8.04 Deck group


Below the video you will find accompanying notes and some pre-class questions.


Deck group

(0.00) Let \(p\colon Y\to X\) be a path-connected covering space. Let \(G=\pi_1(X,x)\), \(H=p_*\pi_1(Y,y)\) (where \(y\in p^{-1}(x)\)) and let \(N_H\subset G\) be the normaliser of \(H\), that is the largest subgroup of \(G\) containing \(H\) for which \(H\subset N_H\) is normal. Then \[Deck(Y,p)\cong N_H/H.\] In particular, if \(Y\) is a normal cover then \(H\) is normal in \(G\), so \(N_H=G\) and \(Deck(Y,p)\cong G/H=\pi_1(X,x)/p_*\pi_1(Y,y)\) as we suspected.

(2.48) We will write down a surjective homomorphism \(\phi\colon N_H\to Deck(Y,p)\) whose kernel is \(H\). The first isomorphism theorem will then imply that \(Deck(Y,p)\cong N_H/H\).

(3.57) We define \(\phi\) as follows. Given an element \(\beta\in N_H\) we have \[\beta p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,y),\] and we also have \(\beta p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,\sigma_\beta(y))\), so by the existence and uniqueness results for deck transformations, there exists a unique deck transformation \(F_\beta\colon Y\to Y\) such that \(F_\beta(y)=\sigma_\beta(y)\). We define \(\phi(\beta)=F_\beta\).

(6.00) We need to check that \(\phi\) is a homomorphism, that \(\phi\) is surjective, and that \(\ker\phi=H\).

To show that \(\phi\) is a homomorphism:

\begin{align*} F_{\beta_1}(F_{\beta_2}(y))&=F_{\beta_1}(\sigma_{\beta_2}(y))\\ &=\sigma_{\beta_1}\sigma_{\beta_2}(y)\\ &=\sigma_{\beta_1\cdot\beta_2}(y)\\ &=F_{\beta_1\cdot\beta_2}(y), \end{align*}

where we used that the monodromy \(\beta\mapsto\sigma_\beta\) is a homomorphism.

(7.46) To see that \(\phi\) is surjective, given a deck transformation \(F\) we want to find \(\beta\in N_H\) such that \(F=F_{\beta}\). Let \(\alpha\) be a path in \(Y\) from \(y\) to \(F(y)\). Let \(\beta=p\circ\alpha\). This \(\beta\) is a loop in \(X\) because \(p(y)=p(F(y))=x\) and \(F(y)=\sigma_\beta(y)\) by definition of monodromy. Therefore \(F=F_\beta\), because these covering transformations agree at \(y\).

(10.38) To see that \(\ker\phi=H\), suppose that \(\beta\in\ker\phi\) (so that \(F_\beta=id_Y\)). Therefore \(y=F_\beta(y)=\sigma_\beta(y)\), so \(y\) is fixed by the monodromy around \(\beta\). This means that the unique lift of \(\beta\) starting at \(y\) is a loop \(\tilde{\beta}\), so \([\beta]=p_*[\tilde{\beta}]\in p_*\pi_1(Y,y)\). This shows that \(\ker\phi\subset p_*\pi_1(Y,y)\). The inclusion \(p_*\pi_1(Y,y)\subset\ker\phi\) is an exercise.

Deck group of the universal cover

(13.23) If \(p\colon Y\to X\) is a simply-connected covering space then \(Deck(Y,p)\cong\pi_1(X,x)\).

For example:

  • the deck group of \(p\colon\mathbf{R}\to S^1\) is \(\mathbf{Z}\), which is also \(\pi_1(S^1)\).
  • the deck group of \(p\colon S^n\to\mathbf{RP}^n\) is \(\mathbf{Z}/2=\pi_1(\mathbf{RP}^n)\).

Pre-class questions

  1. Show that the deck group of a simply-connected covering space of \(X\) is isomorphic to \(\pi_1(X,x)\).
  2. Show that \(p_*\pi_1(Y,y)\subset\ker\phi\).
  3. There is a gap in the proof of surjectivity of \(\phi\). Can you find it? Can you fix it?


CC-BY-SA, Jonny Evans 2017