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First, http://pi.math.cornell.edu/~hatcher/AT/AT.pdf is the site of the book.

Exercise 1.3.27 is :

Exercise 1.3.27. For a universal cover $p : \tilde X \to X $ we have two actions of $\pi_1(X, x_0)$ on the fiber $p^{-1}(x_0)$, namely the action given by lifting loops at $x_0$ and the action given by restricting deck transformations to the fiber. Are these two actions the same when $X=S_1 \vee S_1$ or $X=S_1 \times S_1$? Do the actions always agree when $\pi_1(X,x_0)$ is abelian?

My interpretation of the two actions are as :

(1) The first action is given in p.69 and defined as follows : Let $[\gamma] \in \pi_1(X ,x_0)$ and let $\tilde x \in p^{-1}(x_0)$. Then there is a unique lift $\delta$ of $\bar{\gamma}$ starting at $\tilde x$. We define $[\gamma] \tilde x$ to be the point $\delta (1)$. (Hatcher used $\bar{\gamma}$, not $\gamma$, to make the action a left action. Also, as in the definition of an action in p.71, every action in this book means a left action. )

(2) The second action is : Fix $\tilde x _0 \in p^{-1}(x_0) $. Let $[\gamma] \in \pi_1(X ,x_0)$ and let $\tilde x \in p^{-1}(x_0)$. Since $p$ is a simply-connected covering, there is a unique deck transformation $\tau_\gamma$ sending $\tilde x _0$ to $\tilde \gamma (1)$, where $\tilde \gamma$ is the lift of $\gamma$ starting at $\tilde x _0$ (This is Proposition 1.39. In fact $\pi_1(X, x_0)$ is isomorphic to $G(\tilde X)$, the group of deck transformations, via the isomorphism sending $[\gamma]$ to $\tau_\gamma$.) Now we define $[\gamma]\tilde x$ to be the point $\tau_\gamma (\tilde x)$.

Assuming my understand is correct, what I am confused is these two actions are trivially different. Am I wrong with the definitions of the actions?

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    $\begingroup$ Why do you say they are trivially different? $\endgroup$ Aug 26, 2019 at 3:16
  • $\begingroup$ I don't see the difference, can you explain ? Except that in 1) you are more or less constructing the universal cover as the set of curves in $X$ modulo homotopy (so $\bar{X}$ is a manifold) and in 2) you are assuming you are given a simply connected space with a covering $\endgroup$
    – reuns
    Aug 26, 2019 at 3:25
  • $\begingroup$ @EricWofsey For example, consider $X=S_1 \vee S_1$ as a graph (see p.57 of Hatcher) and the universal covering space of $S_1 \vee S_1$ given in Example 1.45(p.77). Let $\alpha \in \pi_1(X, x_0)$ be the class of the loop $a$ in $X$ (where $x_0$ is the obvious vertex in $X$). Then for the first action, $\alpha e=a^{-1}$ while $\alpha e=a$. This argument is similar for the torus, where the universal covering spcae of the torus is $\Bbb R^2 $ with grids along $\Bbb Z \times \Bbb Z$ $\endgroup$
    – user302934
    Aug 26, 2019 at 3:43
  • $\begingroup$ Note that Hatcher has revised this problem. It now ends with: Show that these two actions are different when 𝑋=𝑆1∨𝑆1 and when 𝑋=𝑆1×𝑆1 and determine when the two actions are the same. [This is a revised version of the original form of this exercise.] $\endgroup$ Dec 18, 2023 at 14:09

2 Answers 2

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These are indeed not the same in most cases (including when $X=S^1\vee S^1$ or $X=S^1\times S^1$) for the rather trivial reason that the first action is defined using $\bar{\gamma}$ and the second is defined using $\gamma$ (so at least when acting on the basepoint $\tilde{x}_0$, the action of $[\gamma]$ by the first definition corresponds to the action of $[\gamma]^{-1}$ by the second). I'm guessing that what Hatcher had in mind, though, is the version of the first action which is defined using a lift of $\gamma$, not a lift of $\bar{\gamma}$. Of course, that makes the first action a right action rather than a left action, so the actions cannot possibly be the same when $\pi_1(X,x_0)$ is nonabelian (given that the actions are faithful). But for an abelian group, right and left actions are the same, and so this still leaves a nontrivial question of whether they coincide when $\pi_1(X,x_0)$ is abelian.

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  • $\begingroup$ How can it not be true for when $X = S^1 \times S^1$? The fundamental group is abelian. $\endgroup$
    – mi.f.zh
    Dec 3, 2020 at 8:14
  • $\begingroup$ Identifying the fiber with $\pi_1(X,x_0)$, one action has $g\in \pi_1(X,x_0)$ act by left translation by $g$ and the other has it act by right translation by $g^{-1}$. $\endgroup$ Dec 3, 2020 at 15:24
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I am three years too late, but there is an interesting solution to this exercise without assuming a mistake on Hatcher's part, as in Eric Wofsey's answer. Here is an outline of the proof:

  1. Prove that both the deck action and the so-called "Monodromy action" are sharply transitive
  2. Prove that if two transitive group actions on the same set agree on a single element, then they are the same
  3. Using their sharp transitivity, prove that $\pi_1(X,x_0)$ is a 2-group if and only if the two actions agree on $\tilde x_0$
  4. Thus these two actions agree if and only if every element of $\pi_1(X,x_0)$ is of order 2

This approach is necessary because a priori, we only know how the deck action acts on $\tilde x_0$, but not the other elements in the fiber.

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