Let $\Sigma$ be a compact surface of genus at least $1$ with one boundary component, equipped with a hyperbolic metric so that the boundary is geodesic. There is a version of the collar lemma that says there is a collar neighbourhood $C$ of the boundary such that no simple closed geodesic on $\Sigma$ enters $C$.


Given a curve, I am interested in measuring what length of it lies outside $C$. More precisely, I am curious about the following: does there exist a constant $d>0$ such that, if $\gamma$ is an essential simple closed curve (not nullhomotopic and not homotopic to the boundary component), then $$\ell(\gamma|\_{\Sigma\backslash C})+d \geq \ell(\tilde{\gamma})$$ Here, $\ell$ is the hyperbolic length, $\tilde{\gamma}$ is the unique simple closed geodesic homotopic to $\gamma$, and, to be clear, $\ell(\gamma|\_{\Sigma\backslash C})$ refers to the length of the restriction of $\gamma$ to $\Sigma\backslash C$ (this is a union of disconnected curves). Here $d$ should depend on the hyperbolic metric and the choice of $C$, but not the homotopy class of the curve.


The reason I add the "$+d$" is that one could cook up the following example: take a curve that enters $C$ once, and then replace the region that enters $C$ with a straight line connecting the boundary points at which the curve crosses. We then get a closed curve homotopic to $\tilde{\gamma}$, and the length is at most $\ell(\gamma|\_{\Sigma\backslash C})+d$, where $d=\ell(\partial C)$.


It seems true for just about any example I draw, but any proof I attempt to write down becomes terribly buggy and complicated.



Hyperbolic length of curve that does not enter a collar
There is no such constant $d$. The "reason" is that geodesic laminations exist, and are approximated by simple closed curves.


Thought process (which can be ignored):



> 
> Suppose that $\tau$ is a train track in $\Sigma$. Let $R$ be a
> rectangle embedded in $\Sigma - \tau$ with one side running along
> $\partial \Sigma$ and one side running along a branch $b$ of $\tau$.
> If $\gamma$ is carried by $\tau$ then we can use $R$ to perform a
> homotopy of $\gamma$; we take all of the arcs of $\gamma$ running
> along $b$ and isotope a subsegment of each, along $R$ into $C$. We
> call the sides of $R$ in $\partial \Sigma$ and in $b$ "horizontal" and
> we call the other two sides "vertical".
> 
> 
> Let $\gamma'$ be the result of the homotopy. After the homotopy we
> have added many copies of the vertical sides (to the length of
> $\gamma' - C$) and subtracted the same number of many copies of the
> segment along $b$.
> 
> 
> So if we can find a surface $\Sigma$ equipped with a track $\tau$
> where $R$ has short verticals and long horizontals, we are done.
> 
> 
> 


The resulting proof:


Suppose that $\Sigma'$ is a closed surface with a very short geodesic curve $\alpha$. Let $\beta$ and $\beta'$ be distinct, disjoint, simple geodesic loops that cross $\alpha$ exactly once. It will be useful to arrange for $\beta$ to be separating, and $\beta'$ to be non-separating.


Let $C\_\alpha$ be a collar of $\alpha$ where $\partial C\_\alpha$ is much smaller than the "height" of $C\_\alpha$. Since $\alpha$ is short, the curves $\beta$ and $\beta'$ fellow travel, very closely, inside of $C\_\alpha$. By Dehn twisting about $\beta'$ we can now produce a family of curves $\gamma\_n$ that are (a) disjoint from $\beta$ and (b) cross $\alpha$ exactly $n$ times. We produce $\Sigma$ by cutting along $\beta$ and throwing away the component that does not contain the $\gamma\_n$.


Let $C\_\beta$ be the given collar of $\partial \Sigma = \beta$. We now perform a homotopy of $\gamma\_n$ (supported in the remains of $C\_\alpha$) so that the $n$ long arcs of $\gamma\_n \cap C\_\alpha$ lie inside of $C\_\beta$. The length of $\gamma\_n - C$ is less than that of $\gamma\_n$ by (approximately) $n$ times the difference between the height of $C\_\alpha$ and the length of $\partial C\_\alpha$.


