March 21, 2013

Mark on Magnitude

Posted by Tom Leinster

Here’s a truly superb talk by Mark Meckes, on The magnitude of metric spaces.

The talk was at Banff last week, as part of a meeting on The interplay of convex geometry and Banach space theory.

Maybe I was predisposed to like the talk, its subject matter being so close to my heart. But I was awed by how well thought-out it was: it’s just incredibly clear. And he goes from nothing to the cutting edge in under half an hour, without appearing to hurry. If you’re on the internet looking for diversion, look no further!

Posted at March 21, 2013 1:12 AM UTC

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Re: Mark on Magnitude

I’ll take the opportunity to make a small correction. In so far as deciding on a name can be called “work” (and I suppose it must be—marketing companies are paid millions for this), the name “magnitude” is joint work with Simon Willerton.

Posted by: Tom Leinster on March 21, 2013 1:44 AM | Permalink | Reply to this

Re: Mark on Magnitude

Nice talk! What are the examples of spaces whose box dimension is not the magnitude dimension? Are they suitably weird?

Posted by: Simon Willerton on March 21, 2013 8:36 AM | Permalink | Reply to this

Re: Mark on Magnitude

Thanks!

Here’s a big class of examples where magnitude dimension and box dimension differ. Let $A$ be a compact subset of $n$-dimensional Euclidean space with positive Lebesgue measure, and write $A^\alpha$ for $A$ equipped with the metric $d(x,y) = \|x - y \|^\alpha$, where $\|\cdot\|$ is the Euclidean norm and $\alpha \in (0,1]$. Then $A^\alpha$ has magnitude dimension $n$ for each $\alpha$ (by the last two theorems in this paper, which generalize results of Tom’s for the case $\alpha = 1$; there are also $\ell_p$ versions of this).

On the other hand, it’s an easy exercise to check that the box dimension of $A^\alpha$ is $1/\alpha$ times the box dimension of $A^1$, thus $n/\alpha$.

Is that weird enough for you? The set $A$ can be as tame as you like, but the metric on $A^\alpha$ isn’t so easy to visualize — distances get stretched a lot at small scales and shrunk at large scales. It’s a nice related puzzle to explicitly isometrically embed an interval, equipped with the square root of the usual metric, into a Hilbert space — but it doesn’t embed into a finite dimensional subspace.

Posted by: Mark Meckes on March 21, 2013 3:30 PM | Permalink | Reply to this

Re: Mark on Magnitude

More recently, inspired by work I learned about in Banff, I proved that (under some hypotheses which are always satisfied in the cases discussed above) box dimension is less than or equal to magnitude dimension. Since Hausdorff dimension is less than or equal to box dimension, this is an improvement of a result mentioned on the slides.

This was somewhat distressing, though, since it contradicts the observation I made in the comment above. Tom helped me find the error behind the above observation. The upshot is that what I actually proved in my paper is that the magnitude dimension of such spaces $A^\alpha$ is $n/\alpha$ — precisely equal to the box dimension.

So as far as I know right now, magnitude dimension may always be equal to box dimension. This would be a big piece of support for the point of view that magnitude measures the “number of points you can see if you squint your eyes”.

Posted by: Mark Meckes on April 7, 2013 9:27 PM | Permalink | Reply to this

Re: Mark on Magnitude

And in fact I can now prove that in Euclidean space, magnitude dimension is always equal to box dimension. In fact box dimension is equal to something else closely related to magnitude dimension. The fact that this other quantity coincides with magnitude dimension in Euclidean space depends on a nontrivial result in potential theory, and I don’t know (yet) how far it generalizes.

Posted by: Mark Meckes on April 17, 2013 6:49 PM | Permalink | Reply to this
Read the post Magnitude of Metric Spaces: A Roundup
Weblog: The n-Category Café
Excerpt: Resources on magnitude of metric spaces.
Tracked: April 7, 2013 9:54 PM

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