A guest post by Dylan McKay, recently graduated from Georgia Tech and soon to be PhD student at Stanford.
[Editor's Note: Turns out the given solution doesn't work and whether a metric group product over the non-negative reals exists remains open. Update 7/10/18: A non-constructive solution was posted on Mathoverflow back in 2010]
[Editor's Note: Turns out the given solution doesn't work and whether a metric group product over the non-negative reals exists remains open. Update 7/10/18: A non-constructive solution was posted on Mathoverflow back in 2010]
Here a cute puzzle motivated by a pair of undergrads and their poor understanding of what the phrase “Algebraic Geometry” really should mean:
Find a function f from the nonnegative reals to the nonnegative reals that satisfies the group axioms and the metric axioms, or prove that there is no such function. That is, find an f:RxR→R such that (R,f) is a group and a metric space. (I am using R to refer to the set of nonnegative real numbers).
As a quick reminder, the group axioms are:- Closure: f(a,b) must be in R
- Identity: There must exist an element e such that for all a, f(e,a)=f(a,e)=a
- Associative: for all a,b,c, f(f(a,b),c) = f(a,f(b,c))
- Inverse: for all a, there must exist a b such that f(a,b)=e
And the metric axioms are:
- f(a,b) = 0 iff a=b
- f(a,b) <= f(a,c) + f(c,b)
- f(a,b) = f(b,a)
One really bizarre thing about this supposed function is that, what this metric does is essentially guarantee that, given some number x, every other number has a distance from x that is unique -- no other number is exactly that distance away! This is quite counterintuitive to how we think about distance. Each number is its own distance from 0, though, which is very much in line with intuition.
(If you want to solve the puzzle yourself, don't read any further until you do! Unless you want some hints, in which case, look at the next paragraph.)
We can, however, make some other observations that point us in the right direction. For instance, f(e,e) = 0 from the metric axioms and f(e,e) = e from the group axioms, so you get that e = 0. From this and the metric axioms, you get that every element must be it’s own inverse!
Here, we are reminded of a pretty familiar and trusty function, this exclusive-or (XOR) function! And indeed, this will lead us a to a solution.
Here is our candidate function:
Consider x and y in their binary representations. If they do not have the same number of bits to the left of the decimal point (or should it be called the binary point?), pad the one with fewer such bits with 0’s so they have the same number of such bits. Then f(x,y) will be the number represented in binary by the bit-wise exclusive or of these two strings (maintaining the same number of bits to the left of the decimal point, of course). And there we have it: our function!
Now of course, we still need to prove that it satisfies the axioms. But if you do not believe that it does, work it out for yourself! Each axiom is pretty simple to check, especially as we are (for the most part) familiar with this as an extension of an idea for some smaller groups. Well, all of them except our good friend the triangle inequality. This one actually isn’t too bad, but if you have trouble, I will include a hint at the bottom of the post.
Anyway, I hope you liked our puzzle!
Thanks!
-Dylan
Hint for triangle inequality: XOR and addition are really similar, but there is a key difference between them that make a XOR b <= a + b.
