TITLE: A Johnson-Lindenstrauss lemma for finite fields?
QUESTION [9 upvotes]: Given $m$ points in $\mathbb{R}^N$, the Johnson-Lindenstrauss lemma guarantees the existence of a linear operator $\mathbb{R}^N\rightarrow\mathbb{R}^n$ that nearly preserves pairwise distances between the points.  Here, we can take $n=\Omega(\log(m)/\varepsilon^2)$, where $\varepsilon$ is the level of distortion.
Is there a similar result for points in a vector space over a finite field, e.g. $\mathbb{F}_2^N$?  I assume a result of this form would be in terms of Hamming distance.

REPLY [2 votes]: The problem with Hamming distance is that it's bounded above by $N$, so if you have a subset of $\mathbb{F}_2^N$ with Hamming distances in that range, you're not going to be able to embed it in $\mathbb{F}_2^n$ for $n$ much smaller than $N$.
Perhaps more natural is to build in a scaling, so that you want to find an embedding $f$ of the subset $S$ of $\mathbb{R}^N$ into $\mathbb{R}^n$ in such a way that
$d(f(x),f(y)) \approx \frac{n}{N}\cdot d(x,y)$.
In other words, given two vectors $x$ and $y$, you want the proportion of coordinates in which they agree to be more or less left alone by the projection.  Then you could try a random projection as in Johnson-Lindenstrauss -- i.e. show that (if indeed this is true) a random choice of one of the $N \choose n$ coordinate projections gives you low distortion in this sense, when $n$ is not too horrifically small.