3
$\begingroup$

Hatcher lists some examples of covers of a figure 8 (page 58). One of them corresponds with the group with two generators $a$ and $b$ and the relations $a^2, b^2, aba^-1, bab^-1$.

Three-fold cover of figure-eight graph

I thought threefold covers had to have three deck transformations, but I can't seem to think of an automorphism of order three. Am I misunderstanding something?

$\endgroup$
2
  • 2
    $\begingroup$ there are threefold covering spaces with trivial deck transformation group. To obtain the isomorphism $Deck \cong \pi_1(X)/\pi_1(\hat X)$ you want your covering space to be normal, i.e. $\pi_1(\hat X) \subset \pi_1(X)$ be a normal subgroup. $\endgroup$
    – Daniel Valenzuela
    Jul 20, 2014 at 9:55
  • $\begingroup$ The figure 8 symbol, not the eighth figure. It's the third figure. $\endgroup$
    – leewz
    Jul 20, 2014 at 9:55

1 Answer 1

Reset to default
3
$\begingroup$

Your statement is only true for normal covering spaces $\hat X\to X$, i.e. $\pi_1(\hat X) \subset \pi_1(X)$ is normal. Then we get the isomorphism $Deck(\hat X) \cong \pi_1(X) / \pi_1(\hat X)$, which is very interesting, have a look at its geometric interpretation.

Furthermore there are threefold coverings which have trivial deck transformation group. Of course this can never happen for abelian fundamentalgroups by the first part of the answer.

$\endgroup$

You must log in to answer this question.

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