Progress on the Curvature Problem

Progress on the Curvature Problem David Futer April 7, 00 Abstract This writeup presents the known results about the curvature problem: the proof for the n = case, and the argument generalizing the R result to R n when w 1 1 The Conjecture The conjecture we are studying comes out of the Hutchings theory of component bounds on double bubbles in general dimensions The basic setup is illustrated in Figure 1 Note: All the figures and other related materials are available on the Web at http://wwwmathmsuedu/ dfuter/research/curvature/ Conjecture 11 In R n with n ), let H 0, H 1, and H, respectively, denote the mean curvature of a sphere of volume w, a sphere of volume w + 1, and the exterior of the second region of the standard double bubble of volumes 1, w Then H > H 0 + H 1 In studying this problem, it is convenient to divide the double bubble of volumes 1, w into four regions separated by pieces of the spheres We call these regions R 0, R 1, R, and R Thus R 0 R 1 is the bubble of volume w and R R is the bubble of volume 1 Regions R and R are of particular interest to us, so we call their volumes A and B, respectively We usually think think of volumes A and B = 1 A as functions A n w) and B n w), where n indicates dimension It turns out to be convenent to rephrase the problem in terms of w and A n w) rather than in terms of curvature To that end, let V n) be the volume of the n-dimensional unit ball This can be calculated in closed form using the Gamma function It is interesting to note, and occasionally relevant to our estimates, that V n) reaches a maximum at n = 5, and decreases steadily thereafter) Because mean curvature is just the reciprocal of the radius, we have 1

H 0 = V n)/w) 1/n H 1 = V n)/w + 1)) 1/n H = V n)/w + A n )) 1/n Thus, substituting this into the conjectured inequality and solving for A n, the conjecture is equivalent to the statement that A n w) < M n w) n ww + 1) w 1/n + w + 1) 1/n ) n w Thus we have an explicitly defined function M n w) expressing the bound that A n w) must satisfy for the conjecture to hold Before we proceed with the task of comparing the two functions, the following lemmas provide an appreciation of how similar they are Lemma 1 For each n, the function A n w) is strictly increasing in w As w 0, it shrinks to 0; as w, it is asymptotic to 1/ Proof: This proof is similar in spirit to Frank Morgan s proof that only one standard double bubble exists for every pair of volumes Let r n w) be the ratio between the radii of the spherical caps outside regions R 0 and R, respectively Now, r n w) must increase monotonically with w otherwise there will be two double bubbles with radii in the same proportion r n w) = r n w )) but disproportional volumes w/1 w /1) This would violate the scaling symmetry of R n Now, let us focus our attention on the the sphere S containing regions R 1, R, and R, and on one particular point where the spherical caps meet at 10 angles Consider what happens to S as we vary the other sphere but keep the intersection tied to that point As w increases, the angle is conserved but the ratio r n w) goes up so regions R 1 and R take up a bigger and bigger portion of this sphere See Figure ) So B n w), the volume of R, is decreasing in w, and thus A n w) = 1 B n w) is increasing As w 0, the bubble of volume w looks like a small lens on a vastly larger sphere See Figure ) Thus the sphere containing this bubble in addition to region R grows smaller and smaller forcing A n w), the volume of R to approach 0 In fact, the ratio A n w)/w approaches a constant more on this later) As w, the picture is reversed: now, the bubble of volume 1 is a lens on a much larger sphere Regions R and R are symmetric with respect to inversion in this sphere; thus, in the limit, they are reflections of one another in a flat plane and have equal volume So A n w) B n w) = 1 A n w), and thus A n w) 1/ Lemma 1 Let m < n Then M m w) < M n w) for all w Proof: For this proof, it is convenient to rewrite M n w) = ww + 1) w 1/n +w+1) 1/n ) n w

Thus the only difference as we vary n is in the averages in the denominator Now, since m < n, fx) = x n/m is a strictly convex function on [0, ) So, for all distinct x, y 0, ) n/m x + y < xn/m + y n/m Let x = w 1/n and y = w + 1) 1/n Then w 1/n + w + 1) 1/n ) n/m < w1/m + w + 1) 1/m, w 1/n + w + 1) 1/n )n w 1/m + w + 1) 1/m < and thus M m w) < M n w) )m, Lemma 14 M n w) has the same limits as A n w): it approaches 0 as w 0 and 1/ as w Proof: Observe that M n w) = w ) n w + 1 w 1/n + w + 1) 1/n ) 1, n and thus clearly vanishes as w 0 Now, to compute the limit of M n w) as w, we will need to squeeze it between two other functions We know, by Lemma 1, that M 1 w) < M n w) for n For an upper bound on M n w), recall that the geometric mean is always lower than the arithmetic Thus w 1/n w+1) 1/n < w1/n + w+1) 1/n Therefore we have, so w 1/n + w+1) 1/n ww+1) < )n ww + 1) M 1 w) = w+w+1) ww + 1) w + 1 w < M n w) < ww + 1) ww + 1) w w < M n w) < w + w w w + w w w w + 1 < M n w) < w + w + 1 4 w w w + 1 < M n w) < w + 1 w w w + 1 < M n w) < 1 Thus lim w M n w) = 1/

Lemma 15 For each n, the function M n w) is strictly increasing and concave down Proof: We need to check that M nw) > 0 and M nw) < 0 M nw) = n w1/n + w+1) 1/n ) n w+1) w 1/n + w+1) 1/n ) n 1 nww+1) nw1/n + w+1) 1/n ) n 1 1 n w1/n 1 + w+1) 1/n 1 ) w 1/n + w+1) 1/n ) n = n w1/n + w + 1) 1/n )w + w + 1)) w 1/n + w + 1) 1/n ) n+1 nww + 1)w1/n 1 + w + 1) 1/n 1 ) w 1/n + w + 1) 1/n ) n+1 1 = n w1+1/n + ww + 1) 1/n + w + 1)w 1/n + w + 1) 1+1/n w 1/n + w + 1) 1/n ) n+1 nw + 1)w1/n ww + 1) 1/n w 1/n + w + 1) 1/n ) n+1 1 = n w1+1/n + w + 1) 1+1/n w 1/n + w + 1) 1/n ) n+1 1 Now, lim M nw) = lim w w n w = n n+1 1 = 0 w+1 )1+1/n + 1 ) n+1 1 w w+1 )1/n + 1 Thus, once we establish that M w) is decreasing, it will follow that it s everywhere positive 4

M n w) = nw1/n + w+1) 1/n ) n+1 1 + 1 n )w1/n + w+1) 1/n ) w 1/n + w+1) 1/n ) n+ nw1+1/n +w+1) 1+1/n ) n+1)w 1/n +w+1) 1/n ) n 1 n w1/n 1 +w+1) 1/n 1 ) w 1/n + w+1) 1/n ) n+ = n 1 + 1 ) w 1/n + w+1) 1/n ) n w 1/n + w+1) 1/n ) n+ n 1 + 1 ) w 1+1/n + w+1) 1+1/n )w 1/n 1 + w+1) 1/n 1 ) n w 1/n + w+1) 1/n ) n+ = n 1 + 1 ) w 1/n + w+1) 1/n ) n w 1/n + w+1) 1/n ) n+ n 1 + 1 ) w 1/n w+1 + w+1)1/n w )w+1)w 1/n + ww+1) 1/n ) n w 1/n + w+1) 1/n ) n+ = n 1 + 1 ) w /n + w 1/n w+1) 1/n + w+1) /n n w 1/n + w+1) 1/n ) n+ ) w /n + w w+1 w1/n w+1) 1/n + w+1 w w1/n w+1) 1/n + w+1) /n n 1 + 1 n w 1/n + w+1) 1/n ) n+ = n 1 + 1 ) w 1/n w+1) 1/n w w+1 w1/n w+1) 1/n w+1 w w1/n w+1) 1/n n w 1/n + w+1) 1/n ) n+ = n 1 + 1 n = n 1 + 1 n = n 1 + 1 n < 0 ) w 1/n w+1) 1/n w w+1 w+1 w ) w 1/n + w+1) 1/n ) n+ ) w 1/n 1/n 1 w+1) ww+1) w 1/n + w+1) 1/n ) n+ ) w 1/n 1 w+1) 1/n 1 w 1/n + w+1) 1/n ) n+ Asymptotic analysis for small w Because the functions A n w) and M n w) are so similar in their asymptotic behavior, we need to analyze carefully the intervals when w is very small and very large In this section, we present the proof that A n w) < M n w) for sufficiently small w; the proof for large w in the case n = will come in the next section Lemma 1 M n w) n 1)w n n 1 w 1+1/n 5

Proof: Taylor s Theorem implies that M n w) = M n 0) + M n 0)w + 1 w 0 M n t)w t)dt From Lemmas 14 and 15, we have M n 0) = 0 and M n0) = n 1 Also, 1 + 1 n n 1 + 1 n M n w) = n ) w 1/n 1 w 1/n + w+1) 1/n ) n+ w+1) 1 1/n ) w 1/n 1, since w 1/n + w+1) 1/n ) n+ w+1) 1 1/n 1 for all w Thus 1 w 0 completing the proof M n t)w t)dt 1 w 0 = n 1 1 + 1 n = n 1n + 1 n n 1 + 1 ) t 1/n 1 w t)dt n ) w t 1/n 1 w t 1/n dt 0 [ nwt 1/n n ] w n + 1 t1+1/n 0 = n 1 n + 1)w 1+1/n w 1+1/n) = n n 1 w 1+1/n, Now that we have a lower bound on M n w), it would help to have an upper bound on A n w) This is obtained with the help of some geometry When w is very small, the bubble enclosing volume w looks like a lens on a nearly flat surface See Figure ) Because the boundary surfaces of a double bubble meet at 10 angles, each of the regions R 0 and R 1 looks like a truncated portion of an n-ball; specifically, the part of the ball when for unit radius) x n 1 Region R, of volume A, fills out the remainder of this n-ball This picture provides an upper bound on the ratio A/w: as w grows, the bubble of volume w fills out a greater and greater portion of the ball Thus, to get an upper bound on the ratio A/w, we need to calculate the volume of this lens in an n-ball and compare it to the volume of the remainder of the ball Lemma ) n 1 A n w) < n w Proof: Because we are calculating the ratio between the volumes of two different portions of a ball, we may suppose that the ball has unit radius We can 6

compute the n-dimensional volume Ln) of the lens by integrating by cylindrical shells See Figure 4) Each of these cylindrical shells is an interval times an n 1)-sphere, whose surface area n )-dimensional measure) for radius r is n 1)V n 1)r n Then / 1 Ln) = n 1)V n 1)r n r 1 ) dr 0 / 1 n 1)V n 1) r n 1 ) r dr = n 1)V n 1) 0 / 0 r n r n 1 dr ] / [ r n 1 = n 1)V n 1) n 1 r n n 0 [ = V n 1) r n 1 n 1 ] / r n n 0 ) n 1 ) n 1 = V n 1) n 1 n = V n 1) n ) n 1 A n w) w V n) Ln) Ln) < V n) Ln) < n ) n 1 n V n) V n 1) ) n 1 when n 6 When n 5, we can compute the constant cn) = V n) Ln))/Ln) explicitly from the integral expression for Ln): n π+ cn) 4π 15575 11 5 = 4π+9 4 8π 9 9499 0 5 5 80 7

Thus A n w) < n/ ) n 1 w for all n Theorem A n w) < M n w) whenever w n 1 ) ) n 1 n 1 n n 1 Proof: The proof is just a computation based on the past two lemmas w n 1 ) ) n 1 n 1 n n 1 w 1/n n 1 1 n n 1 ) n 1 n n 1 w 1/n n 1 n ) n 1 ) n 1 n n 1) n n 1 w 1/n ) n 1 n w n 1)w n n 1 w 1+1/n A n w) < M n w) The two-dimensional case Our proof of Conjecture 11 for n = comes in three pieces First, we know from Theorem that the conjecture holds for sufficiently small volumes Second, asymptotic analysis at the other end will prove that the conjecture holds for sufficiently large volumes And finally, the compact interval in between can be checked numerically, relying on the result Lemmas 1 and 15) that both M n and A n are increasing functions 1 Different parameters and explicit formulae Since infinity is such an unwieldy notion, it would be preferable to replace w with some other parameter that stays finite as w As it happens, not one but two alternate parameters are useful to our purposes here First, we can rescale the double bubble of volumes w and 1 so that instead it has volumes v and 1 v Now, as w varies from 0 to, v varies from 0 to 1 They are related by the equations v = w w + 1 and w = v 1 v 8

Now, we can express the bound M w) in terms of v: M wv) = M wv)) ) w + 1 = w 4 w + w + 1) 1 = = v 1 v v 1 v 1 1 v 4 v 1 v + 4 v + 1) 1 1 1 v ) ) 1 The other parameter, even more useful for this section, is the oriented) angle θ between the chord and the separating cap in Figure 1 When w < 1 and the separating cap bulges into the bubble of volume 1, we say that θ < 0; when w > 1 and the separating cap bulges the other way, we say θ > 0 Thus, as w varies from 0 to and v varies from 0 to 1), θ varies from π/ to π/ Because w, v and θ are all strictly increasing functions of one another, the monotonicity results of Lemmas 1 and 15 apply with any parameter) The advantage of θ as a parameter is that both A and v can be written explicitly in terms of it, allowing us to compare A and M directly The computations hinge on the geometrical formula in the following lemma Definition 1 For an angle θ, define a function ϕθ) by ϕθ) = θ sin θ cosθ sin θ Lemma Consider the sector S of a circle subtended by a chord of length c, where the chord meets the circle at internal angle θ See Figure 5) Then the area of S is aθ, c) = c θ sin θ cosθ ϕθ) = c sin θ Proof: Connect the intersection points of the chord and the circle to the center by a pair of radii; then the length of the radius is r = c/sin θ) See Figure 5) Now, S can be described as the wedge between the two radii, minus the triangle of the chord and the two radii In terms of r, the area of the wedge is r θ and the area of the triangle is r sin θ cosθ The area of S is the difference Lemma Consider a planar double bubble of areas w, 1, where the separating cap meets the chord at oriented angle θ, as above Then A θ) = ϕθ) + ϕ π θ) ϕ ϕθ) + ϕ π and vθ) = θ) + θ) ϕθ) ϕ π + θ) + ϕπ θ) π 9

Proof: We compute the areas using Figure 6 It is evident from the picture that, when θ > 0, A = aθ, c) + a π θ, c) and w = aπ + θ, c) aθ, c) When θ < 0, extending the formula of Lemma gives a negative expression for aθ, c), so in fact the same formulae for A and w still apply To expand them completely, we need an expression for c in terms of θ To that end, observe that 1 = A + B = aθ, c) + a π θ, c) = c ϕθ) + ϕ π θ) ) Thus so c = 1 ϕθ) + ϕ π θ), A = ϕθ) + ϕ π θ) ϕπ + θ) ϕθ) ϕθ) + ϕ π and w = θ) ϕθ) + ϕ π θ) From this, we can compute w + 1 = ϕπ + θ) ϕθ) ϕθ) + ϕπ ϕθ) + ϕ π + θ) ϕπ θ) ϕθ) + ϕ π = + θ) + ϕπ θ) θ) ϕθ) + ϕ π θ), and thus v = w w + 1 = + θ) ϕθ) ϕ π + θ) + ϕπ θ) ϕ π Asymptotic analysis for large w Our plan is to prove that A < M when θ is close to π by bounding their derivatives da dθ and dm dθ In order to do that, we need to know more about the building-block function ϕθ) Lemma 4 ϕθ) is positive, increasing, and concave up on 0, π) Specifically, ϕ θ) = sin θ θ cosθ sin θ 10

Proof: Lemma tells us that ϕθ) is the area of a sector Sθ) of a circle of radius 1/ sinθ cut by a chord of length See Figure 7) So ϕθ) is clearly positive The chord length is fixed, so as θ increases, the sector Sθ) will grow larger Thus ϕθ) is increasing Now consider the change in area between Sθ) and Sθ+ǫ) for some small ǫ The difference between the two sectors is a narrow strip along the circumference Now, keep ǫ fixed and vary θ For larger θ, the strip is both longer and wider, and thus has larger area So ϕ θ) is increasing and ϕθ) is concave up ϕ θ) = sin θ1 + sin θ cos θ) θ sin θ cosθ) sin θ cosθ) sin 4 θ = sin θ sin θ) θ sin θ cosθ + sin θ cos θ sin 4 θ = sin θ θ sinθ cosθ sin 4 θ sin θ θ cosθ = sin θ For computing derivatives of A and M we introduce the following notation: α = π θ, β = π θ, and γ = π + θ With this notaton, we have A θ) = ϕθ) + ϕα) ϕθ) + ϕβ) Lemma 5 When θ 1, π da ), dθ > 1 6 Proof: Let θ 1, π ) Then and vθ) = ϕγ) ϕθ) ϕγ) + ϕβ) da dθ = ϕθ) + ϕβ))ϕ θ) ϕ α)) ϕθ) + ϕα))ϕ θ) ϕ β)) ϕθ) + ϕβ)) = ϕθ) + ϕβ))ϕ θ) ϕ α)) + ϕθ) + ϕα))ϕ β) ϕ θ)) ϕθ) + ϕβ)) > ϕθ) + ϕβ))ϕ θ) ϕ α)) ϕθ) + ϕβ)), since ϕ β) > ϕ θ) > ϕθ)ϕ θ) ϕ α)) ϕβ)), since ϕβ) > ϕθ) > ϕ1)ϕ 1) ϕ π 1)) ϕ π 1) 01749 Lemma 6 When θ 1, π dv ), 0 < dθ < 1 0 11

Proof: Let θ 1, π dv ) Since v is increasing, dθ > 0 Also, dv dθ = ϕγ) + ϕβ))ϕ γ) ϕ θ)) ϕγ) + ϕθ))ϕ γ) ϕ β)) ϕγ) + ϕβ)) = ϕγ)ϕ γ) ϕγ)ϕ θ) + ϕβ)ϕ γ) ϕβ)ϕ θ) ϕγ) + ϕβ)) + ϕγ)ϕ γ) + ϕγ)ϕ β) + ϕθ)ϕ γ) ϕθ)ϕ β) ϕγ) + ϕβ)) = ϕγ)ϕ θ) + ϕβ)ϕ γ) ϕβ)ϕ θ) + ϕγ)ϕ β) + ϕθ)ϕ γ) ϕθ)ϕ β) ϕγ) + ϕβ)) < ϕβ)ϕ γ) + ϕγ)ϕ β) + ϕθ)ϕ γ) ϕγ) + ϕβ)) < ϕβ)ϕ γ) + ϕγ)ϕ β) ϕγ) = ϕ γ) ϕγ) ϕβ) + ϕ β) ϕγ) ) π < ϕ γ) ϕγ) ϕ 1 + ϕ π 1) ϕγ) We now substitute ϕ π 1) 08698 and ϕ π 1) 0667 dv dθ < 18 ϕ γ) ϕγ) + 1 ϕγ) = 18 = 18 sin γ γ cosγ sin γ sin 4 γ γ sin γ cosγ) + sin γ γ sin γ cosγ sin γsin γ γ cosγ) sin γ γ sinγ cosγ) + γ sin γ cosγ < 1 5 sin γ)sin γ γ cosγ) + 1 sin γ, since γ sin γ cosγ > < π 5 sinγ + 1 sin γ, since sin γ γ cosγ < π < π 5 sin π + 1 ) + 1 sin π + 1 ) 0004 Lemma 7 When θ 1, π dm ), 0 < dv < 8 Proof: Recall that M w) = v 1 v 41 + v) 1) Thus 1

dm dv ) 1 v) v 1) 4 = 1 v) 1 + v) 1 + v 1 v 1 = 1 v) 4 1 + v) v 1 + v) 4 1 v 1 1 + v) = + 1 + v)) 1 + v)) 1 v) 1 + 4 v v) 4 1 v)1 + v) 4 + v = 1 v)1 + v) 4 v 4 1 v)1 + v) 4 = = v 1 v)1 + v) 4 1 + v) 4 8 1 + v) ) 1 v Now, when θ > 1, certainly w > 1, and thus v > 1/ Therefore dm dv < 1 + 1 ) 4 05 Theorem 8 When θ > 1, A < M Proof: By the Fundamental Theorem of Calculus, A θ) = A π M wθ) = M w ) π θ π ) da dλ dλ, π θ dm dλ dλ But by Lemmas 1 and 14, A π ) = M w π ) = 1 Also, by the lemmas in this section, dm dλ = dm dv dv dλ < 8 1 0 < 1 6 < da dλ when θ 1, π ) Thus A < M on this interval Proof of the conjecture Theorem 9 Conjecture 11 is true when n = 1

Proof: By Theorem, A w) < M w) when w 1 1 1 ) = 4 1 ) 00981 Using the formula for wθ) from Lemma, we can check that w < 00 when θ < 09 Also, Theorem 8 tells us that A < M when θ > 1 Thus the only remaining task is to check the conjecture when θ [ 09, 1] This is quite easy to do numerically Lemmas 1 and 15 tell us that both functions are increasing, so the graph in Figure 8 is in fact rigorous Alternately, one can find a collection of angles 09 = θ 0 < θ 1 < < θ k = 1 such that A θ i ) < M wθ i 1 ) One such collection is given below θ i Aθ i ) M wθ i ) θ 0 = 09 0016 0059 θ 1 = 084 0045 0046 θ = 075 00441 0084 θ = 06 00817 01484 θ 4 = 04 0156 049 θ 5 = 005 099 0584 θ 6 = 04 0446 04556 θ 7 = 08 04414 0499 θ 8 = 1 04888 04998 4 Centers of mass and inversion in spheres 41 Generalized centers of mass Now that we have proved Conjecture 11 for n =, we must turn our attention to higher dimensions Instead of approaching the general problem from scratch, it would be much easier to somehow make use of what we already know In order to pass from the planar picture to the higher-dimensional one, it becomes important to know the relationship between the volumes of corresponding regions in different dimensions Let G n R n n ) be a measurable set invariant under any rotation preserving the x 1 -axis; that is, a region of revolution We can think of G n as being generated by a subset G of the upper half-plane: G n is what we get when we revolve G about the x-axis More precisely, each point x, y) G corresponds to an n )-sphere of radius y in the n 1)-dimensional crosssection of G n whose first coordinate is x See Figure 9) The surface area n )-dimensional measure) of such a sphere is n 1)V n 1)y n, in terms 14

of the already familiar constant V n 1) describing the volume of the unit n 1 ball Now, this setup allows us to compute the volume of G n : VolG n ) = n 1)V n 1)y n da G When n =, the above setup brings to mind Pappus Theorem: VolG ) = π y AreaG ), where y is the average distance of G from the x-axis, ie the y-coordinate of its center of mass One way to compute the center of mass of a region is with the same integral that we have above: G y = y da AreaG ) This suggests a way to generalize the notion of center of mass to correspond to volumes of higher-dimensional regions of revolution Definition 41 Let G be a bounded, measurable subset of the upper half-plane, and let n We define the n-dimensional center of mass c n G) by G c n G) = yn da AreaG) Our computations above imply Lemma 4 Let G n R n be a bounded, measurable set invariant under rotations about the x 1 -axis Let G be its generating region in the upper half-plane Then VolG n ) = n 1)V n 1)c n G )AreaG ) The following two lemmas are also immediate consequences of the definition of c n G) Lemma 4 Let G and H be disjoint, bounded, measurable subsets of the upper half-plane Then the n-dimensional center of mass of their union is a weighted average of their centers of mass, the weights being the respective areas In other words, c n G H) = AreaG)c ng) + AreaH)c n H) AreaG H) Proof: Obvious from the definition Lemma 44 Let G be a bounded, measurable subset of the upper half-plane, and let λg be the image of G under scaling by some factor λ > 0 Then c n λg) = λ n c n G) 15

Proof: Let G n R n be the region obtained by revolving G around the x-axis Scaling by λ increases the area of G by a factor of λ and the volume of G n by a factor of λ n The result now follows from Lemma 4 4 Centers of mass of circular sectors In our problem, we are specifically concerned with regions bounded between circles in R or spheres in R n The following regions turn out to be fundamental building blocks of double bubbles Definition 45 Consider a circle whose center lies on the x-axis, and a vertical chord of length through the circle Let θ be the internal angle between the chord and the circular arc to the right of the chord See Figure 10) Let Gθ) be the region contained to the right of the chord, inside the circle, and above the x-axis For any θ 0, π), this construction defines Gθ) uniquely up to horizontal translation) For each n, define a function f n on 0, π) by f n θ) = c n Gθ)) Lemma 46 For each n, f n θ) is strictly increasing in θ Proof: In order to make it easier to compute an explicit formula for f n θ), let us rescale the picture so that Gθ) is a sector of the unit circle Since the radius of the original circle is 1/ sinθ), we need to scale by a factor λ = sinθ) After scaling, λgθ) is the subset of the unit circle above the x-axis and to the right of the line x = cosθ) By Lemma, the area of this region is Now, AreaλGθ)) = 1 sin θ ϕθ) = θ sin θ cosθ 1 λgθ) f n θ) = da λ n AreaλGθ)) = 1 1 x cos θ 0 y n dy dx sin n θθ sinθ cosθ) = 1 1 cos θ n 1 1 x n 1 dx = = sin n θθ sin θ cosθ) n 1 n 1 0 θ sinn 1 u sinu)du sin n θθ sin θ cosθ) θ 0 sinn u du sin n θθ sin θ cosθ) To write the derivative of f n θ) explicitly, without any integrals, one needs to use the messy sum formula for θ 0 sinn u du Instead of doing this, we resort to a trick Define a function g n θ) by 16

sin n+1 θ)θ sin θ cosθ) g n θ) = θn )cosθ + n sin θ n )sinθ It is easy to check that g n θ) is related to f nθ) by f n θ) = g nθ) θ 0 sin n u)du n 1 θn )cosθ + n sin θ n )sinθ sin n 1 θ)θ sin θ cosθ) We shall prove that f nθ) > 0 by considering the sign of g n θ) and of the expression multiplying it In the expression above, the denominator is always positive However, the numerator θ n )cosθ + n sin θ n ) sinθ changes sign once on 0, π): it is positive on 0, b n ) and negative on b n, π) for some bound b n > π/ On b n, π), g n θ) is negative and the factor multiplying it is also negative, making f nθ) clearly positive At θ = b n, when this expression is 0, f nb n ) is easily seen to be positive by computing it directly Thus the fact that g n θ) has a discontinuity at b n does not concern us) On 0, b n ), we need to prove that g n θ) > 0; this takes some work For starters, lim g nθ) = lim θ 0 θ 0 = lim θ 0 = 0, sin n+1 θ)θ sinθ cosθ) θ n )cosθ + n sin θ n ) sin θ sin n+1 θ)θ sin θ cosθ) θ n )cosθ + n sin θ n ) sin θ θ 0 sin n u)du because the numerator has a zero of order at least n+ 5 at 0, and the denominator has a zero of order To complete the proof, we show that g n θ) is strictly increasing for as long as it s continuous; then it will have to be positive for θ 0, b n ) ) g nθ) = θn )cosθ + n sin θ n )sinθ)θn+1)sin n θ cosθ + n+)sin n+ θ n+1)sin n+1 θ) θ n )cosθ + n sin θ n ) sin θ) sinn+1 θ)θ sin θ cosθ)n sin θ cosθ θ n )sin θ) θ n )cosθ + n sin θ n ) sin θ) sin n θ Since we are only concerned with the sign of g n θ), we may multiply through by the denominators and factor out sin n θ This gives us 17

h n θ) = θn )cosθ + n sin θ n )sinθ)θn+1)cosθ + n+)sin θ n+1)sinθ) θ sin θ cosθ)n sin θ cosθ θ n )sin θ) θ n )cosθ + n sin θ n ) sinθ) = θn )cosθ + n sin θ n )sinθ)θ cosθ + sin θ sinθ) θ sin θ cosθ)n sin θ cosθ θ n )sin θ) = n 6)θ cos θ + n sin 6 θ + n 6)sin θ + 6n 6)θ sin θ cosθ 6n 1)θ sin θ cosθ 6n 6)sin 4 θ nθ sin θ cosθ + n )θ sin θ + n sin 4 θ cos θ n )θ sin θ cosθ = n 6)θ cos θ + n )θ sin θ + n 4)θ sin θ cosθ 6n 1)θ sin θ cosθ + n 6)sin θ + n sin 6 θ + n sin 4 θ cos θ 6n 6)sin 4 θ = n )θ + n 4)θ cos θ + n 4)θ sin θ cosθ sin θ 1) 4n 8)θ sinθ cosθ + n 6)sin θ + n sin 4 θ 6n 6)sin 4 θ = n )θ n )θ sin θ cosθ + n 4)θ cos θ + n 4)θ sin θ cos θ n 6)θ sinθ cosθ + n 6)sin θ n 6)sin 4 θ = n )θ θ sin θ cosθ) + n )θ cos θ θ sin θ cos θ) n )θ sin θ cosθ sin θ cos θ) = n )θ sin θ cosθ)θ + θ cos θ sinθ cosθ) Now, n )θ sin θ cosθ) > 0 whenever θ > 0 and n, so this term is not a concern To show that kθ) = θ + θ cos θ sinθ cosθ) > 0, observe that this term equals zero at 0 We will show that it increases on 0, π), guaranteeing that it will be positive k θ) = 1 + cos θ 4θ sin θ cosθ cos θ + sin θ = 4 sin θ 4θ sin θ cosθ = 4 sin θ sin θ θ cosθ) > 0 when θ 0, π) Thus h n θ) > 0 for n, implying that g n θ) is increasing and therefore positive This completes the proof 4 Back to double bubbles: inversion in spheres Now we have the tools to prove the result that will be useful in generalizing our results about two-dimensional bubbles to higher dimensions As a motivation for the following theorem, notice that the bubble of volume 1 is the image, under inversion in the separating cap, of the bubble of volume w Regions R 0 and R 1 18

are also symmetric with respect to inversion in a sphere, as are regions R and R Theorem 47 Let D R be a closed disk, centered on the positive x-axis, and not containing the origin Let G be the intersection of the unit disk with the upper half of D Let H be the image of G under inversion in the unit circle Then, for each n, c n G) < c n H) Proof: Suppose first that D is entirely contained in the unit disk Then G is just a half-disk, and its image H is also a half-disk, congruent to λg for some scaling factor λ > 1 Then c n G) < c n H) by Lemma 44 If, on the other hand, D is not contained in the unit disk, we can draw a vertical chord connecting the two points where D intersects the unit circle This chord divides G into two regions, R and S See Figure 11) Let α be the internal angle between the vertical chord and the unit circle, and β the angle between the vertical chord and D Because inversion is conformal, the internal angle of H at the same vertex is α + β The length of the vertical segment between this vertex and the x-axis is λ = sinα), so we see that R, S, and S H are congruent to λgβ), λgα), and λgα + β), respectively By Lemma 46, f n α+β) > f n α), so c n S H) > c n S) Since by Lemma 4) c n S H) is a weighted average of c n S) and c n H), this implies that c n S) < c n S H) < c n H) Applying Lemma 46 again, we see that f n β) < f n α + β), and thus c n R) < c n S H) < c n H) Therefore, since both c n R) < c n H) and c n S) < c n H), we can conclude again by taking weighted averages) that c n G) = c n R S) < c n H) Corollary 48 Let G n R n be the intersection of two closed n-balls B 1 and B, not containing the center of B 1 Let H n the the image of G n under inversion in the boundary of B 1 Let G and H be the generating regions for G n and H n, respectively, in any half-plane through the centers of B 1 and B See Figure 1) Then AreaH ) AreaG ) < VolH n) VolG n ) Proof: Translating, rotating, and scaling the whole picture will not change the ratios of areas and volumes Thus we may assume that B 1 is the unit ball, and that the center of B lies on the positive x 1 -axis Thus c n G ) < c n H ), and the result follows by Lemma 4 19

Given these results, it is natural to ask if Theorem 47 applies in a more general context than regions bounded between circles It turns out that if we let G be any measurable subset of the upper half-disk whose closure excludes the origin, and H its image under inversion, then it no longer follows that c n G) < c n H) Counterexamples exist where G is the union of two half-disks centered on the x-axis For one specific numerical example, let G consist of the upper halves of B07, 000) and B098, 001) Then we have c G) 0009988 c H) 0009158, violating the desired inequality Other counterexamples can be constructed that violate the inequality for c and also for c n with higher values of n However, in all of the counterexamples known to us, G is either a disjoint union of several pieces, or else a region whose parts are connected by very narrow corridors These examples lead us to conjecture that the key sufficient condition is convexity Conjecture 49 Let G be a closed, convex subset of the upper half of the unit disk, that does not contain the origin Let H be the image of G under inversion in the unit circle Then c n G) < c n H) 5 Generalizing from R to R n To show that Conjecture 11 for n = implies the conjecure for n >, we have to consider the cases w 1 and w < 1 separately In the former case, we have a proof using Lemma 1 and Corollary 48; but in the latter case much remains unknown 51 Case 1: w 1 Theorem 51 Conjecture 11 holds for all n when w 1 Proof: Theorem 9 states the result for n =, and we know by Lemma 1) that M n w) > M w) when n > Thus if we can prove that A n w) < A w) when n >, we will have the result for general n This can be shown when w 1 Consider our problem in R, and suppose w > 1 Then the bubble of area 1 can be obtained from the bubble of area w by inversion in the separating cap, with the latter bubble on the outside Call the regions occupied by these two bubbles G and H, respectively Now, consider the double bubble in R n where the three spherical caps have the same radii as the corresponding caps in R, and let G n and H n be the regions that correspond to G and H, respectively Then, by Corollary 48, w 1 < VolH n) VolG n ) 0

Now, scaling the bubbles in R n to make the small bubble have volume 1 will give the big bubble volume w = VolH n )/VolG n ) Thus we know that w < w It is useful to note at this point that if w = 1, then the two bubbles are symmetric with respect to a plane, so w = w = 1 in this case In general, for w 1, we have w w ) Recall also that the bubble G of area 1 is composed of regions R and R, symmetric under inversion in a circle, with R inside the circle Similarly, G n is composed of regions R and R Corollary 48 tells us that B A = VolR ) VolR ) < VolR ) VolR ) = B n A n As usual, we write A = A w) Furthermore, because R and R make up the small bubble that has volume 1 when the big bubble has volume w, we also have A n = A n w ) Thus 1 + B w) A w) A w) + B w) A w) 1 A w) < 1 + B nw ) A n w ) < A nw ) + B n w ) A n w ) 1 < A n w ) A n w ) < A w) But since A n is an increasing function, we have A n w) A n w ) < A w) 5 Case : w < 1 At the moment, very little can be said conclusively about this case Of course, the asymptotic result of Theorem applies in general dimension, and thus the conjecture only needs to be checked in the interval ǫ n, 1) for each n Following the example of section, we could derive explicit formulae for A n θ) and w n θ), using integrals to express volume Then the conjecture could be verified numerically for any desired dimension So if the truth of Conjecture 11 with some specific n is needed for some particular application, this method can furnish the result However, the formulae would get harder and harder to compute for large n, and in any case, one cannot numerically check infinitely many dimensions I am aware of two other approaches that could potentially be fruitful The first is to prove an estimate of just how much M n w) increases with n for small w; the increase seems to become more dramatic the smaller w gets Then, even though A n w) also increases with n, one can hope to prove that it increases less This seems very difficult to carry out 1

The second approach is to somehow use the so far unutilized) symmetry of the double bubble That is, region R of volume A n w)) in the w, 1) double bubble is a scaled copy of region R 1 in the 1 w, 1) double bubble If we rescale the problem to deal with double bubbles of volumes v, 1 v) instead of w, 1), then region R in the v, 1 v) double bubble is an exact copy of region R 1 in the 1 v, v) double bubble Perhaps this symmetry could allow us to use the result about large volumes to prove the corresponding result about small volumes This seems to be the most promising line of thought, although it too has not been easy so far