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Let $(X, d)$ be an ultrametric space. In particular, X satisfies the strong triangle inequality: for any $x, y, z \in X$, we have $$d(x,y) \leq \max\{d(x,z), d(y,z)\}.$$

I want to show that $X$ satisfies the Gromov four-point condition: $$ (x \cdot y)_w \geq \min\{(x\cdot z)_w, (z\cdot y)_w\} $$ where $x,y,z,w \in X$ and $(x \cdot y)_w = \frac{1}{2} (d(x,w) + d(y,w) - d(x,y)) $.

I feel this should be relatively straightforward, but I have not been successful in using the strong triangle inequality to somehow deduce the four-point condition. I would appreciate any hints.

Dalop
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2 Answers2

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This is true, but the proof that I know is tedious.

First, observe that if $x, y, z$ are points in an ultrametric space $(X,d)$ then the largest two of the distances $d(x,y), d(y,z), d(z,x)$ are equal.

Now, consider a quadruple $x, y, z, w$ in an ultrametric space $(X,d)$. WLOG,
$$ d(x,y)=a\le d(x,z)=b \le d(x,w)=c. $$ Our first observation then implies that $$ d(y,z)=b, d(z,w)=c, d(w,y)=c. $$ The rest is a straightforward computation of 12 Gromov-products between points in $\{x, y, z, w\}$ and checking the 4-point inequalities. (With 4 exceptions, these will be equalities.) I leave these computations to you.

Moishe Kohan
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There is a "proof by picture" relying on the intuition that ultrameric spaces are spaces of leafs of trees. This is always true for finite ultrametric spaces and since the $4$-point condition only requires considering $4$ points at a time, we may use this observation. We shall need two observations.

Observation 1 If $X$ is $0$-hyperbolic and $E\subset X$, then $E$ is also $0$-hyperbolic (this is obvious).

Observation 2 If $X$ is a tree (in our case, it suffices to consider metric spaces that arise as combinatorial trees with weighted edges and equipped with the shortest path metric), then $X$ is $0$-hyperbolic.

Observation 2 follows from the fact that $(x\cdot y)_w$ is the distance from $w$ to the edge connecting $x$ to $y$ (here drawing a picture of a tree is a very good idea).

Now, let $X$ be ultrametric and suppose $x,y,z,w\in X$. As in the answer of Moishe Kohan, we may WLOG take $$ d(x,y)=a\leq d(x,z)= b \leq d(x,w)=c. $$ And, as is also stated in their answer, we then have $$ d(y,z)=b;\; d(z,w)=c;\; d(w,y)=c. $$

However, instead of making 12 tedious calculations and then comparing some numbers, we can instead construct a tree $T$ and embedd $(\{x,y,z,w\}, d)$ into $T$. The tree is given by the following picture.

The tree $T$ with leafs $x,y,z,w$

The weight of the two edges without a label can be easily calculated from the remaining weights. Since $\{x,y,z,w\}$ sits isometrically inside the tree $T$ (equipped with the shortest path metric), combination of the observations above yields that $\{x,y,z,w\}$ is hyperbolic and since $x,y,z,w$ were arbitrary, also $X$ is $0$-hyperblic.

kiliroy
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  • You'd have to verify the claim "it suffices to consider...": it's true but requires a proof. And there are two combinatorially different trees you would have to consider and you would also have to argue that there are no other trees to deal with. This is where it becomes tedious. – Moishe Kohan Mar 13 '25 at 14:40