Simplex slicing: an asymptotically-sharp lower bound

We show that for the regular n-simplex, the 1-codimensional central slice that's parallel to a facet will achieve the minimum area (up to a 1-o(1) factor) among all 1-codimensional central slices, thus improving the previous best known lower bound (Brzezinski 2013) by a factor of $\frac{2\sqrt{3}}{e} \approx 1.27$. In addition to the standard technique of interpreting geometric problems as problems about probability distributions and standard Fourier-analytic techniques, we rely on a new idea, mainly \emph{changing the contour of integration} of a meromorphic function.


Introduction
We are broadly interested in the following question: if K is a given convex body in R n , and H is some hyperplane of codimension 1, can we obtain good upper and lower bounds on the area of the intersection H ∩ K? ("Area" meaning the (n − 1)-dimensional volume.)We refer to H ∩ K as a (hyperplane) section of K.If H is constrained to pass through the centroid of K, we call H ∩ K a central section.
Let us consider the case where K is the hypercube Q n := [− 1 2 , 1 2 ] n and H is constrained to pass through the origin (which is the centroid of Q n ).The first sharp result was discovered independently by Hadwiger ([Had72]) and Hensley ([Hen79]), who showed that the area of H ∩Q n is minimized by taking H perpendicular to the vector (1, 0, 0, 0, . . ., 0).(In this case H ∩Q n is isometric to Q n−1 .)Hensley's proof relied on a technique which has become commonplace: converting the original geometric problem into the language of probability distributions.Later in the same paper, Hensley used another technique which has become commonplace, namely the use of the Fourier inversion formula to deal with tricky convolutions. 1Ball used both aforementioned techniques in [Bal86] to prove that the area of H ∩ Q n is maximized by taking H perpendicular to the vector (1, 1, 0, 0, 0, 0, . . ., 0).(In this case H ∩ Q n is a rectangular prism with one side √ 2 and all other sides of length 1.) Incidentally, Ball's result immediately implies that for n ≥ 10, any central sections of Q n will have area less than that of any central section of B(Γ( n 2 + 1) 1/n / √ π), the n-ball of volume 1 (see [Bal88]).This gives a negative answer to the Busemann-Petty problem for all n ≥ 10.Now let us turn our attention to the case where K is the regular simplex ∆ n (of side length √ 2).If H is not constrained to pass through the center of ∆ n , Webb noted in the introduction of [Web96] that some results of Ball (involving maximum-volume inscribed ellipsoids) can be used to show that the area of H ∩ ∆ n is maximized by letting H be a facet of ∆ n .Fradelizi (in [Fra99]) has given a different proof of this same result.However, if H is constrained to pass through the center of ∆ n , Webb showed in the same paper ( [Web96]) that the area of H ∩ ∆ n is maximized by taking H to pass through all but two vertices of ∆ n (a quantitative version of Webb's result is given in [MTTT24]).Webb gave two proofs of this result, one of which used both the technique of probability distributions and the technique of the Fourier inversion formula.This leaves open the following question, which is the focus of this paper: Question 1.If H is constrained to pass through the center of ∆ n , then what is the minimum possible value of the area of H ∩∆ n ?In other words, what is the minimum central section of the regular simplex?Furthermore, which choices of H attain this minimum value?
It has been conjectured that the minimum in Question 1 is attained by taking H to be H facet , where H facet is any hyperplane through the center of ∆ n and parallel to a facet (this is Conjecture 4 in [NT23]).Filliman states this conjecture without proof in Section 3, part (h) of [Fil92].Dirksen showed (Theorem 1.3 (i) in [Dir17]) that H facet is a local minimizer; i.e. for all H sufficiently close to H facet , the area of H ∩ ∆ n is at least that of H facet ∩ ∆ n .In Theorem 1.1 of [Brz13], Brzezinski showed a lower bound that is within a factor of 1.27 of the conjectured minimizer; i.e.Brzezinski proved that for any H, the area of H ∩ ∆ n is at least e 2 √ 3 ≈ 1 1.27 times that of H facet ∩ ∆ n .One reason why the question of minimizing H ∩ ∆ n is harder than the analogous question for Q n is that Hensley was able to utilize the fact that Q n is a symmetric convex body, so the probability distribution resulting from slicing Q n is a symmetric probability distribution.The simplex is not a symmetric convex body, so the resulting probability distribution is in general not symmetric.
In this paper we almost resolve the question of minimizing H ∩ ∆ n ; i.e. we prove a lower bound that is within a factor of 1 − o(1) of the conjectured minimizer (the notation o(1) is with respect to taking n ↗ ∞).
Theorem 2. For each hyperplane H passing through the center of ∆ n , we have Note that 1 e n+1 n n+1 n n−1 ↗ 1 as n ↗ ∞, so Theorem 2 is best possible (asymptotically speaking).
Our main new idea is to move the contour of integration.Following the aforementioned techniques of probability distributions and of the Fourier inversion formula, we rewrite Theorem 2 as a statement that the probability density of a sum of independent exponential random variables takes a value of at least 1 e at zero.Specifically, we show the following, which may have independent interest: . ., Y n+1 be i.i.d.standard exponential random variables (mean 1).For each unit vector u ∈ R n+1 , we have that the density at 0 of the random variable Equality holds if u is of form (1, 0, 0, 0, . . ., 0).Using Fourier inversion, we rewrite Theorem 3 as a statement that the integral of the characteristic function of Z is at least 1 e .However, F is highly oscillatory along the real axis, and is difficult to bound.We simply move the contour of integration to some new contour γ, so that F takes only positive real values along γ; now we can bound the integral more easily.
Our proof appears to be highly dependent on the idiosyncracies of the regular simplex, and do not generalize well to other convex bodies.In particular, there is a curious inequality (Equations ( 17) and ( 18)), and it is not clear if there is an analogue when the simplex is replaced with some other convex body.
A failed approach that is worth mentioning is the following false statement, strengthening Theorem 3: Fake Theorem 4. Let f X be the density function of a log-concave random variable X, such that X has mean 0 and variance 1.Then f X (0) ≥ 1 e , with equality holding if X is equal to Y − 1, where Y is a standard exponential random variable (mean 1).This is indeed a strengthening of Theorem 3, since the random variable Z from Theorem 3 is the sum of independent log-concave random variables, and is hence log-concave.However, the Fake Theorem 4 is indeed false, as can be seen by taking X to be the uniform distribution with mean 0 and variance 1; we end up with f X (0) = 1 √ 12 < 1 e .(In fact, 1 √ 12 is precisely the minimum possible value of f X (0), as X varies over all log-concave random variables with mean 0 and variance 1.) Finally, we will note in passing that, letting K be a convex body and H be a hyperplane as before, people have considered constraining H, not so that it passes through the centroid of K, but so that it is at a distance exactly t from the centroid of K, where t > 0 is a fixed constant.In this case H ∩ K is called a noncentral section.König showed (Theorem 1.1 of [Kö21]) that if t is reasonably large, the area of H ∩ ∆ n is maximized by taking H to be parallel to a facet of ∆ n .Liu and Tkocz investigated the cross-polytope ([LT20]), and Moody, Stone, Zach, and Zvavitch investigated the hypercube ( [MSZZ13]).
In the remainder of this section we will sketch the proof of Theorem 2.
• In Section 3, we first use a result of Webb (Equation (2)) to reduce Theorem 2 to a question about the distribution of a sum of independent centered exponential random variables (Theorem 8, which is the same as Theorem 3).This constitutes a use of the technique of probability distributions.• In Section 4, we take the Fourier transform, and the problem reduces to showing that the integral of an explicit meromorphic function F a along the real line is at least 1 e (Theorem 9).Here, F a is the characteristic function of the probability distribution.This constitutes a use of the technique of the Fourier inversion formula.• In Section 5, we change the contour of integration; this is our main new idea.As noted previously, F a is too difficult to bound along the real axis, so we instead change the contour of integration from the real axis to some path γ a , with the property that F a always attains positive real values on γ a .Since the integral of a meromorphic function does not depend on the contour used2 , the problem reduces to showing that the integral of F a along γ a is at least 1 e (Theorem 13).• In Section 6, we give a lower bound for the values of F a along γ a .Essentially, we let v denote the vector (1) ∈ R 1 , and we show that F a (x) ≥ F v (x) for each x (Theorem 14).We do this by showing that the derivative of F a is at least the derivative of F v (Equation ( 12)).There is a curious inequality involved (Equations ( 17) and ( 18)).We obtain the desired 1 e lower-bound by evaluating the integral of F v .

Definitions
Let n ≥ 2 be a positive integer.Define the standard basis vectors e 1 , e 2 , . . ., e n+1 in R n+1 , such that e j denotes the vector with a 1 in the jth coordinate, and zeroes in all other entries.We use ∆ n to denote the regular n-simplex of side length √ 2, embedded in R n+1 as follows: ∆ n is the convex hull of the points e 1 , e 2 , . . ., e n+1 .Equivalently, and x i ≥ 0 for each i .
For any set Q, we let vol k (Q) denote the k-dimensional Hausdorff measure of Q.Then the volume of ∆ n is given by ⃗ 1 is the center of the regular simplex ∆ n .Furthermore, given any two vectors v, w ∈ R n+1 , we write ⟨v, w⟩ for their dot product, and we say that v ⊥ w if ⟨v, w⟩ = 0, i.e. if v, w are orthogonal.Throughout this paper, let a ∈ R n+1 denote an arbitrary unit vector satisfying a ⊥ ⃗ 1.Let H(a) denote the hyperplane perpendicular to a; i.e.
Note that since a ⊥ ⃗ 1, we have that H(a) contains the point 1 √ n+1 ⃗ 1; i.e.H(a) always passes through the center of ∆ n .
The intersection H(a) ∩ ∆ n is an (n − 1)-dimensional polytope embedded in R n+1 .We call this object the central section of ∆ n in the direction of a, and denote it by T (a).As in Question 1, we will be concerned with finding the minimum possible value of vol n−1 (T (a)) as a varies over all unit vectors perpendicular to ⃗ 1.
For convenience, with a defined as above, we will write the entries of a as (1) a = (a 1 , a 2 , . . ., a n+1 ).
Observe that Observe that H(a facet ) corresponds exactly to the hyperplane H facet from Section 1, which corresponds to the conjectured minimum central section (as in Question 1).We compute that T (a facet ) is the convex hull of the n vertices and thus verifying the equality in the second half of Theorem 2.
Thus Theorem 2 can be rewritten as follows:

The technique of probability distributions
In Section 0 of [Web96] it is proven that if Y 1 , Y 2 , . . ., Y n+1 are i.i.d.standard exponential random variables (mean 1), and G a (x) represents the probability density function of the sum In other words, the volume of T (a) is proportional to the density at 0 of the random variable So Question 1 and Theorem 5 may be converted to the following statements about probability distributions: Question 6.What is the smallest possible value of G a (0), as a varies over all unit vectors in R n+1 which are perpendicular to ⃗ 1?
Theorem 7.For each unit vector a ∈ R n+1 satisfying a ⊥ ⃗ 1, we have G a (0) ≥ 1 e .Throughout this paper, let u ∈ R n+1 denote an arbitrary unit vector (not necessarily satisfying u ⊥ ⃗ 1).Let G u (x) denote the probability density of the random variable u (sum of independent centered exponential random variables), and observe that when u ⊥ ⃗ 1 this agrees with the earlier definition of G a (x).We will prove the following strengthening of Theorem 7, which may be of independent interest: Theorem 8.For each unit vector u ∈ R n+1 , not necessarily satisfying u ⊥ ⃗ 1, we have G u (0) ≥ 1 e .Equality occurs if u is of form (1, 0, 0, 0, . . ., 0).
Note that this theorem is precisely the same as Theorem 3. Note.It is true that equality holds in Theorem 8 if and only if u is of form (1, 0, 0, 0, . . ., 0), though this will not be proven.Since (1, 0, 0, 0, . . ., 0) is not perpendicular to ⃗ 1, there is no vector a which achieves equality in Theorem 7, and hence no central section T (a) achieves equality in Theorem 5.In other words, Theorem 5 is not sharp, even though Theorem 8 is.We lost some sharpness when we considered the space of all unit vectors u (instead of the space of unit vectors a satisfying a ⊥ ⃗ 1).This is why we cannot quite prove that a facet achieves the minimum central section.We can only prove Theorem 2, which says that a facet is within a 1 − o(1) factor of the minimum central section.
Since for u = (1, 0, 0, 0, . . ., 0) we have that G u (x) is the probability density of the random variable Y 1 − 1 with G u (0) = 1 e , we henceforth restrict our attention to proving Theorem 8 when u is not of this form; i.e. when u has at least two nonzero entries.
Let us try to understand the probability distribution G u .For each 1 ≤ j ≤ n + 1, define f j (x) to be the probability density function of the random variable u j (Y j − 1).Note that the u j (Y j − 1) are mutually independent.Since G u is defined to be the probability density function of the sum of these independent random variables, we know that G u is the convolution of the f j ; i.e.
Define the function f : R → [0, +∞) via Note that f is the probability density function of a standard exponential random variable (mean 1), so each Y j has probability density function given by f .It follows that f j (x) = 1 |uj | f ( x uj + 1) for each j, and hence we may write

The technique of the Fourier inversion formula
We will now use the Fourier transform to convert the convolution into a pointwise product.Defining the Fourier transform (Section 2 of [Web96]) From f (t) = 1 1+it , we calculate f j (t) = e iu j t 1+iuj t and thus Note that whenever u has at least two nonzero entries, we have G u ∈ L 1 and hence the Fourier inversion formula is valid, yielding In particular, plugging in x = 0 yields If we want to show a lower bound on G u (0) (as in Theorem 8), it suffices to prove the following inequality: The contour γ u will be chosen such that F u is red along γ u .This image was produced using Wolfram Mathematica Student Edition.

Writing
(3) the inequality that we wish to show becomes the following: Theorem 9.For each unit vector u ∈ R n+1 such that u has at least two nonzero entries, we have

Main idea: Changing the contour of integration
In this section we will showcase our main new idea, which is to change the contour of integration.As noted in Section 1, it is difficult to bound the integral +∞ −∞ F u (t) dt as written, because F u oscillates on the real axis (see Figure 2).We will instead find a contour γ u in the complex plane (the shape of the contour will depend on the choice of u), along which F u takes positive real values (Proposition 10); finally we will argue that the integral +∞ −∞ F u (t) dt equals the contour integral γu F u (t) dt (see Equation (10)), allowing us to change the integral in Theorem 9 to an easier integral.
The rest of this section will be dedicated to finding and defining γ u .
As in Section 3, let u ∈ R n+1 be a fixed unit vector.We want to find a function y u : R → R, depending on u, such that the graph of y u (i.e. the set x + iy u (x) x ∈ R ) is exactly the contour γ u that we seek.In order for this to be the case, we need arg F u (x + iy u (x)) ≡ 0 (mod 2π) for each x, where arg z denotes the argument of a nonzero complex number z.Using the definition of F u (t) (Equation (3)), we can write so all we need to do is to choose a continuous function y u such that holds for all x.For any fixed x, there may be multiple values of y u (x) which satisfy Equation (5) (due to the fact that that equation is modulo 2π), but if we additionally stipulate that y u is continuous and satisfies y u (0) = 0, then there is only one way to choose y u .We will describe how to choose y u below.Let x, y be arbitrary real numbers satisfying x > 0. We will adopt the convention that the arccot function is a continuous function from R to R that has range (0, π).(Some authors define arccot θ to have a jump discontinuity at θ = 0; we will not use their convention here.)Assume that all u j are nonzero (since we can just discard any zero entries).

−y)
x if u j < 0. We will not define α j for u j < 0, nor will we define β j for u j > 0. (We may suppress the dependence on x, y, u, thus writing merely α j for what should technically be written as α j (x, y; u), if it is clear from context.)We define the function and the computation in Equation (4) implies for all (x, y) satisfying x > 0.
The advantage of defining Φ u is that Φ u takes values in R, whereas arg F u (x + iy) is only defined modulo 2π.So we can say that Φ u (x, y) is a real-valued function that is a lifting of the function arg F u (x + iy).In other words, Φ u (x, y) is a real-valued function that always is equivalent to the argument of F u (x + iy) modulo 2π.
We can now say that the pair (x, y) has good angle if Φ u (x, y) = 0, this equality holding in R (instead of just R/2πZ).
Let m + , m − denote the number of indices j such that u j is positive (respectively, negative).So m + +m − = n + 1.Since Φ u is a strictly decreasing function of y, we know that: (1) If x satisfies −m − π < x n+1 j=1 u j < m + π, then there exists a unique y such that (x, y) has good angle.In this case we will set y u (x) = y.
(2) If x does not satisfy −m − π < x n+1 j=1 u j < m + π, then there does not exist y such that (x, y) has good angle.

Re
Im
Thus we have found a function y u defined on the domain (Observe that D u contains the interval (0, π).)By definition, (8) Φ u (x, y u (x)) = 0 for all x ∈ D u .In other words, y u is contained in the zero locus of Φ u .
We can obtain that y u is C ∞ (in fact, real analytic) on D u using the Implicit Function Theorem (Φ u is a real analytic function of (x, y) which is also strictly decreasing with respect to y).
We may also show that lim x↘0 y u (x) = 0. To do this, let ϵ ∈ (0, 1) be arbitrary and fixed.If j is such that u j > 0, we note that 1 uj − ϵ > 1 1 − 1 = 0, so (arccot −ϵ ) goes to 1 as x ↘ 0. Similarly, if j is such that u j < 0, we have (arccot u j , we see that when x is sufficiently close to 0, we have Φ u (x, ϵ) < 0. Using the fact that Φ u is strictly decreasing with respect to y, we obtain y u (x) < ϵ when x is sufficiently close to 0. Similarly, we can show that y u (x) > −ϵ when x is sufficiently close to 0. Since ϵ was arbitrary, this proves lim x↘0 y u (x) = 0, as desired.
Thus y u extends continuously to the point x = 0, by setting y u (0) := 0. Let us extend y u by reflection about the y-axis.Explicitly, we define an expanded domain and extend y u to this expanded domain by setting y u (x) = y u (−x) whenever −x ∈ D u .So y u will be a continuous function on this expanded domain, and is in fact C ∞ away from the point x = 0.
We will now show that y u is C ∞ on the entirety of E u .Observe that for x > 0 and y ∈ (−1, 1), we have The right-hand-side can be extended to a real analytic function ψ u (x, y) on all of R × (−1, 1).Moreover, we compute so by the Implicit Function Theorem we know that near (0, 0), the zero locus of ψ u is the graph of a real analytic function.But this zero locus is exactly the graph of y u , so we obtain that y u is a real analytic function of x at the point x = 0, as desired.
Remark.What happens for other probability distributions (other than the exponential distribution that we have been considering)?Are we still able to define a smooth y u on which F u takes positive real values?
We have our function y u which is C ∞ on E u .By construction, we have that y u satisfies Equation ( 5) for all x ∈ E u .Now if we define γ u := x + iy u (x) x ∈ E u to be the graph of y u , then γ u is exactly the contour that we seek.Indeed, the key property of γ u is the following: Proposition 10.For each t ∈ γ u , we have that F u (t) is a positive real number.This proposition follows from Equation (7).Now that we have our contour γ u , we would like to say the following (in order to change the integral in Theorem 9 to an easier integral): Theorem 11.For each unit vector u ∈ R n+1 such that u has at least two nonzero entries, we have that the integral 1 2π Note that the latter integral is well-defined because the integrand is always a positive real number on the path of integration.
To show Theorem 11, we will use the fact that the integral of a meromorphic function does not depend on the contour, as long as one contour can be moved to the other without passing over any poles.This is the main idea of our paper: moving the contour of integration.We will approximate both integrals in Theorem 11 by integrals over bounded contours, and then we will use Cauchy's Integral Theorem to shift the contour of integration.We must use an upper bound on the magnitude of F u (t); this is because when shifting the contour, we will incur a small error at each of the fringes, and we need to know that this error goes to zero as we take the fringes to be further and further away from the origin.
Proof of Theorem 11.To upper bound the magnitude of F u (t), we now use the fact that u has at least two nonzero entries.Hence there exists a constant C > 0, depending only on u, such that for all pairs (x, y) ∈ R 2 with x ̸ = 0, we have (In particular, the integral 1 2π +∞ −∞ F u (t) dt in Theorem 9 is well-defined.)We split into cases based on the sign of n+1 j=1 u j .Case I: n+1 j=1 u j > 0. In this case, E u = (−m + π/ n+1 j=1 u j , m + π/ n+1 j=1 u j ) is a finite interval.Now let us choose x 1 ∈ D u and x 2 > x 1 .Let K x1,x2 denote the path consisting of the line segment from x 1 + iy u (x 1 ) to x 2 + iy u (x 1 ), concatenated with the line segment from x 2 + iy u (x 1 ) to x 2 ; we will use K x1,x2 as a fringe to relate the two contours (γ u and the real axis).Equation (9) implies that |F u (x + iy)| ≤ C exp −y n+1 j=1 u j /x 2 for all x ̸ = 0.By breaking K x1,x2 into its constituent line segments, we obtain Now note that lim x↗m+π/ n+1 j=1 uj y u (x) = +∞.If x 1 is chosen sufficiently close to the right endpoint of D u , we will have y u (x 1 ) > 0, and then we can write F u (t) dt by Cauchy's Integral Theorem (this step is where we have shifted the contour from the real axis to γ u ).We are not done yet, since the integrand on the left-hand-side of the above equation still contains an extra term of F u (s + iy u (s)) • i dyu(s) ds ds.Since y u (x) is an even function of x, we know that dyu(x) dx is an odd function of x.Additionally, F u (x + iy u (x)) is an even function of x, so x −x F u (s + iy u (s)) dyu (s)  ds ds = 0 and we may delete the extra term from the integrand: from which it follows that (10) This proves Theorem 11, as desired.This concludes Case I.
Case II: n+1 j=1 u j < 0. This case is similar to Case I.
Case III: n+1 j=1 u j = 0.In this case, we have that E u is the entire real line R. Then Equation (9) becomes for all x ̸ = 0.In particular, we have |F u (x + iy)| ≤ C/x 2 .For convenience, let M > 0 be large enough so that 1/u j ∈ (−M, M ) for all j, so that we have For any x > 0, we may define the path K x to be the line segment from x + iy u (x) to x.As before, we will use K x as a fringe to relate the two contours (γ u and the real axis).We obtain where the first term comes from the bound |F u (x + iy)| ≤ C/x 2 for |y| ≤ M + 1, and the other two terms come from the bound i.e. the integral along the fringe K x goes to zero, from which we conclude lim by Cauchy's Integral Theorem (this step is where we have shifted the contour from the real axis to γ u ).Arguing as in Case I, we note that x −x F u (s + iy u (s)) dyu(s) ds ds = 0, so we may delete the ds ds term from the integrand, resulting in lim which proves Theorem 11, as desired.This concludes Case III.□ We have proven that Theorem 11 holds in all cases.This allows us to change the integral in Theorem 9. We have successfully applied our main idea: changing the contour of integration.
It remains to show that 1 2π s∈Eu F u (s + iy u (s)) ds ≥ 1 e for each u with at least two nonzero entries.Defining Fu (x) := F u (x + iy u (x)), we wish to show the following: Theorem 12.For each unit vector u ∈ R n+1 with at least two nonzero entries, we have This is a special case of the following theorem: Theorem 13.For each unit vector u ∈ R n+1 (possibly of form (1, 0, 0, 0, . . ., 0)), we have Equality occurs if u is of form (1, 0, 0, 0, . . ., 0).
We will henceforth focus on proving Theorem 13.
In order to prove the equality case of Theorem 13, we wish to show that Fv (s) ds = 1 e .
We will use Cauchy's Residue Theorem.For x ∈ (0, π), let P x denote the path which is the line segment from −x + iy v (x) to x + iy v (x).We have Now that we know that v achieves the equality case in Theorem 13, we can prove Theorem 13 just by showing s∈Eu Fu (s) ds ≥ s∈Ev Fv (s) ds for each u.Note that for any u, we have E v = (−π, π) ⊆ E u .Thus we can write s∈Eu Fu (s) ds ≥ s∈Ev Fu (s) ds since the integrand is positive.It follows that it suffices to prove s∈Ev Fu (s) ds ≥ s∈Ev Fv (s) ds, which would be implied by the following pointwise bound: Theorem 14.For each unit vector u ∈ R n+1 , and for any x ∈ (−π, π), we have Fu (x) ≥ Fv (x).
The remainder of this section will be devoted to proving Theorem 14.We first note that we only have to prove Theorem 14 when x ∈ (0, π) (since Fu (0) = Fv (0) = 1, and since Fu (x), Fv (x) are both even functions of x).
From now on, we suppress the variable x when writing y u , so we write simply y u and y v instead of y u (x) and y v (x).Now we will give a differential equation that y u must satisfy.For all x ∈ D u , we have Φ u (x, y u ) = 0, which implies Differentiating with respect to x, we obtain where y ′ u denotes d dx y u .We can further simplify to Rearranging terms yields (13) which is our desired differential equation, valid for all x ∈ D u .In particular, by plugging in v for u, we obtain valid for all x ∈ (0, π).From Equation (13) and the fact that each u j ∈ [−1, 1], we obtain We must first prove an auxiliary lemma relating y u and y v : Lemma 15.For each x ∈ (−π, π), we have −y v ≤ y u ≤ y v .
Proof.It suffices to prove that y u ≤ y v for all x ∈ (0, π) (because we can obtain the other cases by replacing x with −x and by replacing u with −u).
Recall the Φ function from Equations ( 6) and ( 7).(As a reminder, Φ u : (0, ∞) × R → R is defined so that Φ u (x, y) always agrees with the argument of F u (x + iy) modulo 2π.)Note that for each u, we have that Φ u is a C ∞ function of (x, y) (in fact, it is a real analytic function).
If we plug in v for u in Equation (6), we obtain for all (x, y) ∈ (0, ∞) × R. Furthermore, we note that by the definition of y v , we have that Φ v vanishes along γ v (see Equation ( 8)).
Our general proof strategy will be to show that Φ v is nondecreasing along γ u .Combined with the fact that Φ v (x, y) is a strictly decreasing function of y, we can conclude that y u ≤ y v for all x ∈ (0, π), as desired.
We will have to take the derivative d dx (Φ v (x, y u (x))) and show that it is nonnegative.
We conclude that when x ∈ D u , the function Φ v (x, y u (x)) is a nondecreasing function of x.Since lim x↘0 y u (x) = 0, we obtain lim x↘0 Φ v (x, y u (x)) = 0, from which it follows that Φ v (x, y u (x)) ≥ 0 for all x ∈ D u .
As observed above, we have Φ v (x, y v (x)) = 0 for all x ∈ (0, π).So for all x ∈ (0, π).Since Φ v (x, y) is a strictly decreasing function of y, we must have which is exactly Equation (12), as desired.This proves Theorem 14.

A final question
We showed that H facet is the smallest central slice, but only up to a factor of 1 − o(1).
Question 16.Can we remove the factor of 1 − o(1), so that we can say that vol n−1 (H facet ∩ ∆ n ) is exactly the minimum value of a central section?

Figure 1 .
Figure 1.The gray triangle represents ∆ 2 , and the black triangle represents H(a).The thick line at the intersection of the two triangles represents the hyperplane section T (a).

Figure 2 .
Figure 2. A plot of F u (t) on the Argand plane when u = ( √ 0.42, √ 0.38, √ 0.20).The horizontal axis denotes the real part of t, and the vertical axis denotes the imaginary part of t.The brightness represents the modulus of F u (t), while the hue represents the argument of F u (t) (see the legends to the right of the plot).Note the poles along the imaginary axis.The contour γ u will be chosen such that F u is red along γ u .This image was produced using Wolfram Mathematica Student Edition.

Figure 3 .
Figure 3.For each j, we define α j := arccot 1 u j −y x if u j > 0, and we define β j :=