Clayton Shonkwiler

Colorado State University

09.15.16

Jason Cantarella

U. of Georgia

Thomas Needham

Ohio State

Gavin Stewart

NYU

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,

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

Text

**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 \(\mathbb{R}^2\).

Three vertices uniformly in the disk:

\(\mathbb{P}(\text{obtuse})=\frac{9}{8}-\frac{4}{\pi^2}\approx 0.72\)

[Edelman–Strang]: Normalize sides so \(a^2+b^2+c^2=1\):

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

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

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

Ideally, this should generalize to \(n\)-gons.

**Spoiler: **\(T\simeq\mathbb{RP}^2=G_2\mathbb{R}^3\)

- The sidelengths \((a,b,c)\) uniquely determine a triangle (remember SSS from high school)
- Obtuseness is scale-invariant

Rescale so that \(a+b+c=2\); then a triangle shape is uniquely specified by a point in the simplex.

Not all points in the simplex correspond to triangles

\(b+c<a\)

\(a+b<c\)

\(a+c<b\)

\(\mathbb{P}(\text{obtuse})=9-12\ln 2 \approx 0.68\)

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

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

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

Let \(s=\frac{1}{2}(a+b+c)\) and define

**Note:** This is why we chose the normalization \(a+b+c=2\)

\(s_a=s-a, \quad s_b = s-b, \quad s_c = s-c\)

Then

\(s_a+s_b+s_c=3s-(a+b+c)=3s-2s=s\)

and the triangle inequalities become

\(s_a>0, \quad s_b > 0, \quad s_c > 0\)

Text

\((1-s_a)^2+(1-s_b)^2=(1-s_c)^2\)

\(\mathbb{P}(\text{obtuse})=9-12\ln 2 \approx 0.68\)

Consider \((x,y,z)\) so that

\(x^2=s_a, \quad y^2 = s_b, \quad z^2 = s_c\)

The unit sphere is a \(2^3\)-fold cover of triangle space

\(SO(3)\) obviously acts transitively on the sphere, and the corresponding action on triangles is natural.

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

\(z\) fixed

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

Since the uniform measure is the unique (up to scale) measure on \(S^2\) invariant under the action of \(SO(3)\)...

**Definition**

The *symmetric measure* on triangle space is the probability measure proportional to the uniform measure on the sphere.

The right triangles are exactly those satisfying

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

Since \(a=1-s_a=1-x^2\), 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.

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

But now \(C\) has the parametrization

And the integral reduces to

By Stokes’ Theorem

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

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

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

**Theorem [w/ Cantarella, Needham, Stewart]**

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

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

The pushforward of the uniform measure on the sphere to the simplex \(s_a+s_b+s_c=1\) has density

\(\frac{1}{2\pi \sqrt{s s_a s_b s_c}}\)

This is the Dirichlet(1/2,1/2,1/2) distribution.

By Heron's formula, the density is proportional to \(\frac{1}{\text{Area}}\).

**Corollary [w/ Cantarella, Needham, Stewart]**

The expected area of a random triangle is \(\frac{1}{4\pi}\).

**Corollary [w/ Cantarella, Needham, Stewart]**

The expected curvature of the circumscribed circle to a random triangle is \(\frac{\pi}{2}\).

For \(n>3\), the sidelengths do not uniquely determine an \(n\)-gon, so the simplex approach doesn‘t obviously generalize.

**Key Observation: **The coordinates \((x,y,z)\) of a point on the sphere are the Plücker coordinates of the perpendicular 2-plane.

\(\vec{p}=\vec{a} \times \vec{b}\)

\(\vec{p}=\vec{a} \times \vec{b} = \begin{pmatrix} \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \\ \begin{vmatrix} a_3 & b_3 \\ a_1 & b_1 \end{vmatrix} \\ \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}\end{pmatrix}\)

In general, \(G_k(\mathbb{R}^n)\) is the *Grassmannian* of \(k\)-dimensional linear subspace of \(\mathbb{R}^n\).

If \(P \in G_k(\mathbb{R}^n)\) and \(\{a_1,\ldots , a_n\}\) is a basis for \(P\), then the map \(G_k(\mathbb{R}^n) \to \mathbb{RP}^{\binom{n}{k}-1}\) recording all \(k \times k\) minors of the matrix \([a_1 \cdots a_k]\) is called the *Plücker embedding* of \(G_k(\mathbb{R}^n)\).

Equivalently, \(P \mapsto a_1 \wedge \ldots \wedge a_k \in \mathbb{P}(\bigwedge^k \mathbb{R}^n) \simeq \mathbb{RP}^{\binom{n}{k}-1}\).

The Plücker coordinates satisfy certain relations; in the case of \(G_2(\mathbb{R}^4)\), these reduce to

\(\Delta_{12}\Delta_{34}-\Delta_{13}\Delta_{24}+\Delta_{14}\Delta_{23}=0\)

Suppose \(T\) is a triangle with edge vectors \(e_1, e_2, e_3\). Let \(p \in S^2\) be any lift of \(T\), and let \((a,b)\) be any orthonormal basis of \(p^\bot\).

\((a, b) = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}\)

Let \(z_k = a_k + i b_k\) for each \(k\).

Up to rotation,

\(e_k = z_k^2\) as complex numbers

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 \(\mathbb{R}^n\).

**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.

**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.

In particular, notice that the symmetric measure is invariant under permutations of the edges.

We can package the Plücker coordinates \(\Delta_{ij} = \begin{vmatrix} a_i & b_i \\ a_j & b_j \end{vmatrix}\) of a point in \(G_2(\mathbb{R}^n)\) as a skew-symmetric matrix

\(\Delta = (\Delta_{ij})_{i,j}\)

In other words, this is the matrix of cross products of the \((a_k,b_k)\).

**Semi-Miraculous Fact: **\(-\Delta^2\) is the Gram matrix (i.e., matrix of dot products) of the \((a_k,b_k)\).

**Definition: **Given \(P \in G_2(\mathbb{R}^n)\), the *Plücker sign matrix* of \(P\) is \(\operatorname{sgn} \Delta(P)\) and the *projection sign matrix* of \(P\) is \(\operatorname{sgn}( -\Delta(P)^2)\).

Since \((a_k + i b_k)^2=e_k\), the Plücker sign matrix and projection sign matrix notice when two edges in a polygon become parallel or anti-parallel.

The sign matrices notice, among other things, when a polygon becomes non-convex or non-embedded.

convex

reflex/reentrant

self-intersecting

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

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

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

**Theorem [Blaschke]**

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

**Theorem [w/ Cantarella, Needham, Stewart]**

With respect to the symmetric measure, each of the three classes of quadrilaterals occurs with equal probability. In particular, \(\mathbb{P}(\text{reflex})=\frac{1}{3}\).

More generally...

**Theorem [w/ Cantarella, Needham, Stewart]**

With respect to *any* permutation-invariant measure on \(n\)-gon space, the probability that a random \(n\)-gon is convex is \(\frac{2}{(n-1)!}\).

There is a version of this story for polygons in \(\mathbb{R}^3\) as well.

**Example Theorem [w/ Cantarella, Grosberg, Kusner]**

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

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

The polygon space is \(G_2(\mathbb{C}^n)\) and the analog of the squaring map is the Hopf map.

What is the manifold of *equilateral* planar \(n\)-gons up to translation and rotation?

Is there a good parametrization of this manifold?

- J. Cantarella, T. Deguchi, and C. Shonkwiler. Probability theory of random polygons from the quaternionic perspective.
*Communications on Pure and Applied Mathematics***67**(2014), 1658–1699 - J. Cantarella, A. Y. Grosberg, R. Kusner, and C. Shonkwiler. Expected total curvature of random polygons.
*American Journal of Mathematics***137**(2015), 411–438 - J. Cantarella, T. Needham, C. Shonkwiler, and G. Stewart. Random triangles and polygons in the plane. Coming soon!