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For any universal cover p of the wedge S1 V S1 is it true that the two actions of π1(X, x_0) on the fiber p^-1(x0) given by lifting loops at x0 and the action given by restricting deck transformations to the fiber coincide. I'm not sure but I think you need for π1(X , x0) to be abelian no?

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  • $\begingroup$ Lifting loops is the right action, since generators get appended to the ends of words. Deck transformations come from the left action. $\endgroup$
    – S. Carnahan
    Nov 19, 2010 at 14:37
  • $\begingroup$ Scott, aren't "the right action" and "the left action" meaningless until you identify the fiber with the fundamental group? (I'm just saying that the definite article could cause confusion.) I've always found the question in this post confusing, and at some point Wikipedia had it wrong... $\endgroup$
    – Dan Ramras
    Nov 22, 2010 at 22:34

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It's indeed true that the action by lifting loops and the action by deck transformations agree exactly when the fundamental group is abelian. This statement is a bit vague, so let me be precise.

Let $X$ be a space with universal cover $Y\stackrel{p}{\longrightarrow} X$, and choose basepoints $x_0 \in X$, $y_0 \in Y$. Then we can identify the fiber $p^{-1} (x_0)$ with $\pi_1 (X, x_0)$ using path lifting: a loop $\gamma$ at $x_0$ lifts to a path $\tilde{\gamma}$ starting at $y_0$, and we identify $[\gamma]$ with $\tilde{\gamma} (0)$. Now the action of $\pi_1 (X, x_0)$ on the fiber, via lifting, corresponds to right multiplication in $\pi_1 (X, x_0)$. On the other hand, the group of deck transformations of $Y$ is also isomorphic to $\pi_1 (X, x_0)$, by sending a loop $\gamma$ to the deck transformation taking $y_0$ to $\tilde{\gamma} (0)$. Now under the identifications of both the deck transformations and the fiber with $\pi_1 (X, x_0)$, the action of deck transformations on the fiber corresponds to left multiplication in $\pi_1 (X, x_0)$.

Checking these statements is a worthwile exercise. In some sense, this is easier to think about if you initially just think of actions as functions that assign group elements to bijections, and then you can later worry about left versus right.

In any event, a group G is abelian if and only if for every element $g\in G$, left multiplication by g and right multiplication by g are the same function $G\to G$.

I'll also note that any left action of a group $G$ on a set $S$ can be converted into a right action by setting $s\cdot g = g^{-1} \cdot s$, but the above discussion shows that if you convert the left action of $\pi_1 (X, x_0)$ on the fiber (via deck transformations) into a right action, you definitely do not get the lifting action.

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  • $\begingroup$ Very nicely explained the last part helped me see it by performing the actual actions you definitely need commutativity. Thanks a lot. $\endgroup$
    – andrewlk
    Nov 29, 2010 at 5:21
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Sorry I'm new to this area it's just self study out of hatcher, one being left and other being right do not prevent them from being the same right? you can always convert a left action onto a right one and vice versa, I'm not sure what it really means for those 2 actions to be the same.

Forget about S1 V S1 that's too complicated what about the simple torus S1 x S1 where the universal cover is R x R.

by definition an action G on X is (G,X) -> X. so here the action group G would be pi_1(X, x0) itself or a subgroup of pi_1(X, x0)? and X would be some point in p^-1(x0) and this "action" gives another point in p^-1(x0) by a lift. So for p:R->S1 by the usual covering p(x) = (cos2pix, sin2pix) has 0,1,2,3,4...all in the fibre of (1,0) in S1 and a lift from [0,1] to R would start at one integer and end at another.

Now deck transformations are defined to be homeomorphisms of the covering space with itself so they also map fibre elements onto themselves. I can kind of see what is going on geometrically but I think there are some definitions/theorems I'm not aware of. Is this correct for the most part how do I tackle the problem.

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  • $\begingroup$ In future please leave updates to your original question either as comments, or as an edit to your original question, not as an answer. (The software rearranges answers depending on votes, meaning that what you write here might end up moving around. This is one way in which the MO setup does not behave like a forum or blog.) $\endgroup$
    – Yemon Choi
    Nov 20, 2010 at 7:44
  • $\begingroup$ ahh ok I didn't know what to do because of the character limit for comments $\endgroup$
    – andrewlk
    Nov 20, 2010 at 9:36

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