The isomorphism and bi-embeddability relations for finitely - PowerPoint PPT Presentation

The Polish space of finitely generated groups Corollary If K is an open subset of G fg , then every G ∈ K is a homomorphic image of a finitely presented group H ∈ K . Proof. Let ( G n ) be a sequence of finitely presented groups such that: G is a homomorphic image of G n . lim n →∞ G n = G . Then G n ∈ K for all but finitely many n ∈ N . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The Main Theorems Theorem (Thomas-Williams 2013) The isomorphism and bi-embeddability relations on the space of Kazhdan groups are both weakly universal. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The Main Theorems Theorem (Thomas-Williams 2013) The isomorphism and bi-embeddability relations on the space of Kazhdan groups are both weakly universal. Theorem (Thomas 2013) The isomorphism and bi-embeddability relations on the space of finitely generated simple groups are both nonsmooth. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The Main Theorems Theorem (Thomas-Williams 2013) The isomorphism and bi-embeddability relations on the space of Kazhdan groups are both weakly universal. Theorem (Thomas 2013) The isomorphism and bi-embeddability relations on the space of finitely generated simple groups are both nonsmooth. Definition The Borel equivalence relation E is smooth if E is Borel reducible to the identity relation ∆ X on some (equivalently every) uncountable Polish space X . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Weakly universal Borel equivalence relations Definition A countable Borel equivalence relation E is weakly universal if for every countable Borel equivalence relation F , there exists a countable-to-one Borel homomorphism from F to E . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Weakly universal Borel equivalence relations Definition A countable Borel equivalence relation E is weakly universal if for every countable Borel equivalence relation F , there exists a countable-to-one Borel homomorphism from F to E . Theorem (Kechris-Miller 2008) A countable Borel equivalence relation E is weakly universal iff there exists a universal countable Borel equivalence relation F ⊆ E. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Weakly universal Borel equivalence relations Definition A countable Borel equivalence relation E is weakly universal if for every countable Borel equivalence relation F , there exists a countable-to-one Borel homomorphism from F to E . Example The Turing equivalence relation ≡ T on 2 N is weakly universal. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Weakly universal Borel equivalence relations Question Does there exist a weakly universal Borel equivalence relation which is not countable universal? Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Weakly universal Borel equivalence relations Question Does there exist a weakly universal Borel equivalence relation which is not countable universal? Theorem (Dougherty & Kechris 1999) If Martin’s Conjecture on degree invariant Borel maps holds, then the Turing equivalence relation ≡ T is weakly universal but not countable universal. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The main theorems Theorem (Thomas-Williams 2013) The isomorphism and bi-embeddability relations on the space of Kazhdan groups are both weakly universal. Theorem (Thomas 2013) The isomorphism and bi-embeddability relations on the space of finitely generated simple groups are both nonsmooth. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The main theorems Theorem (Thomas-Williams 2013) The isomorphism and bi-embeddability relations on the space of Kazhdan groups are both weakly universal. Theorem (Thomas 2013) The isomorphism and bi-embeddability relations on the space of finitely generated simple groups are both nonsmooth. Initial Target The bi-embeddability relation ≈ on the space G fg of finitely generated groups is weakly universal. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The word problem for finitely generated groups Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The word problem for finitely generated groups For each m ≥ 1, fix a recursive enumeration { w k ( x 1 , · · · , x m ) | k ∈ N } of the (not necessarily reduced) words in x 1 , · · · , x m , x − 1 1 , · · · , x − 1 m . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The word problem for finitely generated groups For each m ≥ 1, fix a recursive enumeration { w k ( x 1 , · · · , x m ) | k ∈ N } of the (not necessarily reduced) words in x 1 , · · · , x m , x − 1 1 , · · · , x − 1 m . Definition If ( G , ¯ s ) ∈ G m is a finitely generated group, then Rel ( G ) = { k ∈ N | w k ( s 1 , · · · , s m ) = 1 } Nonrel ( G ) = { k ∈ N | w k ( s 1 , · · · , s m ) � = 1 } . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The word problem for finitely generated groups For each m ≥ 1, fix a recursive enumeration { w k ( x 1 , · · · , x m ) | k ∈ N } of the (not necessarily reduced) words in x 1 , · · · , x m , x − 1 1 , · · · , x − 1 m . Definition If ( G , ¯ s ) ∈ G m is a finitely generated group, then Rel ( G ) = { k ∈ N | w k ( s 1 , · · · , s m ) = 1 } Nonrel ( G ) = { k ∈ N | w k ( s 1 , · · · , s m ) � = 1 } . Question But what if we choose a different generating set G = � t 1 , · · · , t n � ? Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The most obvious Turing reductions Definition If A , B ∈ 2 N , then A is one-one reducible to B , written A ≤ 1 B , if there exists an injective recursive function f : N → N such that for all n ∈ N , n ∈ A ⇐ ⇒ f ( n ) ∈ B . Definition If A ≤ 1 B and B ≤ 1 A , then we write A ≡ 1 B and say that A , B are recursively isomorphic. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The most obvious Turing reductions Definition If A , B ∈ 2 N , then A is one-one reducible to B , written A ≤ 1 B , if there exists an injective recursive function f : N → N such that for all n ∈ N , n ∈ A ⇐ ⇒ f ( n ) ∈ B . Definition If A ≤ 1 B and B ≤ 1 A , then we write A ≡ 1 B and say that A , B are recursively isomorphic. Example s ) , ( H , ¯ s ) ≤ 1 Rel ( H , ¯ If ( G , ¯ t ) ∈ G fg and G ֒ → H , then Rel ( G , ¯ t ) and s ) ≤ 1 Nonrel ( H , ¯ Nonrel ( G , ¯ t ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal finitely generated groups Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal finitely generated groups Definition If G is any group, then the corresponding skeleton is defined to be Sk ( G ) = { H ∈ G fg | H ֒ → G } . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal finitely generated groups Definition If G is any group, then the corresponding skeleton is defined to be Sk ( G ) = { H ∈ G fg | H ֒ → G } . Definition If A ∈ 2 N , then the finitely generated group G is relatively universal of degree A if Sk ( G ) = { H ∈ G fg | Rel ( H ) ≤ T A } . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal finitely generated groups Definition If G is any group, then the corresponding skeleton is defined to be Sk ( G ) = { H ∈ G fg | H ֒ → G } . Definition If A ∈ 2 N , then the finitely generated group G is relatively universal of degree A if Sk ( G ) = { H ∈ G fg | Rel ( H ) ≤ T A } . Observation If A, B ∈ 2 N and G A , G B are relatively universal of degrees A, B, then G A ≈ G B ⇐ ⇒ A ≡ T B . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal finitely generated groups Definition If G is any group, then the corresponding skeleton is defined to be Sk ( G ) = { H ∈ G fg | H ֒ → G } . Definition If A ∈ 2 N , then the finitely generated group G is relatively universal of degree A if Sk ( G ) = { H ∈ G fg | Rel ( H ) ≤ T A } . Theorem For each f.g. group G, there exists a f.g. group H such that: Rel ( H ) ≤ T Rel ( G ) ; and H does not embed into G. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . For each k ∈ N , let r k ( x , y ) be the word ( x k + 1 y k + 1 ) 7 . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . For each k ∈ N , let r k ( x , y ) be the word ( x k + 1 y k + 1 ) 7 . Let H have presentation � b , c | R � , where r k ( b , c ) ∈ R ⇐ ⇒ r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) � = 1 Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . For each k ∈ N , let r k ( x , y ) be the word ( x k + 1 y k + 1 ) 7 . Let H have presentation � b , c | R � , where r k ( b , c ) ∈ R ⇐ ⇒ r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) � = 1 “Clearly” Rel ( H ) ≤ T Rel ( G ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . For each k ∈ N , let r k ( x , y ) be the word ( x k + 1 y k + 1 ) 7 . Let H have presentation � b , c | R � , where r k ( b , c ) ∈ R ⇐ ⇒ r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) � = 1 “Clearly” Rel ( H ) ≤ T Rel ( G ) . → G , then there exists k ∈ N such that If ϕ : H ֒ ϕ ( b ) = u k ( a 1 , · · · , a n ) and ϕ ( c ) = v k ( a 1 , · · · , a n ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . For each k ∈ N , let r k ( x , y ) be the word ( x k + 1 y k + 1 ) 7 . Let H have presentation � b , c | R � , where r k ( b , c ) ∈ R ⇐ ⇒ r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) � = 1 “Clearly” Rel ( H ) ≤ T Rel ( G ) . → G , then there exists k ∈ N such that If ϕ : H ֒ ϕ ( b ) = u k ( a 1 , · · · , a n ) and ϕ ( c ) = v k ( a 1 , · · · , a n ) . Clearly r k ( b , c ) / ∈ R and so r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) = 1. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G = � a 1 , · · · , a n � . Let { � u k ( x 1 , · · · , x n ) , v k ( x 1 , · · · , x n ) � | k ∈ N } be a recursive enumeration of the ordered pairs of words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . For each k ∈ N , let r k ( x , y ) be the word ( x k + 1 y k + 1 ) 7 . Let H have presentation � b , c | R � , where r k ( b , c ) ∈ R ⇐ ⇒ r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) � = 1 “Clearly” Rel ( H ) ≤ T Rel ( G ) . → G , then there exists k ∈ N such that If ϕ : H ֒ ϕ ( b ) = u k ( a 1 , · · · , a n ) and ϕ ( c ) = v k ( a 1 , · · · , a n ) . Clearly r k ( b , c ) / ∈ R and so r k ( u k ( a 1 , · · · , a n ) , v k ( a 1 , · · · , a n )) = 1. But since r k ( b , c ) / ∈ R , it follows that r k ( b , c ) � = 1 in H , which is a contradiction. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Enumeration reducibility Definition If A , B ⊆ N , then A is enumeration reducible to B , written A ≤ e B , if there exists a recursively enumerable subset R ⊆ N × Fin ( N ) such that n ∈ A ⇐ ⇒ ( n , F ) ∈ R for some finite subset F ⊆ B . Remark Intuitively, A ≤ e B if there is a fixed effective procedure which produces an enumeration of A from any enumeration of B . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Enumeration reducibility Definition If A , B ⊆ N , then A is enumeration reducible to B , written A ≤ e B , if there exists a recursively enumerable subset R ⊆ N × Fin ( N ) such that n ∈ A ⇐ ⇒ ( n , F ) ∈ R for some finite subset F ⊆ B . Remark Intuitively, A ≤ e B if there is a fixed effective procedure which produces an enumeration of A from any enumeration of B . Observation It is easily checked that if A ≤ 1 B, then A ≤ e B. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Enumeration reducibility Definition If A , B ⊆ N , then A is enumeration reducible to B , written A ≤ e B , if there exists a recursively enumerable subset R ⊆ N × Fin ( N ) such that n ∈ A ⇐ ⇒ ( n , F ) ∈ R for some finite subset F ⊆ B . Remark Intuitively, A ≤ e B if there is a fixed effective procedure which produces an enumeration of A from any enumeration of B . Remark If G , H ∈ G fg and G ֒ → H , then Rel ( G ) ≤ e Rel ( H ) and Nonrel ( G ) ≤ e Nonrel ( H ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Enumeration reducibility Definition ≡ e is the countable Borel equivalence relation on 2 N defined by A ≡ e B ⇐ ⇒ A ≤ e B and B ≤ e A . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Enumeration reducibility Definition ≡ e is the countable Borel equivalence relation on 2 N defined by A ≡ e B ⇐ ⇒ A ≤ e B and B ≤ e A . Theorem (Folklore) ≡ T is Borel reducible to ≡ e . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Enumeration reducibility Definition ≡ e is the countable Borel equivalence relation on 2 N defined by A ≡ e B ⇐ ⇒ A ≤ e B and B ≤ e A . Theorem (Folklore) ≡ T is Borel reducible to ≡ e . Proof. A ≤ T B ⇐ ⇒ A ⊕ ( N � A ) ≤ e B ⊕ ( N � B ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal f.g. groups revisited Definition If A ∈ 2 N , then the finitely generated group G is relatively universal of e -degree A if Sk ( G ) = { H ∈ G fg | Rel ( H ) ≤ e A } . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal f.g. groups revisited Definition If A ∈ 2 N , then the finitely generated group G is relatively universal of e -degree A if Sk ( G ) = { H ∈ G fg | Rel ( H ) ≤ e A } . Observation If A, B ∈ 2 N and G A , G B are relatively universal of e-degrees A, B, then G A ≈ G B ⇐ ⇒ A ≡ e B . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Relatively universal f.g. groups revisited Definition If A ∈ 2 N , then the finitely generated group G is relatively universal of e -degree A if Sk ( G ) = { H ∈ G fg | Rel ( H ) ≤ e A } . Observation If A, B ∈ 2 N and G A , G B are relatively universal of e-degrees A, B, then G A ≈ G B ⇐ ⇒ A ≡ e B . Theorem (Higman-Scott 1998) For each A ∈ 2 N , there exists a relatively universal group G A of e-degree A. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Two contrasting results Theorem There exists a Borel map A �→ G A from 2 N to G fg such that: Rel ( G A ) ≡ e A, and if A ≡ e B, then G A ≈ G B . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Two contrasting results Theorem There exists a Borel map A �→ G A from 2 N to G fg such that: Rel ( G A ) ≡ e A, and if A ≡ e B, then G A ≈ G B . Theorem (Thomas 2010) If A �→ G A is a Borel map from 2 N to G fg such that Rel ( G A ) ≡ T A, then there exists a Turing degree d 0 such that for all Turing degrees d ≥ T d 0 , there exists an infinite subset { A n | n ∈ N } ⊆ d such that the groups { G A n | n ∈ N } are pairwise incomparable with respect to embeddability. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The bi-embeddability relation for Kazhdan groups Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The bi-embeddability relation for Kazhdan groups Theorem (Thomas-Williams 2013) For each A ∈ 2 N , there exists a finitely generated group K A such that K A is relatively universal of e-degree A; and K A is a Kazhdan group. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The bi-embeddability relation for Kazhdan groups Theorem (Thomas-Williams 2013) For each A ∈ 2 N , there exists a finitely generated group K A such that K A is relatively universal of e-degree A; and K A is a Kazhdan group. Corollary The bi-embeddability relation for Kazhdan groups is weakly universal. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G A = � s 1 , · · · , s n � be relatively universal of e -degree A . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G A = � s 1 , · · · , s n � be relatively universal of e -degree A . Let H be an infinite hyperbolic Kazhdan group. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G A = � s 1 , · · · , s n � be relatively universal of e -degree A . Let H be an infinite hyperbolic Kazhdan group. Then H has a finite presentation H = � x 1 , · · · , x m | T � . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G A = � s 1 , · · · , s n � be relatively universal of e -degree A . Let H be an infinite hyperbolic Kazhdan group. Then H has a finite presentation H = � x 1 , · · · , x m | T � . By Ol’shanskii, H has a free subgroup F = � a 1 , · · · , a n � such that for every N � F , there exists M � H such that N = M ∩ F . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G A = � s 1 , · · · , s n � be relatively universal of e -degree A . Let H be an infinite hyperbolic Kazhdan group. Then H has a finite presentation H = � x 1 , · · · , x m | T � . By Ol’shanskii, H has a free subgroup F = � a 1 , · · · , a n � such that for every N � F , there exists M � H such that N = M ∩ F . Let a i = u i (¯ x ) for 1 ≤ i ≤ n . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Sketch proof of Theorem Let G A = � s 1 , · · · , s n � be relatively universal of e -degree A . Let H be an infinite hyperbolic Kazhdan group. Then H has a finite presentation H = � x 1 , · · · , x m | T � . By Ol’shanskii, H has a free subgroup F = � a 1 , · · · , a n � such that for every N � F , there exists M � H such that N = M ∩ F . Let a i = u i (¯ x ) for 1 ≤ i ≤ n . Let K A = � x 1 , · · · , x m | R � , where R = T ∪ { w ( u 1 (¯ x ) , · · · , u n (¯ x ) ) | w ( a 1 , · · · , a n ) ∈ Rel ( G A ) } . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

How about f.g. simple groups? Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

How about f.g. simple groups? Definition A finitely generated group G ∈ G fg is recursively presented if Rel ( G ) is recursively enumerable. Theorem (Kuznetsov 1958) If G is a recursively presented simple group, then G has a solvable word problem. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

How about f.g. simple groups? Definition A finitely generated group G ∈ G fg is recursively presented if Rel ( G ) is recursively enumerable. Theorem (Kuznetsov 1958) If G is a recursively presented simple group, then G has a solvable word problem. Extended Kuznetsov Theorem If G ∈ G fg is simple, then Nonrel ( G ) ≤ e Rel ( G ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

How about f.g. simple groups? Definition A finitely generated group G ∈ G fg is recursively presented if Rel ( G ) is recursively enumerable. Theorem (Kuznetsov 1958) If G is a recursively presented simple group, then G has a solvable word problem. Extended Kuznetsov Theorem If G ∈ G fg is simple, then Nonrel ( G ) ≤ e Rel ( G ) . Theorem (Thomas-Williams 2013) If A ∈ 2 N and G A is relatively universal of e-degree A, then G A is not simple. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Let G = � a 1 , · · · , a n � be a finitely generated simple group. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Let G = � a 1 , · · · , a n � be a finitely generated simple group. Let F n be the free group on x 1 , · · · , x n and let W be the set of all (not necessarily reduced) words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Let G = � a 1 , · · · , a n � be a finitely generated simple group. Let F n be the free group on x 1 , · · · , x n and let W be the set of all (not necessarily reduced) words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . Fix some word u ∈ W such that u ( a 1 , · · · , a n ) � = 1 in G . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Let G = � a 1 , · · · , a n � be a finitely generated simple group. Let F n be the free group on x 1 , · · · , x n and let W be the set of all (not necessarily reduced) words in x 1 , · · · , x n , x − 1 1 , · · · , x − 1 n . Fix some word u ∈ W such that u ( a 1 , · · · , a n ) � = 1 in G . Since G is simple, if w ∈ W , then the following are equivalent: (i) w ∈ Nonrel ( G ) . (ii) The group G w with presentation � x 1 , · · · , x n | Rel G ∪ { w } � is trivial. (iii) u is in the normal closure of Rel ( G ) ∪ { w } in F n . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Consider the set R ⊆ W × Fin ( W ) defined by ( w , F ) ∈ R ⇐ ⇒ u is in the normal closure of F ∪ { w } in F n . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Consider the set R ⊆ W × Fin ( W ) defined by ( w , F ) ∈ R ⇐ ⇒ u is in the normal closure of F ∪ { w } in F n . Then R is recursively enumerable and the following are equivalent: (i) w ∈ Nonrel ( G ) . (ii) ( w , F ) ∈ R for some finite subset F ⊆ Rel ( G ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

Proof of the Extended Kuznetsov Theorem Consider the set R ⊆ W × Fin ( W ) defined by ( w , F ) ∈ R ⇐ ⇒ u is in the normal closure of F ∪ { w } in F n . Then R is recursively enumerable and the following are equivalent: (i) w ∈ Nonrel ( G ) . (ii) ( w , F ) ∈ R for some finite subset F ⊆ Rel ( G ) . Thus Nonrel ( G ) ≤ e Rel ( G ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

A characterization of the total enumeration degrees Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

A characterization of the total enumeration degrees Definition A set A ∈ 2 N is total if A ≡ e A ⊕ ( N � A ) . Example If G ∈ G fg is simple, then Rel ( G ) is total. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

A characterization of the total enumeration degrees Definition A set A ∈ 2 N is total if A ≡ e A ⊕ ( N � A ) . Example If G ∈ G fg is simple, then Rel ( G ) is total. Definition An enumeration degree e is total if it contains a total set. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

A characterization of the total enumeration degrees Definition A set A ∈ 2 N is total if A ≡ e A ⊕ ( N � A ) . Example If G ∈ G fg is simple, then Rel ( G ) is total. Definition An enumeration degree e is total if it contains a total set. Theorem (Thomas 2016) An enumeration degree e is total if and only if there exists a finitely generated simple group G such that Rel ( G ) ∈ e . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The obvious conjectures ... Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The obvious conjectures ... Conjecture The isomorphism relation for finitely generated simple groups is not weakly universal. Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The obvious conjectures ... Conjecture The isomorphism relation for finitely generated simple groups is not weakly universal. Conjecture There does not exist an isomorphism-invariant Borel map φ : G fg → G fg such that for all G ∈ G fg , φ ( G ) is a finitely generated simple group; and G ֒ → φ ( G ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016

The obvious conjectures ... Conjecture The isomorphism relation for finitely generated simple groups is not weakly universal. Conjecture There does not exist an isomorphism-invariant Borel map φ : G fg → G fg such that for all G ∈ G fg , φ ( G ) is a finitely generated Kazhdan group; and G ֒ → φ ( G ) . Simon Thomas (Rutgers University) Turin Logic Seminar 1st June 2016