Mixing
Lifts of 2 curves in the same homotopy class
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
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 $p:tildeS to S$ is the universal covering. Let $tau_alpha$ be the deck transformation which sends $tildex_0 in tildealpha(0)$ to $tildealpha(1)$ where $tildealpha$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. Let $tau_beta$ be the deck transformation which sends $tildey_0 in tildebeta(0)$ to $tildebeta(1)$ where $tildebeta$ is the lift of $beta$ to $tildeS$ starting at $tildey_0$.
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.
I'm thinking about the following picture, but I couldn't find a rigorous way to explain the idea.
hyperbolic-geometry covering-spaces
asked Jul 31 at 8:09
Haitrung
237
 |Â
show 2 more comments
up vote
2
down vote
favorite
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 $p:tildeS to S$ is the universal covering. Let $tau_alpha$ be the deck transformation which sends $tildex_0 in tildealpha(0)$ to $tildealpha(1)$ where $tildealpha$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. Let $tau_beta$ be the deck transformation which sends $tildey_0 in tildebeta(0)$ to $tildebeta(1)$ where $tildebeta$ is the lift of $beta$ to $tildeS$ starting at $tildey_0$.
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.
I'm thinking about the following picture, but I couldn't find a rigorous way to explain the idea.
hyperbolic-geometry covering-spaces
asked Jul 31 at 8:09
Haitrung
237
I'm not clear on what you're trying to prove. Could you state the full lemma?
– Daniel Mroz
Jul 31 at 9:33
Is $tilde alpha$ without the subscript $x_0$ the same thing as $tildealpha_x_0$? And the same question for $tildebeta$. If so, it would help if you fixed up that notation in your question.
– Lee Mosher
Jul 31 at 15:48
Yes they are the same, I will fix the notations now. Thank you.
– Haitrung
Jul 31 at 15:53
Let me re-state it in the following way: Given a closed curve $alpha$ in $S$. Let $x_0 = alpha(0)$ and $tildex_0 in p^-1(x_0)$. Let $tau$ be the deck transformation which sends $tildex_0$ to $tildealpha_0(1)$ where $tildealpha_0$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. 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 $tildeS$ conjugacy to $tau$. We claim that the map $psi: [alpha] mapsto [tau]$ is well defined and it is a bijection.
– Haitrung
Jul 31 at 17:05
1
Let me be more explicit: $Stildealpha$ is a lift of $alpha$ starting at $S(tildex)$. Thus the endpoint is $Stildealpha(1)=ST(tildex)$. Thus the unique transformation you care about sends $S(tildex)$ to $ST(tildex)$, which is of course $STS^-1$.
– Steve D
Jul 31 at 23:07
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 $p:tildeS to S$ is the universal covering. Let $tau_alpha$ be the deck transformation which sends $tildex_0 in tildealpha(0)$ to $tildealpha(1)$ where $tildealpha$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. Let $tau_beta$ be the deck transformation which sends $tildey_0 in tildebeta(0)$ to $tildebeta(1)$ where $tildebeta$ is the lift of $beta$ to $tildeS$ starting at $tildey_0$.
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.
I'm thinking about the following picture, but I couldn't find a rigorous way to explain the idea.
hyperbolic-geometry covering-spaces
asked Jul 31 at 8:09
Haitrung
237
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 $p:tildeS to S$ is the universal covering. Let $tau_alpha$ be the deck transformation which sends $tildex_0 in tildealpha(0)$ to $tildealpha(1)$ where $tildealpha$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. Let $tau_beta$ be the deck transformation which sends $tildey_0 in tildebeta(0)$ to $tildebeta(1)$ where $tildebeta$ is the lift of $beta$ to $tildeS$ starting at $tildey_0$.
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.
I'm thinking about the following picture, but I couldn't find a rigorous way to explain the idea.
hyperbolic-geometry covering-spaces
asked Jul 31 at 8:09
Haitrung
237
edited Jul 31 at 15:56
asked Jul 31 at 8:09
Haitrung
237
asked Jul 31 at 8:09
Haitrung
237
asked Jul 31 at 8:09
Haitrung
237
237
I'm not clear on what you're trying to prove. Could you state the full lemma?
– Daniel Mroz
Jul 31 at 9:33
Is $tilde alpha$ without the subscript $x_0$ the same thing as $tildealpha_x_0$? And the same question for $tildebeta$. If so, it would help if you fixed up that notation in your question.
– Lee Mosher
Jul 31 at 15:48
Yes they are the same, I will fix the notations now. Thank you.
– Haitrung
Jul 31 at 15:53
Let me re-state it in the following way: Given a closed curve $alpha$ in $S$. Let $x_0 = alpha(0)$ and $tildex_0 in p^-1(x_0)$. Let $tau$ be the deck transformation which sends $tildex_0$ to $tildealpha_0(1)$ where $tildealpha_0$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. 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 $tildeS$ conjugacy to $tau$. We claim that the map $psi: [alpha] mapsto [tau]$ is well defined and it is a bijection.
– Haitrung
Jul 31 at 17:05
1
Let me be more explicit: $Stildealpha$ is a lift of $alpha$ starting at $S(tildex)$. Thus the endpoint is $Stildealpha(1)=ST(tildex)$. Thus the unique transformation you care about sends $S(tildex)$ to $ST(tildex)$, which is of course $STS^-1$.
– Steve D
Jul 31 at 23:07
 |Â
show 2 more comments
I'm not clear on what you're trying to prove. Could you state the full lemma?
– Daniel Mroz
Jul 31 at 9:33
Is $tilde alpha$ without the subscript $x_0$ the same thing as $tildealpha_x_0$? And the same question for $tildebeta$. If so, it would help if you fixed up that notation in your question.
– Lee Mosher
Jul 31 at 15:48
Yes they are the same, I will fix the notations now. Thank you.
– Haitrung
Jul 31 at 15:53
Let me re-state it in the following way: Given a closed curve $alpha$ in $S$. Let $x_0 = alpha(0)$ and $tildex_0 in p^-1(x_0)$. Let $tau$ be the deck transformation which sends $tildex_0$ to $tildealpha_0(1)$ where $tildealpha_0$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. 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 $tildeS$ conjugacy to $tau$. We claim that the map $psi: [alpha] mapsto [tau]$ is well defined and it is a bijection.
– Haitrung
Jul 31 at 17:05
1
Let me be more explicit: $Stildealpha$ is a lift of $alpha$ starting at $S(tildex)$. Thus the endpoint is $Stildealpha(1)=ST(tildex)$. Thus the unique transformation you care about sends $S(tildex)$ to $ST(tildex)$, which is of course $STS^-1$.
– Steve D
Jul 31 at 23:07
I'm not clear on what you're trying to prove. Could you state the full lemma?
– Daniel Mroz
Jul 31 at 9:33
I'm not clear on what you're trying to prove. Could you state the full lemma?
– Daniel Mroz
Jul 31 at 9:33
Is $tilde alpha$ without the subscript $x_0$ the same thing as $tildealpha_x_0$? And the same question for $tildebeta$. If so, it would help if you fixed up that notation in your question.
– Lee Mosher
Jul 31 at 15:48
Is $tilde alpha$ without the subscript $x_0$ the same thing as $tildealpha_x_0$? And the same question for $tildebeta$. If so, it would help if you fixed up that notation in your question.
– Lee Mosher
Jul 31 at 15:48
Yes they are the same, I will fix the notations now. Thank you.
– Haitrung
Jul 31 at 15:53
Yes they are the same, I will fix the notations now. Thank you.
– Haitrung
Jul 31 at 15:53
Let me re-state it in the following way: Given a closed curve $alpha$ in $S$. Let $x_0 = alpha(0)$ and $tildex_0 in p^-1(x_0)$. Let $tau$ be the deck transformation which sends $tildex_0$ to $tildealpha_0(1)$ where $tildealpha_0$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. 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 $tildeS$ conjugacy to $tau$. We claim that the map $psi: [alpha] mapsto [tau]$ is well defined and it is a bijection.
– Haitrung
Jul 31 at 17:05
Let me re-state it in the following way: Given a closed curve $alpha$ in $S$. Let $x_0 = alpha(0)$ and $tildex_0 in p^-1(x_0)$. Let $tau$ be the deck transformation which sends $tildex_0$ to $tildealpha_0(1)$ where $tildealpha_0$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. 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 $tildeS$ conjugacy to $tau$. We claim that the map $psi: [alpha] mapsto [tau]$ is well defined and it is a bijection.
– Haitrung
Jul 31 at 17:05
1
1
Let me be more explicit: $Stildealpha$ is a lift of $alpha$ starting at $S(tildex)$. Thus the endpoint is $Stildealpha(1)=ST(tildex)$. Thus the unique transformation you care about sends $S(tildex)$ to $ST(tildex)$, which is of course $STS^-1$.
– Steve D
Jul 31 at 23:07
Let me be more explicit: $Stildealpha$ is a lift of $alpha$ starting at $S(tildex)$. Thus the endpoint is $Stildealpha(1)=ST(tildex)$. Thus the unique transformation you care about sends $S(tildex)$ to $ST(tildex)$, which is of course $STS^-1$.
– Steve D
Jul 31 at 23:07
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
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 $tildex in pi^-1(x)$. There is a unique lift $tildealpha$ of $alpha$ such that $tildealpha(0) = tildex$. This determines a unique deck transformations $T_tildex,alpha$ such that $T_tildex,alpha(tildex) = tildealpha(1)$. The following facts are well-known:
(a) $langle T_tildex,alpha rangle$ does not depend on $tildex$.
(b) $T_tildex,alpha$ depends only on $[alpha] in pi_1(S,x)$. This is true because if $alpha$ is homotopic rel. $ 0, 1 $ to $alpha'$ and $tildealpha'$ is a lift of $alpha'$ such that $tildealpha'(0) = tildex$, then $tildealpha(1) = tildealpha'(1)$.
Now let $beta : [0,1] to S$ be a path with $beta(0) = x, beta(1) = y$. It lifts uniquely to a path $tildebeta$ such that $tildebeta(0) = tildex$. Let $tildey = tildebeta(1) in pi^-1(y)$. 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 $tildedelta$ such that $tildedelta(0) = tildey$. This lift can be composed by lifts of its three components. $beta^-1$ lifts to $tildebeta^-1$ which begins at $tildey$ and ends at $tildex$. $alpha$ lifts to $tildealpha$ which begins at $tildex$ and ends at $T_tildex,alpha(tildex)$. Finally $T_tildex,alpha circ tildebeta$ is a lift of $beta$ (because $p circ T_tildex,alpha = p$) which begins at $T_tildex,alpha(tildex)$. Therefore $tildedelta = tildebeta^-1 ast tildealpha ast (T_tildex,alpha circ tildebeta)$. But we have $tildedelta(1) = (T_tildex,alpha circ tildebeta)(1) = T_tildex,alpha(tildebeta(1)) = T_tildex,alpha(tildey)$. This shows that $T_tildey,delta = T_tildex,alpha$.
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 $T_tildey,gamma = T_tildey,delta = T_tildex,alpha$.
Remark:
We work with a universal covering $p : tildeS to S$. 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 $tildex in p^-1(x(alpha))$ the element $ell(alpha,tildex) = T_alpha,tildex$ of $Deck(p)$ = set of deck transformations for $p$.
Here are some known facts:
(1) If $tildex, tildex' in p^-1(x(alpha))$, then $T_alpha,tildex, T_alpha,tildex'$ are conjugate.
(2) Let $x in S$. For each $tildex in p^-1(x)$ we obtain a group homomorphism
$$l_tildex : pi_1(S,x) to Deck(p) .$$
It is injective because only loops which are equivalent in $pi_1(S,x)$ to the trivial loop produce a lift ending at $tildex$.
(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 $l_tildey circ h_beta = l_tildex$.
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 $l_tildex$ maps conjugacy classes in $pi_1(S,x)$ to conjugacy classes in $Deck(p)$. Hence we obtain a well-defined function $l^free_tildex : pi^free_1(S,x) to Conj(Deck(p))$. From (1) we conclude that $l^free_tildex$ does not depend on the choice of $tildex$ 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 $i_x : pi^free_1(S,x) tolambda_1(S), i_x([alpha]^free) = [[alpha]]$. 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 $psi([[gamma]]) = l^free_x(gamma)([gamma]^free) = langle T_gamma,tildexrangle$ where $tildex$ is any point of $p^-1(x(gamma))$.
answered Jul 31 at 23:01
Paul Frost
3,553420
1
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
1
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 $tildex in pi^-1(x)$. There is a unique lift $tildealpha$ of $alpha$ such that $tildealpha(0) = tildex$. This determines a unique deck transformations $T_tildex,alpha$ such that $T_tildex,alpha(tildex) = tildealpha(1)$. The following facts are well-known:
(a) $langle T_tildex,alpha rangle$ does not depend on $tildex$.
(b) $T_tildex,alpha$ depends only on $[alpha] in pi_1(S,x)$. This is true because if $alpha$ is homotopic rel. $ 0, 1 $ to $alpha'$ and $tildealpha'$ is a lift of $alpha'$ such that $tildealpha'(0) = tildex$, then $tildealpha(1) = tildealpha'(1)$.
Now let $beta : [0,1] to S$ be a path with $beta(0) = x, beta(1) = y$. It lifts uniquely to a path $tildebeta$ such that $tildebeta(0) = tildex$. Let $tildey = tildebeta(1) in pi^-1(y)$. 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 $tildedelta$ such that $tildedelta(0) = tildey$. This lift can be composed by lifts of its three components. $beta^-1$ lifts to $tildebeta^-1$ which begins at $tildey$ and ends at $tildex$. $alpha$ lifts to $tildealpha$ which begins at $tildex$ and ends at $T_tildex,alpha(tildex)$. Finally $T_tildex,alpha circ tildebeta$ is a lift of $beta$ (because $p circ T_tildex,alpha = p$) which begins at $T_tildex,alpha(tildex)$. Therefore $tildedelta = tildebeta^-1 ast tildealpha ast (T_tildex,alpha circ tildebeta)$. But we have $tildedelta(1) = (T_tildex,alpha circ tildebeta)(1) = T_tildex,alpha(tildebeta(1)) = T_tildex,alpha(tildey)$. This shows that $T_tildey,delta = T_tildex,alpha$.
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 $T_tildey,gamma = T_tildey,delta = T_tildex,alpha$.
Remark:
We work with a universal covering $p : tildeS to S$. 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 $tildex in p^-1(x(alpha))$ the element $ell(alpha,tildex) = T_alpha,tildex$ of $Deck(p)$ = set of deck transformations for $p$.
Here are some known facts:
(1) If $tildex, tildex' in p^-1(x(alpha))$, then $T_alpha,tildex, T_alpha,tildex'$ are conjugate.
(2) Let $x in S$. For each $tildex in p^-1(x)$ we obtain a group homomorphism
$$l_tildex : pi_1(S,x) to Deck(p) .$$
It is injective because only loops which are equivalent in $pi_1(S,x)$ to the trivial loop produce a lift ending at $tildex$.
(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 $l_tildey circ h_beta = l_tildex$.
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 $l_tildex$ maps conjugacy classes in $pi_1(S,x)$ to conjugacy classes in $Deck(p)$. Hence we obtain a well-defined function $l^free_tildex : pi^free_1(S,x) to Conj(Deck(p))$. From (1) we conclude that $l^free_tildex$ does not depend on the choice of $tildex$ 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 $i_x : pi^free_1(S,x) tolambda_1(S), i_x([alpha]^free) = [[alpha]]$. 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 $psi([[gamma]]) = l^free_x(gamma)([gamma]^free) = langle T_gamma,tildexrangle$ where $tildex$ is any point of $p^-1(x(gamma))$.
answered Jul 31 at 23:01
Paul Frost
3,553420
1
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
1
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
add a comment |Â
up vote
2
down vote
accepted
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 $tildex in pi^-1(x)$. There is a unique lift $tildealpha$ of $alpha$ such that $tildealpha(0) = tildex$. This determines a unique deck transformations $T_tildex,alpha$ such that $T_tildex,alpha(tildex) = tildealpha(1)$. The following facts are well-known:
(a) $langle T_tildex,alpha rangle$ does not depend on $tildex$.
(b) $T_tildex,alpha$ depends only on $[alpha] in pi_1(S,x)$. This is true because if $alpha$ is homotopic rel. $ 0, 1 $ to $alpha'$ and $tildealpha'$ is a lift of $alpha'$ such that $tildealpha'(0) = tildex$, then $tildealpha(1) = tildealpha'(1)$.
Now let $beta : [0,1] to S$ be a path with $beta(0) = x, beta(1) = y$. It lifts uniquely to a path $tildebeta$ such that $tildebeta(0) = tildex$. Let $tildey = tildebeta(1) in pi^-1(y)$. 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 $tildedelta$ such that $tildedelta(0) = tildey$. This lift can be composed by lifts of its three components. $beta^-1$ lifts to $tildebeta^-1$ which begins at $tildey$ and ends at $tildex$. $alpha$ lifts to $tildealpha$ which begins at $tildex$ and ends at $T_tildex,alpha(tildex)$. Finally $T_tildex,alpha circ tildebeta$ is a lift of $beta$ (because $p circ T_tildex,alpha = p$) which begins at $T_tildex,alpha(tildex)$. Therefore $tildedelta = tildebeta^-1 ast tildealpha ast (T_tildex,alpha circ tildebeta)$. But we have $tildedelta(1) = (T_tildex,alpha circ tildebeta)(1) = T_tildex,alpha(tildebeta(1)) = T_tildex,alpha(tildey)$. This shows that $T_tildey,delta = T_tildex,alpha$.
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 $T_tildey,gamma = T_tildey,delta = T_tildex,alpha$.
Remark:
We work with a universal covering $p : tildeS to S$. 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 $tildex in p^-1(x(alpha))$ the element $ell(alpha,tildex) = T_alpha,tildex$ of $Deck(p)$ = set of deck transformations for $p$.
Here are some known facts:
(1) If $tildex, tildex' in p^-1(x(alpha))$, then $T_alpha,tildex, T_alpha,tildex'$ are conjugate.
(2) Let $x in S$. For each $tildex in p^-1(x)$ we obtain a group homomorphism
$$l_tildex : pi_1(S,x) to Deck(p) .$$
It is injective because only loops which are equivalent in $pi_1(S,x)$ to the trivial loop produce a lift ending at $tildex$.
(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 $l_tildey circ h_beta = l_tildex$.
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 $l_tildex$ maps conjugacy classes in $pi_1(S,x)$ to conjugacy classes in $Deck(p)$. Hence we obtain a well-defined function $l^free_tildex : pi^free_1(S,x) to Conj(Deck(p))$. From (1) we conclude that $l^free_tildex$ does not depend on the choice of $tildex$ 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 $i_x : pi^free_1(S,x) tolambda_1(S), i_x([alpha]^free) = [[alpha]]$. 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 $psi([[gamma]]) = l^free_x(gamma)([gamma]^free) = langle T_gamma,tildexrangle$ where $tildex$ is any point of $p^-1(x(gamma))$.
answered Jul 31 at 23:01
Paul Frost
3,553420
1
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
1
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 $tildex in pi^-1(x)$. There is a unique lift $tildealpha$ of $alpha$ such that $tildealpha(0) = tildex$. This determines a unique deck transformations $T_tildex,alpha$ such that $T_tildex,alpha(tildex) = tildealpha(1)$. The following facts are well-known:
(a) $langle T_tildex,alpha rangle$ does not depend on $tildex$.
(b) $T_tildex,alpha$ depends only on $[alpha] in pi_1(S,x)$. This is true because if $alpha$ is homotopic rel. $ 0, 1 $ to $alpha'$ and $tildealpha'$ is a lift of $alpha'$ such that $tildealpha'(0) = tildex$, then $tildealpha(1) = tildealpha'(1)$.
Now let $beta : [0,1] to S$ be a path with $beta(0) = x, beta(1) = y$. It lifts uniquely to a path $tildebeta$ such that $tildebeta(0) = tildex$. Let $tildey = tildebeta(1) in pi^-1(y)$. 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 $tildedelta$ such that $tildedelta(0) = tildey$. This lift can be composed by lifts of its three components. $beta^-1$ lifts to $tildebeta^-1$ which begins at $tildey$ and ends at $tildex$. $alpha$ lifts to $tildealpha$ which begins at $tildex$ and ends at $T_tildex,alpha(tildex)$. Finally $T_tildex,alpha circ tildebeta$ is a lift of $beta$ (because $p circ T_tildex,alpha = p$) which begins at $T_tildex,alpha(tildex)$. Therefore $tildedelta = tildebeta^-1 ast tildealpha ast (T_tildex,alpha circ tildebeta)$. But we have $tildedelta(1) = (T_tildex,alpha circ tildebeta)(1) = T_tildex,alpha(tildebeta(1)) = T_tildex,alpha(tildey)$. This shows that $T_tildey,delta = T_tildex,alpha$.
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 $T_tildey,gamma = T_tildey,delta = T_tildex,alpha$.
Remark:
We work with a universal covering $p : tildeS to S$. 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 $tildex in p^-1(x(alpha))$ the element $ell(alpha,tildex) = T_alpha,tildex$ of $Deck(p)$ = set of deck transformations for $p$.
Here are some known facts:
(1) If $tildex, tildex' in p^-1(x(alpha))$, then $T_alpha,tildex, T_alpha,tildex'$ are conjugate.
(2) Let $x in S$. For each $tildex in p^-1(x)$ we obtain a group homomorphism
$$l_tildex : pi_1(S,x) to Deck(p) .$$
It is injective because only loops which are equivalent in $pi_1(S,x)$ to the trivial loop produce a lift ending at $tildex$.
(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 $l_tildey circ h_beta = l_tildex$.
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 $l_tildex$ maps conjugacy classes in $pi_1(S,x)$ to conjugacy classes in $Deck(p)$. Hence we obtain a well-defined function $l^free_tildex : pi^free_1(S,x) to Conj(Deck(p))$. From (1) we conclude that $l^free_tildex$ does not depend on the choice of $tildex$ 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 $i_x : pi^free_1(S,x) tolambda_1(S), i_x([alpha]^free) = [[alpha]]$. 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 $psi([[gamma]]) = l^free_x(gamma)([gamma]^free) = langle T_gamma,tildexrangle$ where $tildex$ is any point of $p^-1(x(gamma))$.
answered Jul 31 at 23:01
Paul Frost
3,553420
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 $tildex in pi^-1(x)$. There is a unique lift $tildealpha$ of $alpha$ such that $tildealpha(0) = tildex$. This determines a unique deck transformations $T_tildex,alpha$ such that $T_tildex,alpha(tildex) = tildealpha(1)$. The following facts are well-known:
(a) $langle T_tildex,alpha rangle$ does not depend on $tildex$.
(b) $T_tildex,alpha$ depends only on $[alpha] in pi_1(S,x)$. This is true because if $alpha$ is homotopic rel. $ 0, 1 $ to $alpha'$ and $tildealpha'$ is a lift of $alpha'$ such that $tildealpha'(0) = tildex$, then $tildealpha(1) = tildealpha'(1)$.
Now let $beta : [0,1] to S$ be a path with $beta(0) = x, beta(1) = y$. It lifts uniquely to a path $tildebeta$ such that $tildebeta(0) = tildex$. Let $tildey = tildebeta(1) in pi^-1(y)$. 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 $tildedelta$ such that $tildedelta(0) = tildey$. This lift can be composed by lifts of its three components. $beta^-1$ lifts to $tildebeta^-1$ which begins at $tildey$ and ends at $tildex$. $alpha$ lifts to $tildealpha$ which begins at $tildex$ and ends at $T_tildex,alpha(tildex)$. Finally $T_tildex,alpha circ tildebeta$ is a lift of $beta$ (because $p circ T_tildex,alpha = p$) which begins at $T_tildex,alpha(tildex)$. Therefore $tildedelta = tildebeta^-1 ast tildealpha ast (T_tildex,alpha circ tildebeta)$. But we have $tildedelta(1) = (T_tildex,alpha circ tildebeta)(1) = T_tildex,alpha(tildebeta(1)) = T_tildex,alpha(tildey)$. This shows that $T_tildey,delta = T_tildex,alpha$.
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 $T_tildey,gamma = T_tildey,delta = T_tildex,alpha$.
Remark:
We work with a universal covering $p : tildeS to S$. 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 $tildex in p^-1(x(alpha))$ the element $ell(alpha,tildex) = T_alpha,tildex$ of $Deck(p)$ = set of deck transformations for $p$.
Here are some known facts:
(1) If $tildex, tildex' in p^-1(x(alpha))$, then $T_alpha,tildex, T_alpha,tildex'$ are conjugate.
(2) Let $x in S$. For each $tildex in p^-1(x)$ we obtain a group homomorphism
$$l_tildex : pi_1(S,x) to Deck(p) .$$
It is injective because only loops which are equivalent in $pi_1(S,x)$ to the trivial loop produce a lift ending at $tildex$.
(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 $l_tildey circ h_beta = l_tildex$.
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 $l_tildex$ maps conjugacy classes in $pi_1(S,x)$ to conjugacy classes in $Deck(p)$. Hence we obtain a well-defined function $l^free_tildex : pi^free_1(S,x) to Conj(Deck(p))$. From (1) we conclude that $l^free_tildex$ does not depend on the choice of $tildex$ 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 $i_x : pi^free_1(S,x) tolambda_1(S), i_x([alpha]^free) = [[alpha]]$. 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 $psi([[gamma]]) = l^free_x(gamma)([gamma]^free) = langle T_gamma,tildexrangle$ where $tildex$ is any point of $p^-1(x(gamma))$.
answered Jul 31 at 23:01
Paul Frost
3,553420
edited Aug 2 at 9:52
answered Jul 31 at 23:01
Paul Frost
3,553420
answered Jul 31 at 23:01
Paul Frost
3,553420
answered Jul 31 at 23:01
Paul Frost
3,553420
3,553420
1
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
1
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
add a comment |Â
1
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
1
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
1
1
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
You've dropped a conjugacy class along the way, because what you're saying is certainly not true at the level of equality.
– Steve D
Aug 1 at 0:19
1
1
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
@SteveD It is true for the special $tildey$ which is the end point of $tildebeta$.For other elements of $p^-1(y)$ certainly not. Obviously there must exist some some $y' in p^-1(y)$ such that $T_y',gamma = T_tildex,alpha$, otherwise the theorem would be wrong. The question is only how to find it. This was done in my answer.
– Paul Frost
Aug 1 at 8:26
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867788%2flifts-of-2-curves-in-the-same-homotopy-class%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I'm not clear on what you're trying to prove. Could you state the full lemma?
– Daniel Mroz
Jul 31 at 9:33
Is $tilde alpha$ without the subscript $x_0$ the same thing as $tildealpha_x_0$? And the same question for $tildebeta$. If so, it would help if you fixed up that notation in your question.
– Lee Mosher
Jul 31 at 15:48
Yes they are the same, I will fix the notations now. Thank you.
– Haitrung
Jul 31 at 15:53
Let me re-state it in the following way: Given a closed curve $alpha$ in $S$. Let $x_0 = alpha(0)$ and $tildex_0 in p^-1(x_0)$. Let $tau$ be the deck transformation which sends $tildex_0$ to $tildealpha_0(1)$ where $tildealpha_0$ is the lift of $alpha$ to $tildeS$ starting at $tildex_0$. 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 $tildeS$ conjugacy to $tau$. We claim that the map $psi: [alpha] mapsto [tau]$ is well defined and it is a bijection.
– Haitrung
Jul 31 at 17:05
1
Let me be more explicit: $Stildealpha$ is a lift of $alpha$ starting at $S(tildex)$. Thus the endpoint is $Stildealpha(1)=ST(tildex)$. Thus the unique transformation you care about sends $S(tildex)$ to $ST(tildex)$, which is of course $STS^-1$.
– Steve D
Jul 31 at 23:07