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Let $S$ be a closed oriented surface of genus $g\geq 2$. Let $\mathcal{T}$ be the corresponding Teichmuller space. Given a free homotopy class of closed curve $[\gamma]$ we can define the length function $l_{\gamma}$ on $\mathcal{T}$.

It is well known that there exist $[\alpha]\neq [\beta]$ such that $l_\alpha=l_\beta$.

Q) Does there exist free homotopy classes $[\alpha]\neq [\beta]$ such that $l_\alpha-l_\beta=c\neq 0$ for some $c\in\mathbb{R}$?

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  • $\begingroup$ What do you mean exactly ? You fix $c$ and then you look for $\alpha$ and $\beta$ with $l_{\alpha}-l_{\beta}=c$ ? $\endgroup$ – M. Dus Nov 11 '18 at 20:22
  • $\begingroup$ What/where is the argument for the existence of curves with the same length functions? $\endgroup$ – Paul Plummer Nov 11 '18 at 20:54
  • $\begingroup$ @M.Dus I am looking for the existence of such a relation among the lengths. To be precise "Does there exist $c$, $\alpha$ and $\beta$ such that the above relation hold?" $\endgroup$ – tessellation Nov 12 '18 at 8:44
  • $\begingroup$ @PaulPlummer The argument comes from the trace relations in $SL_2(\mathbb{R})$. See arxiv.org/pdf/math/0302280.pdf $\endgroup$ – tessellation Nov 12 '18 at 8:47
  • $\begingroup$ No, such loops do not exist. Are you still interested in a proof? $\endgroup$ – Misha Dec 31 '18 at 3:22
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Let me first prove an easier statement, namely that there are no loops $c_1, c_2$ on $S$ such that $tr(\rho(c_1)) - tr(\rho(c_2))=a$ for some nonzero constant $a$ and all discrete and faithful representations $\rho: \pi_1(S)\to SL(2, {\mathbb R})$. Consider the representation variety $$ Rep(S)=Hom(\pi_1(S), SL(2, {\mathbb C})). $$ This variety is known to be connected (proven by Bill Goldman in his PhD thesis and published sometime in 1980s). This variety is even smooth away from "reducible representations", i.e. representations whose images have centralizers in $SL(2, {\mathbb C}$ of positive dimension.

Also, the subset $$ F(S)=Hom_{df}(\pi_1(S), SL(2, {\mathbb R})) $$ is Zariski dense (over complex numbers) in $Rep(S)$. This follows from the fact that it is an open nonempty subset in the set of real points in $Rep(S)$. Now, if the identity $tr(\rho(c_1))- tr(\rho(c_2))=a$ holds for all $\rho\in F(S)$, then it holds on the entire $Rep(S)$. On the other hand, $Rep(S)$ contains the trivial representation $\rho_0$ and for all loops $c$ on $S$, $tr(\rho_0(c))=2$. A contradiction.

The proof for the length functions is similar, except we have to use holomorphic functions and not quite on $Rep(S)$, but on its 2-fold branched cover.

Recall that for a nontrivial loop $c$ on a hyperbolic surface $S={\mathbb H}^2/\Gamma$ represented by a matrix $C\in \Gamma$ under a discrete representation $\rho: \pi_1(S)\to \Gamma< SL(2, {\mathbb R})$, the hyperbolic length of $c$ equals $$ \log \lambda^2_C, $$ where $\lambda_C>1$ is the largest eigenvalue of $C$.

Therefore, the identity $l(c_1)-l(c_2)=a\ne 0$ for all hyperbolic structures on the surface $S$ translates into the identity $$ \lambda^2_{C_1}= e^a \lambda^2_{C_2}. $$ Now, let $Rep'(S)$ denote the Zariski subset of $Rep(S)$ consisting of representations such that both matrices $\rho(c_1), \rho(c_2)$ have eigenvalues of the multiplicity one, i.e. are different from $\pm 1$. On $Rep(S')$ we have multivalued (actually, each function takes only two values) holomorphic functions $\lambda^2_1, \lambda^2_2$ obtained by analytic continuation of the real-analytic functions $\lambda^2_{C_1}, \lambda^2_{C_2}$ defined on $F(S)$. To make these functions single-valued we pass to a 4-fold analytic cover $Rep''(S)\to Rep'(S)$. We obtain single-valued lifts $\mu_k$ of $\lambda_k, k=1,2$ to $Rep''(S)$. Now, it follows that the identity $$ \mu_1= e^a \mu_2 $$ still holds on $Rep''(S)$. The trouble is that the trivial representation is not in $Rep'(S)$. However, there is a real curve $\alpha(t)$, $t\in (0,1]$ in $Rep'(S)$ such that $\lim_{t\to 0}\alpha(t)=\rho_0$, the trivial representation.

Then, lifting $\alpha$ to a curve $\beta$ in $Rep''(S)$, we again obtain a contradiction: $$ \lim_{t\to 0} \mu_k \beta(t)= 1, k=1, 2, $$ contradicting the identity $\mu_1= e^a \mu_2$, $a\ne 0$.

I can chase references if anybody cares.

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  • $\begingroup$ Thank you for such a beautiful answer. I have the following doubts. 1) Is Rep′(S) Zariski dense? 2) Does Resp′(S) contains reducible representations? If yes, then how to analytically continue the functions? 3)Why can't we analytically continue the functions on whole of Rep(S)? $\endgroup$ – tessellation Jan 6 '19 at 5:17
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    $\begingroup$ @tessellation: 1) It is even dense in the classical topology. 2) Yes, $Rep'(S)$ contains reducible representations, but you can remove reducibles, since the complement is still dense. 3) The eigenvalue functions are genuinely multivalued on $Rep(S)$. You can already see this on the level of $Hom({\mathbb Z}, SL(2, {\mathbb C}))$. $\endgroup$ – Misha Jan 6 '19 at 16:58

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