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Scaling limits and influence of the seed graph in preferential attachment trees Ioan Manolescu joint work with Nicolas Curien, Thomas Duquesne and Igor Kortchemski University of Geneva 9th December 2014 Journ ee cartes al eatoires - IHES

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SLIDE 1

Scaling limits and influence of the seed graph in preferential attachment trees

Ioan Manolescu joint work with Nicolas Curien, Thomas Duquesne and Igor Kortchemski

University of Geneva

9th December 2014 Journ´ ee cartes al´ eatoires - IHES

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 1 / 14

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SLIDE 2

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 3

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 4

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 5

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 6

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 7

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 8

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 9

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 10

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 11

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 12

Linear preferential attachment: the model

Initial tree: T (S)

n0

= S (where n0 = |S|) T (S)

n+1 is obtained from T (S) n

by adding an edge to a random vertex v ∈ T (S)

n

, chosen proportionally to its degree. Questions: Does the process mix? For S1 = S2, does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? How does T (S)

n

look like when n is very large? Scaling limit?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 2 / 14

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SLIDE 13

Recognise the seed

Question: For S1 = S2, does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 14

Recognise the seed

Question: For S1 = S2, does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? It may. . . Example:

T (S2)

3

= S2 T (S1)

2

= S1

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 15

Recognise the seed

Question: For S1 = S2, does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? It may. . . Example:

T (S1)

3

= S2 T (S2)

3

= S2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 16

Recognise the seed

Question: For S1 = S2, does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? It may. . . Example:

T (S1)

3

= S2 T (S2)

3

= S2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 17

Recognise the seed

Question: For S1 = S2, with |S1| =|S2| ≥ 3 does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? NO!

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 18

Recognise the seed

Question: For S1 = S2, with |S1| =|S2| ≥ 3 does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? NO! Idea of Bubeck, Mossel, R´ acz: Study the degree sequence of Tn: Deg(Tn) = {deg(v) : v ∈ Tn}.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 19

Recognise the seed

Question: For S1 = S2, with |S1| =|S2| ≥ 3 does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? NO! Idea of Bubeck, Mossel, R´ acz: Study the degree sequence of Tn: Deg(Tn) = {deg(v) : v ∈ Tn}. Theorem (Bubeck, Mossel, R´ acz) For seeds S1 = S2 with |S1| = |S2| but Deg(S1) = Deg(S2), dTV(Deg(T (S1)

n

); Deg(T (S2)

n

)) / − → 0.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 20

Recognise the seed

Question: For S1 = S2, with |S1| =|S2| ≥ 3 does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? NO! Idea of Bubeck, Mossel, R´ acz: Study the degree sequence of Tn: Deg(Tn) = {deg(v) : v ∈ Tn}. Theorem (Bubeck, Mossel, R´ acz) For seeds S1 = S2 with |S1| = |S2| but Deg(S1) = Deg(S2), dTV(Deg(T (S1)

n

); Deg(T (S2)

n

)) / − → 0. Proof: The degree sequence is given by a P´

  • lya urn.

The tail of max Deg(T (S)

n

) depends on Deg(S).

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 21

Recognise the seed

Question: For S1 = S2, with |S1| =|S2| ≥ 3 does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? NO! Idea of Bubeck, Mossel, R´ acz: Study the degree sequence of Tn: Deg(Tn) = {deg(v) : v ∈ Tn}. Theorem (Bubeck, Mossel, R´ acz) For seeds S1 = S2 with |S1| = |S2| but Deg(S1) = Deg(S2), dTV(Deg(T (S1)

n

); Deg(T (S2)

n

)) / − → 0. Proof: The degree sequence is given by a P´

  • lya urn.

The tail of max Deg(T (S)

n

) depends on Deg(S). Problem: This strategy can not distinguish between seeds with same degree sequences. S1 S2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 22

Recognise the seed

Question: For S1 = S2, with |S1| =|S2| ≥ 3 does dTV(T (S1)

n

; T (S2)

n

) → 0 as n → ∞? NO! Idea of Bubeck, Mossel, R´ acz: Study the degree sequence of Tn: Deg(Tn) = {deg(v) : v ∈ Tn}. Theorem (Bubeck, Mossel, R´ acz) For seeds S1 = S2 with |S1| = |S2| but Deg(S1) = Deg(S2), dTV(Deg(T (S1)

n

); Deg(T (S2)

n

)) / − → 0. Proof: The degree sequence is given by a P´

  • lya urn.

The tail of max Deg(T (S)

n

) depends on Deg(S). Conclusion: Need some geometric observable to distinguish T (S1)

n

form T (S2)

n

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 3 / 14

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SLIDE 23

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 24

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Diameter of Tn: log n. Maximal degree of Tn: √n. No non-trivial compact scaling limit!

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 25

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Solution: consider the loop tree Loop(Tn)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 26

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Solution: consider the loop tree Loop(Tn)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 27

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Solution: consider the loop tree Loop(Tn) − → diameter: √n.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 28

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Solution: consider the loop tree Loop(Tn) − → diameter: √n. Theorem n−1/2 · Loop(T (S)

n

)

a.s. for G.H.

− − − − − − − →

n→∞

L(S), where L(S) is a random compact metric space called the ”Bownian looptree”.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 29

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Solution: consider the loop tree Loop(Tn) − → diameter: √n. Theorem n−1/2 · Loop(T (S)

n

)

a.s. for G.H.

− − − − − − − →

n→∞

L(S), where L(S) is a random compact metric space called the ”Bownian looptree”. The loop tree is well defined for plane trees. How do we embed T (S)

n

?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 30

How to define a limit of Tn? Convergence in Gromov–Hausdorff topology? Solution: consider the loop tree Loop(Tn) − → diameter: √n. Theorem n−1/2 · Loop(T (S)

n

)

a.s. for G.H.

− − − − − − − →

n→∞

L(S), where L(S) is a random compact metric space called the ”Bownian looptree”. The loop tree is well defined for plane trees. How do we embed T (S)

n

?

  • Uniformly. . .

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 4 / 14

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SLIDE 31

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 32

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 33

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 34

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 35

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

slide-36
SLIDE 36

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

slide-37
SLIDE 37

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

slide-38
SLIDE 38

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

slide-39
SLIDE 39

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 40

The plane LPAM and R´ emy’s algorithm

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 41

The plane LPAM and R´ emy’s algorithm

Gromov - Hausdorff CRT Rn + ordered leaves

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 42

The plane LPAM and R´ emy’s algorithm

X0 = root X1 X2 X3 X4 X5 Gromov - Hausdorff + points X0, X1, . . . CRT + uniform points Rn + ordered leaves

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 43

The plane LPAM and R´ emy’s algorithm

X0 = root X1 X2 X3 X4 X5 Gromov - Hausdorff + points X0, X1, . . . CRT + uniform points Gromov - Hausdorff Glue points L = Glue(CRT; X0, X1, . . . ) Glue infinitely many points Rn + ordered leaves Glue(Rn)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 5 / 14

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SLIDE 44

Rn = nth step in R´ emy’s algorithm. X n

0 , . . . , X n n = leaves in order of appearance.

Theorem (R´ emy ’85; Curien & Haas ’13) Then Rn is a uniform tree with n edges and X n

0 , . . . , X n n is a uniform ordering of

its leaves. Moreover, for any k fixed, n−1/2 · (Rn; X n

0 , . . . , X n k ) a.s.for k−pointed G.H.

− − − − − − − − − − − − − →

n→∞

2 √ 2 · (CRT; X0, . . . , Xk), where X0, X1, . . . are i.i.d. points in the CRT, chosen according to its mass measure.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 6 / 14

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SLIDE 45

Rn = nth step in R´ emy’s algorithm. X n

0 , . . . , X n n = leaves in order of appearance.

Theorem (R´ emy ’85; Curien & Haas ’13) Then Rn is a uniform tree with n edges and X n

0 , . . . , X n n is a uniform ordering of

its leaves. Moreover, for any k fixed, n−1/2 · (Rn; X n

0 , . . . , X n k ) a.s.for k−pointed G.H.

− − − − − − − − − − − − − →

n→∞

2 √ 2 · (CRT; X0, . . . , Xk), where X0, X1, . . . are i.i.d. points in the CRT, chosen according to its mass measure. Consequence: n−1/2 · Glue(Rn; X n

0 , . . . , X n k ) a.s. for G.H.

− − − − − − − →

n→∞

2 √ 2 · Glue(CRT; X0, . . . , Xk).

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 6 / 14

slide-46
SLIDE 46

Rn = nth step in R´ emy’s algorithm. X n

0 , . . . , X n n = leaves in order of appearance.

Theorem (R´ emy ’85; Curien & Haas ’13) Then Rn is a uniform tree with n edges and X n

0 , . . . , X n n is a uniform ordering of

its leaves. Moreover, for any k fixed, n−1/2 · (Rn; X n

0 , . . . , X n k ) a.s.for k−pointed G.H.

− − − − − − − − − − − − − →

n→∞

2 √ 2 · (CRT; X0, . . . , Xk), where X0, X1, . . . are i.i.d. points in the CRT, chosen according to its mass measure. Consequence: n−1/2 · Glue(Rn; X n

0 , . . . , X n k ) a.s. for G.H.

− − − − − − − →

n→∞

2 √ 2 · Glue(CRT; X0, . . . , Xk). Theorem n−1/2 · Loop(T ⊸

n ) a.s. for G.H.

− − − − − − − →

n→∞

2 √ 2 · L, where L is the limit of Glue(CRT; X0, . . . , Xk) as k → ∞.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 6 / 14

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SLIDE 47

The plane LPAM and R´ emy’s algorithm

Gromov - Hausdorff CRT Rn + ordered leaves Gromov - Hausdorff Glue k points Glue k points Glue(Rn; Xn

1 . . . , Xn k )

L = Glue(CRT; X0, . . . , Xk) X0 = root X1 X2 X3 X4 X5 + points X0, . . . , Xk + (k + 1) uniform points

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 7 / 14

slide-48
SLIDE 48

The plane LPAM and R´ emy’s algorithm

X0 = root X1 X2 X3 X4 X5 Gromov - Hausdorff + points X0, X1, . . . CRT + uniform points Gromov - Hausdorff Glue points L = Glue(CRT; X0, X1, . . . ) Glue infinitely many points Rn + ordered leaves Glue(Rn)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 7 / 14

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SLIDE 49

Properties of the loop tree

Theorem L has a.s. Hausdorff dimension 2. Big faces touch each other! (In Tn the vertices of large degree are at finite distance.)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 8 / 14

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SLIDE 50

In the previous, we looked at T ⊸

n

with seed ⊸. For general seeds S, with N corners: T (S)

n

is obtained by: sample N variables αn

1, . . . , αn N with P´

  • lya urn distribution,

sample N independent Ln

1, . . . , Ln N of LPAMs started from ⊸ with resp.

αn

1, . . . , αn N vertices;

attach each Ln

i in a corner of S.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 9 / 14

slide-51
SLIDE 51

In the previous, we looked at T ⊸

n

with seed ⊸. For general seeds S, with N corners: T (S)

n

is obtained by: sample N variables αn

1, . . . , αn N with P´

  • lya urn distribution,

sample N independent Ln

1, . . . , Ln N of LPAMs started from ⊸ with resp.

αn

1, . . . , αn N vertices;

attach each Ln

i in a corner of S.

αn

1

αn

2

αn

N

. . . . . . . . .

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 9 / 14

slide-52
SLIDE 52

In the previous, we looked at T ⊸

n

with seed ⊸. For general seeds S, with N corners: T (S)

n

is obtained by: sample N variables αn

1, . . . , αn N with P´

  • lya urn distribution,

sample N independent Ln

1, . . . , Ln N of LPAMs started from ⊸ with resp.

αn

1, . . . , αn N vertices;

attach each Ln

i in a corner of S.

αn

1

αn

2

αn

N

. . . . . . . . .

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 9 / 14

slide-53
SLIDE 53

In the previous, we looked at T ⊸

n

with seed ⊸. For general seeds S, with N corners: n−1/2 · Loop(T (S)

n

)

a.s. for G.H.

− − − − − − − →

n→∞

2 √ 2 · L(S). where L(S) is obtained by: sample N variables α1, . . . , αN with Dirichlet distribution (1/2, . . . , 1/2), sample N i.i.d. instances of L: L1, . . . , LN, glue each loop tree αiLi on an edge of Loop(S).

αn

1

αn

2

αn

N

. . . . . . . . .

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 9 / 14

slide-54
SLIDE 54

In the previous, we looked at T ⊸

n

with seed ⊸. For general seeds S, with N corners: n−1/2 · Loop(T (S)

n

)

a.s. for G.H.

− − − − − − − →

n→∞

2 √ 2 · L(S). where L(S) is obtained by: sample N variables α1, . . . , αN with Dirichlet distribution (1/2, . . . , 1/2), sample N i.i.d. instances of L: L1, . . . , LN, glue each loop tree αiLi on an edge of Loop(S).

αn

1

αn

2

αn

N

. . . . . . . . .

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 9 / 14

slide-55
SLIDE 55

Back to distinguishing the seeds

Which observable to use?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 10 / 14

slide-56
SLIDE 56

Back to distinguishing the seeds

Which observable to use? Number of embeddings of a given (small) tree: A tree τ may be embedded in Dτ(T (S)

n

) ways in T (S)

n

.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 10 / 14

slide-57
SLIDE 57

Back to distinguishing the seeds

Which observable to use? Number of embeddings of a given (small) tree: A tree τ may be embedded in Dτ(T (S)

n

) ways in T (S)

n

. Do Dτ(T (S1)

n

) and Dτ(T (S2)

n

) have different asymptotics? n−? · Dτ(T (S)

n

) → d(S)?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 10 / 14

slide-58
SLIDE 58

For technical conditions, we will work with decorated trees.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-59
SLIDE 59

For technical conditions, we will work with decorated trees.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-60
SLIDE 60

For technical conditions, we will work with decorated trees.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-61
SLIDE 61

For technical conditions, we will work with decorated trees.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-62
SLIDE 62

For technical conditions, we will work with decorated trees. Dτ(T) = number of embeddings.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-63
SLIDE 63

For technical conditions, we will work with decorated trees. Dτ(T) = number of embeddings. We may expect: n−|τ|/2 · Dτ(T (S)

n

) → d(S), with d(S) a random variable that depends on S. Because of the small stubs, there may be logarithmic cor- rections.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-64
SLIDE 64

For technical conditions, we will work with decorated trees. Dτ(T) = number of embeddings. Proposition For τ a decorated tree there exist constants {cn(τ, τ ′) : τ ′ τ, n ≥ 2} such that Mτ(T (S)

n

) =

  • τ ′τ

cn(τ, τ ′) · Dτ ′(T (S)

n

) is a martingale for any seed S, and is bounded in L2.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-65
SLIDE 65

For technical conditions, we will work with decorated trees. Dτ(T) = number of embeddings. Proposition For τ a decorated tree there exist constants {cn(τ, τ ′) : τ ′ τ, n ≥ 2} such that Mτ(T (S)

n

) =

  • τ ′τ

cn(τ, τ ′) · Dτ ′(T (S)

n

) is a martingale for any seed S, and is bounded in L2.

1 1 2 2

For S1 = S2 with n0 = |S1| = |S2| ≥ 3, there exists a decorated tree τ such that Mτ(T (S1)

n0

) = Mτ(T (S2)

n0

) .

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

slide-66
SLIDE 66

For technical conditions, we will work with decorated trees. Dτ(T) = number of embeddings. Proposition For τ a decorated tree there exist constants {cn(τ, τ ′) : τ ′ τ, n ≥ 2} such that Mτ(T (S)

n

) =

  • τ ′τ

cn(τ, τ ′) · Dτ ′(T (S)

n

) is a martingale for any seed S, and is bounded in L2.

1 1 2 2

For S1 = S2 with n0 = |S1| = |S2| ≥ 3, there exists a decorated tree τ such that E[Mτ(T (S1)

n

)] = E[Mτ(T (S2)

n

)]. Theorem For S1 = S2 with |S1|, |S2| ≥ 3, dTV(T (S1)

n

; T (S2)

n

) stays bounded away from 0. In other words d(S1, S2) = lim dTV(T (S1)

n

; T (S2)

n

) is a distance on seeds with at least 3 vertices.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 11 / 14

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SLIDE 67

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

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SLIDE 68

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-69
SLIDE 69

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-70
SLIDE 70

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-71
SLIDE 71

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-72
SLIDE 72

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2 1 2 2

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-73
SLIDE 73

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2 1 2 1

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-74
SLIDE 74

Idea of proof: recurrence for Dτ(Tn)

For any τ, there exist constants {c(τ, τ ′) : τ ′ ≺ τ} such that E

  • Dτ
  • T (S)

n+1

  • T (S)

n

  • =
  • 1 +

|τ| 2n − 2

  • Dτ
  • T (S)

n

  • +

1 2n − 2

  • τ ′≺τ

c(τ, τ ′)Dτ ′ T (S)

n

  • .

When τ =

1 we have D 1 (T (S)

n

) = 2n − 2.

1 1 2 2 1 2 1

Using this recurrence formula, we show the existence of the martingales Mτ(T (S)

n

)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 12 / 14

slide-75
SLIDE 75

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-76
SLIDE 76

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ).

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-77
SLIDE 77

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-78
SLIDE 78

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ. Choose red edges with prob. 1 + α and the other with prob 1 − α; α = 1/(2 + δ)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-79
SLIDE 79

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ. Choose red edges with prob. 1 + α and the other with prob 1 − α; α = 1/(2 + δ)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-80
SLIDE 80

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ. Choose red edges with prob. 1 + α and the other with prob 1 − α; α = 1/(2 + δ)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-81
SLIDE 81

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ. Choose red edges with prob. 1 + α and the other with prob 1 − α; α = 1/(2 + δ)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-82
SLIDE 82

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ. Choose red edges with prob. 1 + α and the other with prob 1 − α; α = 1/(2 + δ)

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-83
SLIDE 83

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ. Choose red edges with prob. 1 + α and the other with prob 1 − α; α = 1/(2 + δ) L⊸,δ should come from a fragmentation tree of Hausdorff dimension 2 + δ.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-84
SLIDE 84

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ.

  • dTV(T (S1),δ

n

; T (S2),δ

n

) / − → 0 for S1 = S2.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-85
SLIDE 85

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ.

  • dTV(T (S1),δ

n

; T (S2),δ

n

) / − → 0 for S1 = S2. In our proof the exchangeability of the corners played an essential role! We expect a similar result. For δ = ∞ (vertex chosen uniformly) - result obtained by Bubeck, Eldan, Mossel, R´ acz.

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-86
SLIDE 86

Related model

Affine reinforcement: For δ > −1, T (S),δ

n+1

is obtained from T (S),δ

n

by adding an edge to a vertex v chosen with probability proportional to deg(v) + δ. Maximal degree: n1/(2+δ). Conjectures: • n−1/(2+δ) · Loop(T (S),δ

n

) → L(S),δ.

  • dTV(T (S1),δ

n

; T (S2),δ

n

) / − → 0 for S1 = S2.

  • Is all the asymptotic information on T (S)

n

contained in L(S)?

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 13 / 14

slide-87
SLIDE 87

Thank you!

Ioan Manolescu (University of Geneva) LPAM 9th December 2014 14 / 14