Fix a prime p. For integers x, let v

_{p}(x) denote the largest power of p dividing x (to save some time typing this, I'll drop the p from the notation, and just write v(x)). Note that v(xy)=v(x)+v(y), which looks logarithmic (as well it should, we're talking about powers...). So lets extend the definition to include rationals, via v(a/b)=v(a)-v(b) (and check that this is well defined). I should comment that we set v(0)=∞, with usual conventions for working with ∞. Anyway, our v has another interesting property: v(x+y)≥min(v(x),v(y)), with equality if v(x)≠v(y). To see this, let a=v(x) and b=v(y) and we can then write x=p

^{a}m, y=p

^{b}n where p doesn't divide m or n (since p is prime, we could say this as "p doesn't divide mn"). Now v(x+y)=v(p

^{a}m+p

^{b}n), so we factor our the least power of p, and we see that v(x+y) is at least that power. Furthermore, suppose a<b. Then x+y=p

^{a}(m+p

^{b-a}n), and since p doesn't divide m, p also doesn't divide m+p

^{b-a}n (here we use b-a≠0), so v(x+y) is exactly the min of v(x) and v(y), as claimed.

We now create a new absolute value on the rationals, via |x|

_{p}=p

^{-v(x)}(I will, again, drop the p from the notation and just write |x|). Translating the two properties we now know about v(x), we see that |xy|=|x|*|y| and |x+y|≤max(|x|,|y|) (which is, in turn, no larger than |x|+|y|). This new absolute value is stronger than the normal absolute value, in the sense that it satisfies the strong triangle inequality (the second property above). This new absolute value is an example of a "non-Archimedean" norm. A norm is Archimedean if it satisfies the following property: if x is smaller than y, there is a positive integer n such that nx is bigger than y (stick enough small line segments together to make as large a number as you want). Notice that our new absolute value cannot possibly have this property, because for integers n, v(n)≥1, so |n|≤1, and then the multiplicative property tells us |nx|≤|x|, so we've actually made (potentially) our length smaller by taking more copies of it.

Given an absolute value, the thing to do is define a new metric. So let d(x,y)=|x-y|. Keep in mind this is our new absolute value, so I should write d

_{p}(x,y)=|x-y|

_{p}(but it takes longer). Anyway, this metric satisfies the "ultrametric" property: For any z, d(x,y)≤max(d(x,z),d(z,y)) (this is stronger than the normal triangle inequality). What's more: if d(x,z) is different from d(z,y), then d(x,y)=max(d(x,z),d(z,y)). This can be interpreted as saying that "all triangles are isosceles".

With a metric you get the associated topology. Let b(a,r) denote the set of all points whose distance from a is less than r, and B(a,r) be those points whose distance is less than or equal to r. These balls (as they are called) have some interesting properties, different from balls in the normal topology on the reals. I think I'll leave it for you to prove that if x is in b(a,r), then b(a,r)=b(x,r) (I think about this as saying that every point is the center). Using this, we can see that if b(a,r) intersects b(s,t), then one is contained in the other: if x is in the intersection, then b(a,r)=b(x,r), and b(s,t)=b(x,t) and the ball with larger radius contains (possibly with equality) the ball of smaller radius. Now every ball is open in the topology (that's how the topology is defined), but what is new with this topology is that all the balls b(a,r) are also closed.

I'd like to leave you with one last fact about these balls: B(0,1) is the disjoint union of b(0,1), b(1,1),... b(p-1,1). The proof I'll summarize here is from Gouvea's book: Suppose a/b is in B(0,1). Taking a/b to be in lowest terms, we may assume p doesn't divide b (since its distance from 0 is no more than 1...). Consider the p integers a, a-b, a-2b,... a-(p-1)b. It is not too hard to see that these are all distinct mod p, and since there are p of them, exactly 1 of them, say a-ib, is equivalent to 0 mod p. Now check that a/b is in b(i,1). Anyway, that's a start. Perhaps you'd rather thing about it algebraically. If I tell you B(0,1) is a subring of the rationals with maximal ideal b(0,1), can you guess what the b(i,1) are?

So anyway, those are the low hanging fruits of the new metric. Fun things to see, I hope. If you want to learn more, the books by Gouvea and Koblitz were the main ones I looked at when preparing my talk.

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