A professional chef notes that the usual method will do: First, she slices the garlic one-thousand times along its length and one-thousand times across its width. This gives

\[ 1000\times 1000 = 1\,000\,000\]

tiny little garlic pieces with only two-thousand cuts. (Our characters remain intentionally two-dimensional—a three-dimensional chef would rotate the garlic to slice along its height, mincing the garlic into one-million pieces with only three-hundred cuts of her knife.)

A naïve chef claims a better way. “First,” he says, “I slice the garlic in two. Then, I cut each of those pieces in half with a single slice.” He continues along, carefully rearranging the garlic so that each run of his blade divides every piece in two again.

“See?” he points. “Each of my slices doubles the number of garlic bits.”

“Two! Four! Eight!” he shouts. “With only twenty slices, I’ll mince it to a million minuscule morsels.”

Then there’s us: sloppy, lazy cooks. Unlike the professional chef, we can’t reliably cut a clove of garlic into one-hundred thin slices without risking our fingers. And, we note wryly, it’s only a matter of time before the cocksure chef discovers a problem with his approach: the final cut requires half a million fragments of garlic arranged into a perfect line!

Instead of risking our fingers or sanity, we’re going to use a technique known to professionals and amateurs alike: we will chop the garlic at random, over and over, until we generate one million pieces. How many chops does our simple approach require?

The answer: Our rustic chop will mince almost as quickly as the professional’s honed technique. Put mathematically, the number of cells in our random chop also grows quadratically with the number of slices we make. With \(n\) slices, our technique produces \(n + n(n-1)/ 4\pi\) bits of garlic when the garlic is a circle. Taking \(n=3540\) yields, on average, just over one million pieces of garlic. Neat!

Let’s assume that our garlic \(G\subset \mathbb{R}^2\) is convex with perimeter \(P\) and area \(A\). We will model our random chopping process as random chords across our garlic, and then counting the number of pieces that this random process creates. These models are known in the mathematical literature as “hyperplane tessellations,” and they have a rich theory associated with them.

Before we resolve the thorny question of “what do you mean by random chord“, let’s take a foray into the basic geometric problem. Suppose we draw the \(n-1\) lines across our garlic. When the \(n\)th line is drawn across the garlic, it will create one new cell for every cell that it crosses on its way across the garlic. This observation forms the core of the analysis below.

What, precisely do we mean by “random line” or “random chord”? This question lies at the heart of Bertrand’s Paradox, where several seemingly reasonable interpretations of “random chord” result in very different distributions of lines. The most common resolution of this paradox involves *invariance*: Generates lines so that the location and orientation of the garlic doesn’t matter—the distribution of the resulting intersections are precisely the same.

Invariant line processes are well-studied in the academic literature. From here on out, we will assume that all of our random lines are drawn from this unique probability measure. The measure has some very nice properties. In particular, the probability that a random line that intersects our garlic also intersects a convex subset of the garlic—denoted by \(K\)—is invariant to the location of the subset. To be precise, we have the following fact.

Fact 1.Let \(K’ \subset K\) be convex. Then the probability that a random line that intersects \(K\) also intersects \(K’\) is equal to \(p(K’) / p(K)\), where \(p(\cdot)\) denotes the perimeter of the set.

With this, we can get in to some computations. Let \(M(i)\) denote the random variable that counts the number of cells of garlic after the \(i\)th chop, \(N(i) := M(i) – M(i-1)\) be the number of new cells of garlic added by the \(i\)th chop, and \(L_i\) be the line generated by the \(i\)th chop. Then using linearity of expectation and the conditioning property, we find

\[\mathbb{E}[M(n)] = 1 + \sum_{i=1}^n \mathbb{E} [ N(i) ] = 1 + \sum_{i=1}^n \mathbb{E}[\mathbb{E}[ N(i) \mid L_1, \dotsc, L_{i-2}, L_{i-1}]].\]

Each fixed arrangement of lines \(L_1, \dotsc, L_{i-2}, L_{i-1}\) generates a cell complex \(\{C_1, \dotsc, C_{R-1}, C_R\}\), where the number \(R\) is a function of the lines. With this notation, the inner expectation takes the form

\[\mathbb{E}[N(i) \mid L_1, \dotsc, L_{n-2}, L_{n-1}] = \sum_{j=1}^R \mathbb{P}\{L_n \text{ intersects } C_j \}= \frac{1}{p(K)} \sum_{j=1}^R p(C_j), \]

where the second equality is Fact 1. The right-hand side is the total perimeter of all cells inside of the garlic. This is equal to the perimeter of the garlic plus twice the length \(\ell\) of every line in the garlic:

\[\frac{1}{p(K)} \sum_{j=1}^R p(C_j) = 1 + \frac{2}{p(K)} \sum_{k=1}^{i-1} \ell(L_k).\]

Combining the three equations above and using the fact that the lines are identically distributed, we find that

\[\mathbb{E}[M(n)] = n + \frac{2}{p(K)}\sum_{i=1}^n \sum_{k=1}^{i-1}\mathbb{E}[\ell(L_i)] = n + \frac{\mathbb{E}[\ell(L_1)]}{p(K)} n(n-1). \]

Thus, the average number of lines grows quadratically, and the only thing that remains to compute is the average length of the chord \(\ell(L_1)\). This is given by [SW, Theorem 8.4.1]:

\[\mathbb{E}[\ell(L_1)] = \frac{v(K)}{p(K)},\]

where \(v(K)\) is the volume of the garlic \(K\). Putting this all together, we find

\[\mathbb{E}[M(n)] = n + n (n-1) \frac{v(K)}{p^2(K)}.\]

The ration \(v(K)/p^2(K)\) is maximized when \(K\) is a circle, where \(v(K)/p^2(K) = 1 / 4\pi\). In this case

\[\mathbb{E}[M(n)] = n + \frac{n(n-1)}{4\pi}.\]

]]>More precisely, when does a function \(\Lambda\) that maps an outward-pointing normal vector \(u \in \mathbb{S}^2\) to a pair of curvatures \(\lambda_1(u), \lambda_2(u)\) always define a convex surface in \(\mathbb{R}^3\)? This question is closely related to deep problems in geometry, notably the Christoffel–Minkowski problem. In this post, we derive a condition on every curvature function \(\Lambda\) that defines a convex surface that limits the available functions.

To state the result, let’s make up a word.

Definition.A point \(x\) on a convex surface \(S\) anequicurved points of \(S\)if all principal curvatures of \(S\) at \(x\) are equal.

In a neighborhood of an equicurved point \(x\), the surface \(S\) looks spherical up to second order. Indeed, spheres are characterized by the fact that every point is a point of equicurvature. Surprisingly, every sufficiently regular three-dimensional convex surface has at least one point of equicurvature.

Theorem 1.Let \(S\subset \mathbb{R}^3\) be a strictly convex twice-differentiable surface. Then \(S\) has at least one point of equicurvature.

The proof requires a few preliminaries.

Fact 2 [Schneider, Chapter 2.5].Let \(S\) be as above, and let \(h\colon \mathbb{S}^2\to \mathbb{R}\) be the support function of \(S\). Then:1. The support function \(h\) is of class \(C^2\), that is, \(h\) is twice continuously differentiable.

2. The eigenvalues of the Hessian \(\nabla^2 h(u)\) of the support function are the principal radii of curvature of \(S\) in the direction \(u\in \mathbb{S}^2\). (Note that this parameterization uses the outward-pointing normal to determine the radius of curvature, as opposed to the base point.)

3. The eigenvectors of \(\nabla^2 h(u)\) corresponding to nonzero eigenvalues are perpendicular to \(u\).

*Proof of Theorem 1.* For a contradiction, suppose that there is no point of equicurvature on \(S\). By Fact 2(1), that the Hessian map \(\nabla^2 h \colon \mathbb{S}^2 \to \mathbb{R}^{3×3}\) is continuous. Moreover, since \(S\) has no point of equicurvature, \(\nabla^2 h(u)\) has distinct eigenvalues everywhere by Fact 2(2).

This implies that the principal unit eigenvector \(\xi_i(u) \in \mathbb{S}^2\) of \(\nabla^2 h(u)\) is also well defined, continuous, and perpendicular to \(u\) by Fact 2(3). In other words, \(\xi \colon \mathbb{S}^2 \to \mathbb{S}^2\) is a non vanishing continuous map from the unit sphere onto it’s tangent space, which is impossible by the hairy ball theorem.

###

Fascinating! We’ve used a classical theorem in differential geometry to find a special point on every smooth, strictly convex surface.

Moreover, the proof doesn’t work for surfaces embedded in even dimensions, and even in odd dimensions, the straightforward extension of the method results in a far less satisfying conclusion (no point of equicurvature, but you must have points where the principal curvatures intertwine).

Of course, the theorem is almost vacuously true in one and two dimensions, so we seem to have stumbled upon a truly “three dimensional” fact.

]]>\[h(\theta) + h”(\theta) \ge 0 \text{ for all } 0 \le \theta < 2\pi,\]

where \(h(\theta):= h(\cos(\theta), \sin(\theta)\) represents evaluation of \(h\) on the unit circle. Since one-homogeneous convex functions are support functions, the equation above characterizes convex bodies in two dimensions.

The non-standard proof below provides a more modern route to the result than classical approaches. I expect that this technique extends to higher dimensions more directly than the classical arguments available, e.g., in Groemer’s text.

Before moving forward, let’s make our claim a theorem. As above, we use the notation \(h(\theta):= h(\cos(\theta),\sin(\theta))\).

Theorem.Support that \(h\colon \mathbb{R}^2\to\mathbb{R}\) is one-homogeneous and that the Hessian \(\nabla^2 h\) exists on \(\mathbb{R}\setminus \{0\}\). Then \(h\) is convex if and only if\[h(\theta) + h”(\theta) \ge 0.\tag{1}\]

We will break the proof into a series of lemmas. We begin with a lemma about the Hessian of one-homogeneous functions.

Lemma 1.The Hessian of a one-homogeneous function \(h\) on the unit circle takes the form\[\nabla^2 h (\cos \theta, \sin\theta) = r(\theta) \begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta \\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix}.\tag{2}\]

*Proof.* Since \(h\) is one-homogeneous, it exhibits no curvature in the radial direction, or in other words, \((x, y)^t\) is in the null space of \(\nabla^2h (x,y)\) for all \((x,y)\ne (0,0)\). Therfore,

\[\nabla^2 h (x,y) = \lambda(x, y) \begin{pmatrix} -y \\ x \end{pmatrix} (-y, x) = \lambda(x,y) \begin{pmatrix} y^2 & -xy \\ -xy & x^2\end{pmatrix},\]

where \(\lambda\) provides the other eigenvalue of \(\nabla^2h\). The conclusion follows by parameterizing \(x= \cos \theta\) and \(y = \sin \theta\). ###

As a corollary, we characterize the Hessian of convex functions.

Corollary 2.The function \(h\) in Lemma 1 is convex if and only if \(r(\theta) \ge 0\) for all \(\theta \in [0,2\pi)\).

*Proof. *The function \(h\) is convex if and only if it’s Hessian is positive definite, which occurs if and only if \(r(\theta) \ge 0\). ###

Below, we define the first derivatives

\[ h_x := \frac{\mathrm{d}h}{\mathrm{d}{x}}, \quad h_y:= \frac{\mathrm{d}h}{\mathrm{d}{y}}, \quad \text{and}\quad h’ := \frac{\mathrm{d}h}{\mathrm{d}\theta} = – h_x \sin\theta + h_y \cos\theta .\]

The second derivatives \(h”\), \(h_{xx}\), \(h_{xy}\), and \(h_{yy}\) are defined analogously. Then we have the following result.

Lemma 3.For any one-homogeneous function \(h\), we have\[ h(\theta) + h”(\theta) = r(\theta) \]

where \(r(\theta)\) is given in (2).

*Proof. *First, recall that Euler’s relation for one-homogeneous functions implies \(h(\theta) = h_x \cos\theta + h_y \sin\theta\) on the unit circle. Moreover,

\[h”(\theta) = h_{xx} \sin^2 \theta – 2 h_{xy} \sin \theta \cos\theta + h_{yy} \cos^2\theta -h_x \cos\theta – h_y\sin\theta.\]

Applying (2), we find

\[h”(\theta) = r(\theta)\underbrace{ \left(\sin^4\theta +2 \sin^2\theta \cos^2\theta + \cos^4\theta\right)}_{= (sin^2\theta + \cos^2\theta)^2 = 1} -h(\theta),\]

as claimed.

###

The proof of Theorem 1 is completed upon combining Corollary 2 and Lemma 3.

]]>Take, for example, the famous Meissner body (Fig. 1). This shape, first described by the Swiss Mathematician Ernst Meissner around 1911, arises from a bit of a mathematical hack. Meissner rounded three edges of a Relaux tetrahedron—a body of *almost* constant width—to create a body of precisely constant width. (In this game, close doesn’t count for much.) Despite a nearly one-hundred year investigation, Meissner’s construction still yields the smallest volume body of constant width known. While some mathematicians suspect that Meissner’s body is the smallest volume shape of constant width, a proof of this conjecture remains dreadfully elusive.

In this post, we do not attempt to address that old and apparently difficult conjecture. Rather, we consider another remarkable fact, due to Wilhelm Blaschke in 1915: The surface area of a body of constant width is proportional to its volume. The proof below is relatively elementary, and apparently new. This exposition may be of interest for anyone looking for a deeper understanding of these wonderful shapes.

Let’s start by making things a bit more precise. We say that a convex set \(K\subset \mathbb{R}^3\) has* constant width* \(c\) if every two parallel planes touching the edges of the set are the same distance apart, no matter how the shape is rotated. The surface area and volume of each of these bodies satisfy a remarkable relationship.

Theorem 1.(Blaschke, 1915).Let \(K\subset \mathbb{R}^3\) be a body of constant width \(c\). Then the volume \(v(K)\) and the surface area \(s(K)\) satisfy\[v(K) = c\frac{1}{2} s(K) – c^3\frac{\pi}{3}.\tag{1}\]

The original proof of this result appears in an article [1] that I’ve been unable to locate online, so I can’t compare the proof below to the approach taken by Blaschke. However, a much later article of Santaló provides an analog in arbitrary dimensions that requires more advanced techniques. The proof below, while restricted to three dimensions, extends to readily to higher dimensions, and yields Santaló’s results with relative ease.

Our proof relies on a few ideas from the theory of convex bodies, in particular, the theory of mixed volumes. The volume \(v(K + \lambda L)\) of the Minkowski sum of the convex bodies \(K\) and \(\lambda L\) is a polynomial in the scale parameter \(\lambda \ge 0\):

\[v(K + \lambda L) = V_0(K,L) + \lambda V_1(K,L) + \lambda^2 V_2(K,L) + \lambda^3 V_3(K,L). \tag{2}\]

From this equation, it’s not hard to show that the coefficients \(V_i(K,L)\), called *mixed volumes*, satisfy:

- \(V_{i}(K,L) = V_{3-i}(L,K)\) for each \(i\); and
- \(V(K, L) = V(-K,-L)\).

If we set \(L= \mathbb{B}_3\) the 3-ball, then the coefficients in (2) take on special values:

\[v(K + \lambda \mathbb{B}_3) = v(K) + s(K) \lambda + 2\pi\bar{w}(K)\lambda^2 + \frac{4 \pi}{3}\lambda^3,\tag{3}\]

where \(\bar{w}(K)\) is the *mean width* of the convex body \(K\). (In particular, when \(K\) has constant width \(c\), we have \(\bar{w}(K) = c\).) The result (3) goes by the name *Steiner’s formula,* after the Swiss mathematician Jakob Steiner who discovered it in 1840. Although Steiner was illiterate until he was 14 and did not attend school until after he turned 18, he managed to make enormous contributions to the theory of analytic geometry during his life.

Finally, we will need another characterization of sets of constant width. A convex set \(K\) in 3-space has constant width \(c\) if and only if

\[ K – K = c \mathbb{B}_3,\tag{4}\]

where \(K – K\) denotes the Minkowski sum of \(K\) and \(-K\). For completeness, we prove this relationship at the end of the post.

At this point, you might already sense our strategy: we will compute the volume of \(K + \lambda (-K)\) using both the mixed volume formula (2) and the Steiner formula (3), and then compare coefficients.

We may assume without loss that the width \(c=1\), since the general \(c\) case follows in (1) by homogeneity. Let \(\lambda \in [0,1]\). Using the mixed volume expansion (2), we find

\begin{align}v(K + \lambda(-K)) &= V_0(K,-K) + 3\lambda V_1(K, -K) + 3\lambda^2 V_2(K, -K) + \lambda^3 V_3(K,-K) \tag{5}\end{align}

On the other hand, we also have \(K + \lambda (-K) = (1-\lambda) K + \lambda \mathbb{B}_3\) by Equation (4). Applying the Steiner formula (3), we find

\[v(K + \lambda(-K)) =(1-\lambda)^3 v(K) +\lambda(1-\lambda)^2 s(K) + \lambda^2 (1-\lambda)2\pi +\lambda^3\frac{4\pi}{3},\tag{6}\]

where we’ve used the fact that the mean width \(\bar w (K) = 1\) because \(K\) has constant width \(c=1\). Grouping like powers of \(\lambda\) in (5) and (6), we arrive at the following system of equations:

\begin{align}V_0(K,-K)&=v(K)\\ 3V_1(K,-K) &= -3 v(K) + s(K) \\ 3V_2(K,-K) &= 3v(K) – 2 s(K) +2\pi\\ V_3(K,-K) &= -v(K) + s(K) – \frac{2\pi}{3}.\end{align}

The properties of the mixed volumes listed above imply that \(V_0(K,-K) = V_3(K,-K)\) and \(V_1(K,-K) = V_2(K,-K)\), so the four equations above reduce to the following two independent statements:

\begin{align}v(k) &= \frac{1}{2}s(K) – \frac{\pi}{3}\tag{7}\\ V_1(K,-K) &= \frac{\pi}{3} – \frac{1}{2} s(K).\end{align}

Equation (7) is precisely Blaschke’s identity (1) for \(c=1\). The general case follows by homogeneity.

#

We make use of properties of the support function \(h_K\), all of which appear in the linked Wikipedia article. First, it follows from the definition of the support function \(h_K\) that \(K\) has constant width \(c\) if and only if

\[ h_K(u) + h_K(-u) = c \quad \text{for all} \quad u \in \mathbb{S}^{d-1}.\]

Since \(h_K(-u) = h_{-K}(u)\) and we have the linear relationship \(h_{K – K} = h_K + h_{-K}\), we see that \(h_{K -K}(u) = c\) for all \(u\in \mathbb{S}^{d-1}\). This latter equation characterizes the scaled ball \(c\mathbb{B}_d\), so we conclude that \(K-K = c \mathbb{B}_d\).

#

[1] W. Blaschke, Einige Bemerkungen über Kurven und Flächen von konstanter Breite, Ber. Verh. Sächs. Akad. Leipzig, 67 (1915), 290–297.

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