# 8.04 Deck group

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

# Notes

## 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?