4
$\begingroup$

This is exercise 12-6 from J. Lee's Introduction to Topological Manifolds.
Let $E=\{(x,y)\in\mathbb{R}^3\times\mathbb{R}^3:x\neq y\}$, and define an equivalence relation by $(x,y)\sim(y,x)$ for all $(x,y)\in E$. Compute the fundamental group of $E/\sim$.
This is the first time I'm asked to compute the fundamental group of something that I can't seem to visualise, and I am not sure how to approach it. Seifert-van Kampen does not seem useful for such a space. What I think I want to do is find a group $G$ such that $q:F\to F/G=E/\sim$ is a covering map for a simply connected $F$, so that the fundamental group is isomorphic to $Aut_q(F)=G$. But even then, I am not quite sure how to do this. Can this space be visualised somehow? If not, how would one generally approach such a problem where one's visualisation is insufficient, as I guess it is in most worthwhile problems?
I would appreciate some advice on this.

$\endgroup$
2
  • 1
    $\begingroup$ Maybe, not sure $\Bbb Z_2$-action on the simply-connected space $E$. $\endgroup$
    – Sumanta
    Commented Sep 21, 2020 at 9:15
  • $\begingroup$ @0-thUser As in, the automorphism group would just be flipping the decks, and $E$ is a double covering of $E/\sim$? $\endgroup$ Commented Sep 21, 2020 at 10:20

1 Answer 1

2
$\begingroup$

Consider the map $$L : E\rightarrow \mathbb R^3\times \mathbb R^3-\{ 0\}$$ $$(x,y)\rightarrow \left (x+y,y-x\right )$$ This is clearly a homeomorphism. So your space $E$ is homotopy equivalent to $\mathbb R^3-\{ 0\}\simeq \mathbb S^2$ which is simply connected.

This already gives you the fundamental group to be $\mathbb Z_2$ but I shall attempt to identify the homotopy type of this space.

$E/_\sim $ is harder to see. Using the isomorphism $L$ we see that $E/_\sim$ is homeomorphic to $X=\mathbb R^3\times \mathbb R^3-\{ 0\}/_{(v,x)\sim (v,-x)}$

[This should remind you of the tangent bundle of $\mathbb {RP}^2$ upto a homotopy which I shall try to make precise]

$\mathbb R^3\times \mathbb R^3-\{ 0\}\xrightarrow{\text{projection }} 0\times \mathbb R^3-\{0 \}$ is a strong deformatrion retract. Let $A= 0\times \mathbb R^3-\{0 \} /_{(0,x)\sim (0,-x)}$

We have the weak retract $$r: X\rightarrow A$$ $$[(v,x)]\mapsto [(0,x)]$$

Consider the homotopy $$H_t : X\rightarrow X$$ $$[(v,x)]\mapsto [(tv,x)]$$ This clearly shows $r$ is a strong deformation retract and hence a homotopy equivalnce. So $E/_\sim \simeq \mathbb {RP}^2$ and once again we can see that $\pi_1(\mathbb {RP}^2)=\mathbb Z_2$

$\endgroup$
1
  • $\begingroup$ Very nice solution, thank you! $\endgroup$ Commented Sep 21, 2020 at 19:21

You must log in to answer this question.

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