewed any of this lately, and have never taught or taken this course in over 30 years.

 

[dan131m]

Consider the torii as CW complexes, and think of the identification as happening along their 1-skeletons. (This technique should let you calculate anything along these lines almost mechanically.)

EDIT: Looks like this is essentially the idea between mathwonk's solution.

 

[quasar987]

I'm pretty sure you got it.. the key being that the torus with diagonal identified with itself is actually homeomorphic to the square aaa^-1a^-1 which apparently is just a wedge of a circle and a sphere, and so has fundamental group Z.

That was the only missing piece.

But originally we thought that the fundamental group of the torus with diagonal identified with itself was still Z². But actually, the only nontrivial loop is the one that follows (half) the diagonal. In particular, there is actually a way to homotope the usual generators of pi_1(T²) (the 2 "sides" of the square aba^-1b^-1) to that diagonal generator of the torus mod identification.

It would be interesting to see how. After thinking about it 5 minutes I still don't see it... :\

P.S. For my defense: originally, I thought that for some obscure reason the torus with diagonal identified with itself was not homeomorphic to the square aaa^-1a^-1 because I wrongfully thought this was a sphere, which seemed absurd since the torus with diagonal identified with itself seemed obviously non simply connected. :p