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when we have a covering $p:Y\to X$ choose $x\in X$ and we have two actions on the fiber $p^{-1}(x)$:

1) The monodromy action of $\pi_1(X,x)$ defined as follows: $[\gamma].\tilde x=\tilde \gamma (1)$ that is; the action gives the endpoint of the unique lift of $\gamma$ starting at $\tilde x$.

2)The Deck group action of the group of Deck transformations $Deck(Y,X)$ on $p^{-1}(x)$ defined as follows: $\phi. \tilde x=\phi(\tilde x)$

When $Y=\tilde X$ is the universal covering we have that the two groups $Deck(\tilde X,X)$ and $\pi_1(X,x)$ are isomorphic.

Now my problem is the following : when defining homology of $X$ with local coefficients in a $\mathbb Z[\pi_1(X)]-$module $A$, many authors say that $\pi_1(X,x)$ acts on $\tilde X$ by Deck transformations and set $C_n(X,A):=C_n(\tilde X,\mathbb Z)\otimes A$

so first why they don't say $\pi_1(X)$ acts on $\tilde X$ by monodromy action as defined above instead of saying that it acts by Deck transformatioins and second why taking the universal cover when they could take any cover of $X$ and we still have monodromy and Deck actions.

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  • $\begingroup$ For you second question: The action will fail to be proper so the quotient might not be Hausdorff. For the first question: You get the same answer as you noted. $\endgroup$ – Moishe Kohan Jul 15 at 14:48
  • $\begingroup$ @MoisheKohan where does the quotient $\tilde X/\pi_1(X,x)$ appear in this construction ? $\endgroup$ – palio Jul 15 at 15:46
  • $\begingroup$ It appears as the base of a fibration. But what I wrote refers to a different quotient. As an extreme case, take $X=\tilde{X}$. $\endgroup$ – Moishe Kohan Jul 15 at 15:56
  • $\begingroup$ @MoisheKohan In Hatcher page 72 : the action of the Deck transformation group is properly discontinuous. $\endgroup$ – palio Jul 15 at 16:08
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    $\begingroup$ Sure, but you are proposing to use the action of $\pi_1(X)$ which is not the same as the deck group, unless you have the universal covering space. Incidentally, your monodromy does not even define an action of $\pi_1(X)$ unless you have a regular covering. I suggest that you work out some examples when $\tilde{X}$ is not simply connected and see what you get. $\endgroup$ – Moishe Kohan Jul 15 at 16:17
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Following @MoisheKohan comments, I think I see the answer to my questions:

Given a covering map $p:Y\to X$, the Deck transformations group $Deck(Y,X)$ acts on $Y$ and this action restricts to an action on each fiber $p^{-1}(x)$. The monodromy action however is an action of $\pi_1(X,x)$ on the fiber $p^{-1}(x)$ and this action need not be defined on the whole space $Y$ unless the covering $p:Y \to X$ is a regular covering.

When the covering $Y$ is the universal covering $\tilde X$, we have that $\pi_1(X,x_0)$ is isomorphic to $Deck(\tilde X,X)$ and that's why we talk about the action of $\pi_1(X,x)$ on $\tilde X$ by Deck transformations.

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