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.


1 Answer 1


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 $\partial C_\alpha$.


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