Poisson Brackets for the Grassmann Pentagram Map

Poisson Brackets

for the

Grassmann Pentagram Map

Nick Ovenhouse

University of Minnesota

October, 2019

Notre Dame

Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 1 / 29

Outline

1 Background

2 Grassmann Version

3 Non-Commutative Integrability

4 Combinatorial Models

5 Recovering the Lax Invariants


The Basic Idea

Draw a polygon in the (projective) plane

Label the vertices 1, . . . , nDraw diagonals connecting i, i + 2

Label intersections of diagonals by 1′, . . . , n′

1

2

3

4

5

6

1′

2′

3′

4′

5′

6′

De�nition [Schwartz]1

The pentagram map sends the original polygon to the new polygon.

1Richard Schwartz. “The Pentagram Map”. In: Experimental Mathematics 1.1 (1992), pp. 71–81


The Basic Idea




1

2

3

4

5

6

1′

2′

3′

4′

5′

6′





The Basic Idea


Label the vertices 1, . . . , n

Draw diagonals connecting i, i + 2


1

2

3

4

5

6

1′

2′

3′

4′

5′

6′





The Basic Idea




1

2

3

4

5

6

1′

2′

3′

4′

5′

6′





The Basic Idea




1

2

3

4

5

6

1′

2′

3′

4′

5′

6′





The Basic Idea




1

2

3

4

5

6

1′

2′

3′

4′

5′

6′





The Basic Idea




1

2

3

4

5

6

1′

2′

3′

4′

5′

6′





Twisted Polygons

More generally, we consider twisted polygons.

De�nition

A twisted n-gon is a bi-in�nite sequence (pi)i∈Z of points in P2, such that pi+n = Mpi

for some M ∈ PGL3. The matrix M is called the monodromy of the polygon.

De�nition

PGL3 acts on the set of twisted polygons by A · (pi) = (Api). Two polygons in the

same orbit are called projectively equivalent. Denote by Pn the set of projective

equivalence classes of twisted n-gons.


Twisted Polygons


De�nition



De�nition





Twisted Polygons


De�nition



De�nition





Twisted Polygons


De�nition



De�nition





Integrability

De�nition

A Poisson mapping T : M → M on a manifold M is called completely integrable if

there are su�ciently many independent conserved quantities f1, . . . , fN so that all

{fi, fj} = 0.

Theorem [OST]1[GSTV]

2

The pentagram map on Pn is completely integrable.

1Valentin Ovsienko, Richard Schwartz, and Serge Tabachnikov. “The Pentagram Map: a Discrete

Integrable System”. In: Communications in Mathematical Physics 299.2 (2010), pp. 409–446

2Michael Gekhtman et al. “Integrable cluster dynamics of directed networks and pentagram maps”.

In: Advances in Mathematics 300 (2016), pp. 390–450


Integrability

De�nition



{fi, fj} = 0.


2







Integrability

De�nition



{fi, fj} = 0.


2







Outline

1 Background

2 Grassmann Version





The Basic Idea

“Draw a polygon” (choose n points P1, . . . , Pn ∈ GrN (3N ))

“Draw diagonals” (consider Li, the span of Pi, Pi+2)

“Intersect diagonals” (take intersection Qi = Li ∩ Li+1)

De�nition [Marí-Be�a, Felipe]1

The Grassmann pentagram map sends P1, . . . , Pn to Q1, . . . ,Qn.

Theorem [Marí-Be�a, Felipe]1

The Grassmann pentagram map has a Lax representation. That is, there is a matrix

whose spectral invariants are conserved quantities.

1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint

arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


The Basic Idea










arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


The Basic Idea










arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


The Basic Idea










arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


The Basic Idea










arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


The Basic Idea










arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


The Basic Idea










arXiv:1507.04765 (2015)


arXiv:1507.04765 (2015)


Twisted Grassmann Polygons

Again, we consider more generally twisted polygons:

De�nition

A twisted Grassmann n-gon is a bi-in�nite sequence (Pi)i∈Z of points in GrN (3N ),

together with some M ∈ PGL3N so that Pi+n = MPi for all i.

De�nition

Let GPn,N denote the moduli space of twisted n-gons in GrN (3N ), up to the action

of PGL3N .




De�nition



De�nition


of PGL3N .




De�nition



De�nition


of PGL3N .




De�nition



De�nition


of PGL3N .


Coordinates

If P1, . . . , Pn ∈ GrN (3N ) are vertices, choose a lift V1, . . . ,Vn ∈ Mat3N×N .

If the polygon is “generic”, then the combined columns of Vi,Vi+1,Vi+2 are a

basis of R3N.

Then there are Ai,Bi,Ci ∈ GLN so that Vi+3 = Vi+1Ai + ViBi + Vi+2Ci

Lemma

The lift can be chosen so that Ci = IdN for all i. So then for all i,

Vi+3 = Vi+1Xi + ViYi + Vi+2,

and there is a Z ∈ GLN so that Xi+n = ZXiZ−1and Yi+n = ZYiZ−1

.


Coordinates



basis of R3N.


Lemma




.


Coordinates



basis of R3N.


Lemma




.


Coordinates



basis of R3N.


Lemma




.


Coordinates



basis of R3N.


Lemma




.


Coordinates



basis of R3N.


Lemma




.


Expression for the Map

The Grassmann pentagram map transforms the matrices Xi, Yi by

Xi 7→ (Xi + Yi+1)−1 Xi (Xi+2 + Yi+3)

Yi 7→ (Xi + Yi+1)−1 Yi+1 (Xi+2 + Yi+3)

In what follows, we will consider this as a transformation on a set of formal

noncommutative variables.




Xi 7→ (Xi + Yi+1)−1 Xi (Xi+2 + Yi+3)

Yi 7→ (Xi + Yi+1)−1 Yi+1 (Xi+2 + Yi+3)






Xi 7→ (Xi + Yi+1)−1 Xi (Xi+2 + Yi+3)

Yi 7→ (Xi + Yi+1)−1 Yi+1 (Xi+2 + Yi+3)






Xi 7→ (Xi + Yi+1)−1 Xi (Xi+2 + Yi+3)

Yi 7→ (Xi + Yi+1)−1 Yi+1 (Xi+2 + Yi+3)




Outline

1 Background

2 Grassmann Version





Non-Commutative Poisson Structures

If A is any associative algebra, then the commutator bracket [a, b] = ab − ba is

always a Poisson bracket.

Theorem [Farkas, Letzter]1

Let A be an associative, but non-commutative, prime ring. Then the only Poisson

bracket on A (up to scalar multiple) is the commutator bracket [a, b].

In the commutative case, this reduces to the trivial Poisson bracket.

Question: What is the right notion of Poisson structure for a non-commutative

algebra?

1Daniel R Farkas and Gail Letzter. “Ring theory from symplectic geometry”. In: Journal of Pure and

Applied Algebra 125.1-3 (1998), pp. 155–190










algebra?












algebra?












algebra?












algebra?




H0-Poisson Structures

Let A be an associative algebra.

Denote A\ := A/[A,A], the cyclic space.Elements are cyclic words in A, since x1 · · · xn = xnx1 · · · xn−1 mod [A,A].

Each a\ ∈ A\ gives a GLn-invariant function tr(a) on Hom(A,Matn), and hence

a function on Repn(A) := Hom(A,Matn)/GLn.

De�nition [Crawley-Boevey]1

An H0-Poisson structure on A is a Lie bracket [−,−] on A\ such that each [a\,−] is

induced by a derivation of A.

Theorem [Crawley-Boevey]1

An H0-Poisson structure on A induces a Poisson bracket on Repn(A) given by

{tr(a), tr(b)} = tr[a\, b\]

1William Crawley-Boevey. “Poisson Structures on Moduli Spaces of Representations”. In: Journal of

Algebra 325.1 (2011), pp. 205–215. doi: 10.1016/j.jalgebra.2010.09.033


http://dx.doi.org/10.1016/j.jalgebra.2010.09.033











{tr(a), tr(b)} = tr[a\, b\]







Denote A\ := A/[A,A], the cyclic space.

Elements are cyclic words in A, since x1 · · · xn = xnx1 · · · xn−1 mod [A,A].








{tr(a), tr(b)} = tr[a\, b\]















{tr(a), tr(b)} = tr[a\, b\]















{tr(a), tr(b)} = tr[a\, b\]















{tr(a), tr(b)} = tr[a\, b\]















{tr(a), tr(b)} = tr[a\, b\]


Algebra 325.1 (2011), pp. 205–215. doi: 10.1016/j.jalgebra.2010.09.033Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29












{tr(a), tr(b)} = tr[a\, b\]


Algebra 325.1 (2011), pp. 205–215. doi: 10.1016/j.jalgebra.2010.09.033Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29


Non-Commutative Integrability

By a non-commutative integrable system in A, we mean a map T : A→ A such that

There is an in�nite family of invariants t\1, t\

2, . . . , t\i , . . . ∈ A\

(i.e. ti = T(ti) mod [A,A])

There is an H0-Poisson structure so that [t\i , t\j ] = 0 for all i, j.

Consider the expression for the Grassmann pentagram map in the X , Y matrices

formally as a map on the free skew �eld in the indeterminates Xi, Yi.

Theorem [Ovenhouse]1

The Grassmann pentagram map is a non-commutative integrable system in the free

skew �eld.

1Nicholas Ovenhouse. “Non-Commutative Integrability of the Grassmann Pentagram Map”. In:

arXiv:1810.11742 (2018)





2, . . . , t\i , . . . ∈ A\







skew �eld.


arXiv:1810.11742 (2018)





2, . . . , t\i , . . . ∈ A\







skew �eld.


arXiv:1810.11742 (2018)





2, . . . , t\i , . . . ∈ A\







skew �eld.


arXiv:1810.11742 (2018)





2, . . . , t\i , . . . ∈ A\







skew �eld.


arXiv:1810.11742 (2018)





2, . . . , t\i , . . . ∈ A\







skew �eld.


arXiv:1810.11742 (2018)





2, . . . , t\i , . . . ∈ A\







skew �eld.


arXiv:1810.11742 (2018)


Outline

1 Background

2 Grassmann Version





A Combinatorial Model for GPn,N

Conisder the following graph embedded on a torus:

3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z

Recall the relations Vi+3 = Vi+1Ai + ViBi + Vi+2Ci.

If we change the lift Vi 7→ ViG...

Do this repeatedly to cancel the C’s...




3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z







3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z







3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z







3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z







3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z







3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z







3

2

1

1

3

2

V1

V3

V5

V4

V2

A3

B5

A5

B3

B1

A2 C3

C2

C1

C5

C4

B2

A1

B4 A4

V2G

B4G

C4G

A4G

G−1A1

G−1B2G−1C5

X4

X5

X1

X2

X3

Y3

Y4

Y5

Y1

Y2

Z

Z

Z





Van den Bergh’s Double Brackets

De�nition [Van den Bergh]1

Let A be an associative algebra. A double bracket on A is a bilinear operation

{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)

{{a, bc}} = {{a, b}} (1⊗ c) + (b ⊗ 1) {{a, c}}{{ab, c}} = {{a, c}} (b ⊗ 1) + (1⊗ a) {{b, c}}

1Michel Van Den Bergh. “Double Poisson Algebras”. In: Transactions of the American Mathematical

Society 360.11 (2008), pp. 5711–5769





{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)



Society 360.11 (2008), pp. 5711–5769





{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)



Society 360.11 (2008), pp. 5711–5769





{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)

{{a, bc}} = {{a, b}} (1⊗ c) + (b ⊗ 1) {{a, c}}

{{ab, c}} = {{a, c}} (b ⊗ 1) + (1⊗ a) {{b, c}}


Society 360.11 (2008), pp. 5711–5769





{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)



Society 360.11 (2008), pp. 5711–5769





{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)



Society 360.11 (2008), pp. 5711–5769





{{−,−}} : A⊗ A→ A⊗ A

such that:

{{b, a}} = −{{a, b}}τ ((x ⊗ y)τ = y ⊗ x)



Society 360.11 (2008), pp. 5711–5769


Double Brackets from Networks

Let Q be a network drawn on a cylinder, so that

Boundary vertices are univalent.

One boundary component has only sources, the other only sinks.

All internal vertices are trivalent, and neither sources nor sinks:

x

z

y ab

c

Let A be the algebra generated by the arrows.

De�ne a double bracket on A by {{y, z}} = y ⊗ z and {{b, c}} = c ⊗ b.

By composing with multiplication and the quotient map, we get an operation

on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\







x

z

y ab

c




on A\ = A/[A,A]:〈a, b〉 := µ({{a, b}})\


An H0 Poisson Bracket

Let L ⊂ A the subspace spanned by loops.

Choosing a vertex of the graph •, let L• be the space of loops based at •. It is a

subalgebra.

If f , g ∈ L intersect at a point p, then fpgp represents the loop which follows f ,

then g, based at p.

Theorem

The induced bracket 〈−,−〉 on L \• is a Lie bracket, and it is given by:

〈f , g〉 =∑

p∈f∩g

εp(f , g) fpgp

The coe�cients εp(f , g) are given by:

· · ·

εp(f , g) = 1

· · ·

εp(f , g) = 0





subalgebra.


then g, based at p.

Theorem


〈f , g〉 =∑

p∈f∩g

εp(f , g) fpgp


· · ·

εp(f , g) = 1

· · ·

εp(f , g) = 0





subalgebra.


then g, based at p.

Theorem


〈f , g〉 =∑

p∈f∩g

εp(f , g) fpgp


· · ·

εp(f , g) = 1

· · ·

εp(f , g) = 0





subalgebra.


then g, based at p.

Theorem


〈f , g〉 =∑

p∈f∩g

εp(f , g) fpgp


· · ·

εp(f , g) = 1

· · ·

εp(f , g) = 0





subalgebra.


then g, based at p.

Theorem


〈f , g〉 =∑

p∈f∩g

εp(f , g) fpgp


· · ·

εp(f , g) = 1

· · ·

εp(f , g) = 0





subalgebra.


then g, based at p.

Theorem


〈f , g〉 =∑

p∈f∩g

εp(f , g) fpgp


· · ·

εp(f , g) = 1

· · ·

εp(f , g) = 0


The X , Y Coordinates

3

2

1

1

3

2

Xi

Yi

The Xi and Yi variables represent the pictured cycles.

Their brackets are given by

〈Xi+1,Xi〉 = Xi+1Xi 〈Yi+k, Yi〉 = Yi+kYi for k = 1, 2

〈Yi+k,Xi〉 = Yi+kXi for k = 0, 1 〈Xi+k, Yi〉 = Xi+kYi for k = 1, 2

〈X3,X2〉 = X3Z−1X2Z 〈Y3, Y2〉 = Y3Z−1Y2Z



3

2

1

1

3

2

Xi

Yi





〈X3,X2〉 = X3Z−1X2Z 〈Y3, Y2〉 = Y3Z−1Y2Z



3

2

1

1

3

2

Xi

Yi





〈X3,X2〉 = X3Z−1X2Z 〈Y3, Y2〉 = Y3Z−1Y2Z



3

2

1

1

3

2

Xi

Yi





〈X3,X2〉 = X3Z−1X2Z 〈Y3, Y2〉 = Y3Z−1Y2Z


Boundary Measurements

Given a path p in the network,

Let wt(p) be the product of the edge weights (in order).

Let WT(p) := wt(p)λdp , where dp is the “winding number” of the path.

The boundary measurement matrix B(λ) = (bij(λ)) is given by

bij(λ) =∑

p : i→j

WT(p)







bij(λ) =∑

p : i→j

WT(p)







bij(λ) =∑

p : i→j

WT(p)







bij(λ) =∑

p : i→j

WT(p)







bij(λ) =∑

p : i→j

WT(p)







bij(λ) =∑

p : i→j

WT(p)


The Pentagram Map via Networks

We perform the following sequence of local transformations (“Postnikov moves”) of

the network, considered on a torus:

“Square Move”

“White-swap”

“Black-swap”

a

b

c

d

dc∆−1

∆

∆−1ad

dc∆−1bc−1

(∆ := b + adc)

b

a

c

b

b−1a

bc

b

a

c

b

ab

cb−1





“Square Move”

“White-swap”

“Black-swap”

a

b

c

d

dc∆−1

∆

∆−1ad

dc∆−1bc−1

(∆ := b + adc)

b

a

c

b

b−1a

bc

b

a

c

b

ab

cb−1





“Square Move”

“White-swap”

“Black-swap”

a

b

c

d

dc∆−1

∆

∆−1ad

dc∆−1bc−1

(∆ := b + adc)

b

a

c

b

b−1a

bc

b

a

c

b

ab

cb−1





“Square Move”

“White-swap”

“Black-swap”

a

b

c

d

dc∆−1

∆

∆−1ad

dc∆−1bc−1

(∆ := b + adc)

b

a

c

b

b−1a

bc

b

a

c

b

ab

cb−1





“Square Move”

“White-swap”

“Black-swap”

a

b

c

d

dc∆−1

∆

∆−1ad

dc∆−1bc−1

(∆ := b + adc)

b

a

c

b

b−1a

bc

b

a

c

b

ab

cb−1



After this sequence, we end up with the same network, but with weights

Xi = (Xi + Yi)−1Xi(Xi+2 + Yi+2)

Yi = (Xi+1 + Yi+1)−1Yi+1(Xi+3 + Yi+3)

These are the expressions for the pentagram map (up to a shift of indices Yi 7→ Yi+1).




Xi = (Xi + Yi)−1Xi(Xi+2 + Yi+2)

Yi = (Xi+1 + Yi+1)−1Yi+1(Xi+3 + Yi+3)





Xi = (Xi + Yi)−1Xi(Xi+2 + Yi+2)

Yi = (Xi+1 + Yi+1)−1Yi+1(Xi+3 + Yi+3)



Integrability

The sequence of moves we performed do not change the boundary measurements.

But moving the edges/vertices around on the torus can change the matrix up to

conjugation.

So although the entries of B(λ) are not invariants, the spectral invariants are.

Let tr(B(λ)i) =∑

tijλj .

Theorem [Ovenhouse]⟨t\ik, t

\j`

⟩= 0 for all i, j, k, `.

The proof is combinatorial and topological in nature. It is done by enumerating the

paths with given homology classes (on the torus), and examining their intersections.


Integrability



conjugation.


Let tr(B(λ)i) =∑

tijλj .


\j`

⟩= 0 for all i, j, k, `.




Integrability



conjugation.


Let tr(B(λ)i) =∑

tijλj .


\j`

⟩= 0 for all i, j, k, `.




Integrability



conjugation.


Let tr(B(λ)i) =∑

tijλj .


\j`

⟩= 0 for all i, j, k, `.




Integrability



conjugation.


Let tr(B(λ)i) =∑

tijλj .


\j`

⟩= 0 for all i, j, k, `.




Integrability



conjugation.


Let tr(B(λ)i) =∑

tijλj .


\j`

⟩= 0 for all i, j, k, `.




Outline

1 Background

2 Grassmann Version





Another View of the Moduli Space

A polygon, and a choice of lift Vi, determines the 2n + 1 matrices Xi, Yi,Z .

However, simultaneously changing all Vi by Vi 7→ ViG for some �xed G ∈ GLN ,

induces Xi 7→ G−1XiG, Yi 7→ G−1YiG, and Z 7→ G−1ZG. So the matrices Xi, Yi,Z are

only well-de�ned up to simultaneous conjugation.

The moduli space is identi�ed with GPn,N ∼= GL2n+1

N /GLN .

If A is the group algebra of the free group on 2n + 1 generators, then this is

RepN (A). So by Crawley-Boevey’s theorem, the H0-Poisson bracket induces a

Poisson bracket on GPn,N so that

{tr(a), tr(b)} = tr(〈a, b〉)








N /GLN .




{tr(a), tr(b)} = tr(〈a, b〉)








N /GLN .




{tr(a), tr(b)} = tr(〈a, b〉)








N /GLN .




{tr(a), tr(b)} = tr(〈a, b〉)








N /GLN .




{tr(a), tr(b)} = tr(〈a, b〉)


Interpreting the Invariants

Since the invariants tij are (noncommutative) polynomials in the Xi’s and Yi’s, we

can interpret the traces tr(tij) as functions on GPn,N , and by Crawley-Boevey’s

theorem, {tr(tij), tr(tk`)} = 0.

Marí-Be�a and Felipe’s Lax matrix was constructred as follows. Form the matrix Viwhose columns are Vi, Vi+1, and Vi+2.

Consider the “shift” matrix

Li =

0 0 YiIdN 0 Xi0 IdN IdN

Then Vi+1 = ViLi, and the action of the monodromy matrix is given by

MVi = ViL1L2 · · · Ln








Li =











Li =











Li =











Li =





Scaling Parameter

Consider the modi�ed matrices, with a scaling parameter:

Li(λ) =

0 0 λYiIdN 0 λXi0 IdN IdN

The product L(λ) = L1(λ) · · · Ln(λ) is the Lax matrix of Marí-Be�a and Felipe.

Proposition

The product L(λ) is conjugate to the boundary measurement matrix B(λ).


Scaling Parameter


Li(λ) =



Proposition



Scaling Parameter


Li(λ) =



Proposition



Thank You!


Documents

Poisson Brackets for the Grassmann Pentagram Map