Post on 14-Jul-2022
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 2 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
The Basic Idea
Draw a polygon in the (projective) plane
Label the vertices 1, . . . , n
Draw 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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 3 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 4 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 4 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 4 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 4 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 5 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 5 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 5 / 29
Outline
1 Background
2 Grassmann Version
3 Non-Commutative Integrability
4 Combinatorial Models
5 Recovering the Lax Invariants
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 6 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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)
1Raúl Felipe and Gloria Marí Be�a. “The Pentagram Map on Grassmannians”. In: arXiv preprint
arXiv:1507.04765 (2015)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 7 / 29
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 .
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 8 / 29
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 .
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 8 / 29
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 .
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 8 / 29
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 .
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 8 / 29
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
.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 9 / 29
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
.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 9 / 29
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
.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 9 / 29
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
.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 9 / 29
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
.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 9 / 29
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
.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 9 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 10 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 10 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 10 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 10 / 29
Outline
1 Background
2 Grassmann Version
3 Non-Commutative Integrability
4 Combinatorial Models
5 Recovering the Lax Invariants
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 11 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 12 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 12 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 12 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 12 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 12 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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.2010.09.033Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 13 / 29
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.2010.09.033Nick 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 14 / 29
Outline
1 Background
2 Grassmann Version
3 Non-Commutative Integrability
4 Combinatorial Models
5 Recovering the Lax Invariants
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 15 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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...
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 16 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 17 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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}})\
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 18 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 19 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 19 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 19 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 19 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 19 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 19 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 20 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 20 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 20 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 20 / 29
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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 21 / 29
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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 21 / 29
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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 21 / 29
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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 21 / 29
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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 21 / 29
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)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 21 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 22 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 22 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 22 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 22 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 22 / 29
The Pentagram Map via Networks
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).
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 23 / 29
The Pentagram Map via Networks
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).
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 23 / 29
The Pentagram Map via Networks
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).
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 23 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 24 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 24 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 24 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 24 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 24 / 29
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.
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 24 / 29
Outline
1 Background
2 Grassmann Version
3 Non-Commutative Integrability
4 Combinatorial Models
5 Recovering the Lax Invariants
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 25 / 29
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〉)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 26 / 29
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〉)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 26 / 29
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〉)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 26 / 29
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〉)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 26 / 29
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〉)
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 26 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 27 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 27 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 27 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 27 / 29
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
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 27 / 29
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).
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 28 / 29
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).
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 28 / 29
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).
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 28 / 29
Thank You!
Nick Ovenhouse (UMN) Pentagram Map October, 2019 Notre Dame 29 / 29