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Embeddability between the right-angled Artin groups of surfaces Takuya Katayama Hiroshima University (JSPS Research Fellow PD) Mathematics of Knots II Nihon University, December 19, 2019 Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 1

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Embeddability between the right-angled Artin groups of surfaces

Takuya Katayama

Hiroshima University (JSPS Research Fellow PD)

Mathematics of Knots II Nihon University, December 19, 2019

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 1 / 23

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I. Introduction
II. Main Theorem

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 2 / 23

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Right-angled Artin groups

Γ: a finite (simplicial) graph V (Γ) = {v1, v2, . . . , vn}: the vertex set of Γ E(Γ): the edge set of Γ

Definition

The right-angled Artin group (RAAG) A(Γ) on Γ is the group given by the following presentation: A(Γ) = ⟨v1, v2, . . . , vn | [vi, vj] = 1 if {vi, vj} ∈ E(Γ)⟩. A(Γ1) ∼ = A(Γ2) if and only if Γ1 ∼ = Γ2. A( ) ∼ = F3 A( ) ∼ = Z ∗ Z2 A( ) ∼ = Z × F2 A( ) ∼ = Z3

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 3 / 23

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Previous studies and motivation

Problem (Crisp-Sageev-Sapir, 2006)

For given two finite graphs Λ and Γ, decide whether A(Λ) can be embedded into A(Γ). Graph theory ⇝ Embeddability between RAAGs

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 4 / 23

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Full graph embedding of the defining graph A graph embedding ι: Λ ֒ → Γ is said to be full if {u, v} ∈ E(Λ) ⇔ {u, v} ∈ E(Γ) for all u, v ∈ V (Λ). We denote by Λ ≤ Γ if Λ has a full graph embedding into Γ. Λ is called full subgraph of Γ.

Theorem (van der Lek, 1983)

Λ, Γ: finite graphs If Λ ≤ Γ, then A(Λ) ֒ → A(Γ). Natural map v

→ ι(v) extends to a homomorphism, and is injective (non-trivial).

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 5 / 23

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Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 6 / 23

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Extension graph The extension graph Γe of a finite graph Γ is a graph that V (Γe) = {v g | v ∈ V (Γ), g ∈ A(Γ)} and E(Γe) = {{uh, v g} | [uh, v g] = 1 in A(Γ)}. v g := g −1vg E.g. P4 = ⇒ Pe

4 is a locally infinite tree. valence ∞

valence 1

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 7 / 23

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valence ∞ valence 1

Note: If [u, v] = uvu−1v −1 = 1, then [uw, v w] = (w −1uw)(w −1vw)(w −1u−1w)(w −1v −1w) = w −1uvu−1v −1w = 1.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 8 / 23

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Theorem (Kim–Koberda, 2013)

Λ, Γ: finite graphs If Λ ≤ Γe, then A(Λ) ֒ → A(Γ).

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 9 / 23

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The existence of embeddings not derived from the extension graph

Theorem (Kim–Koberda, 2013)

C5 ̸≤ (Pc

8)e.

Theorem (Casals-Ruiz, Duncan, Kazachkov, 2015)

A(C5) ֒ → A(Pc

8).

Kim–Koberda (2015) gave infinitely many examples.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 10 / 23

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Some results on non-existence of embeddings Kim–Koberda (2013): Every embedding A(Λ) ֒ → A(Γ) has a “normal form” described by Λ and Γe.

Theorem (Kim–Koberda, 2013)

Λ, Γ: finite graphs w/o triangles If A(Λ) ֒ → A(Γ), then Λ ≤ Γe.

Theorem (K., 2018)

Λ: the complement graph of a liner forest Γ: a finite graph If A(Λ) ֒ → A(Γ), then Λ ≤ Γe.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 11 / 23

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Problem (recall)

For given two finite graphs Λ and Γ, decide whether A(Λ) can be embedded into A(Γ). Graph theory ⇝ Embeddability between RAAGs How about topology??

Motivation

Topology ⇝

?? Embeddability between RAAGs

Theorem (Kim–Koberda, 2013)

Λ: a circle Γ: a finite graph If A(Λ) ֒ → A(Γ), then Γ contains a circle.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 12 / 23

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In order to consider manifolds together with graphs, we need the following concept.

Definition

Λ: a finite graph The flag complex of Λ is a simplicial complex that {n-simplex in XΛ}

1:1

↔ {complete subgraph on (n + 1) vertices in Λ} X (1)

= Λ.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 13 / 23

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E.g. A( ) ∼ = Z ∗ Z2 A( ) ∼ = Z3

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 14 / 23

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Main Theorem

Main Thereom

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ ∼ = S2, then XΓ contains a subcomplex ∼ = S2.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 15 / 23

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Main Thereom (recall)

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ ∼ = S2, then XΓ contains a subcomplex ∼ = S2. Main Theorem and the fact that a 2-sphere S2 is not contained in any other surfaces imply the following.

Corollary

Λ, Γ: finite graphs Suppose that XΛ ∼ = S2 and XΓ ∼ =“a surface other than S2”. Then A(Λ) ̸֒ → A(Γ).

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 16 / 23

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Main Thereom (recall)

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ ∼ = S2, then XΓ contains a subcomplex ∼ = S2. We divide the proof into the following two steps. Step A. Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). Assuming XΛ ∼ = S2, prove that XΓe contains a subcomplex ∼ = S2. Here Γe is the extension graph of Γ. Step B. Assuming XΓe contains a subcomplex ∼ = S2, prove that XΓ contains a subcomplex ∼ = S2.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 17 / 23

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Step B. Assuming XΓe contains a subcomplex ∼ = S2, prove that XΓ contains a subcomplex ∼ = S2.

Lemma B

Γ: a finite graph If XΓe contains a subcomplex ∼ = S2, then so does XΓ. In order to understand the extension graph well, we use the “double”

f graphs.

Definition (recall)

Let Γ be a graph and v a vertex of Γ. The star subgraph St(v, Γ) of Γ is a full subgraph induced by {u ∈ V (Γ) | {u, v} ∈ E(Γ)} ∪ {v}. The double D(v, Γ) of Γ is Γ ∪St(v,Γ) Γ.

=

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 18 / 23

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Lemma B (recall)

Γ: a finite graph If XΓe contains a subcomplex ∼ = S2, then so does XΓ.

Lemma (Kim–Koberda)

Λ, Γ: finite graphs If Λ ⊂ Γe, there is a sequence of doubles Γ = D1 ≤ D2 ≤ . . . ≤ Dn such that Λ ⊂ Dn and that Di+1 is a double of Di. By the above Lemma, it is enough to prove the following.

Lemma B’

Γ: a finite graph, v ∈ V (Γ) If XD(v,Γ) contains a subcomplex ∼ = S2, then so does XΓ. S ⊂ XΓe w/ S ∼ = S2 ⇒ Lemma (KK) XΓ = XD1 ≤ ∃XD2 ≤ . . . ≤ ∃XDn; S ⊂ XDn.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 19 / 23

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Lemma B’ (recall)

Γ: a finite graph, v ∈ V (Γ) If XD(v,Γ) contains a subcomplex ∼ = S2, then so does XΓ. Idea of the proof) Note: XD(v,Γ) = XΓ ∪XSt(v,Γ) XΓ.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 20 / 23

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How to find a disk V in the sphere with ∂V ⊂ XSt and V ⊂ XΓ?

Claim

Suppose that S2 ̸⊂“copies of XΓ”. Then S2 ∩ XSt contains a circle.

Claim

V : a disk in S2 with ∂V ⊂ XSt and V ̸⊂“copies of XΓ”. For any edge-path P joining opposite points, ∃C a circle; C ⊂ V ∩ XSt and C ∩ P ̸= ∅.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 21 / 23

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Future work

Main Thereom (recall)

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ has a subcomplex ∼ = S2, then XΓ has a subcomplex ∼ = S2.

Asphericity Question

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). dim(XΓ) = 2. Then does π2(XΛ) ̸= 1 imply π2(XΓ) ̸= 1? Note: For 2-complexes, π2 = 1 ⇔ asphericity.

Whitehead’s Asphericity Conjecture for simp.cpx., 1941

X: an aspherical 2-dim simplicial complex Every connected subcomplex of X is aspherical.

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 22 / 23

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Thank you very much for your attention!

Takuya Katayama (Hiroshima Univ.) RAAGs of surfaces 23 / 23

Embeddability between the right-angled Artin groups of surfaces

Takuya Katayama

Hiroshima University (JSPS Research Fellow PD)

Mathematics of Knots II Nihon University, December 19, 2019

Table of contents

Right-angled Artin groups

Γ: a finite (simplicial) graph V (Γ) = {v1, v2, . . . , vn}: the vertex set of Γ E(Γ): the edge set of Γ

Definition

The right-angled Artin group (RAAG) A(Γ) on Γ is the group given by the following presentation: A(Γ) = ⟨v1, v2, . . . , vn | [vi, vj] = 1 if {vi, vj} ∈ E(Γ)⟩. A(Γ1) ∼ = A(Γ2) if and only if Γ1 ∼ = Γ2. A( ) ∼ = F3 A( ) ∼ = Z ∗ Z2 A( ) ∼ = Z × F2 A( ) ∼ = Z3

Previous studies and motivation

Problem (Crisp-Sageev-Sapir, 2006)

For given two finite graphs Λ and Γ, decide whether A(Λ) can be embedded into A(Γ). Graph theory ⇝ Embeddability between RAAGs

Full graph embedding of the defining graph A graph embedding ι: Λ ֒ → Γ is said to be full if {u, v} ∈ E(Λ) ⇔ {u, v} ∈ E(Γ) for all u, v ∈ V (Λ). We denote by Λ ≤ Γ if Λ has a full graph embedding into Γ. Λ is called full subgraph of Γ.

Theorem (van der Lek, 1983)

Λ, Γ: finite graphs If Λ ≤ Γ, then A(Λ) ֒ → A(Γ). Natural map v

→ ι(v) extends to a homomorphism, and is injective (non-trivial).

Extension graph The extension graph Γe of a finite graph Γ is a graph that V (Γe) = {v g | v ∈ V (Γ), g ∈ A(Γ)} and E(Γe) = {{uh, v g} | [uh, v g] = 1 in A(Γ)}. v g := g −1vg E.g. P4 = ⇒ Pe

Note: If [u, v] = uvu−1v −1 = 1, then [uw, v w] = (w −1uw)(w −1vw)(w −1u−1w)(w −1v −1w) = w −1uvu−1v −1w = 1.

Theorem (Kim–Koberda, 2013)

Λ, Γ: finite graphs If Λ ≤ Γe, then A(Λ) ֒ → A(Γ).

The existence of embeddings not derived from the extension graph

Theorem (Kim–Koberda, 2013)

C5 ̸≤ (Pc

Theorem (Casals-Ruiz, Duncan, Kazachkov, 2015)

A(C5) ֒ → A(Pc

Kim–Koberda (2015) gave infinitely many examples.

Some results on non-existence of embeddings Kim–Koberda (2013): Every embedding A(Λ) ֒ → A(Γ) has a “normal form” described by Λ and Γe.

Theorem (Kim–Koberda, 2013)

Λ, Γ: finite graphs w/o triangles If A(Λ) ֒ → A(Γ), then Λ ≤ Γe.

Theorem (K., 2018)

Λ: the complement graph of a liner forest Γ: a finite graph If A(Λ) ֒ → A(Γ), then Λ ≤ Γe.

Problem (recall)

For given two finite graphs Λ and Γ, decide whether A(Λ) can be embedded into A(Γ). Graph theory ⇝ Embeddability between RAAGs How about topology??

Motivation

Topology ⇝

Theorem (Kim–Koberda, 2013)

Λ: a circle Γ: a finite graph If A(Λ) ֒ → A(Γ), then Γ contains a circle.

In order to consider manifolds together with graphs, we need the following concept.

Definition

Λ: a finite graph The flag complex of Λ is a simplicial complex that {n-simplex in XΛ}

↔ {complete subgraph on (n + 1) vertices in Λ} X (1)

= Λ.

E.g. A( ) ∼ = Z ∗ Z2 A( ) ∼ = Z3

Main Theorem

Main Thereom

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ ∼ = S2, then XΓ contains a subcomplex ∼ = S2.

Main Thereom (recall)

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ ∼ = S2, then XΓ contains a subcomplex ∼ = S2. Main Theorem and the fact that a 2-sphere S2 is not contained in any other surfaces imply the following.

Corollary

Λ, Γ: finite graphs Suppose that XΛ ∼ = S2 and XΓ ∼ =“a surface other than S2”. Then A(Λ) ̸֒ → A(Γ).

Main Thereom (recall)

Step B. Assuming XΓe contains a subcomplex ∼ = S2, prove that XΓ contains a subcomplex ∼ = S2.

Lemma B

Γ: a finite graph If XΓe contains a subcomplex ∼ = S2, then so does XΓ. In order to understand the extension graph well, we use the “double”

Definition (recall)

Let Γ be a graph and v a vertex of Γ. The star subgraph St(v, Γ) of Γ is a full subgraph induced by {u ∈ V (Γ) | {u, v} ∈ E(Γ)} ∪ {v}. The double D(v, Γ) of Γ is Γ ∪St(v,Γ) Γ.

=

Lemma B (recall)

Γ: a finite graph If XΓe contains a subcomplex ∼ = S2, then so does XΓ.

Lemma (Kim–Koberda)

Λ, Γ: finite graphs If Λ ⊂ Γe, there is a sequence of doubles Γ = D1 ≤ D2 ≤ . . . ≤ Dn such that Λ ⊂ Dn and that Di+1 is a double of Di. By the above Lemma, it is enough to prove the following.

Lemma B’

Γ: a finite graph, v ∈ V (Γ) If XD(v,Γ) contains a subcomplex ∼ = S2, then so does XΓ. S ⊂ XΓe w/ S ∼ = S2 ⇒ Lemma (KK) XΓ = XD1 ≤ ∃XD2 ≤ . . . ≤ ∃XDn; S ⊂ XDn.

Lemma B’ (recall)

Γ: a finite graph, v ∈ V (Γ) If XD(v,Γ) contains a subcomplex ∼ = S2, then so does XΓ. Idea of the proof) Note: XD(v,Γ) = XΓ ∪XSt(v,Γ) XΓ.

How to find a disk V in the sphere with ∂V ⊂ XSt and V ⊂ XΓ?

Claim

Suppose that S2 ̸⊂“copies of XΓ”. Then S2 ∩ XSt contains a circle.

Claim

V : a disk in S2 with ∂V ⊂ XSt and V ̸⊂“copies of XΓ”. For any edge-path P joining opposite points, ∃C a circle; C ⊂ V ∩ XSt and C ∩ P ̸= ∅.

Future work

Main Thereom (recall)

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). If XΛ has a subcomplex ∼ = S2, then XΓ has a subcomplex ∼ = S2.

Asphericity Question

Λ, Γ: finite graphs w/ A(Λ) ֒ → A(Γ). dim(XΓ) = 2. Then does π2(XΛ) ̸= 1 imply π2(XΓ) ̸= 1? Note: For 2-complexes, π2 = 1 ⇔ asphericity.

Whitehead’s Asphericity Conjecture for simp.cpx., 1941

X: an aspherical 2-dim simplicial complex Every connected subcomplex of X is aspherical.

Thank you very much for your attention!