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Local P-Adic Analysis Local p-adic analysis Peter Schneider Course at MPI MiS Leipzig, August 30 - Sept. 2, 2021 Contents 1 Ultrametric spaces 2 2 Nonarchimedean fields 4 3 Convergent series 8 4 Differentiability 9 5 Power series 11 6 Locally analytic functions 17 7 Charts and atlases 20 8 Manifolds 22 9 The tangent space 26 10 Reminder: Locally convex K-vector spaces 33 11 The topological vector space Can(M; E) 35 References 36 1 1 Ultrametric spaces We recall the very basic notions. Definition. A metric space (X; d) is called ultrametric if the strict triangle inequality d(x; z) ≤ max(d(x; y); d(y; z)) for any x; y; z 2 X is satisfied. Remark. i. If (X; d) is ultrametric then (Y; d jY × Y ), for any subset Y ⊆ X, is ultrametric as well. ii. If (X1; d1);:::; (Xm; dm) are ultrametric spaces then the cartesian pro- duct X1 × ::: × Xm is ultrametric with respect to d((x1; : : : ; xm); (y1; : : : ; ym)) := max(d1(x1; y1); : : : ; dm(xm; ym)) : Let (X; d) be an ultrametric space. Lemma 1.1. For any three points x; y; z 2 X such that d(x; y) 6= d(y; z) we have d(x; z) = max(d(x; y); d(y; z)) : Proof. Suppose that d(x; y) < d(y; z). Then d(x; y) < d(y; z) ≤ max(d(y; x); d(x; z)) = max(d(x; y); d(x; z)) = d(x; z) : Hence d(x; y) < d(y; z) ≤ d(x; z) and then d(x; z) ≤ max(d(x; y); d(y; z)) ≤ d(x; z) : Let a 2 X be a point and " > 0 be a real number. We call − B"(a) := fx 2 X : d(a; x) ≤ "g and B" (a) := fx 2 X : d(a; x) < "g the closed and the open ball, respectively, around a of radius ". But be careful! Lemma 1.2. i. Every ball is open and closed in X. − − ii. For b 2 B"(a), resp. b 2 B" (a), we have B"(b) = B"(a), resp. B" (b) = − B" (a). 2 Proof. Observe that the open, resp. closed, balls are the equivalence classes of the equivalence relation x ∼ y, resp. x ≈ y, on X defined by d(x; y) < ", resp. by d(x; y) ≤ ". So any point of a ball can serve as its midpoint. Exercise. The radius of a ball is not well determined. Corollary 1.3. For any two balls B and B0 in X such that B \ B0 6= ; we have B ⊆ B0 or B0 ⊆ B. Remark. If the ultrametric space X is connected then it is empty or consists of one point. Lemma 1.4. Let U ⊆ X be an open subsets and let "1 > "2 > : : : > 0 be a strictly descending sequence of positive real numbers which converges to zero; then any open covering of U can be refined into a DECOMPOSITION of U into balls of the form B"i (a). As usual the metric space X is called complete if every Cauchy sequence in X is convergent. Lemma 1.5. A sequence (xn)n2N in X is a Cauchy sequence if and only if limn!1 d(xn; xn+1) = 0. There is a stronger concept than completeness. For a subset A ⊆ X call d(A) := supfd(x; y): x; y 2 Ag the diameter of A. Corollary 1.6. Let B ⊆ X be a ball with " := d(B) > 0 and pick any point − a 2 B; we then have B = B" (a) or B = B"(a). Consider now a descending sequence of balls B1 ⊇ B2 ⊇ ::: ⊇ Bn ⊇ ::: in X. Suppose that X is complete and that limn!1 d(Bn) = 0. We pick points xn 2 Bn and obtain the Cauchy sequence (xn)n2 . Since each Bn is N T closed we must have x := limn!1 xn 2 Bn and therefore x 2 n Bn. Hence \ Bn 6= ; : n2N T But without the condition on the diameters the intersection n Bn can be empty (see the next section for an important example). This motivates the following definition. 3 Definition. The ultrametric space (X; d) is called spherically complete if any descending sequence of balls B1 ⊇ B2 ⊇ ::: in X has a nonempty intersection. Lemma 1.7. i. If X is spherically complete then it is complete. ii. Suppose that X is complete; if 0 is the only accumulation point of the set d(X × X) ⊆ R+ of values of the metric d then X is spherically complete. Lemma 1.8. Suppose that X is spherically complete; for any family (Bi)i2I of closed balls in X such that Bi \ Bj 6= ; for any i; j 2 I we then have T i2I Bi 6= ;. 2 Nonarchimedean fields Let K be any field. Definition. A nonarchimedean absolute value on K is a function j j : K −! R which satisfies: (i) jaj ≥ 0, (ii) jaj = 0 if and only if a = 0, (iii) jabj = jaj · jbj, (iv) ja + bj ≤ max(jaj; jbj). Exercise. i. jn · 1j ≤ 1 for any n 2 Z. × × ii. j j : K −! R+ is a homomorphism of groups; in particular, j1j = j − 1j = 1. iii. K is an ultrametric space with respect to the metric d(a; b) := jb − aj; in particular, we have ja + bj = max(jaj; jbj) whenever jaj 6= jbj. iv. Addition and multiplication on the ultrametric space K are continuous maps. Definition. A nonarchimedean field (K; j j) is a field K equipped with a nonarchimedean absolute value j j such that: 4 (i) j j is non-trivial, i. e., there is an a 2 K with jaj 6= 0; 1, (ii) K is complete with respect to the metric d(a; b) := jb − aj. Examples. { Fix a prime number p. Then −r r m jajp := p if a = p n with r; m; n 2 Z and p 6 j mn is a nonarchimedean absolute value on the field Q of rational numbers. The corresponding completion Qp is called the field of p-adic numbers. Note that jQpjp = pZ[f0g. Hence Qp is spherically complete by Lemma 1.7.ii. { Let K=Qp be any finite extension of fields. Then [K:Qqp] jaj := jNormK=Qp (a)jp is the unique extension of j jp to a nonarchimedean absolute value on K. The corresponding ultrametric space K is complete and spherically complete and, in fact, locally compact. (See [Ser] Chap. II §§1-2.) alg { The algebraic closure Qp of Qp is not complete. Its completion Cp is algebraically closed but not spherically complete (see [Sch] §17 and Cor. 20.6). In the following we fix a nonarchimedean field (K; j j). By the strict triangle inequality the closed unit ball oK := B1(0) is a subring of K, called the ring of integers in K, and the open unit ball − mK := B1 (0) × is an ideal in oK . Because of oK = oK nmK this ideal mK is the only maximal ideal of oK . The field oK =mK is called the residue class field of K. Exercise 2.1. i. If the residue class field oK =mK has characteristic zero then K has characteristic zero as well and we have jaj = 1 for any nonzero a 2 Q ⊆ K. ii. If K has characteristic zero but oK =mK has characteristic p > 0 then we have log jpj − log p jaj = jajp for any a 2 Q ⊆ K; in particular, K contains Qp. 5 A nonarchimedean field K as in the second part of Exercise 2.1 is called a p-adic field. Lemma 2.2. If K is p-adic then we have n−1 jnj ≥ jn!j ≥ jpj p−1 for any n 2 N. Proof. We may obviously assume that K = Qp. Then the reader should do this as an exercise but also may consult [B-LL] Chap. II §8.1 Lemma 1. Now let V be any K-vector space. Definition. A (nonarchimedean) norm on V is a function k k : V −! R such that for any v; w 2 V and any a 2 K we have: (i) kavk = jaj · kvk, (ii) kv + wk ≤ max(kvk; kwk), (iii) if kvk = 0 then v = 0. Moreover, V is called normed if it is equipped with a norm. Exercise. i. kvk ≥ 0 for any v 2 V and k0k = 0. ii. V is an ultrametric space with respect to the metric d(v; w) := kw−vk; in particular, we have kv + wk = max(kvk; kwk) whenever kvk 6= kwk. iii. Addition V × V −−!+ V and scalar multiplication K × V −! V are continuous. Lemma 2.3. Let (V1; k k1) and (V2; k k2) let two normed K-vector spaces; a linear map f : V1 −! V2 is continuous if and only if there is a constant c > 0 such that kf(v)k2 ≤ c · kvk1 for any v 2 V1 : Definition. The normed K-vector space (V; k k) is called a K-Banach space if V is complete with respect to the metric d(v; w) := kw − vk. n Examples. 1) K with the norm k(a1; : : : ; an)k := max1≤i≤n jaij is a K-Banach space. 6 2) Let I be a fixed but arbitrary index set. A family (ai)i2I of elements in K is called bounded if there is a c > 0 such that jaij ≤ c for any i 2 I. The set 1 ` (I) := set of all bounded families (ai)i2I in K with componentwise addition and scalar multiplication and with the norm k(ai)ik1 := sup jaij i2I is a K-Banach space.
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