Show that this quasi-geodesic ray is not Gromov hyperbolic?

Question

Consider the spiral (t, log(1+t)) (given in polar coordinates); it inherits the Euclidean metric from the plane. I have to show that this spiral (a quasi-geodesic) is not Gromov hyperbolic. In other words, it isn't hyperbolic for any > 0. I've been stuck on this problem for a while now. My professor suggested that I should use the 4-point condition, but I haven't gotten anywhere with that either. Any ideas?

Answer 1

Hints:

Step 1. Show that your spiral is Hausdorff-close to the spiral $S$ given in polar coordinates by the parameteric equation $$ (r, \log(r)), r\ge 1. $$

Step 2. Verify that the spiral $S$ is not 0-hyperbolic.

Step 3. Verify that the spiral $S$ maps into itself under the sequence of dilations $$ f_n: (r, \theta)\mapsto (e^{2\pi n} r, \theta), n\in {\mathbb N}, $$ i.e. $f_n(S)\subset S$ for each $n$.

Step 4. Combine steps 2 and 3 to conclude that $S$ is not $\delta$-hyperbolic for any $\delta<\infty$.

Step 5. Make the same conclusion for the original spiral using Step 1 and your knowledge (I assume you know this staff) that the 4-point hyperbolicity is preserved under $(1,C)$-quasi-isometries.