Wednesday, February 09, 2011

it's cold, and i'm thinking about snowflakes.

lately it seems like all i write about is teaching. i suppose that's where the recent frustration comes from, especially with this class that is new to me.

as for research frustrations, i can handle those. this line of work is so full of setbacks (and occasionally, advances) that i've developed a sense of humor about it.

that said, before the teaching rants begin anew, here's a little problem that came to mind today. it's nothing serious (not for research or anything) but right now i can't seem to answer it.
given a metrιc space $(X,d)$ and a number $0 \leq \epsilon \leq 1$, we call $X^\epsilon := (X,d^\epsilon)$ the $\epsilon$-snowflake of $X$.

it's not hard to show that $X^\epsilon$ is also a metrιc space; usually one finds such a problem in textbooks.

question: is the real line the snowflake of some metrιc space? what about the plane, or other Euclιdean spaces? for those experts out there: if the answer is yes, then can one choose the metrιc space to be doubling?

(epilogue: the answer is easy. thanks for the comments, guys.)
as for how it came up, i was thinking about embeddings of metric spaces, including the recent result discussed in a preprint of naοr and neιman. apparently assοuad's embedding theorem now works with a fixed dimension, regardless of the snowflaking parameter $\epsilon$.

roughly speaking, the doubling condition on a metrιc space is a finite-dimensιonality condition. so the naοr-neιman result suggests (to me) that one shouldn't have to push too far from the dimension of the given space in order to embed it.


Leonid said...

Real line is not a snowflake of anything (unless you really want to allows epsilon=1...). Indeed, otherwise |x-y|^p would be comparable to a metric for some p>1, which it is not (the triangle inequality, applied to fine partitions, would collapse such a metric). Laakso's appendix to the Tyson-Wu snowflake paper gives a very nice answer to the "snowflake or not" question in general.

janus said...

thanks, L. it sounded like this sort of thing has been investigated before. i'll check out the reference.

Anonymous said...

Also, If it were a snowflake of something, then it would be homeomorphic to space with Hausdorff dimension less than one, which would therefore have topological dimension less than one. Same applies for R^n, or more generally, any space whose Hausdorff and topological dimensions coincide.

Anonymous said...

The above, by the way, is one way to see that a snowflake space has no rectifiable curves.