# When is a metric space a snowflake?

Let $$(X,d)$$ be a metric space. For any $$0<\epsilon<1$$, we call the metric space $$(X,d^{\epsilon})$$; where $$d^{\epsilon}(x,y)\triangleq (d(x,y))^{\epsilon}$$ the $$\epsilon$$-snowflake of $$(X,d)$$.

My question is, given $$(X,d)$$ and some $$\epsilon\in(0,1)$$, under what conditions can we deduce that there exists some other metric space $$(Z,\rho)$$ such that $$(X,d)$$ is the $$\epsilon$$-snowflake of $$(Z,\rho)$$; i.e.: $$(X,d)=(Z,\rho^{\epsilon})?$$

• I suppose you don't consider “there exists $e>1$ such that $d^e$ satisfies the triangle inequality” a satisfactory answer to your question. 😁 Mar 15, 2021 at 12:14
• Haha, nope also note that I require that $0<\epsilon<1$ ;). Mar 15, 2021 at 12:21
• @Gro-Tsen In the question, $\epsilon$ is fixed, so instead of "there exists $e>1$", you want $e=1/\epsilon$. Mar 15, 2021 at 18:27
• @Bernard_Karkanidis Why not? I can hardly imagine a simpler condition than the one Gro-Tsen suggests. If there is a triple satisfying $d^{1/\epsilon}(x,z) > d^{1/\epsilon}(x,y) + d^{1/\epsilon}(y,z)$ then the answer is no, otherwise it is yes. Mar 17, 2021 at 12:59

The paper contains a number of equivalent characterizations of the property that a metric space $$X$$ is bi-Lipschitz equivalent to a snowflake. Some of the characterizations in the paper require that the space is embedded in a certain Banach space, and some do not.