The Decomposability Problem for Torsion-Free Abelian Groups is Analytic-Complete
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1 The Decomposability Problem for Torsion-Free Abelian Groups is Analytic-Complete Kyle Riggs Indiana University April 29, 2014 Kyle Riggs Decomposability Problem 1/ 22
2 Computable Groups background A group (G, +) is computable if there is an enumeration ν : N G such that the set {(a, b, c) : ν(a) + ν(b) = ν(c)} is computable. Such an enumeration is called a computable presentation of G. This means statements of the form x + y = z and nx = y are computable for x, y, z G, n N. Kyle Riggs Decomposability Problem 2/ 22
3 Computable Groups background An abelian group G is decomposable if it has two nontrivial subgroups A and B such that G = A B. If not, it is indecomposable. Kyle Riggs Decomposability Problem 3/ 22
4 Computable Groups background An abelian group G is decomposable if it has two nontrivial subgroups A and B such that G = A B. If not, it is indecomposable. We would like to write a group as the direct sum of indecomposable subgroups. Can we classify the indecomposable groups? Kyle Riggs Decomposability Problem 3/ 22
5 Computable Groups background The only indecomposable torsion groups are the cocyclic groups (Z(p k ) for 1 k ). Kyle Riggs Decomposability Problem 4/ 22
6 Computable Groups background The only indecomposable torsion groups are the cocyclic groups (Z(p k ) for 1 k ). Every mixed group is decomposable. Kyle Riggs Decomposability Problem 4/ 22
7 Computable Groups background The only indecomposable torsion groups are the cocyclic groups (Z(p k ) for 1 k ). Every mixed group is decomposable. Torsion-free groups of rank 1 (subgroups of Q) are indecomposable. There is no other well-known class of indecomposable torsion-free abelian groups. This talk will explain why. Kyle Riggs Decomposability Problem 4/ 22
8 Computable Groups background The only indecomposable torsion groups are the cocyclic groups (Z(p k ) for 1 k ). Every mixed group is decomposable. Torsion-free groups of rank 1 (subgroups of Q) are indecomposable. There is no other well-known class of indecomposable torsion-free abelian groups. This talk will explain why. For computable torsion-free groups of finite rank (> 1), being decomposable is a Σ 0 3-complete property. Among computable groups of infinite rank, the property is Σ 1 1 -complete. Kyle Riggs Decomposability Problem 4/ 22
9 Here s an example of an indecomposable group of rank 2 (from Abelian Groups, L. Fuchs) : Kyle Riggs Decomposability Problem 5/ 22
10 Here s an example of an indecomposable group of rank 2 (from Abelian Groups, L. Fuchs) : Start with the free abelian group generated by two elements: x 1 and x 2. Add to the group the following elements: x 1 3, x 2 5, x 1 7, x 2 11, x 1 13, x Kyle Riggs Decomposability Problem 5/ 22
11 Here s an example of an indecomposable group of rank 2 (from Abelian Groups, L. Fuchs) : Start with the free abelian group generated by two elements: x 1 and x 2. Add to the group the following elements: x 1 3, x 2 5, x 1 7, x 2 11, x 1 13, x Finally, add the element x 1+x 2 2 to the group. Denote the group generated by these elements as G. Kyle Riggs Decomposability Problem 5/ 22
12 {x 1, x 2 } is still a basis for G because for every g G, there exist m, m 1, m 2 Z such that mg = m 1 x 1 + m 2 x 2 We can also say that there exist q 1, q 2 Q such that g = q 1 x 1 + q 2 x 2 However, this does not imply that the elements q 1 x 1 and q 2 x 2 exist. For example, x 1+x 2 2 = 1 2 x x 2. Kyle Riggs Decomposability Problem 6/ 22
13 Note that χ G (x 1 ) = (0, 1, 0, 1, 0, 1, 0,...) and χ G (x 2 ) = (0, 0, 1, 0, 1, 0, 1,...) But for any g = q 1 x 1 + q 2 x 2 with q 1, q 2 nonzero, t(g) = (0, 0, 0, 0, 0, 0,...) Kyle Riggs Decomposability Problem 7/ 22
14 If there are nontrivial subgroups A, B such that G = A B, then x 1 must be contained in a direct summand of this decomposition. This also holds for x 2. Kyle Riggs Decomposability Problem 8/ 22
15 If there are nontrivial subgroups A, B such that G = A B, then x 1 must be contained in a direct summand of this decomposition. This also holds for x 2. Because we added the element x 1+x 2 2 to the group, but neither x 1 2 nor x 2 2, then x 1 and x 2 must be in the same direct summand. Kyle Riggs Decomposability Problem 8/ 22
16 If there are nontrivial subgroups A, B such that G = A B, then x 1 must be contained in a direct summand of this decomposition. This also holds for x 2. Because we added the element x 1+x 2 2 to the group, but neither x 1 2 nor x 2 2, then x 1 and x 2 must be in the same direct summand. Thus, G = A (or B), so the group is indecomposable. We say the element x 1+x 2 2 serves as a link connecting x 1 and x 2. Kyle Riggs Decomposability Problem 8/ 22
17 In this example, x 1 and x 2 are elements of strictly maximal type. An element x has strictly maximal type if, for any y linearly independent from x, t(x) t(y) (there are infinitely many pairs p, k P N such that p k x but p k y). Kyle Riggs Decomposability Problem 9/ 22
18 In this example, x 1 and x 2 are elements of strictly maximal type. An element x has strictly maximal type if, for any y linearly independent from x, t(x) t(y) (there are infinitely many pairs p, k P N such that p k x but p k y). As we have seen, any element of strictly maximal type in a decomposable group must be contained in one of the direct summands. Kyle Riggs Decomposability Problem 9/ 22
19 In this example, x 1 and x 2 are elements of strictly maximal type. An element x has strictly maximal type if, for any y linearly independent from x, t(x) t(y) (there are infinitely many pairs p, k P N such that p k x but p k y). As we have seen, any element of strictly maximal type in a decomposable group must be contained in one of the direct summands. A group is indecomposable if it has a basis of elements of strictly maximal type with a link (or a chain of links) between every pair of basis elements. The converse of this statement does not hold. Kyle Riggs Decomposability Problem 9/ 22
20 If G is a torsion-free group of finite rank and G = A B, then for any bases ā, b of A, B (resp.) the set ā b is a basis for G with the following property: For every g G, if g = i q i a i + j r j b j, then there are elements a, b such that a = i q i a i and b = j r j b j Kyle Riggs Decomposability Problem 10/ 22
21 If G is a torsion-free group of finite rank and G = A B, then for any bases ā, b of A, B (resp.) the set ā b is a basis for G with the following property: For every g G, if g = i q i a i + j r j b j, then there are elements a, b such that a = i q i a i and b = j r j b j This allows us to talk about decomposability in terms of finite bases instead of subgroups. Kyle Riggs Decomposability Problem 10/ 22
22 For a computable group G there is a Π 0 2 sets such that formula BASIS on finite BASIS( x) x is a basis of G Kyle Riggs Decomposability Problem 11/ 22
23 For a computable group G there is a Π 0 2 sets such that formula BASIS on finite BASIS( x) x is a basis of G This gives us the following Σ 0 3 formula for decomposable torsion-free groups of finite rank: G is decomposable iff ( ā, b [G] <ω ) { BASIS(ā b) ā ø b ø ( y G) ( q Q <ω )( w G) [ ( q = ā + b y = q i a i + q j b j ) w = ] } q i a i i j i Kyle Riggs Decomposability Problem 11/ 22
24 To see this property is Σ 0 3-complete, we describe a computable function f : ω {torsion-free abelian groups of rank 2} such that G n is decomposable iff W n is cofinite. Kyle Riggs Decomposability Problem 12/ 22
25 Start with the group G in the previous example. G = x 1 + x 2, x 1 2 3, x 2 5, x 1 7, x 2 11,... Kyle Riggs Decomposability Problem 13/ 22
26 Start with the group G in the previous example. G = x 1 + x 2, x 1 2 3, x 2 5, x 1 7, x 2 11,... To create the group G n, we add to G the element such that Φ n (k). x 2 p 2k+1 for every k Kyle Riggs Decomposability Problem 13/ 22
27 Start with the group G in the previous example. G = x 1 + x 2, x 1 2 3, x 2 5, x 1 7, x 2 11,... To create the group G n, we add to G the element such that Φ n (k). x 2 p 2k+1 for every k If W n is coinfinite, there are still infinitely many primes dividing x 1 but not x 2, so the group remains indecomposable. Kyle Riggs Decomposability Problem 13/ 22
28 Start with the group G in the previous example. G = x 1 + x 2, x 1 2 3, x 2 5, x 1 7, x 2 11,... To create the group G n, we add to G the element such that Φ n (k). x 2 p 2k+1 for every k If W n is coinfinite, there are still infinitely many primes dividing x 1 but not x 2, so the group remains indecomposable. If W n is cofinite, then x 1 no longer has strictly maximal type, and a decomposition of G n exists. Kyle Riggs Decomposability Problem 13/ 22
29 Start with the group G in the previous example. G = x 1 + x 2, x 1 2 3, x 2 5, x 1 7, x 2 11,... To create the group G n, we add to G the element such that Φ n (k). x 2 p 2k+1 for every k If W n is coinfinite, there are still infinitely many primes dividing x 1 but not x 2, so the group remains indecomposable. If W n is cofinite, then x 1 no longer has strictly maximal type, and a decomposition of G n exists. Thus, the set of decomposable torsion-free groups of finite rank (> 1) is Σ 0 3 -complete. Kyle Riggs Decomposability Problem 13/ 22
30 If we adapt our previous formula to describe decomposable groups of infinite rank, we get the following: ( (ā b) [G] ω ) [BASIS(ā b) ā ø b ø ( y G) ( q Q <ω )( w G) (y = q i a i + q j b j w = q i a i )] i j i This is a Σ 1 1 formula, and in fact we will show that this property is Σ 1 1-complete for computable groups, and analytic-complete for groups in general. Kyle Riggs Decomposability Problem 14/ 22
31 We describe a (computable) function from (computable) trees in ω <ω to torsion-free abelian groups of infinite rank that gives us a group G T which is decomposable iff there is an infinite path through the tree T. Kyle Riggs Decomposability Problem 15/ 22
32 We start with a countable basis {x i } i 1, {x σ } σ ω <ω, {y j } j 1 and make each element of this basis infinitely divisible by a unique prime. Kyle Riggs Decomposability Problem 16/ 22
33 We start with a countable basis {x i } i 1, {x σ } σ ω <ω, {y j } j 1 and make each element of this basis infinitely divisible by a unique prime. Then, for each pair of x-elements we create a link using a unique prime. We do the same for each pair of y-elements. Kyle Riggs Decomposability Problem 16/ 22
34 We start with a countable basis {x i } i 1, {x σ } σ ω <ω, {y j } j 1 and make each element of this basis infinitely divisible by a unique prime. Then, for each pair of x-elements we create a link using a unique prime. We do the same for each pair of y-elements. For every n we also add an element of the form y 1 + y y n p f (n) Kyle Riggs Decomposability Problem 16/ 22
35 At this point, the group decomposes as A B, where A is the pure subgroup containing all the x-elements, and B the pure subgroup containing all the y-elements. Then, for every i we create a link between x i and y i. The group (denoted G) is now indecomposable. Kyle Riggs Decomposability Problem 17/ 22
36 Idea: Given a tree T, we add elements to G to create a group G T. If T has no infinite paths, G T is still indecomposable. If T has an infinite path π, G T = A T B π, where A T is the pure subgroup containing all the x-elements and B π is the pure subgroup containing all the elements of the form y n + x π n B π = y 1 + x σ1, y 2 + x σ2,... where σ n = n and σ n σ n+1 for all n. Kyle Riggs Decomposability Problem 18/ 22
37 Sketch: Let σ be a string of length n, and let p be the prime that infinitely divides y n. Whenever a string τ σ is found on T, we make x σ divisible by another power of p. Kyle Riggs Decomposability Problem 19/ 22
38 Sketch: Let σ be a string of length n, and let p be the prime that infinitely divides y n. Whenever a string τ σ is found on T, we make x σ divisible by another power of p. If there is a string of length n with infinitely many strings above it, then y n no longer has strictly maximal type. However, the set {y n } {x σ : σ = n ( τ T ) τ σ} does have strictly maximal type. This means that if y n decomposes as y n = a + b, then a and b must be in the span of this set. Kyle Riggs Decomposability Problem 19/ 22
39 When we find σ on T ( σ = n), then we break some links to allow y n + x σ to be contained in a direct summand. For example: x n + y n q = x n x σ q + y n + x σ q Kyle Riggs Decomposability Problem 20/ 22
40 When we find σ on T ( σ = n), then we break some links to allow y n + x σ to be contained in a direct summand. For example: x n + y n q = x n x σ q + y n + x σ q y 1 + y y n p f (n) = x σ 1 + x σ x σ p f (n) + (y 1 + x σ 1 ) + (y 2 + x σ 2 ) (y n + x σ ) p f (n) Kyle Riggs Decomposability Problem 20/ 22
41 We denote the finished group G T. If there is an infinite path π through T, G T = A T B π. If G T is decomposable, we can use the increasing sequence of links to find an infinite path through T. Kyle Riggs Decomposability Problem 21/ 22
42 We denote the finished group G T. If there is an infinite path π through T, G T = A T B π. If G T is decomposable, we can use the increasing sequence of links to find an infinite path through T. Thus, the set of computable decomposable torsion-free abelian groups of infinite rank is Σ 1 1-complete, and the set of decomposable torsion-free abelian groups of infinite rank is Σ 1 1 -complete. Kyle Riggs Decomposability Problem 21/ 22
43 Thank you! Kyle Riggs Decomposability Problem 22/ 22
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