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I'm reading the proof of Lemma B.4.3 page 64, Lectures on Hyperbolic Geometry. There is a part of the proof I didn't understand. Suppose $\alpha$ and $\beta$ are two freely homotopic curves in the hyperbolic surface $S$ and $p:\tilde{S} \to S$ is the universal covering. Let $\tau_{\alpha}$ be the deck transformation which sends $\tilde{x}_0 \in \tilde{\alpha}(0)$ to $\tilde{\alpha}(1)$ where $\tilde{\alpha}$ is the lift of $\alpha$ to $\tilde{S}$ starting at $\tilde{x}_0$. Let $\tau_{\beta}$ be the deck transformation which sends $\tilde{y}_0 \in \tilde{\beta}(0)$ to $\tilde{\beta}(1)$ where $\tilde{\beta}$ is the lift of $\beta$ to $\tilde{S}$ starting at $\tilde{y}_0$. How to prove that $\tau_{\beta} = S \circ \tau_{\alpha} \circ S^{-1}$ for some deck transformation $S$.

Dear Daniel Mroz, this is the full statement. enter image description here I'm thinking about the following picture, but I couldn't find a rigorous way to explain the idea.enter image description here

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  • $\begingroup$ I'm not clear on what you're trying to prove. Could you state the full lemma? $\endgroup$ – Daniel Mroz Jul 31 '18 at 9:33
  • $\begingroup$ Is $\tilde \alpha$ without the subscript $x_0$ the same thing as $\tilde\alpha_{x_0}$? And the same question for $\tilde\beta$. If so, it would help if you fixed up that notation in your question. $\endgroup$ – Lee Mosher Jul 31 '18 at 15:48
  • $\begingroup$ Yes they are the same, I will fix the notations now. Thank you. $\endgroup$ – user578196 Jul 31 '18 at 15:53
  • $\begingroup$ Let me re-state it in the following way: Given a closed curve $\alpha$ in $S$. Let $x_0 = \alpha(0)$ and $\tilde{x}_0 \in p^{-1}(x_0)$. Let $\tau$ be the deck transformation which sends $\tilde{x}_0$ to $\tilde{\alpha}_0(1)$ where $\tilde{\alpha}_0$ is the lift of $\alpha$ to $\tilde{S}$ starting at $\tilde{x}_0$. Let us denote $[\alpha]$ the class of all closed curves in $S$ freely homotopic to $\alpha$ and $[\tau]$ the class of all deck transformations of $\tilde{S}$ conjugacy to $\tau$. We claim that the map $\psi: [\alpha] \mapsto [\tau]$ is well defined and it is a bijection. $\endgroup$ – user578196 Jul 31 '18 at 17:05
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    $\begingroup$ Let me be more explicit: $S\tilde{\alpha}$ is a lift of $\alpha$ starting at $S(\tilde{x})$. Thus the endpoint is $S\tilde{\alpha}(1)=ST(\tilde{x})$. Thus the unique transformation you care about sends $S(\tilde{x})$ to $ST(\tilde{x})$, which is of course $STS^{-1}$. $\endgroup$ – Steve D Jul 31 '18 at 23:07
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Let us write down once more how deck transformations are associated to loops (this has also been done in one of Haitrung's comments).

Consider a loop $\alpha : [0,1] \to S$ at $x \in S$. Choose $\tilde{x} \in \pi^{-1}(x)$. There is a unique lift $\tilde{\alpha}$ of $\alpha$ such that $\tilde{\alpha}(0) = \tilde{x}$. This determines a unique deck transformations $T_{\tilde{x},\alpha}$ such that $T_{\tilde{x},\alpha}(\tilde{x}) = \tilde{\alpha}(1)$. The following facts are well-known:

(a) $\langle T_{\tilde{x},\alpha} \rangle$ does not depend on $\tilde{x}$.

(b) $T_{\tilde{x},\alpha}$ depends only on $[\alpha] \in \pi_1(S,x)$. This is true because if $\alpha$ is homotopic rel. $\{ 0, 1 \}$ to $\alpha'$ and $\tilde{\alpha'}$ is a lift of $\alpha'$ such that $\tilde{\alpha'}(0) = \tilde{x}$, then $\tilde{\alpha}(1) = \tilde{\alpha'}(1)$.

Now let $\beta : [0,1] \to S$ be a path with $\beta(0) = x, \beta(1) = y$. It lifts uniquely to a path $\tilde{\beta}$ such that $\tilde{\beta}(0) = \tilde{x}$. Let $\tilde{y} = \tilde{\beta}(1) \in \pi^{-1}(y)$. Define $\delta = \beta^{-1} \ast \alpha \ast \beta$, where $\ast$ denotes the product of paths. This is a closed path at $y$. It lifts to a path $\tilde{\delta}$ such that $\tilde{\delta}(0) = \tilde{y}$. This lift can be composed by lifts of its three components. $\beta^{-1}$ lifts to $\tilde{\beta}^{-1}$ which begins at $\tilde{y}$ and ends at $\tilde{x}$. $\alpha$ lifts to $\tilde{\alpha}$ which begins at $\tilde{x}$ and ends at $T_{\tilde{x},\alpha}(\tilde{x})$. Finally $T_{\tilde{x},\alpha} \circ \tilde{\beta}$ is a lift of $\beta$ (because $p \circ T_{\tilde{x},\alpha} = p$) which begins at $T_{\tilde{x},\alpha}(\tilde{x})$. Therefore $\tilde{\delta} = \tilde{\beta}^{-1} \ast \tilde{\alpha} \ast (T_{\tilde{x},\alpha} \circ \tilde{\beta})$. But we have $\tilde{\delta}(1) = (T_{\tilde{x},\alpha} \circ \tilde{\beta})(1) = T_{\tilde{x},\alpha}(\tilde{\beta}(1)) = T_{\tilde{x},\alpha}(\tilde{y})$. This shows that $T_{\tilde{y},\delta} = T_{\tilde{x},\alpha}$.

Now let $\gamma$ be a loop at $y$ freely homotopic to $\alpha$ via a homotopy $H : [0,1] \times [0,1] \to S$ (which has the property $H(0,s) = H(1,s)$ for all $s$). Let $\beta$ be the path from $x$ to $y$ given by $\beta(t) = H(0,t) = H(1,t)$. Then $\gamma$ and $\delta = \beta^{-1} \ast \alpha \ast \beta$ represent the same element in $\pi_1(S,y)$. Our above considerations show that $T_{\tilde{y},\gamma} = T_{\tilde{y},\delta} = T_{\tilde{x},\alpha}$.

Remark:

We work with a universal covering $p : \tilde{S} \to S$. Universal coverings are only defined for connected base spaces $S$.

Since we have a hyperbolic surface $S$ (which is locally path connected), we conclude that $S$ must be path connected to admit a universal covering.

Let $\Lambda_1(S)$ denote the set of all loops $\alpha : [0,1] \to S$ in $S$. Given $\alpha$ we say it is a loop at $x(\alpha) = \alpha(0) = \alpha(1)$.

We have constructed a function $\ell$ assigning to each $\alpha \in \Lambda_1(S)$ and each $\tilde{x} \in p^{-1}(x(\alpha))$ the element $\ell(\alpha,\tilde{x}) = T_{\alpha,\tilde{x}}$ of $Deck(p)$ = set of deck transformations for $p$.

Here are some known facts:

(1) If $\tilde{x}, \tilde{x}' \in p^{-1}(x(\alpha))$, then $T_{\alpha,\tilde{x}}, T_{\alpha,\tilde{x}'}$ are conjugate.

(2) Let $x \in S$. For each $\tilde{x} \in p^{-1}(x)$ we obtain a group homomorphism $$l_{\tilde{x}} : \pi_1(S,x) \to Deck(p) .$$ It is injective because only loops which are equivalent in $\pi_1(S,x)$ to the trivial loop produce a lift ending at $\tilde{x}$.

(3) Each path $\beta$ from $x$ to $y$ induces a group isomorphism $$h_\beta : \pi_1(S,x) \to \pi_1(S,y), h_\beta([\alpha]) = [\beta^{-1} \ast \alpha \ast \beta] .$$

What we have shown above is that $l_{\tilde{y}} \circ h_\beta = l_{\tilde{x}}$.

Now let $\pi^{free}_1(S,x)$ denote the set of free homotopy classes of loops at $x$. We have a surjection $f_x : \pi_1(S,x) \to \pi^{free}_1(S,x)$ and it is well-known (although not completely trivial) that the preimage $f_x^{-1}([\alpha]^{free})$ is nothing else than the conjugacy classes of $[\alpha]$ in $\pi_1(S,x)$.

As a homomorphism $l_{\tilde{x}}$ maps conjugacy classes in $\pi_1(S,x)$ to conjugacy classes in $Deck(p)$. Hence we obtain a well-defined function $l^{free}_{\tilde{x}} : \pi^{free}_1(S,x) \to Conj(Deck(p))$. From (1) we conclude that $l^{free}_{\tilde{x}}$ does not depend on the choice of $\tilde{x}$ so that we actually have a function $$l^{free}_x : \pi^{free}_1(S,x) \to Conj(Deck(p)) .$$ As above each path $\beta$ from $x$ to $y$ induces a bijection $$h^{free}_\beta : \pi^{free}_1(S,x) \to \pi^{free}_1(S,y), h_\beta([\alpha]^{free}) = [\beta^{-1} \ast \alpha \ast \beta]^{free}$$ and our results imply that $l^{free}_y \circ h^{free}_\beta = l^{free}_x$.

Now let $\lambda_1(S)$ denote the set of all free homotopy classes in $\Lambda_1(S)$ (which will denoted by $[[\alpha]]$). For each $x \in S$ w obtain an obvious map $i_x : \pi^{free}_1(S,x) \to\lambda_1(S), i_x([\alpha]^{free}) = [[\alpha]]$. It is clearly injective. Moreover, if $\beta$ is a path from $x$ to $y$, we can easily verify that $i_y \circ h^{free}_\beta = i_x$. This shows that each $i_x$ is surjective: For each $[[\gamma]]$ we have $[[\gamma]] = i_{x(\gamma)}([\gamma^{free}]) = i_x((h^{free}_\beta)^{-1}([\gamma^{free}]))$, where $\beta$ is any path from $x$ to $x(\gamma)$.

This shows that we get a well-defined function

$$\psi_x = l^{free}_x \circ i_x^{-1} : \lambda_1(S) \to Conj(Deck(p)) .$$

But it does not depend on $x$: Given $x,y \in S$ and a path $\beta$ from $x$ to $y$, we get $l^{free}_x \circ i_x^{-1} = l^{free}_x \circ (h^{free}_\beta)^{-1} \circ i_y^{-1} = l^{free}_y \circ i_y^{-1}$. This map

$$\psi : \lambda_1(S) \to Conj(Deck(p))$$

is given by $\psi([[\gamma]]) = l^{free}_{x(\gamma)}([\gamma]^{free}) = \langle T_{\gamma,\tilde{x}}\rangle$ where $\tilde{x}$ is any point of $p^{-1}(x(\gamma))$.

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    $\begingroup$ You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality. $\endgroup$ – Steve D Aug 1 '18 at 0:19
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    $\begingroup$ @SteveD It is true for the special $\tilde{y}$ which is the end point of $\tilde{\beta}$.For other elements of $p^{-1}(y)$ certainly not. Obviously there must exist some some $y' \in p^{-1}(y)$ such that $T_{y',\gamma} = T_{\tilde{x},\alpha}$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer. $\endgroup$ – Paul Frost Aug 1 '18 at 8:26

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