# Shortcutting quasigeodesics

Let $$\Gamma$$ be a connected graph, let $$\lambda \ge 1$$ and $$c \ge 0$$ be some constants. Recall that a combinatorial path $$p$$ in $$\Gamma$$ is said to be $$(\lambda,c)$$-quasigeodesic if for every combinatorial subpath $$q$$ of $$p$$ one has $$\ell(q) \le \lambda d(q_-,q_+)+c,$$ where $$\ell(q)$$ is the length of $$q$$, $$q_-$$ and $$q_+$$ are the endpoints of $$q$$, and $$d(\cdot,\cdot)$$ is the standard metric on $$\Gamma$$.

Question 1: suppose that a path $$s$$ in $$\Gamma$$ has been obtained from a $$(\lambda,c)$$-quasigeodesic path $$p$$ by replacing some (combinatorial) subpaths of $$p$$ with geodesics. Is it true that this new path $$s$$ is again $$(\lambda,c)$$-quasigeodesic?

Intuitively, the answer should be "Yes", because replacing subpaths with geodesics ("shortcutting") should only improve the quasigeodesicity constants. However, I do not see how to prove this.

If the answer to Question 1 is negative, the natural next question is the following:

Question 2: suppose that a path $$s$$ in $$\Gamma$$ has been obtained from a $$(\lambda,c)$$-quasigeodesic path $$p$$ by replacing some subpaths of $$p$$ with geodesics. Is it true that this new path $$s$$ is $$(\lambda',c')$$-quasigeodesic, where the constants $$\lambda' \ge 1$$, $$c' \ge 0$$ depend only on $$\Gamma$$, $$\lambda$$ and $$c$$?

It's not hard to show that the answer to Question 2 is positive when the graph $$\Gamma$$ is $$\delta$$-hyperbolic, but I do not have a proof or a counter-example for more general graphs.

• How about ${\mathbb Z}^2$? There are many geodesics with the same endpoints there. Oct 12, 2021 at 14:20
• here is an easy counterexample in $\mathbb Z^2$: the path [$k$ steps up, $k$ steps right, $k$ steps down] is a $(3,0)$ quasi-geodesic, but replacing the first $2k$ steps by the geodesic [$(k-1)$ steps right, $k$ steps up, one step right] does not preserve this. Oct 12, 2021 at 14:31
• @FlorianLehner: thanks, that's a great counter-example for both questions! I have thought about the same path, but treated it as a $(1,2k)$-quasigeodesic, instead of changing the multiplicative constant! What if I require $\lambda=1$? Oct 12, 2021 at 14:43
• @AshotMinasyan: If I'm not mistaken, then a path is a $(1,c)$-quasigeodesic if and only if its length exceeds the distance between its endpoints by at most $c$, i.e. we only need to check the whole path, not all sub-paths. This property is certainly preserved under replacing a subpath by a shorter (or equally long) one. Oct 12, 2021 at 15:26
• @FlorianLehner: you are right, thanks for your comments. If you decide to put them as an answer I would be happy to accept it. Oct 12, 2021 at 21:14

As suggested, I am turning my comments into an answer. The answer to both questions is negative for any $$\lambda > 1$$, and positive for $$\lambda = 1$$.
For $$\lambda = 1+\epsilon$$ note that in $$\mathbb Z^2$$, the concatenation of the (unique) geodesics from $$(0,0)$$ to $$(0,k)$$, from $$(0,k)$$ to $$(n,k)$$ and from $$(n,k)$$ to $$(n,0)$$ is $$(1+\epsilon,0)$$-quasigeodesic given $$\frac{2k}{n} < \epsilon$$.
If we replace the segment between $$(0,0)$$ and $$(n-1,k)$$ by the geodesic from $$(0,0)$$ via $$(n-1,0)$$ to $$(n-1,k)$$ then the points $$(0,n-1)$$ and $$(0,n)$$ have distance $$1$$ in $$\mathbb Z^2$$, but distance $$2k+1$$ on the path. So this path can't be $$(\lambda',c')$$-quasigeodesic unless $$\lambda'+c' \geq 2k+1$$.
For $$\lambda = 1$$, note that a path $$p$$ is $$(1,c)$$-quasigeodesic if and only if $$\ell(p) \leq d(p_-,p_+)+c$$; in other words, it suffices to check the condition for the whole path rather than for all sub-paths. This property is clearly preserved when replacing any subpath by a shorter or equally long one.