joint work with joscha diehl, terry lyons and jeremy reizenstein … of... · theorem...

Areas of Areas generate the shuffle algebraJoint work with Joscha Diehl, Terry Lyons and Jeremy Reizenstein

Rosa Preiß

Technische Universität Berlin

Algebra, Combinatorics and Applications, Zoom, April 17, 2020

Rosa Preiß (Technische Universität Berlin) Areas of Areas Algebra, Comb. & Appl. 1 / 23

Introduction

✓ SO(2) invariant


Introduction

The Planimeter: a mechanical device to calculate the area

http://persweb.wabash.edu/facstaff/footer/Planimeter/Polar&Linear.htm


http://persweb.wabash.edu/facstaff/footer/Planimeter/Polar&Linear.htm

Introduction

For functions X, Y ∶ [0, T] → R, let Area(X, Y)t denote two times thesigned area enclosed by the two-dimensional path (X, Y) up to time t. Wehave

Area(X, Y)t = ∫ t

0XsdYs − ∫ t

0YsdXs

= ∫ t

0XsYsds − ∫ t

0YsXsds.


Introduction

✓ SO(2) invariant

✓ iterateable: Area(Area(X, Y), Z), Area(Area(X, Y), Area(Y, Z)), . . .

✓ behaves nicely with respect to discretization andItô-Stratonovich-integration for semimartingals


Introduction

✓ SO(2) invariant✓ iterateable: Area(Area(X, Y), Z), Area(Area(X, Y), Area(Y, Z)), . . .

✓ behaves nicely with respect to discretization andItô-Stratonovich-integration for semimartingals


The iterated integral signature

We consider the signature σ(X) of a path X as an element of T((Rd)),the dual space of the space of words T(Rd), i.e.

⟨σ(X), i1⋯in⟩ = ∫ T

0∫ rn

0⋯∫ r2

0dXi1

r1 . . . dXinrn .

Both T(Rd) and T((Rd)) are endowed with an associative algebraproduct: concatenation

i1⋯in • in+1⋯im = i1⋯im, e • w = w • e = w

Example1422 • 432 = 1422432



Definition

Let w, w1 and w2 be three words and a and b two letters. We define theshuffle product of two words recursively by

e� w = w� e = w, and(w1 • a)� (w2 • b) = (w1 � (w2 • b)) • a + ((w1 • a)� w2) • b.

The shuffle product is associative, i.e.(x� y)� z = x� (y� z).

Example1� 2 = 12 + 21, 12� 3 = 123 + 132 + 312



Definition

Let w, w1 and w2 be three words and a and b two letters. We define theshuffle product of two words recursively by

e� w = w� e = w, and(w1 • a)� (w2 • b) = (w1 � (w2 • b)) • a + ((w1 • a)� w2) • b.

The shuffle product is associative, i.e.(x� y)� z = x� (y� z).Example1� 2 = 12 + 21, 12� 3 = 123 + 132 + 312



Proposition (Shuffle identity, Ree ’58)

Let X ∶ [0, L] ⟶ Rd be a regular enough path. Then, for everyu, v ∈ T(Rd), ⟨σ(X), u⟩ ⟨σ(X), v⟩ = ⟨σ(X), u� v⟩ .

Proposition (Chen’s identity, Chen ’57)

Let X, Y ∶ [0, L] ⟶ Rd be two regular enough paths and consider theirconcatenation X ⊔ Y ∶ [0, 2L] ⟶ Rd. Then, σ(X ⊔ Y) = σ(X) • σ(Y)


Halfshuffle identity

Definition (Eilenberg-MacLane ’54, Schützenberger ’58)

The right half-shuffle ≻ ∶ T≥1(Rd) × T≥1(Rd) → T≥1(Rd) is recursivelygiven on words as

w ≻ i ∶= wi,

w ≻ vi ∶= (w ≻ v + v ≻ w) • i,

where w, v are words and i is a letter.

Therefore, for any non-empty words w, v

w� v = w ≻ v + v ≻ w.



By Johnny Blood - Flickr, CC BY-SA 2.0, https://commons.wikimedia.org/w/index.php?curid=752624


https://commons.wikimedia.org/w/index.php?curid=752624


Theorem (Eilenberg-MacLane ’54, Schützenberger ’58)(T≥1(Rd),≻) is a Zinbiel algebra, i.e. for any non-empty words w, v, andu,

w ≻ (v ≻ u) = (w ≻ v + v ≻ w) ≻ u. (1)



Theorem (Schützenberger ’58)

Indeed, (T≥1(Rd),≻) is the free Zinbiel algebra over Rd.

I.e., for any Zinbiel algebra (Z,⋟) and any linear map L ∶ Rd → Z, there isa unique homomorphism ΛL ∶ (T≥1(Rd),≻) → (Z,⋟) such that

Rd (T≥1(Rd),≻)(Z,⋟)

ι

LΛL



Theorem (This has been around for a while...)

∫ s

0Xa

t dXbt = Xa≻b

s

for any a, b ∈ T≥1(Rd), where Xat ∶= ⟨σ(X↾[0,t]), a⟩



For functions X, Y ∶ [0, T] → R, let Area(X, Y)t denote two times thesigned area enclosed by the two-dimensional path (X, Y) up to time t. Wehave

Area(X, Y)t = ∫ t

0XsdYs − ∫ t

0YsdXs.

Thus,

Area(Xa, Xb)T = ⟨σ(X), area(a, b)⟩.


Areas & Tortkara

Definearea(a, b) ∶= a ≻ b − b ≻ a

area is non-associative,

area(area(1, 2), 3) = 123 − 132 + 213 − 231 − 312 + 321≠ 123 − 132 − 213 + 231 − 312 + 321 = area(1, area(2, 3))

area is anticommutative, but doesn’t satisfy the Jacobi identity,

vol(1, 2, 3) ∶= area(1, area(2, 3)) + area(2, area(3, 1)) + area(3, area(1, 2))= −123 + 132 + 213 − 231 − 312 + 321 ≠ 0


Areas & Tortkara

However, area satisfies the so-called Tortkara identity introduced byDzhumadil’daev in 2007:

area(area(a, b), area(c, d)) + area(area(a, d), area(c, b))= area(vol(a, b, c), d) + area(vol(a, d, c), b),

where

vol(x, y, z) ∶= area(area(x, y), z)+area(area(y, z), x)+area(area(z, x), y).vol corresponds to the signed volume:

Volume(Xa, Xb

, Xc) = ⟨σ(X), vol(a, b, c)⟩is six times the signed volume enclosed by the three-dimensional path(Xa

, Xb, Xc), where Xz

t = ⟨σ(X↾[0,t]), z⟩.Rosa Preiß (Technische Universität Berlin) Areas of Areas Algebra, Comb. & Appl. 16 / 23

Areas & Tortkara

Let A(Rd) be the smallest Tortkara subalgebra of (T≥1(Rd), area) thatcontains the letters.Theorem (Dzhumadil’daev-Ismailov-Mashurov 2018)

A(Rd) = Rd ⊕⨁i<j

T(Rd) • (ij − ji)


Areas & Tortkara

Theorem (Diehl-Lyons-P.-Reizenstein 2019)

T≥1(Rd) is shuffle generated by A(Rd), i.e.

span{a1 � ..� an ∶ n ≥ 1, ai ∈ A(Rd)} = T≥1(Rd).This means that for a given path, knowledge of the values of all areas ofareas is equivalent to knowledge of the full signature!

The result is a corollary of the following more general fact.


Let Xn ⊆ Tn(Rd) and X = ⋃n Xn. Then,

For all n ≥ 1, for all nonzero L ∈ gn there is an x ∈ Xn such that ⟨x, L⟩ /= 0if and only if

X shuffle generates the shuffle algebra T(Rd).


Areas & Tortkara


T≥1(Rd) is shuffle generated by A(Rd), i.e.

span{a1 � ..� an ∶ n ≥ 1, ai ∈ A(Rd)} = T≥1(Rd).This means that for a given path, knowledge of the values of all areas ofareas is equivalent to knowledge of the full signature!The result is a corollary of the following more general fact.


Let Xn ⊆ Tn(Rd) and X = ⋃n Xn. Then,

For all n ≥ 1, for all nonzero L ∈ gn there is an x ∈ Xn such that ⟨x, L⟩ /= 0if and only if

X shuffle generates the shuffle algebra T(Rd).Rosa Preiß (Technische Universität Berlin) Areas of Areas Algebra, Comb. & Appl. 18 / 23

Areas & Tortkara

Define a left bracketing of area, i.e.←−−−area(i1 . . . in) ∶= area(. . . area(area(area(i1, i2), i3), i4), . . . , in)Conjecture(A(Rd), area) is the free Tortkara algebra. A linear basis is given by theunion of the letters and (←−−−area(wij))(w,i,j), where w runs over all words ind letters and (i, j) over all letters such that i < j.

This was proven for d = 2 in Dzhumadil’daev-Ismailov-Mashurov 2018. Itremains an open problem for d ≥ 3.


Areas & Tortkara

Weird...Theorem (Diehl-Lyons-P.-Reizenstein 2019)We have

←−−−area(vx) = ←−−−area(v) •←−−−area(x)for any T≥1(Rd) and any Lie polynomial x without a first order term.


Areas & Tortkara

Let X in R be a piecewise linear curve through the points0, x1, . . . , xn ∈ R. Then: for every tree (bracketing) τ and every word w,

⟨areaτ (w) , S(X)0,n⟩ = discreteAreaτ (w, x)n ,

where

discreteArea(a, b)ℓ ∶= Corr1(a, b)ℓ − Corr1(b, a)ℓ

∶=ℓ−1∑i=0

ai+1bi −ℓ−1∑i=0

bi+1ai, ℓ = 0, dots, n,


Areas & Tortkara

Open Problems

• Find computationally useful examples of free shuffle generating sets interms of areas of areas,

• Understand Tortkara identity in geometric terms,• Show freeness of the area algebra as a Tortkara algebra in any

dimension.


Areas & Tortkara

Thank you.arXiv preprint: https://arxiv.org/abs/2002.02338My website: http://page.math.tu-berlin.de/~preiss/

Btw: I’m looking for a postdoc position starting latest in september 2020.


https://arxiv.org/abs/2002.02338

http://page.math.tu-berlin.de/~preiss/

Areas & Tortkara

Thank you.arXiv preprint: https://arxiv.org/abs/2002.02338My website: http://page.math.tu-berlin.de/~preiss/Btw: I’m looking for a postdoc position starting latest in september 2020.


https://arxiv.org/abs/2002.02338

http://page.math.tu-berlin.de/~preiss/

joint work with joscha diehl, terry lyons and jeremy reizenstein … of... · theorem...

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