Zair Ibragimov CSUF. Talk at Fullerton College July 14, Geometry of p-adic numbers. Zair Ibragimov CSUF. Valuations on Rational Numbers

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1 integers Talk at Fullerton College July 14, 2011

2 integers Let Z = {..., 2, 1, 0, 1, 2,... } denote the integers. Let Q = {a/b : a, b Z and b > 0} denote the rational. We can add and multiply rational : a/b + c/d = (ad + bc)/(bd) and (a/b) (c/d) = (ac)/(bd) By a valuation on Q we mean a function v : Q [0, + ) satisfying the following three properties: 1 v(x) 0, and v(x) = 0 if and only if x = 0; 2 v(x + y) v(x) + v(y) (the triangle inequality); 3 v(xy) = v(x)v(y). We say that a valuation v is non-archimedean if it satisfies the strong triangle inequality: v(x + y) max{v(x), v(y)}.

3 The standard valuation on Q integers The real absolute value function on Q is a well-known example of a valuation. Recall that x = x if x 0 and x = x if x < 0. Properties (1) and (3) are easily verified. To show property (2), enough to consider three cases: Case 1: x, y 0. Then x + y = x + y = x + y. Case 2: x, y 0. Then x + y = x y = x + y. Case 3: x 0 and y 0. Then x + y is either x + y or x y and x + y = x + y. Hence x + y x + y. The function 0 (defined by x 0 = 0 if x = 0 and x 0 = 1 if x 0) is easily seen to be a valuation on Q. It is referred to as the trivial valuation on Q.

4 Fundamental Theorem of Arithmetics integers Theorem (Fundamental Theorem of Arithmetics) Any integer can be written as a product of primes. That is, n = p a 1 1 pa 2 2 pa k k, where p 1, p 2,..., p k are primes and a 1, a 2,..., a k are positive integers. For example, 360 = and = If we allow a 1, a 2,..., a k to be any integers (positive or negative), then we can write any rational number as a product of primes. That is, x = p a 1 1 pa 2 2 pa k k. For example, =

5 valuation on Q integers Let p be a fixed prime number, i.e., p = 2, 3, 5, 7, 11, 13,.... Any non-zero rational number x can be written uniquely as x = p n a b, where neither a not b is divisible by p and n Z. absolute value on Q is defined by { 0 if x = 0 x p = p n if x 0. For example, since 360/93555 = , 360/ = 2 2, 360/ = 3 2, 360/ = 7 and 360/93555 p = p 0 = 1 for all p 2, 3, 5, 7, 11. Show that the absolute value is a non-archimedean valuation on Q. Show that x p 1 if and only if x Z.

6 Ostrowski s Theorem integers Two valuations v 1 and v 2 on Q is said to be equivalent if there exists c > 0 such that Theorem (Ostrowski) v 2 (x) = v 1 (x) c for all x Q. Any valuation on Q is equivalent to either the trivial valuation or the real absolute value or a absolute value p for some prime p. The classical Analysis, Algebra and Geometry is based on the standard valuation on Q. valuation, which is the topic of my talk, has been introduced in 1897 by Kurt Hensel. A systematic study of began in the late 1960 s.

7 Metrics and Ultrametrics on Q integers A distance function d : Q Q [0, + ) is called a metric on Q if (1) d(x, y) 0 and d(x, y) = 0 if and only if x = y, (2) d(x, y) = d(y, x) and (3) d(x, y) d(x, z) + d(z, y). It is called an ultrametric if it satisfies (1), (2) and (3 ): d(x, y) max{d(x, z), d(z, y)}. Each valuation v on Q induces a metric d to measure a distance between points in Q by d(x, y) = v(x y). The real absolute value induces Euclidean metric on Q, given by d(x, y) = x y. As we mentioned above, much of classical analysis and geometry is based on Euclidean metric. Show that the distance function d(x, y) = x y p on Q induced by the absolute value p is an ultrametric.

8 of Q integers The rational Q, equipped with Euclidean metric or ultrametric, is not adequate to study problems in Algebra or Analysis. For example, the polynomial equation x 2 2 = 0 in Q is not solvable in Q because 2 is not a rational number. Similarly, the series k=0 1 k! in Q does not converge in Q because its sum is equal to e, which is not a rational number. Overcoming these problems require a completion of Q. In other words, we need to add more points to Q. of Q depend on the metric used and is based on the notion of Cauchy sequences.

9 Completion of Q with respect to Euclidean metric integers We say that a sequence {x n } of in Q is a Cauchy sequence if x n x m 0 as n, m. For example, a trivial sequence x, x,... is a Cauchy sequence. So is the sequence 1, 1/2, 1/3,.... Two Cauchy sequences {x n } and {y n } are said to be equivalent if x n y n 0 as n. Let [{x n }] denote the set of all Cauchy sequences in Q that are equivalent to {x n }. The set of all equivalent Cauchy sequences, equipped with the metric d([{x n }], [{y n }]) = lim n x n y n, is called the metric completion of Q. This space can be identified with the real R. Show that a sequence converging to 2 is a Cauchy sequence. Show that all such sequences are equivalent.

10 Completion of Q with respect to ultrametric integers, denoted by Q p, is obtained as the completion of the rational Q with respect to the ultrametric. Since every rational x can be identified with a Cauchy sequence {x, x,... }, every rational number is a number. That is, Q Q p. But there are much more that cannot be identified with rational. In fact, while the rational Q are countable, the Q p are uncountable. There are significant differences between the two completions of Q due to the fact that Euclidean metric satisfies only the triangle inequality while the metric satisfies the strong triangle inequality. For example, R is a connected space while Q p is totally disconnected like the middle-third Cantor set.

11 Completion of Z with respect to ultrametric integers The integers Z, equipped with Euclidean metric n m, is complete. This is because there are no Cauchy sequences in Z except for trivial ones like {1, 1, 1,... }, {2, 2, 2,... } and so on. So the completion of Z with respect to Euclidean metric is the same as Z. That is, this completion does not add any new points to Z. On the other hand, the same integers Z, equipped with the metric n m p, is NOT complete! For example, the following sequence of integers is a nontrivial Cauchy sequence 1, 1 + p, 1 + p + p 2, 1 + p + p 2 + p 3,... integers, denoted by Z p, is the completion of Z with respect to the ultrametric.

12 integers and integers integers Z p can be identified with the formal series a k p k, where a k {0, 1, 2,..., p 1}. k=0 Q p can be identified with the formal series a k p k, where m Z and a k {0, 1, 2,..., p 1}. k=m Show that the sequence of partial sums of each series above is a Cauchy sequence. Use the series representations above to show that Z p = {x Q p : x p 1}.

13 Decomposition of Z integers Our goal is to obtain the integers as the end space of a metric tree. For each k {1, 2, 3,..., p} consider the function f k on Z, defined by f k (z) = pz + (k 1). Show that f k (z) f k (w) p = p 1 z w p for all z, w Z. We have the following decompositions of (Z, p ): Z = B 1 B 2 B p, where B k = f k (Z). For each k = 1, 2,... p, show that B k = {pn + (k 1): n Z}. Show that B k is a ball in (Z, p ) of radius 1/p. Show that the diameter of B k is also equal to 1/p.

14 Decomposition of Z (continued) integers For each fixed n {1, 2,..., p} we have B n = p k=1 B kn, B kn = f k (B n ) = (f k f n )(Z). Show that diam(b kn ) = p 2 and dist(b k1 n, B k2 n) = p 1. Show that each B kn is a ball in (Z, p ) of radius 1/p 2. For each fixed pair n, m {1, 2,..., p} we have B nm = p k=1 B knm, B knm = f k (B nm ) = (f k f n f m )(Z). Show that diam(b knm ) = p 3 and dist(b k1 nm, B k2 nm) = p 2. Show that each B knm is a ball in (Z, p ) of radius 1/p 3. The process continues...

15 Collection of balls in (Z, n m p ) integers Let Z p be the collection of all the balls in (Z, p ) described above. Equip Z p with the distance function h p, h p (A, B) = 2 log Show that h p is a metric on Z p. diam(a B) diam(a) diam(b). Next, we define a metric tree associated with the above decomposition. A metric space X is called a metric tree if each pair of points in it can be joined by a unique arc and this arc is a geodesic segment.

16 Metric tree associated with Z p integers Consider the set Z p as the set of vertices and connect each ball B to its children f 1 (B), f 2 (B),..., f p (B) by an edge of length log p. Note that h p (B, f k (B)) = log p. For each pair (B 1, B 2 ) of distinct vertices there is a unique vertex B of smallest diameter containing B 1 and B 2. Let B, B 1, B 2 be as above. Show that diam(b) = diam(b 1 B 2 ). Let γ k, k = 1, 2, be the arc connecting B to B k. Hence length(γ k ) = h p (B, B k ). Then γ = γ 1 γ 2 is the unique arc joining B 1 and B 2. Since h p (B 1, B 2 ) = h p (B, B 1 ) + h p (B, B 2 ) = length(γ), we conclude that γ is a geodesic segment.

17 End space of T p integers The resulting tree, denoted by T p, is a metric tree. The unit ball Z in Z p is the root of T p and the pair (T p, Z) is referred to as the rooted tree. The end space End(T p, Z) of the pair (T p, Z) is defined to be the set of (infinite) geodesic rays emanating from the root Z. If γ 1 and γ 2 are two such rays, let b(γ 1, γ 2 ) be their bifurcation point, i.e., the vertex where they split and let l(γ 1, γ 2 ) be the length of their common part. The distance between γ 1 and γ 2 is defined to be ρ(γ 1, γ 2 ) = e l(γ 1,γ 2 ). We have l(γ 1, γ 2 ) = h d (Z, b(γ 1, γ 2 )) = n log p, where n is the degree of separation of b(γ 1, γ 2 ) from Z.

18 Z p is the end space of T p integers We conclude that ρ(γ 1, γ 2 ) = e l(γ 1,γ 2 ) = e n log p = p n. Show that ρ is a metric on End(F p, Z). Theorem (Ibragimov, 2010) The space ( End(F p, Z), ρ ) is isometric to (Z p, p ).