Suppose $p:E\longrightarrow B$ is a covering projection. I have a general question, on how to find the group of all deck transformations $\Delta(p)$. Is there a common way to do this, or what could be a good approach?
Thanks in advance.
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$\begingroup$ Your question is very unclear. What is $p$? $\endgroup$– Lee MosherCommented Jun 22, 2016 at 21:02
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$\begingroup$ $p$ is supposed to be a function, a covering space to be more specific. Should I give an example? $\endgroup$– Mr.TopologyCommented Jun 22, 2016 at 21:06
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$\begingroup$ You should define $\Delta p$. $\endgroup$– Pedro ♦Commented Jun 22, 2016 at 21:07
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$\begingroup$ $\Delta(p)$ is the group of all deck transformations, of a given covering space $p$. $\endgroup$– Mr.TopologyCommented Jun 22, 2016 at 21:08
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$\begingroup$ So $p$ is a covering projection, and $\Delta(p)$ is the group of all deck transformations. $\endgroup$– Pedro ♦Commented Jun 22, 2016 at 21:09
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1 Answer
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Recall what is the subgroup of $\pi_1(B)$ associated to the covering $p$. To this end, recall this prop:
Thm 1.38 [Hatcher page 67] Let $B$ be path-connected, locally path-connected, and semilocally simply-connected. Then there is a bijection between the set of basepoint-preserving isomorphism classes of path-connected covering spaces $p \colon (\tilde{B} , \tilde{b_0})\to (B, b_0)$ and the set of subgroups of $\pi_1(B, b_0)$, obtained by associating the subgroup $p_* \pi_1(\tilde{B}, \tilde{b_0})$ to the covering space $(\tilde{B}, \tilde{b_0})$. If basepoints are ignored, this correspondence gives a bijection between isomorphism classes of path-connected covering spaces and conjugacy classes of subgroups of $\pi_1$.
With that in mind, let's try to solve your doubt:
The key-point is Prop 1.39 on Hatcher which briefly tells you that if $p_* \pi_1(\tilde{B}, \tilde{b_0})$ is normal in $\pi_1(B, b_0)$, then the associated covering space is normal, i.e. the for each $b\in B$, and each pair of lifts $\tilde{b},\tilde{b}'$ of $b$, there is a deck transformation taking $\tilde{b}\to \tilde{b}'$. (and therefore it's uniquely determined by this). Moreover the proposition tells you that the group $\Delta(p)$ you are interested in, is isomorphic to the quotient $N(H)/H$, where $N(H)$ is the normaliser of $H$ in $\pi_1(B,b_0)$.
tl;dr: Your first step should be to determine the groups $H$ and $\pi_1(B,b_0)$, and then apply prop 1.39.
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$\begingroup$ Thank you. I am going to edit an example in the starting post, and hope we can discuss this example. $\endgroup$ Commented Jun 23, 2016 at 9:05
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