A note on “The Cartan-Hadamard conjecture and the Little Prince” S. Michalakis Institute for Quantum Information and Matter, Caltech, Pasadena, CA 91125 WeprovideelementaryproofsofLemmas7.1and7.4appearingin”TheCartan-Hadamardconjec- tureandtheLittlePrince”,byB.KloecknerandG.Kuperberg. TheLemmasplayanimportantrole in the derivation of novel isoperimetric inequalities. The original proofs relied on Sage, a symbolic algebra package, to factor certain algebraic varieties intoirreducible components. I. THE TWO LEMMAS 7 1 0 The followinglemmas canbe found in[1]as Lemmas7.1and7.4. Elementaryversionsoftheir proofsfollow below. 2 Lemma 1 (Lemma 7.1). Show that for x,y ≥0 and θ ∈[0,π], n a J (sinθ)3xy+ (cosθ)3−3cosθ+2 (x+y)−(sinθ)3−6sinθ−6θ+6π−6arctan(x)+ 2x −6arctan(y)+ 2y ≥0, 1 1+x2 1+y2 (cid:0) (cid:1) 3 (I.1) with equality at x=y =cot(θ/2). ] G Proof. We begin by setting D 2x h. G(θ,x,y)=(sinθ)3xy+ (cosθ)3−3cosθ+2 (x+y)−(sinθ)3−6sinθ−6θ+6π−6arctan(x)+ 1+x2 t (cid:0) (cid:1) a 2y m −6arctan(y)+ . (I.2) 1+y2 [ In the proof of Lemma 7.1 in [1], the authors begin by showing that the negative values of G(θ,x,y) are confined in 1 a compact subset of [0,2π]×R2. Hence, we can use the derivative and boundary-value test to prove positivity. v 2x 2y 7 It is easy to see that G(0,x,y) = 6π−6arctan(x)+ −6arctan(y)+ ≥ 0, with equality only when 7 1+x2 1+y2 0 x=y =cot(0)=∞, so we may assume that θ >0. Noting that the following identity holds (by differentiating both 0 sides): 0 2. 4x−6arctan(x)+2x/(1+x2)=4 x s4 ds, 0 Z (1+s2)2 0 7 1 (and similarly for y), we see that the condition G(θ,x,y)≥0, is equivalent to: : v i x s4 y s4 X (sinθ)3xy+((cosθ)3−3cosθ−2)(x+y)−(sinθ)3−6sinθ−6θ+6π+4 ds+4 ds≥0. (I.3) Z (1+s2)2 Z (1+s2)2 r 0 0 a Setting the partial derivatives of G(θ,x,y) to zero, yields in turn: ∂ G(θ,x,y)=0 =⇒ sin2(θ)(cos(θ)xy+sin(θ)(x+y))=2(1+cos(θ))+sin2(θ)cos(θ), (I.4) θ ∂ G(θ,x,y)=0 =⇒ sin3(θ)y+4x4/(1+x2)2 =(1+cosθ)2+sin2θ(1+cosθ), (I.5) x ∂ G(θ,x,y)=0 =⇒ sin3(θ)x+4y4/(1+y2)2 =(1+cosθ)2+sin2θ(1+cosθ), (I.6) y where we used (cosθ)3−3cosθ−2=−(1+cosθ)2(2−cosθ)=−(1+cosθ)2−sin2θ(1+cosθ), to get the last two equations. Now, noting that 2(1 + cos(θ)) + sin2(θ)cos(θ) = sin2(θ) cosθcot2(θ/2)+2sinθcot(θ/2) , where we used, cot(θ/2)=(1+cosθ)/sin(θ), we get from Eqn (I.4): (cid:0) (cid:1) ∂ G(θ,x,y)=0 =⇒ sin2(θ) cos(θ)(xy−cot2(θ/2))+sin(θ)(x+y−2cot(θ/2)) =0. θ (cid:2) (cid:3) Hence, θ =π, or cos(θ)(cot2(θ/2)−xy)=sin(θ)(x+y−2cot(θ/2)), θ ∈(0,π). (I.7) 2 A quick check shows that: G(π,x,y) = 4 x s4 ds + 4 y s4 ds ≥ 0, since x,y ≥ 0, with equality when 0 (1+s2)2 0 (1+s2)2 x=y =cot(π/2)=0. To treat the remaininRg case, we makeRthe following substitutions: x=αcot(θ/2), y =βcot(θ/2). (I.8) Substituting the above in equation (I.7), and using cot(θ/2)=(1+cosθ)/sinθ, we get: (1−αβ)cosθ =((α−1)+(β−1))(1−cosθ), which can be further simplified to: (1−α)(1−β)cosθ =(1−α)+(1−β). Now, if α = 1, then β = 1 (and vice versa), and equation (I.8) implies that x = y = cot(θ/2), which is what we set out to prove. So, we may assume that (1−α)(1−β)6=0 and that: cosθ =(1−α)−1+(1−β)−1, θ ∈(0,π). The rest of the proof verifies that the above conditions on α,β and θ are incompatible with (I.5) and (I.6), so that the only global minimum is attained at x=y =cot(θ/2), at which point G(θ,x,y)=0. We begin with (I.5). Using equation (I.8) and the trigonometric identities cot(θ/2) = (1 + cosθ)/sinθ and cot2(θ/2)=(1+cosθ)/(1−cosθ), we see that ∂ G(θ,x,y)=0, is equivalent to the following equality: x 4α4 = (α2−1)cosθ+α2+1 2(1+(1−β)(1−cosθ)). (I.9) (cid:0) (cid:1) Substituting the expression we derived above for cosθ, we get: 4α4(1−β)=α(α−1)(1+α+α(1−β))2, (I.10) whereweused1+(1−β)(1−cosθ)=α(1−β)(α−1)−1and(α2−1)cosθ+α2+1=(α+1)((α−1)cosθ+1)+α(α−1)= (α−1)(1−β)−1(2α+1−αβ). If α = 0, then equation (I.10) with α and β exchanged (derived by taking the partial w.r.t. y instead of x, and noting that G(θ,x,y) = G(θ,y,x)) yields 4β4 = β(β −1)(1+2β)2, which is equivalent to β(3β +1) = 0. Since β ≥0, we have that α =0 implies β =0, which leads to the contradiction cosθ = 2 (note that x =y = 0 minimizes G(θ,x,y) when θ = π, which was treated separately above). By symmetry, we also know that if β = 0, then α = 0, since otherwise 3α+1=0, which is a contradiction. Hence, we can assume that α,β >0. Dividing through by α and re-writing 4α3(1−β)=4α(1+α)(α−1)(1−β)+4α(1−β), equation (I.10) yields: 4α(1−β)=(α−1)(1+αβ)2, (I.11) which may be further simplified to: 4α+(1−αβ)2 =α(1+αβ)2. Moreover,by exchanging α and β (due to the symmetry of G discussed above), we also have the equation: 4β+(1−αβ)2 =β(1+αβ)2. Subtracting the two equations yields: 4(α−β)=(α−β)(1+αβ)2, which implies that α=β and/or αβ =1. If α=β, the (I.11) becomes 4α(1−α)=(α−1)(1+α2)2, which immediately implies α = 1 (since the left and right side of the equality have opposite signs otherwise). This contradicts our earlier assumption that (1−α)(1−β) 6= 0. If αβ = 1, then cosθ = (1−α)−1 +(1−1/α)−1 = 1, which contradicts the assumption that θ ∈(0,π). This completes the proof. 3 Lemma 2 (Lemma 7.4). Let 1 2x F(ℓ,x,y)=sinh(ℓ)3xy− cosh(ℓ)3−3cosh(ℓ)+2 (x+y)+sinh(ℓ)3−6sinh(ℓ)−6ℓ+6arctanh + (cid:18)x(cid:19) x2−1 (cid:0) (cid:1) 1 2y +6arctanh + . (I.12) (cid:18)y(cid:19) y2−1 The function F(ℓ,x,y) is non-negative for ℓ≥0 and x,y ≥1, vanishing only when x=y =coth(ℓ/2). Proof. TheprooffollowsthestepsoftheargumentgivenaboveforLemma1almostidentically(theargumentconfining thenegativevaluesofF inacompactsetcanbefoundinthebeginningoftheproofofLemma7.4in[1]). Inparticular, applying similar reasoning as before and using coth(ℓ)=(1+coshℓ)/sinhℓ, one can show that: ∂ F(ℓ,x,y)=0 =⇒ sinh2(ℓ) cosh(ℓ)(xy−coth2(ℓ))−sinh(ℓ)(x+y−2coth(ℓ)) =0, (I.13) ℓ ∂ F(ℓ,x,y)=0 =⇒ sinh3(ℓ)y(cid:2)−4x4/(x2−1)2 =−(1+coshℓ)2+sinh2ℓ(1+cos(cid:3)hℓ), (I.14) x and similarly for ∂ F(ℓ,x,y), since F(ℓ,x,y) is symmetric in x and y. It is trivial to check that F(0,x,y)≥ 0, with y equalityonlywhenx=y =coth(0)=∞. Hence,wewillassumethatℓ>0fromnowon. Settingx=αcoth(ℓ/2),y = βcoth(ℓ/2) and using cosh2(ℓ)−sinh2(ℓ)=1, we get: ∂ F(ℓ,x,y)=0 =⇒ (1−αβ)cosh(ℓ)=((α−1)+(β−1))(1−coshℓ), (I.15) ℓ ∂ F(ℓ,x,y)=0 =⇒ 4α4 = (α2−1)coshℓ+α2+1 2(1+(1−β)(1−coshℓ)), (I.16) x ∂ F(ℓ,x,y)=0 =⇒ 4β4 = (cid:0)(β2−1)coshℓ+β2+1(cid:1)2(1+(1−α)(1−coshℓ)). (I.17) y (cid:0) (cid:1) Noting that the above conditions for extrema of F(ℓ,x,y) are identical to the ones we derived for G(θ,x,y) in the proofof Lemma 1 (if we swapcosθ with coshℓ), we see that the only values of α and β satisfying allthree conditions are α = β = 1 (one still needs to check that α = β = 0 is not allowed, which follows from the requirement that x,y ≥1). Hence, the global minimum of F(ℓ,x,y) is attained at x=y =cot(ℓ/2), at which point it is each to check that F(ℓ,x,y)=0. ACKNOWLEDGMENTS S.MichalakiswouldliketoacknowledgesupportprovidedbytheInstituteforQuantumInformationandMatter,an NSFPhysicsFrontiersCenterwithsupportofthe GordonandBetty MooreFoundationthroughGrant#GBMF1250 and by the AFOSR Grant #FA8750-12-2-0308. [1] B. Kloecknerand G. Kuperberg, “The Cartan-Hadamard conjecture and The Little Prince”, arXiv:1303.3115 [math.DG] (2014)