The Gauss congruences and the Nielsen numbers

Transcription

1 The Gauss congruences and the Nielsen numbers Sogang University, Seoul, KOREA Department of Mathematics Sookmyung Women s University October 8, 2015

2 Begin with Thanks for invitation!

3 Outline I would like to give a survey talk on based on the papers Gauss congruences + Nielsen numbers A. Fel shtyn and J. B. Lee, The Nielsen and Reidemeister numbers of maps on infra-solvmanifolds of type (R), Topology Appl., 181 (2015), A. Fel shtyn and J. B. Lee, The Nielsen numbers of iterations of maps on infra-solvmanifolds of type (R) and periodic orbits, arxiv:

4 Famous Congruences in Number Theory FERMAT S LITTLE THEOREM For any a Z and p prime with gcd(a, p) = 1, a p 1 1 mod p For any a Z and p prime, a p a mod p (FC)

5 Pierre de Fermat Figure: Pierre de Fermat,

6 Famous Congruences in Number Theory EULER S THEOREM For any a Z and n N with gcd(a, n) = 1, a ϕ(n) 1 mod n The Euler s phi function: ϕ(n) = #{k 1 k n, gcd(n, k) = 1}. If p is a prime then ϕ(p) = p 1. Hence Euler s Theorem Fermat s Little Theorem

7 Proof of Euler s theorem Recall Consider ϕ(n) = #{k 1 k n, gcd(n, k) = 1}. K := {k 1,, k ϕ(n) 1 k i n, gcd(n, k i ) = 1} ak = {ak 1,, ak ϕ(n) } Since gcd(a, n) = 1, two sets K and ak are the same modulo n. Thus (ak 1 ) (ak ϕ(n) ) k 1 k ϕ(n) mod n hence a ϕ(n) 1 mod (n)

8 Leonhard Euler Figure: Leonhard Euler,

9 Euler Congruences EULER S THEOREM For a Z and n N with gcd(a, n) = 1, a ϕ(n) 1 mod n (ET) EULER CONGRUENCES For a Z and p, r N with p prime and gcd(a, p) = 1, a pr a pr 1 mod p r (EC) Since ϕ(p r ) = p r p r 1, we have Euler s Theorem Euler Congruences

10 Gauss Congruence in Number Theory GAUSS CONGRUENCE For any a Z and n N, µ(d)a n d 0 mod n (GC) Here, µ is the Möbius function.

11 Carl Friedrich Gauss Figure: Carl Friedrich Gauss,

12 Möbius function µ Möbius function is µ : N { 1, 0, 1} defined by 1 if n is square-free with an EVEN number of prime factors, µ(n) = 1 if n is square-free with an ODD number of prime factors, 0 if n has a squared prime factor. Remark that µ(n) = sum of primitive nth roos of unity.

13 Möbius function µ For example, µ(4) = 0 ( 4 = 2 2, squared) µ(1) = 1 ( square-free and 0 prime factor) µ(2) = 1 ( square-free and 1 prime factor) µ(3) = 1 ( square-free and 1 prime factor) On the other hand, note that x 3 = 1 has three roots 1, ω and ω 2 = ω. Hence µ(3) = ω + ω = 1.

14 Möbius Inversion Formula Möbius Inversion Formula g(n) = f (d) n 1 f (n) = = ( n ) µ(d)g d ( n ) µ g (d) n 1 d

15 Equivalences We have seen that (ET) (EC), and (ET) (FC). In (GC) µ(d)a n d = ( n ) µ a d 0 d mod n take n = p r, then µ(n/d) = 1 when d = p r, 1 when d = p r 1, and 0 otherwise. Hence (GC) (EC) Fact is that the above congruences are equivalent. (GC) (EC) (ET) (FC)

16 Question Equivalent congruences: (GC): µ(d)a n d 0 mod n (EC): a pr a pr 1 mod p r QUESTION: Generalize these congruences from integers a to some other mathematical integer invariants.

17 Regarding the question The traces of all integer matrices A, tr(a), by µ(d)tr(a n d ) 0 mod n, tr(a pr ) tr(a pr 1 ) mod p r A. V. Zarelua, On congruences for the traces of powers of some matrices, Proc. Steklov Inst. Math., 263 (2008), In fact, two congruences are equivalent.

18 Regarding the question The Lefschetz numbers of continuous maps f, L(f ), by µ(d)l(f n d ) 0 mod n, L(f pr ) L(f pr 1 ) mod p r A. Dold, Fixed point indices of iterated maps, Invent. Math., 74 (1983),

19 Lefschetz number Let f : X X be a continuous map. Then f induces a vector space maps (f ) i : H i (X; Q) H i (X; Q) i 0. The Lefschetz number of f is L(f ) = i 0( 1) i tr((f ) i ). LEFSCHETZ FIXED POINT THEOREM If L(f ) 0, then any map homotopic to f has a fixed point.

20 Fixed Points A point x X is a fixed point of f if f (x) = x(= id(x)). Example by graphs of f with X = [0, 1]: fixed points, G(f ) G(id) homotopy; if f g then L(f ) = L(g) existence, cardinality of the number of fixed points L(f ) = 1 0 existence of a fixed point Another example: Coffee cup Any f : D 2 D 2 has a fixed point.

21 Euler s Polyhedron formula Let X be a polyhedron. Then χ(x) = V E + F For example, χ(s 1 ) = 0 and χ(s 2 ) = 2. It is well-known that χ(x) = L(id). That is, the Lefschetz number is a generalization of Euler number. Hence L(id S 1) = 0 and L(id S 2) = 2 0 Remark that the identity map on S 1 fixes every point of S 1 but a little perturbation is fixed point-free. But the Lefschetz Fixed Point Theorem tells that since L(id S 2) = 2 0, any perturbation of the identity map on S 2 has a fixed point.

22 Two fundamental Fixed Point Theorems BANACH THEOREM If X is a complete metric space and f is a contraction, then f has a unique fixed point. BROUWER THEOREM If f is a map of X = D n (unit ball) then f has a fixed point Remark that Banach theorem requires a metric of X in the crucial assumption that f is a contraction Metrical fixed point theory In Brouwer theorem, a metric on R n is used in defining D n and the continuity of f. But the essence is a topological property of the unit ball, i.e., compact and contractible. Topological fixed point theory

23 Brief History History of topological fixed point theory: 1880 s H. Poincaré 1910 s L. E. J. Brouwer 1920 s S. Lefschetz, H. Hopf, J. Nielsen Fixed point theory became a major branch of topology in 1920 s K. Reidemeister and his student F. Wecken 1970 s W. Thurston theory on surfaces -present; many researchers, diverse areas

24 Nielsen theory Let f : X X be a map of a compact space X. Let Fix(f ) = {x X f (x) = x}. Interested in N(f ) : = min{#fix(g) g f } called the Nielsen number of f. The computation of N(f ) is EXTREMELY difficult! So, a main objective of Topological fixed point theory is to compute the Nielsen numbers

25 Example: computation of N(f ) Let X = S 1 and f : S 1 S 1 be a continuous map. Example by a picture. wild map winding map z = e 2πθi z k = e 2πkθi By Hopf, f is homotopic to a winding map of S 1. Hence in computing N(f ), we may assume f (z) = z k. If k = 1 then f = id and N(f d ) = N(id) = 0. Then µ(d)n(f n/d ) = µ(d) 0 = 0, so µ(d)n(f n/d ) 0 mod n.

26 Example: computation of N(f ) continued Let f (z) = z k with k 1. Then N(f ) = #{θ [0, 1) e 2πkθi = e 2πθi } Hence N(f ) = k 1. = #{θ [0, 1) e 2π(k 1)θi = 1} = #{θ [0, 1) (k 1)θ = 0, ±1, ±2, } = #{0, 1 k 1, 2 k 1,, k 2 k 1, k 1 k 1 = 0}. In conclusion, N(f r ) = k r 1.

27 Example: Gauss congruence for Nielsen numbers Let f (z) = z k on S 1 with k 1. Hence N(f r ) = k r 1. Consider Gauss congruences: µ(d)a n/d 0 mod n n µ(d)a n/d. and replace the integers a by the Nielsen numbers N(f ): µ(d)n(f n/d ) = µ(d) k n/d 1 is divisible by n?

28 Example: Gauss congruence for Nielsen numbers Now, µ(d) k n/d 1 = ± µ(d)( k n/d 1) = ± µ(d) k n/d µ(d) = ± µ(d) k n/d is divisible by n? For example, for n = 3 with k > 0, µ(d) k n/d = µ(d)k 3/d = µ(1)k 3 + µ(3)k d 3 = k 3 k = (k 1)k(k + 1) is divisible by 3!

29 Main Result I Let X be an infra-solvmanifold of type (R). For example, X is the torus, the Klein bottle, and etc. Let f : X X be any map. Then the Gauss congruences for the Nielsen numbers hold. Namely, for all n N, µ(d)n(f n/d ) 0 mod n (GN)

30 Main Result II Let X be an infra-solvmanifold of type (R). Let f : X X be any map. Then the Euler congruences for the Nielsen numbers hold. Namely, for all r N and primes p, N(f pr ) N(f pr 1 ) mod p r (EN) In fact, (GN) (EN). My project is to know if the converse holds.

31 End Thanks folk for attending and listening my talk.