Automorphism orbit membership for right-angled Artin groups Matthew - - PowerPoint PPT Presentation

Automorphism orbit membership for right-angled Artin groups

Matthew B. Day University of Arkansas matthewd@uark.edu GAGTA7@CCNY, May 30, 2013 Arχiv reference: 1211:0078, Full-featured peak reduction in right-angled Artin groups

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 1 / 22

Right-angled Artin groups

Let Γ be a finite simplicial graph.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 2 / 22

Right-angled Artin groups

Let Γ be a finite simplicial graph. We define a finitely presented group based on Γ:

Definition

The right-angled Artin group (RAAG) AΓ is the group with presentation: AΓ = Vertices(Γ)|{xy = yx|Γ has an edge from x to y}. These are also called “graph groups” or “partially commutative groups” or “PC groups”.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 2 / 22

Right-angled Artin groups

Let Γ be a finite simplicial graph. We define a finitely presented group based on Γ:

Definition

The right-angled Artin group (RAAG) AΓ is the group with presentation: AΓ = Vertices(Γ)|{xy = yx|Γ has an edge from x to y}. These are also called “graph groups” or “partially commutative groups” or “PC groups”. So Γ encodes the commuting between generators of AΓ.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 2 / 22

Right-angled Artin groups

Let Γ be a finite simplicial graph. We define a finitely presented group based on Γ:

Definition

The right-angled Artin group (RAAG) AΓ is the group with presentation: AΓ = Vertices(Γ)|{xy = yx|Γ has an edge from x to y}. These are also called “graph groups” or “partially commutative groups” or “PC groups”. So Γ encodes the commuting between generators of AΓ. These groups are simple to define but have interesting properties, especially concerning their subgroup structure, automorphism groups, and connections to geometry and topology.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 2 / 22

Examples of RAAGs

Example: Γ edgeless implies AΓ ∼ = Fn and Aut(AΓ) ∼ = Aut(Fn).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 3 / 22

Examples of RAAGs

Example: Γ edgeless implies AΓ ∼ = Fn and Aut(AΓ) ∼ = Aut(Fn). Example: Γ complete implies AΓ ∼ = Zn and Aut(AΓ) ∼ = GL(n, Z).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 3 / 22

Examples of RAAGs

Example: Γ edgeless implies AΓ ∼ = Fn and Aut(AΓ) ∼ = Aut(Fn). Example: Γ complete implies AΓ ∼ = Zn and Aut(AΓ) ∼ = GL(n, Z). Free products of finitely many RAAGs are RAAGs (take disjoint unions of graphs).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 3 / 22

Examples of RAAGs

Example: Γ edgeless implies AΓ ∼ = Fn and Aut(AΓ) ∼ = Aut(Fn). Example: Γ complete implies AΓ ∼ = Zn and Aut(AΓ) ∼ = GL(n, Z). Free products of finitely many RAAGs are RAAGs (take disjoint unions of graphs). Direct products of finitely many RAAGs are RAAGs (take graph joins).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 3 / 22

Examples of RAAGs

Example: Γ edgeless implies AΓ ∼ = Fn and Aut(AΓ) ∼ = Aut(Fn). Example: Γ complete implies AΓ ∼ = Zn and Aut(AΓ) ∼ = GL(n, Z). Free products of finitely many RAAGs are RAAGs (take disjoint unions of graphs). Direct products of finitely many RAAGs are RAAGs (take graph joins). The group a, b, c, d|[a, b] = [b, c] = [c, d] = 1 is the RAAG of the following graph.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 3 / 22

Automorphisms of RAAGs

Suppose Γ contains vertices x, y, z with x adjacent to y but y not adjacent to z.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 4 / 22

Automorphisms of RAAGs

Suppose Γ contains vertices x, y, z with x adjacent to y but y not adjacent to z. Then there is no automorphism φ ∈ Aut(AΓ) with φ(x) = zx and φ(v) = v if v = z for vertices v of Γ.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 4 / 22

Automorphisms of RAAGs

Suppose Γ contains vertices x, y, z with x adjacent to y but y not adjacent to z. Then there is no automorphism φ ∈ Aut(AΓ) with φ(x) = zx and φ(v) = v if v = z for vertices v of Γ. Why? Because [x, y] = 1 but [φ(x), φ(y)] = 1.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 4 / 22

Automorphisms of RAAGs

Suppose Γ contains vertices x, y, z with x adjacent to y but y not adjacent to z. Then there is no automorphism φ ∈ Aut(AΓ) with φ(x) = zx and φ(v) = v if v = z for vertices v of Γ. Why? Because [x, y] = 1 but [φ(x), φ(y)] = 1. Compare to Aut(Fn) and GL(n, Z) where such automorphisms always exist and nearly generate the group (they are called Nielsen moves or transvections or elementary matrices).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 4 / 22

Automorphisms of RAAGs

Suppose Γ contains vertices x, y, z with x adjacent to y but y not adjacent to z. Then there is no automorphism φ ∈ Aut(AΓ) with φ(x) = zx and φ(v) = v if v = z for vertices v of Γ. Why? Because [x, y] = 1 but [φ(x), φ(y)] = 1. Compare to Aut(Fn) and GL(n, Z) where such automorphisms always exist and nearly generate the group (they are called Nielsen moves or transvections or elementary matrices). Laurence’s theorem (1995) gives a concrete finite generating set for Aut(AΓ) that can be read off of Γ.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 4 / 22

Automorphism orbits

Let G be a group and let CG (or just C) denote the set of conjugacy classes of G.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 5 / 22

Automorphism orbits

Let G be a group and let CG (or just C) denote the set of conjugacy classes of G. Then Aut(G) acts diagonally on both G k = G × · · · × G and Ck

G = CG × · · · × CG (for any k).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 5 / 22

Automorphism orbits

Let G be a group and let CG (or just C) denote the set of conjugacy classes of G. Then Aut(G) acts diagonally on both G k = G × · · · × G and Ck

G = CG × · · · × CG (for any k).

Question (Whitehead problem): For a given group G, is there an algorithm that tells us whether two elements U and V in G k (or in Ck

G) are in the same orbit?

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 5 / 22

Automorphism orbits

Let G be a group and let CG (or just C) denote the set of conjugacy classes of G. Then Aut(G) acts diagonally on both G k = G × · · · × G and Ck

G = CG × · · · × CG (for any k).

Question (Whitehead problem): For a given group G, is there an algorithm that tells us whether two elements U and V in G k (or in Ck

G) are in the same orbit?

(That is, how can we tell if there is φ ∈ Aut(G) with φ · U = V ?)

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 5 / 22

Automorphism orbits

Let G be a group and let CG (or just C) denote the set of conjugacy classes of G. Then Aut(G) acts diagonally on both G k = G × · · · × G and Ck

G = CG × · · · × CG (for any k).

Question (Whitehead problem): For a given group G, is there an algorithm that tells us whether two elements U and V in G k (or in Ck

G) are in the same orbit?

(That is, how can we tell if there is φ ∈ Aut(G) with φ · U = V ?) If we determine that U and V are in the same orbit, then we also want to produce φ ∈ Aut(G) with φ · U = V .

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 5 / 22

Orbits of Matrix Groups,1

As a warm up, consider G = Qn and consider the action of Aut(G) = GL(n, Q) on k-tuples of elements of G: G k = (Qn)k = Mn,k(Q), the set of n-row, k-column matrices.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 6 / 22

Orbits of Matrix Groups,1

As a warm up, consider G = Qn and consider the action of Aut(G) = GL(n, Q) on k-tuples of elements of G: G k = (Qn)k = Mn,k(Q), the set of n-row, k-column matrices. The action is given by multiplying matrices from GL(n, Q) on the left.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 6 / 22

Orbits of Matrix Groups,1

As a warm up, consider G = Qn and consider the action of Aut(G) = GL(n, Q) on k-tuples of elements of G: G k = (Qn)k = Mn,k(Q), the set of n-row, k-column matrices. The action is given by multiplying matrices from GL(n, Q) on the left. Then U, V ∈ Mn,k(Q) are in the same orbit if and only if they have the same reduced row-echelon form. If so, we can find A ∈ GL(n, Q) with A · U = V by row-reducing each.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 6 / 22

Orbits of Matrix Groups,2

For G = Zn, the situation is trickier. Then Aut(G) = GL(n, Z) and G k = Mn,k(Z).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 7 / 22

Orbits of Matrix Groups,2

For G = Zn, the situation is trickier. Then Aut(G) = GL(n, Z) and G k = Mn,k(Z). A matrix is in Hermite Normal Form (HNF) if it is in row-echelon form, the pivots are positive, and the entries above pivots are reduced modulo the pivots.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 7 / 22

Orbits of Matrix Groups,2

For G = Zn, the situation is trickier. Then Aut(G) = GL(n, Z) and G k = Mn,k(Z). A matrix is in Hermite Normal Form (HNF) if it is in row-echelon form, the pivots are positive, and the entries above pivots are reduced modulo the pivots. for example:   3 −2 1 −7 1 3 2 1 3 −2  

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 7 / 22

Orbits of Matrix Groups,2

For G = Zn, the situation is trickier. Then Aut(G) = GL(n, Z) and G k = Mn,k(Z). A matrix is in Hermite Normal Form (HNF) if it is in row-echelon form, the pivots are positive, and the entries above pivots are reduced modulo the pivots. for example:   3 −2 1 −7 1 3 2 1 3 −2   The HNF of a matrix can be found by a row-reduction procedure.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 7 / 22

Orbits of Matrix Groups,2

For G = Zn, the situation is trickier. Then Aut(G) = GL(n, Z) and G k = Mn,k(Z). A matrix is in Hermite Normal Form (HNF) if it is in row-echelon form, the pivots are positive, and the entries above pivots are reduced modulo the pivots. for example:   3 −2 1 −7 1 3 2 1 3 −2   The HNF of a matrix can be found by a row-reduction procedure. Matrices A and B ∈ Mn,k(Z) are in the same GL(n, Z)–orbit iff they have the same HNF.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 7 / 22

Orbits of Matrix Groups,2

For G = Zn, the situation is trickier. Then Aut(G) = GL(n, Z) and G k = Mn,k(Z). A matrix is in Hermite Normal Form (HNF) if it is in row-echelon form, the pivots are positive, and the entries above pivots are reduced modulo the pivots. for example:   3 −2 1 −7 1 3 2 1 3 −2   The HNF of a matrix can be found by a row-reduction procedure. Matrices A and B ∈ Mn,k(Z) are in the same GL(n, Z)–orbit iff they have the same HNF. If A and B are in the same orbit, we can find Q ∈ GL(n, Z) with A = QB by keeping track of the moves used in putting each in HNF.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 7 / 22

Peak reduction for free groups, 1

For G = Fn, membership checking for Aut(Fn)–orbits in Ck is a part

• f a bigger story, that of peak reduction.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 8 / 22

Peak reduction for free groups, 1

For G = Fn, membership checking for Aut(Fn)–orbits in Ck is a part

• f a bigger story, that of peak reduction.

For a sequence of elements of Z, a peak is a subsequence of length 3

• f one of the following forms:

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 8 / 22

Peak reduction for free groups, 1

For G = Fn, membership checking for Aut(Fn)–orbits in Ck is a part

• f a bigger story, that of peak reduction.

For a sequence of elements of Z, a peak is a subsequence of length 3

• f one of the following forms:

A sequence is peak-reduced if no peaks appear in it:

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 8 / 22

Peak reduction for free groups,2

Let X be a fixed basis for the free group Fn. The Whitehead automorphisms Ω of Fn with respect to X are

all α ∈ Aut(Fn) s.t. there is a ∈ X ±1 with α(x) ∈ {x, ax, xa−1, axa−1} and all α ∈ Aut(Fn) that restrict to a permutation of X ±1

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 9 / 22

Peak reduction for free groups,2

Let X be a fixed basis for the free group Fn. The Whitehead automorphisms Ω of Fn with respect to X are

all α ∈ Aut(Fn) s.t. there is a ∈ X ±1 with α(x) ∈ {x, ax, xa−1, axa−1} and all α ∈ Aut(Fn) that restrict to a permutation of X ±1

Ω is a finite set.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 9 / 22

Peak reduction for free groups,3

The length |W | of a tuple of conjugacy classes W ∈ Ck is the sum of the lengths of the shortest representatives (with respect to X).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 10 / 22

Peak reduction for free groups,3

The length |W | of a tuple of conjugacy classes W ∈ Ck is the sum of the lengths of the shortest representatives (with respect to X).

Theorem (Whitehead, 1930’s)

Let W ∈ Ck. For α ∈ Aut(Fn), there is a factorization α = βm ◦ . . . ◦ β1 with β1, . . . , βm ∈ Ω and such that the sequence of lengths i → |βi ◦ . . . ◦ β1 · W | is peak reduced.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 10 / 22

Peak reduction for free groups,3

The length |W | of a tuple of conjugacy classes W ∈ Ck is the sum of the lengths of the shortest representatives (with respect to X).

Theorem (Whitehead, 1930’s)

Let W ∈ Ck. For α ∈ Aut(Fn), there is a factorization α = βm ◦ . . . ◦ β1 with β1, . . . , βm ∈ Ω and such that the sequence of lengths i → |βi ◦ . . . ◦ β1 · W | is peak reduced. J.H.C. Whitehead originally proved this using topological methods, but a combinatorial version was developed by Rapaport (1958) and Higgins–Lyndon (1974).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 10 / 22

Peak reduction for free groups,3

The length |W | of a tuple of conjugacy classes W ∈ Ck is the sum of the lengths of the shortest representatives (with respect to X).

Theorem (Whitehead, 1930’s)

Let W ∈ Ck. For α ∈ Aut(Fn), there is a factorization α = βm ◦ . . . ◦ β1 with β1, . . . , βm ∈ Ω and such that the sequence of lengths i → |βi ◦ . . . ◦ β1 · W | is peak reduced. J.H.C. Whitehead originally proved this using topological methods, but a combinatorial version was developed by Rapaport (1958) and Higgins–Lyndon (1974). There is also a version for the action of Aut(Fn) on F k

n . Working with

Ck is easier.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 10 / 22

Corollaries of peak reduction,1

Corollary (Whitehead)

It is possible to check whether two elements of Ck are in the same orbit under the action of Aut(Fn).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 11 / 22

Corollaries of peak reduction,1

Corollary (Whitehead)

It is possible to check whether two elements of Ck are in the same orbit under the action of Aut(Fn). Proof ideas: peak reduction implies that if W ∈ Ck can be shortened by an automorphism, then W can be shortened by a Whitehead

• automorphism. Use finiteness of Ω to check whether W is minimal

length in Aut(Fn) · W .

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 11 / 22

Corollaries of peak reduction,1

Corollary (Whitehead)

It is possible to check whether two elements of Ck are in the same orbit under the action of Aut(Fn). Proof ideas: peak reduction implies that if W ∈ Ck can be shortened by an automorphism, then W can be shortened by a Whitehead

• automorphism. Use finiteness of Ω to check whether W is minimal

length in Aut(Fn) · W . If U, V ∈ Ck are minimal length in their orbits and the same length, then peak reduction implies they are in the same orbit iff there is a sequence of Whitehead automorphisms taking U to V through the finite set of elements of length |U|.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 11 / 22

Corollaries of peak reduction,1

Corollary (Whitehead)

It is possible to check whether two elements of Ck are in the same orbit under the action of Aut(Fn). Proof ideas: peak reduction implies that if W ∈ Ck can be shortened by an automorphism, then W can be shortened by a Whitehead

• automorphism. Use finiteness of Ω to check whether W is minimal

length in Aut(Fn) · W . If U, V ∈ Ck are minimal length in their orbits and the same length, then peak reduction implies they are in the same orbit iff there is a sequence of Whitehead automorphisms taking U to V through the finite set of elements of length |U|.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 11 / 22

Corollaries of peak reduction,2

Corollary (McCool, 1975)

For W ∈ Ck, the stabilizer Aut(Fn)W is finitely presented.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 12 / 22

Corollaries of peak reduction,2

Corollary (McCool, 1975)

For W ∈ Ck, the stabilizer Aut(Fn)W is finitely presented. Proof ideas: We can use the finite set of minimal-length reps of Aut(Fn) · W and Ω to build a finite graph whose fundamental group must surject on Aut(Fn)W by peak reduction.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 12 / 22

Corollaries of peak reduction,2

Corollary (McCool, 1975)

For W ∈ Ck, the stabilizer Aut(Fn)W is finitely presented. Proof ideas: We can use the finite set of minimal-length reps of Aut(Fn) · W and Ω to build a finite graph whose fundamental group must surject on Aut(Fn)W by peak reduction. By peak reducing with respect to a more complicated tuple built out

• f W and conjugacy classes of length 2, we can factor expressions for

1 into products of simpler relations that “respect” W .

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 12 / 22

Peak reduction for RAAGs,1

Theorem (Day, 2008)

For any RAAG AΓ, there are subgroups L and S of Aut(AΓ) such that L has peak reduction with respect to a finite generating set Ωℓ, S has an ‘obvious’ faithful linear representation, and Aut(AΓ) = SL and S ∩ L = 1. Generally neither S nor L is normal.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 13 / 22

Peak reduction for RAAGs,1

Theorem (Day, 2008)

For any RAAG AΓ, there are subgroups L and S of Aut(AΓ) such that L has peak reduction with respect to a finite generating set Ωℓ, S has an ‘obvious’ faithful linear representation, and Aut(AΓ) = SL and S ∩ L = 1. Generally neither S nor L is normal. Ideas from proof: Ωℓ is defined just like Ω, except that α ∈ Ωℓ cannot map x to xy if x is adjacent to y in Γ. Follow Higgins–Lyndon proof outline.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 13 / 22

Peak reduction for RAAGs,1

Theorem (Day, 2008)

For any RAAG AΓ, there are subgroups L and S of Aut(AΓ) such that L has peak reduction with respect to a finite generating set Ωℓ, S has an ‘obvious’ faithful linear representation, and Aut(AΓ) = SL and S ∩ L = 1. Generally neither S nor L is normal. Ideas from proof: Ωℓ is defined just like Ω, except that α ∈ Ωℓ cannot map x to xy if x is adjacent to y in Γ. Follow Higgins–Lyndon proof outline. L is generated by adjacent transvections so the “homology representation” is faithful.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 13 / 22

Peak reduction for RAAGs,1

Theorem (Day, 2008)

For any RAAG AΓ, there are subgroups L and S of Aut(AΓ) such that L has peak reduction with respect to a finite generating set Ωℓ, S has an ‘obvious’ faithful linear representation, and Aut(AΓ) = SL and S ∩ L = 1. Generally neither S nor L is normal. Ideas from proof: Ωℓ is defined just like Ω, except that α ∈ Ωℓ cannot map x to xy if x is adjacent to y in Γ. Follow Higgins–Lyndon proof outline. L is generated by adjacent transvections so the “homology representation” is faithful. There is a rewriting procedure to prove the third part.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 13 / 22

Peak reduction for RAAGs,2

Theorem (Day, 2008)

For any Γ, Aut(AΓ) is finitely presented.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 14 / 22

Peak reduction for RAAGs,2

Theorem (Day, 2008)

For any Γ, Aut(AΓ) is finitely presented. Proof ideas: Prove separately that S and L are finitely presented, and carefully use these to rewrite identities in Aut(AΓ).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 14 / 22

Peak reduction for RAAGs,2

Theorem (Day, 2008)

For any Γ, Aut(AΓ) is finitely presented. Proof ideas: Prove separately that S and L are finitely presented, and carefully use these to rewrite identities in Aut(AΓ). Apparently no easy way to use these theorems to get other classic corollaries!

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 14 / 22

Peak reduction failure for RAAGs

Big problem for peak reduction in RAAGs:

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 15 / 22

Peak reduction failure for RAAGs

Big problem for peak reduction in RAAGs:

Theorem (Day, 2008)

For the following graph Γ, for any finite generating set Ω ⊂ Aut(AΓ) there is a conjugacy class w such that Ω does not have the peak-reduction property with respect to w for Aut(AΓ). AΓ = a, b, c, d|[a, b] = [b, c] = [c, d] = 1

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 15 / 22

Peak reduction failure for RAAGs

Big problem for peak reduction in RAAGs:

Theorem (Day, 2008)

For the following graph Γ, for any finite generating set Ω ⊂ Aut(AΓ) there is a conjugacy class w such that Ω does not have the peak-reduction property with respect to w for Aut(AΓ). AΓ = a, b, c, d|[a, b] = [b, c] = [c, d] = 1 Proof idea: if w = adk and φ(a) = ack and φ(d) = c−1d (others fixed) then φ fixes w but no “simpler” nontrivial automorphism fixes w, contradicting peak reduction for a set Ω whose elements have “complexity” less than k.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 15 / 22

Related results

Collins–Zieschang (1984) proved a relative Whitehead theorem for free products.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 16 / 22

Related results

Collins–Zieschang (1984) proved a relative Whitehead theorem for free products. Charney–Stambaugh–Vogtmann (2012) proved that for any AΓ, there are certain conjugacy classes that such that peak reduction works for these classes using a certain finite Ω.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 16 / 22

Related results

Collins–Zieschang (1984) proved a relative Whitehead theorem for free products. Charney–Stambaugh–Vogtmann (2012) proved that for any AΓ, there are certain conjugacy classes that such that peak reduction works for these classes using a certain finite Ω. This was used in part 1 of their

• ngoing project to create a completely general outer space for RAAGs.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 16 / 22

Related results

Collins–Zieschang (1984) proved a relative Whitehead theorem for free products. Charney–Stambaugh–Vogtmann (2012) proved that for any AΓ, there are certain conjugacy classes that such that peak reduction works for these classes using a certain finite Ω. This was used in part 1 of their

• ngoing project to create a completely general outer space for RAAGs.

Delgado–Ventura (2013) solved endomorphism and monomorphism decision problems in the special case AΓ = Zm × Fn.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 16 / 22

Infinite peak reducing sets for RAAGs

One approach to overcome the problem is to use infinite generating sets.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 17 / 22

Infinite peak reducing sets for RAAGs

One approach to overcome the problem is to use infinite generating sets. For a in Γ, define [a] to be the set of vertices in Γ adjacent to a and with the same star as a. (star(b) is the set of vertices adjacent to b.)

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 17 / 22

Infinite peak reducing sets for RAAGs

One approach to overcome the problem is to use infinite generating sets. For a in Γ, define [a] to be the set of vertices in Γ adjacent to a and with the same star as a. (star(b) is the set of vertices adjacent to b.)

Definition

For a ∈ Γ, the generalized Whitehead automorphisms with multiplier set [a], denoted Ω[a], is the set of automorphisms α such that:

α(b) ∈ [a] if b ∈ [a] and α(b) = ubv for some u, v ∈ [a] if b / ∈ [a].

Let P be the set of automorphisms restricting to permutations of Γ±1.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 17 / 22

Strong peak reduction for RAAGs,2

Theorem (Day, 2012)

Fix Γ. For any W ∈ Ck, Aut(AΓ) has a peak reduction algorithm with respect to W using the infinite generating set P ∪

• a∈Γ

Ω[a]. Further, for any a ∈ Γ, there are algorithms to check whether elements U, V ∈ Ck are in the same Ω[a]–orbit and produce a finite presentation for the stabilizer (Ω[a])W for W ∈ Ck.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 18 / 22

Corollaries of strong peak reduction

Corollary (Day, 2012)

For any Γ, there is an algorithm that takes in pairs of elements U, V ∈ Ck, determines whether they are in the same Aut(AΓ)-orbit, and returns φ ∈ Aut(AΓ) with φ · U = V if they are.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 19 / 22

Corollaries of strong peak reduction

Corollary (Day, 2012)

For any Γ, there is an algorithm that takes in pairs of elements U, V ∈ Ck, determines whether they are in the same Aut(AΓ)-orbit, and returns φ ∈ Aut(AΓ) with φ · U = V if they are.

Corollary (Day, 2012)

For any Γ, there is an algorithm that takes in an element W ∈ Ck and returns a finite presentation for the stabilizer (Aut(AΓ))W .

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 19 / 22

Proof ideas for strong peak reduction

The actual peak reduction algorithm is similar in style to the Higgins–Lyndon argument, but is more involved.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 20 / 22

Proof ideas for strong peak reduction

The actual peak reduction algorithm is similar in style to the Higgins–Lyndon argument, but is more involved. Fix a. There is an isomorphism Ω[a] ∼ = GL(l, Z) ⋉ Ml,m(Z), i.e. Ω[a] is isomorphic to the group of block upper triangular matrices of the form A B O I

• ,

where A ∈ GL(l, Z), B ∈ Ml,m(Z), l = |[a]| and m is determined by a and Γ.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 20 / 22

Proof ideas for strong peak reduction

The actual peak reduction algorithm is similar in style to the Higgins–Lyndon argument, but is more involved. Fix a. There is an isomorphism Ω[a] ∼ = GL(l, Z) ⋉ Ml,m(Z), i.e. Ω[a] is isomorphic to the group of block upper triangular matrices of the form A B O I

• ,

where A ∈ GL(l, Z), B ∈ Ml,m(Z), l = |[a]| and m is determined by a and Γ. There is a finite-to-one, Ω[a]–equivariant set map:

• k

Ck →

• k

Ml+m,k(Z)

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 20 / 22

Proof ideas for strong peak reduction

The actual peak reduction algorithm is similar in style to the Higgins–Lyndon argument, but is more involved. Fix a. There is an isomorphism Ω[a] ∼ = GL(l, Z) ⋉ Ml,m(Z), i.e. Ω[a] is isomorphic to the group of block upper triangular matrices of the form A B O I

• ,

where A ∈ GL(l, Z), B ∈ Ml,m(Z), l = |[a]| and m is determined by a and Γ. There is a finite-to-one, Ω[a]–equivariant set map:

• k

Ck →

• k

Ml+m,k(Z) The theorem reduces to orbit-checking and finding stabilizer presentations for the action of GL(l, Z) ⋉ Ml,m(Z) on Ml+m,k(Z).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 20 / 22

Algorithm for checking GL(l, Z) ⋉ Ml,m(Z) orbits

The trick here is to use rational row moves in the off-diagonal block.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 21 / 22

Algorithm for checking GL(l, Z) ⋉ Ml,m(Z) orbits

The trick here is to use rational row moves in the off-diagonal block. For A ∈ Ml+m,k(Z), put A into a modified Hermite Normal Form:

add rational multiples of the bottom m rows to the top l rows to zero

• ut as many columns as possible,

do integer row reduction on top l rows to put the top l rows in (a multiple of) Hermite Normal Form.

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 21 / 22

Algorithm for checking GL(l, Z) ⋉ Ml,m(Z) orbits

The trick here is to use rational row moves in the off-diagonal block. For A ∈ Ml+m,k(Z), put A into a modified Hermite Normal Form:

add rational multiples of the bottom m rows to the top l rows to zero

• ut as many columns as possible,

do integer row reduction on top l rows to put the top l rows in (a multiple of) Hermite Normal Form.

For A, B ∈ Ml+m,k(Z) with the same normal form N in Ml+m,k(Q):

find a common denominator d such that N ∈ Ml+m,k( 1

d Z) and there

are P, Q ∈ GL(l, Z) ⋉ Ml,m( 1

d Z) with N = PA = QB;

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 21 / 22

Algorithm for checking GL(l, Z) ⋉ Ml,m(Z) orbits

The trick here is to use rational row moves in the off-diagonal block. For A ∈ Ml+m,k(Z), put A into a modified Hermite Normal Form:

add rational multiples of the bottom m rows to the top l rows to zero

• ut as many columns as possible,

do integer row reduction on top l rows to put the top l rows in (a multiple of) Hermite Normal Form.

For A, B ∈ Ml+m,k(Z) with the same normal form N in Ml+m,k(Q):

find a common denominator d such that N ∈ Ml+m,k( 1

d Z) and there

are P, Q ∈ GL(l, Z) ⋉ Ml,m( 1

d Z) with N = PA = QB;

find a finite generating set S for the stabilizer of N in GL(l, Z) ⋉ Ml,m( 1

d Z);

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 21 / 22

Algorithm for checking GL(l, Z) ⋉ Ml,m(Z) orbits

The trick here is to use rational row moves in the off-diagonal block. For A ∈ Ml+m,k(Z), put A into a modified Hermite Normal Form:

add rational multiples of the bottom m rows to the top l rows to zero

• ut as many columns as possible,

do integer row reduction on top l rows to put the top l rows in (a multiple of) Hermite Normal Form.

For A, B ∈ Ml+m,k(Z) with the same normal form N in Ml+m,k(Q):

find a common denominator d such that N ∈ Ml+m,k( 1

d Z) and there

are P, Q ∈ GL(l, Z) ⋉ Ml,m( 1

d Z) with N = PA = QB;

find a finite generating set S for the stabilizer of N in GL(l, Z) ⋉ Ml,m( 1

d Z);

then A and B are in the same GL(l, Z) ⋉ Ml,m(Z)–orbit if and only if the cosets of P and Q are in the same connected component of the Schreier graph of GL(l, Z) ⋉ Ml,m(Z) in GL(l, Z) ⋉ Ml,m( 1

d Z) with

respect to S (a finite check).

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 21 / 22

Thank you!

Matthew Day (U of Arkansas) Aut orbit membership for RAAGs May 30,2013 22 / 22