Geometry of Random Polygons, Knots, and Biopolymersclayton/research/talks/... · 2015. 2. 9. · CCM and Discrete Seminar February 2, 2015. A pillow problem 1 Introduction ... and

Geometry of Random Polygons, Knots, andBiopolymers

Clayton Shonkwiler

Colorado State University

University of Colorado DenverCCM and Discrete Seminar

February 2, 2015

A pillow problem

1 Introduction

Triangles live on a hemisphere and are linked to 2 by 2 matrices. The familiar triangle is seen in a di↵erent light.New understanding and new applications come from its connections to the modern developments of random matrixtheory. You may never look at a triangle the same way again.

We began with an idle question: Are most random triangles acute or obtuse ? While looking for an answer, anote was passed in lecture. (We do not condone our behavior !) The note contained an integral over a region in R6.The evaluation of that integral gave us a number – the fraction of obtuse triangles. This paper will present severalother ways to reach that number, but our real purpose is to provide a more complete picture of “triangle space.”

Later we learned that in 1884 Lewis Carroll (as Charles Dodgson) asked the same question. His answer for theprobability of an obtuse triangle (by his rules) was

3

8 � 6

⇡

p3⇡ 0.64.

Variations of interpretation lead to multiple answers (see [11, 33] and their references). Portnoy reports that in thefirst issue of The Educational Times (1886), Woolhouse reached 9/8�4/⇡2 ⇡ 0.72. In every case obtuse triangles arethe winners – if our mental image of a typical triangle is acute, we are wrong. Probably a triangle taken randomlyfrom a high school geometry book would indeed be acute. Humans generally think of acute triangles, indeed nearlyequilateral triangles or right triangles, in our mental representations of a generic triangle. Carroll’s answer is shortof our favorite answer 3/4, which is more mysterious than it seems. There is no paradox, just di↵erent choices ofprobability measure.

The most developed piece of the subject is humbly known as “Shape Theory.” It was the last interest of the firstprofessor of mathematical statistics at Cambridge University, David Kendall [21, 26]. We rediscovered on our ownwhat the shape theorists knew, that triangles are naturally mapped onto points of the hemisphere. It was a thrill todiscover both the result and the history of shape space.

We will add a purely geometrical derivation of the picture of triangle space, delve into the linear algebra point ofview, and connect triangles to random matrix theory.

We hope to rejuvenate the study of shape theory !

Figure 1: Lewis Carroll’s Pillow Problem 58 (January 20, 1884). 25 and 83 are page numbers for his answer and hismethod of solution. He specifies the longest side AB and assumes that C falls uniformly in the region where ACand BC are not longer than AB.

3

What does it mean to take a random triangle?

QuestionWhat does it mean to choose a random triangle?

Statistician’s Answer

The issue of choosing a “random triangle” isindeed problematic. I believe the difficulty is explainedin large measure by the fact that there seems to be nonatural group of transitive transformations acting onthe set of triangles.

–Stephen Portnoy, 1994(Editor, J. American Statistical Association)



Statistician’s Answer

The issue of choosing a “random triangle” isindeed problematic. I believe the difficulty is explainedin large measure by the fact that there seems to be nonatural group of transitive transformations acting onthe set of triangles.

–Stephen Portnoy, 1994(Editor, J. American Statistical Association)



Applied Mathematician’s Answer

We will add a purely geometrical derivation of thepicture of triangle space, delve into the linear algebrapoint of view, and connect triangles to random matrixtheory.

–Alan Edelman and Gil Strang, 2012(and you all know who Strang is, at least!)



Applied Mathematician’s Answer

We will add a purely geometrical derivation of thepicture of triangle space, delve into the linear algebrapoint of view, and connect triangles to random matrixtheory.

–Alan Edelman and Gil Strang, 2012(and you all know who Strang is, at least!)



Differential Geometer’s Answer

Pick by the measure defined by the volume form ofthe natural Riemannian metric on the manifold of3-gons, of course. That ought to be a special case ofthe manifold of n-gons.

But what’s the manifold of n-gons?And what makes one metric natural?Don’t algebraic geometers understand this?

–Jason Cantarella to me, a few years ago



Differential Geometer’s Answer

Pick by the measure defined by the volume form ofthe natural Riemannian metric on the manifold of3-gons, of course. That ought to be a special case ofthe manifold of n-gons.

But what’s the manifold of n-gons?And what makes one metric natural?Don’t algebraic geometers understand this?

–Jason Cantarella to me, a few years ago

The Algebraic Geometer’s Answer

(Seemingly) Trivial Observation 1Given z1 = a1 + ib1, . . . , zn = an + ibn

∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2



∑z2

i = (a21 − b2

1) + i(2a1b1) + · · ·+ (a2n − b2

n) + i(2anbn)

= (a21 + · · ·+ a2

n)− (b21 + · · ·+ b2

n) + i 2(a1b1 + · · ·+ anbn)

= (|a|2 − |b|2) + i 2〈a,b〉


|z21 |+ · · ·+ |z2

n | = |z1|2 + · · ·+ |zn|2

= (a21 + b2

1) + · · ·+ (a2n + b2

n)

= (a21 + · · ·+ a2

n) + (b21 + · · ·+ b2

n)

= |a|2 + |b|2

The algebraic geometer’s answer, continued

Theorem (Hausmann and Knutson, 1997)The space of closed planar n-gons with length 2, up totranslation and rotation, is identified with the Grassmannmanifold G2(Rn) of 2-planes in Rn.

Proof.Take an orthonormal frame ~a, ~b for the plane, let

~z = (z1 = a1 + ib1, . . . , zn = an + ibn), vi =i∑

j=1

z2i

By the observations, v0 = vn = 0,∑ |vi+1 − vi | = 2. Rotating

the frame a,b in their plane rotates the polygon in the complexplane.

The algebraic geometer’s answer, continued

Theorem (Hausmann and Knutson, 1997)The space of closed planar n-gons with length 2, up totranslation and rotation, is identified with the Grassmannmanifold G2(Rn) of 2-planes in Rn.

Proof.Take an orthonormal frame ~a, ~b for the plane, let

~z = (z1 = a1 + ib1, . . . , zn = an + ibn), vi =i∑

j=1

z2i

By the observations, v0 = vn = 0,∑ |vi+1 − vi | = 2. Rotating

the frame a,b in their plane rotates the polygon in the complexplane.

Our answer

Theorem (with Cantarella and Deguchi)The volume form of the standard Riemannian metric on G2(Rn)defines the natural probability measure on closed, planarn-gons of length 2 up to translation and rotation. It has a(transitive) action by isometries given by the action of SO(n) onG2(Rn).

So random triangles are points selected uniformly on S2 since

random triangle→ random point in G2(R3)

→ random point in G1(R3)

→ random point in RP2

→ random point in S2

Putting the pillow problem to bed

Acute triangles (gold) turn out tobe defined by natural algebraicconditions on the sphere.

Proposition (with Cantarella, Chapman, and Needham,2015)The fraction of obtuse triangles is

32− log 8

π∼ 83.8%

Transitioning to topology

Theorem (with Cantarella and Deguchi)The volume form of the standard Riemannian metric on G2(Cn)defines the natural probability measure on closed space n-gonsof length 2 up to translation and rotation. It has a (transitive)action by isometries given by the action of U(n) on G2(Cn).

Proof.Again, we use an identification due to Hausmann and Knutsonwhere• instead of combining two real vectors to make one complex

vector, we combine two complex vectors to get onequaternionic vector

• instead of squaring complex numbers, we apply the Hopfmap to quaternions

Random Polygons (and Polymer Physics)

Statistical Physics Point of ViewA polymer in solution takes on an ensemble of random shapes,with topology (knot type!) as the unique conserved quantity.

Protonated P2VPRoiter/MinkoClarkson University

Plasmid DNAAlonso-Sarduy, Dietler LabEPF Lausanne

Random Polygons (and Polymer Physics)

Statistical Physics Point of ViewA polymer in solution takes on an ensemble of random shapes,with topology (knot type!) as the unique conserved quantity.

Physics SetupModern polymer physics is based on the analogy

between a polymer chain and a random walk.—Alexander Grosberg, NYU.

Our (geometers and topologists) jobUnderstand the probability that a closed random walk is knottedand the distribution of knot types that result from differentconditions on the walk (walk model, number of segments,confinement, self-avoidance, fixed bond angles, etc).

A theorem about random knots

DefinitionThe total curvature of a space polygon is the sum of its turningangles.

Theorem (with Cantarella, Grosberg, and Kusner)The expected total curvature of a random n-gon of length 2sampled according to the measure on G2(Cn) is

π

2n +

π

42n

2n − 3.

Corollary (with Cantarella, Grosberg, and Kusner)At least 1/3 of hexagons and 1/11 of heptagons are unknots.




π

2n +

π

42n

2n − 3.





π

2n +

π

42n

2n − 3.


Responsible sampling algorithms

We’ll need to sample polygons and check knot types: how?

Proposition (classical?)The natural measure on G2(Cn) is obtained by generatingrandom complex n-vectors with independent Gaussiancoordinates and applying (complex) Gram-Schmidt.

In[9]:= RandomComplexVector@n_D := Apply@Complex,Partition@ð, 2D & �� RandomVariate@NormalDistribution@D, 81, 2 n<D, 82<D@@1DD;

ComplexDot@A_, B_D := Dot@A, Conjugate@BDD;ComplexNormalize@A_D := H1 � Sqrt@Re@ComplexDot@A, ADDDL A;

RandomComplexFrame@n_D := Module@8a, b, A, B<,8a, b< = 8RandomComplexVector@nD, RandomComplexVector@nD<;A = ComplexNormalize@aD;B = ComplexNormalize@b - Conjugate@ComplexDot@A, bDD AD;

8A, B<D;

Using this, we can generate ensembles of random polygonsand distributions of knot types . . .

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Random 2,000-gons

Equilateral Random Walks

For 40 years, physicists have been modeling polymers withequilateral random walks in R3; i.e., walks consisting of nunit-length steps up to translation. The moduli space of suchwalks up to translation is just S2(1)× . . .× S2(1)︸︷︷︸

n

.

Let ePol(n) be the submanifold of closed equilateral randomwalks (or random equilateral polygons); i.e., those walks whichsatisfy both ‖~ei‖ = 1 for all i and

n∑

i=1

~ei = ~0.

Equilateral Random Walks

For 40 years, physicists have been modeling polymers withequilateral random walks in R3; i.e., walks consisting of nunit-length steps up to translation. The moduli space of suchwalks up to translation is just S2(1)× . . .× S2(1)︸︷︷︸

n

.

Let ePol(n) be the submanifold of closed equilateral randomwalks (or random equilateral polygons); i.e., those walks whichsatisfy both ‖~ei‖ = 1 for all i and

n∑

i=1

~ei = ~0.

The Triangulation Polytope

DefinitionA abstract triangulation T of the n-gon picks out n − 3nonintersecting chords. The lengths of these chords obeytriangle inequalities, so they lie in a convex polytope in Rn−3

called the triangulation polytope P.

d1

d2

00

1

2

1 2




d1

d2 d1 + d2 ≥ 1

d1 ≤ 2

d2 ≤ 2

d1 ≤ d2 + 1

d2 ≤ d1 + 1

00

1

2

1 2




d1

d2

d3

(2,3,2)

(0,0,0)(2,1,0)

Action-Angle Coordinates

DefinitionIf P is the triangulation polytope and T n−3 = (S1)n−3 is thetorus of n − 3 dihedral angles, then there are action-anglecoordinates:

α : P × T n−3 → Pol(n)/SO(3)

14

d1d2

✓1

✓2

FIG. 2: This figure shows how to construct an equilateral pentagon in cPol(5;~1) using the action-angle map.First, we pick a point in the moment polytope shown in Figure 3 at center. We have now specified diagonalsd1 and d2 of the pentagon, so we may build the three triangles in the triangulation from their side lengths,as in the picture at left. We then choose dihedral angles ✓1 and ✓2 independently and uniformly, and jointhe triangles along the diagonals d1 and d2, as in the middle picture. The right hand picture shows the finalspace polygon, which is the boundary of this triangulated surface.

Arm3(n;~r) admits a Hamiltonian action by the Lie group SO(3) given by rotating the polygonalarm in space (this is the diagonal SO(3) action on the product of spheres) whose moment mapµ gives the vector joining the ends of the polygon. The closed polygons Pol3(n;~r) are the fiberµ�1(~0) of this map. While the group action does not generally preserve fibers of this moment map,it does preserve µ�1(~0) = Pol3(n;~r) and in this situation, we can perform what is known as asymplectic reduction (or Marsden–Weinstein–Meyer reduction [49, 50]) to produce a symplecticstructure on the quotient of the fiber µ�1(~0) by the group action. This yields a symplectic structureon the (2n � 6)-dimensional moduli space cPol3(n;~r). The symplectic measure induced by thissymplectic structure is equal to the standard measure given by pushing forward the subspace mea-sure on Pol3(n;~r) to cPol3(n;~r) because the “parent” symplectic manifold Arm3(n;~r) is a Kahlermanifold [33].

The polygon space cPol3(n;~r) is singular if

"I(~r) :=X

i2I

ri �X

j /2I

rj

is zero for some I ⇢ {1, . . . , n}. Geometrically, this means it is possible to construct a linearpolygon with edgelengths given by ~r. Since linear polygons are fixed by rotations around theaxis on which they lie, the action of SO(3) is not free in this case and the symplectic reductiondevelops singularities. Nonetheless, the reduction cPol3(n;~r) is a complex analytic space withisolated singularities; in particular, the complement of the singularities is a symplectic (in factKahler) manifold to which Theorem 13 applies.

Both the volume and the cohomology ring of cPol3(n;~r) are well-understood from this sym-plectic perspective [11, 32, 36, 38, 39, 46, 66]. For example:

Polygons and Polytopes, Together

Theorem (with Cantarella)α pushes the standard probability measure on P × T n−3

forward to the correct probability measure on ePol(n)/SO(3).

Ingredients of the Proof.Kapovich–Millson toric symplectic structure on polygon space +Duistermaat–Heckman theorem + Hitchin’s theorem oncompatibility of Riemannian and symplectic volume onsymplectic reductions of Kähler manifolds +Howard–Manon–Millson analysis of polygon space.

Polygons and Polytopes, Together

Theorem (with Cantarella)α pushes the standard probability measure on P × T n−3

forward to the correct probability measure on ePol(n)/SO(3).

Ingredients of the Proof.Kapovich–Millson toric symplectic structure on polygon space +Duistermaat–Heckman theorem + Hitchin’s theorem oncompatibility of Riemannian and symplectic volume onsymplectic reductions of Kähler manifolds +Howard–Manon–Millson analysis of polygon space.

A change of coordinates

If we introduce a fake chordlength d0 = 1, and make the lineartransformation

si = di − di+1, for 0 ≤ i ≤ n − 4, sn−3 = dn−3 − d0

then our inequalities

0 ≤ d1 ≤ 21 ≤ di + di+1|di − di+1| ≤ 1

0 ≤ dn−3 ≤ 2

become

−1 ≤ si ≤ 1,∑

si = 0︸︷︷︸

|di−di+1|≤1

, 2i−1∑

j=0

sj + si ≤ 1

︸︷︷︸di+di+1≥1

A change of coordinates

If we introduce a fake chordlength d0 = 1, and make the lineartransformation

si = di − di+1, for 0 ≤ i ≤ n − 4, sn−3 = dn−3 − d0

then our inequalities

0 ≤ d1 ≤ 21 ≤ di + di+1|di − di+1| ≤ 1

0 ≤ dn−3 ≤ 2

become

−1 ≤ si ≤ 1,∑

si = 0︸︷︷︸

easy conditions

, 2i−1∑

j=0

sj + si ≤ 1

︸︷︷︸hard conditions

Basic Idea

DefinitionThe n-dimensional cross-polytope C is the slice of thehypercube [−1,1]n+1 by the plane x1 + · · ·+ xn+1 = 0.

→

0 1 2

0

1

2

0 1 2

0

1

2

IdeaSample points in the cross polytope, which all obey the “easyconditions”, and reject any samples which fail to obey the “hardconditions”.

Sampling the Cross Polytope

DefinitionThe hypersimplex ∆k ,n is the slab of the cube [0,1]n−1 withk − 1 ≤∑n−1

i=1 xi ≤ k .

Theorem (Stanley)There is a unimodular triangulation1 of ∆k ,n in [0,1]n−1 indexedby permutations of (1, . . . ,n − 1) with k − 1 descents.

Standard triangulation

ψ−1

−→

Stanley triangulation1a decomposition into disjoint simplices of equal volume



i=1 xi ≤ k .



ψ−1

−→




i=1 xi ≤ k .



ψ−1

−→




i=1 xi ≤ k .



ψ−1

−→




i=1 xi ≤ k .



ψ−1

−→




i=1 xi ≤ k .



ψ−1

−→


Parity Issues

Recall that we’re interested in the cross polytope determined by

s0 + s1 + . . .+ sn−3 = 0

in the cube [−1,1]n−2. After translation, scaling, and droppingthe last coordinate, this corresponds to the slab

n − 22− 1 ≤ x0 + x1 + . . .+ xn−3 ≤

n − 22

in the cube [0,1]n−3.

For even n this is just the hypersimplex ∆(n−2)/2,n−2 andStanley’s triangulation applies.

For odd n this is covered by the union∆ n−2

2 −12 ,n−2 ∪∆ n−2

2 + 12 ,n−2, which we can rejection sample.

The Algorithm (for n even)

Moment Polytope Sampling Algorithm (with Cantarellaand Uehara, 2015)

1 Generate a permutation of n − 3 numbers with n/2− 2descents. O(n2) time .

2 Generate a point in the corresponding simplex of theStanley triangulation for ∆(n−2)/2,n−2.

3 Project up to the cross polytope in [0,1]n−2, over to thecross polytope in [−1,1]n−2 and down to a set of diagonals.

4 Test the proposed set of diagonals against the “hard”conditions. acceptance ratio > 1/n

5 Generate dihedral angles from T n−3.6 Build sample polygon in action-angle coordinates.

An Equilateral Polygon with 3,500 Steps

Looking Forward

• Computations! Diameter, diffusion constant, cyclizationenergies, knot probabilities, etc.

• Compare distributions of knot types obtained by differentmodels for random polygons. Is there evidence foruniversal properties of random knot distributions?

• Stronger theoretical results about the topology of randompolygons.

• Extend this approach to directly sample polygons withmore restricted geometry: confinement, excluded volume,fixed turning angle, etc.

• The spaces Pol(n) correspond to the Finite Unit NormTight Frames (FUNTFs) of C2 which arise in signalprocessing . . . do these techniques generalize to otherFUNTF spaces?

Thank you!

Thank you for listening!

References

• Probability Theory of Random Polygons from theQuaternionic ViewpointJason Cantarella, Tetsuo Deguchi, and Clayton ShonkwilerCommunications on Pure and Applied Mathematics 67(2014), no. 10, 658–1699.

• The Expected Total Curvature of Random PolygonsJason Cantarella, Alexander Y Grosberg, Robert Kusner,and Clayton ShonkwilerAmerican Journal of Mathematics, to appear.

• The Symplectic Geometry of Closed Equilateral RandomWalks in 3-SpaceJason Cantarella and Clayton ShonkwilerAnnals of Applied Probability, to appear.

http://arxiv.org/a/shonkwiler_c_1

http://arxiv.org/a/shonkwiler_c_1

Documents

Geometry of Random Polygons, Knots, and Biopolymersclayton/research/talks/... · 2015. 2. 9. · CCM and Discrete Seminar February 2, 2015. A pillow problem 1 Introduction ... and