The Geometry of Polygon Space

Acute Triangles, Convex Quadrilaterals, Flag Means, and More

Clayton Shonkwiler

Colorado State University

http://shonkwiler.org

08.08.17

/tokyo17

this talk!

Collaborators

Jason Cantarella

U. of Georgia

Tom Needham

Ohio State

Gavin Stewart

NYU

Laney Bowden

CSU

Andrea Haynes

CSU

Aaron Shukert

CSU

Lewis Carroll’s Pillow Problem #58

Earlier versions

W. S. B. Woolhouse, Educational Times 18 (1865), p. 189

J. J. Sylvester, Educational Times 18 (1865), p. 68

W. S. B. Woolhouse, The Lady's and Gentleman's Diary 158 (1861), p. 76

Carroll’s answer

Suppose \(AB\) is the longest side. Then

\(\mathbb{P}(\text{obtuse})=\frac{\pi/8}{\pi/3-\sqrt{3}/4} \approx 0.64\)

But if \(AB\) is the second longest side,

\(\mathbb{P}(\text{obtuse}) = \frac{\pi/2}{\pi/3+\sqrt{3}/2} \approx 0.82\)

A probabilist’s answer

Proposition [Portnoy]: If the distribution of \((x_1,y_1,x_2,y_2,x_3,y_3)\in\mathbb{R}^6\) is spherically symmetric (for example, a standard Gaussian), then

\(\mathbb{P}(\text{obtuse}) = \frac{3}{4}\)

Consider the vertices \((x_1,y_1),(x_2,y_2),(x_3,y_3)\) as determining a single point in \(\mathbb{R}^6\).

For example, when the vertices of the triangle are chosen from independent, identically-distributed Gaussians on \(\mathbb{R}^2\).

Choose three vertices uniformly in the disk:

\(\mathbb{P}(\text{obtuse})=\frac{9}{8}-\frac{4}{\pi^2}\approx 0.7197\)

Restricted domain?

Choose three vertices uniformly in the square:

\(\mathbb{P}(\text{obtuse})=\frac{97}{150}-\frac{\pi}{40}\approx 0.7252\)

reentrant

J.J. Sylvester, Educational Times, April 1864

Sylvester’s four point problem

Sylvester and Cayley’s solution

J.J. Sylvester, Phil. Trans. R. Soc. London 154 (1864), p. 654, footnote 64(b)

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VII (1867), p. 81

A. De Morgan’s solution

A. De Morgan, Trans. Cambridge Phil. Soc. XI (1871), pp. 147–148

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VI (1866), p. 52

C.M. Ingleby, Mathematical Questions with Their Solutions V (1865), p. 82

C.M. Ingleby’s solution

G. C. De Morgan’s solution

G.C. De Morgan, Mathematical Questions with Their Solutions V (1865), p. 109

W.S.B. Woolhouse’s solution

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VI (1866), p. 52

W.S.B. Woolhouse’s solutions

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VIII (1868), p. 105

Quadrilaterals in convex bodies

\(\mathbb{P}(\text{reflex})=\frac{1}{3}\)

\(\mathbb{P}(\text{reflex})=\frac{35}{12\pi^2}\approx 0.296\)

Theorem [Blaschke, 1917]

\(\frac{35}{12\pi^2}\leq\mathbb{P}(\text{reflex})\leq\frac{1}{3}\)

J.M. Wilson’s solution

J.M. Wilson, Mathematical Questions with Their Solutions V (1866), p. 81

We need measure theory!

W.A. Whitworth, Mathematical Questions with Their Solutions VIII (1868), p. 36

Report on J.J. Sylvester’s presentation of his paper “On a Special Class of Questions on the Theory of Probabilities” to the British Association for the Advancement of Science, 1865

We need geometry!

Random polygons?

Are these questions really about choosing random points, or are they actually about choosing random polygons?

How would you choose a polygon “at random”?

— Stephen Portnoy, Statistical Science 9 (1994), 279–284

Geometric idea

The space of all \(n\)-gons should be a (preferably compact) manifold \(P_n\) with a transitive isometry group. We should use the left-invariant metric on \(P_n\), scaled so vol\((P_n)=1\). Then the Riemannian volume form induced by this metric is a natural probability measure on \(P_n\), and we should compute the volume of the subset of \(n\)-gons satisfying our favorite condition.

Spoiler: \(P_n \simeq G_2\mathbb{R}^n\)

Polygons and Stiefel manifolds

Let \(e_1, \ldots , e_n\) be the edges of a planar \(n\)-gon with total perimeter 2. Choose \(z_1, \ldots , z_n\) so that \(z_k^2 = e_k\). Let \(z_k = a_k + i b_k\).

The polygon is closed \(\Leftrightarrow e_1 + \ldots + e_n = 0\)

\(\sum e_k =\sum z_k^2 = \left(\sum a_k^2 - \sum b_k^2\right) + 2i \sum a_k b_k\)

The polygon is closed \(\Leftrightarrow \|a\|=\|b\|\) and \(a \bot b\)

Since \(\sum |e_k| = \sum a_k^2 + \sum b_k^2 = \|a\|^2 + \|b\|^2\), we see that \((a,b) \in V_2(\mathbb{R}^n)\), the Stiefel manifold of 2-frames in \(\mathbb{R}^n\).

Polygons and Grassmannians

Proposition: Rotating \((a,b)\) in the plane it spans rotates the corresponding \(n\)-gon twice as fast.

Corollary [Hausmann–Knutson]

The Grassmannian \(G_2(\mathbb{R}^n)\) is (almost) a \(2^n\)-fold covering of the space of planar \(n\)-gons of perimeter 2.

The symmetric measure

Definition [w/ Cantarella & Deguchi]

The symmetric measure on \(n\)-gons of perimeter 2 up to translation and rotation is the pushforward of Haar measure on \(G_2(\mathbb{R}^n)\).

Therefore, \(SO(n)\) acts transitively on \(n\)-gons and preserves the symmetric measure.

Triangles

A triangle corresponds to an element of \(G_2(\mathbb{R}^3)\), which is a plane in \(\mathbb{R}^3\) with ON basis \((a,b)\)...

or, equivalently, the line through

\(p=a\times b\)

\begin{bmatrix} | & | \\ a & b \\ | & | \end{bmatrix} = \begin{bmatrix} - \sqrt{e_1} - \\ - \sqrt{e_2} - \\ - \sqrt{e_3} - \end{bmatrix}

\begin{bmatrix} x \\ y \\ z \end{bmatrix} = p = a \times b = \begin{bmatrix} \sqrt{e_2} \times \sqrt{e_3} \\ \sqrt{e_3} \times \sqrt{e_1} \\ \sqrt{e_1} \times \sqrt{e_2} \end{bmatrix} = \begin{bmatrix} \sqrt{\|e_2\| \|e_3\|}\sin\frac{\theta_{23}}{2} \\ \sqrt{\|e_3\| \|e_1\|}\sin\frac{\theta_{31}}{2} \\ \sqrt{\|e_1\| \|e_2\|}\sin\frac{\theta_{12}}{2} \end{bmatrix}

Translating to side lengths

\begin{bmatrix} x \\ y \\ z \end{bmatrix}

z^2 = a b \cos^2\frac{\gamma}{2}= \frac{1}{2} a b (1+\cos \gamma)

= \frac{1}{2} ab \left(1+\frac{a^2+b^2-c^2}{2ab}\right) = \frac{(a+b)^2-c^2}{4}

= \frac{(2-c)^2-c^2}{4} = \frac{4-4c}{4} = 1-c

= \begin{bmatrix} \sqrt{\|e_2\| \|e_3\|}\sin\frac{\theta_{23}}{2} \\ \sqrt{\|e_3\| \|e_1\|}\sin\frac{\theta_{31}}{2} \\ \sqrt{\|e_1\| \|e_2\|}\sin\frac{\theta_{12}}{2} \end{bmatrix}

= \begin{bmatrix} \sqrt{bc} \cos \frac{\alpha}{2} \\ \sqrt{ca} \cos \frac{\beta}{2} \\ \sqrt{ab} \cos \frac{\gamma}{2} \end{bmatrix}

Law of Cosines

Perimeter \(2\)

Translating to side lengths

\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1-a \\ 1-b \\ 1-c \end{bmatrix}

The transitive group

The rotations are natural transformations of the sphere, and the corresponding action on triangles is natural.

\(c=1-z^2\) fixed

\(z\) fixed

\(C(\theta) = (\frac{z^2+1}{2}\cos 2\theta, z \sin 2\theta)\)

The equal-area-in-equal-time parametrization of the ellipse

A more complicated rotation

Right triangles

The right triangles are exactly those satisfying

\(a^2+b^2=c^2\) & permutations

Since \(a=1-x^2\), etc., the right triangles are determined by the quartic

\((1-x^2)^2+(1-y^2)^2=(1-z^2)^2\) & permutations

\(x^2 + x^2y^2 + y^2 = 1\), etc.

Obtuse triangles

\(\mathbb{P}(\text{obtuse})=\frac{1}{4\pi}\text{Area} = \frac{24}{4\pi} \int_R d\theta dz\)

But now \(C\) has the parametrization

And the integral reduces to

Solution to the pillow problem

By Stokes’ Theorem

\(\frac{6}{\pi} \int_R d\theta dz=\frac{6}{\pi}\int_{\partial R}z d\theta = \frac{6}{\pi}\left(\int_{z=0} zd\theta + \int_C zd\theta \right)\)

\(\left(\sqrt{\frac{1-y^2}{1+y^2}},y,y\sqrt{\frac{1-y^2}{1+y^2}}\right)\)

\(\frac{6}{\pi} \int_0^1 \left(\frac{2y}{1+y^4}-\frac{y}{1+y^2}\right)dy\)

Our Answer

Theorem [with Cantarella, Needham, Stewart]

With respect to the symmetric measure on triangles, the probability that a random triangle is obtuse is

\(\frac{3}{2}-\frac{3\ln 2}{\pi}\approx0.838\)

Sylvester’s Four Point Problem

convex

reflex/reentrant

self-intersecting

Modern Reformulation: What is the probability that all vertices of a random quadrilateral lie on its convex hull?

A discrete group action

A polygon corresponds to \(\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \vdots & \vdots \\ a_n & v_n \end{bmatrix}\), where \((a_k + i b_k)^2 = e_k\).

\((a_k,b_k)\mapsto(-a_k,-b_k)\) doesn’t change the polygon, so \((\mathbb{Z}/2\mathbb{Z})^n\) acts trivially on polygons.

\(S_n \le SO(n)\) permutes rows and hence edges.

The hyperoctahedral group \(B_n = (\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n = S_2 \wr S_n\) acts by isometries on \(G_2(\mathbb{R}^n)\) and permutes edges.

\(|B_n| = 2^n n!\); e.g., \(|B_3| = 2^3 3! = 48\).

\(S_n \le SO(n)\) permutes rows and hence edges.

Proposition [with Cantarella, Needham, Stewart]

The fundamental domain \(\mathcal{D}_n\) of the \(B_n\) action on \(G_2(\mathbb{R}^n)\) consists of (certain lifts of) the convex \(n\)-gons in which all pairs of adjacent or opposite sides form oriented bases in the same class.
The stabilizer of \(\mathcal{D}_4\) has order 4, so \(G_2(\mathbb{R}^4)\) is built from 96 isometric copies of \(\mathcal{D}_4\).
Each of these 96 cells consists of all convex, all reflex, or all self-intersecting quadrilaterals.
\(\mathcal{D}_4\) is path-connected.

Upshot: We can count how many elements of the permutation orbit of a single element of \(\mathcal{D}_4\) are convex, reflex, and self-intersecting.

Volumes via algebra

The average convex quadrilateral

The (empirical) flag mean of \(\mathcal{D}_4\).

\left\{\begin{bmatrix} a_{11} & b_{11} \\ a_{12} & b_{12} \\ \vdots & \vdots \\ a_{1n} & b_{1n} \end{bmatrix},\ldots , \begin{bmatrix} a_{k1} & b_{k1} \\ a_{k2} & b_{k2} \\ \vdots & \vdots \\ a_{kn} & b_{kn} \end{bmatrix}\right\}\subset G_2(\mathbb{R}^n)

The flag mean of

is (the plane spanned by) the first two left singular vectors of the matrix

\begin{bmatrix} a_{11} & b_{11} & a_{21} & b_{21} & \cdots & a_{k1} & b_{k1} \\ a_{12} & b_{12} & a_{22} & b_{22} & \cdots & a_{k2} & b_{k2} \\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ a_{1n} & b_{1n} & a_{2n} & b_{2n} & \cdots & a_{kn} & b_{kn} \end{bmatrix}

The symmetric group orbit

Theorem [with Cantarella, Needham, Stewart]

Convex, reflex, and self-intersecting quadrilaterals are all equiprobable.

A partial generalization

Theorem [with Cantarella, Needham, Stewart]

With respect to any measure on \(n\)-gon space which is invariant under permuting edges, the fraction of convex \(n\)-gons is exactly \(\frac{2}{(n-1)!}\).

What is the least symmetric triangle?

Symmetric triangles

Isosceles

triangles

Degenerate

triangles

The least symmetric triangle

Theorem [with Bowden, Haynes, Shukert]

The least symmetric triangle has side length ratio

1 : 3-\frac{1}{\sqrt{2}} : 3

The least symmetric obtuse triangle

Theorem [with Bowden, Haynes, Shukert]

The least symmetric obtuse triangle has side length ratio

\sim 1 : 2.125 : 2.898

The least symmetric acute triangle

Theorem [with Bowden, Haynes, Shukert]

The least symmetric acute triangle has side length ratio

\sim 1 : 1.221 : 1.408

Polygons in space

Example Theorem [w/ Cantarella, Grosberg, Kusner]

The expected total curvature of a random space \(n\)-gon is exactly

\(\frac{\pi}{2}n + \frac{\pi}{4} \frac{2n}{2n-3}\)

Polygons in \(\mathbb{R}^3\) correspond to points in \(G_2(\mathbb{C}^n)\) and the analog of the squaring map is the Hopf map.

Questions

What is the average \(n\)-edge trefoil?

Observation [Needham]: The flag mean of the equilateral six-edge trefoils is

Theorem*: The flag mean of the positive Grassmannian is the regular \(n\)-gon.

What is the most knotted \(n\)-gon?

What is the flag mean good for in a polygons context?

Observation [Needham]: The flag mean of the equilateral six-edge positive trefoils is

Theorem [Baxter]

The expected number of vertices on the convex hull of a random planar \(n\)-gon is

2\sum_{k=1}^{n-1} \frac{1}{k}

2\sum_{k=1}^{n-1}\frac{\sqrt{\pi } \Gamma \left(\frac{k+1}{2}\right) \Gamma \left(\frac{n-1}{2}\right) \Gamma \left(\frac{n-k+1}{2} \right)}{k \Gamma \left(\frac{k}{2}\right) \Gamma \left(\frac{n}{2}+1\right) \Gamma \left(\frac{n-k}{2}\right)}