Equality cases as diagnostics for all comparison signs

Elementary scalar estimates. [F3] For $x>0$ one has $\sin x<x$: the function $x-\sin x$ vanishes at $0$ and has derivative $1-\cos x\ge 0$, which is positive except at multiples of $2\pi$, so it is strictly increasing on $[0,2\pi ]$ and then visibly positive; for $x>0$ one has $\sinh x>x$, since $\sinh x-x$ vanishes at $0$ and has derivative $\cosh x-1\ge 0$, positive for $x>0$. For $0<x<\pi$ one has $\cot x<1\mathrm{/}x$: for $0<x\le \pi \mathrm{/}2$ the function $\sin x-x\cos x$ vanishes at $0$ and has derivative $x\sin x>0$, so $\sin x>x\cos x$ and $\cot x=\cos x\mathrm{/}\sin x<1\mathrm{/}x$; for $\pi \mathrm{/}2<x<\pi$ one has $\cot x<0<1\mathrm{/}x$. For $y>0$ one has $\coth y>1\mathrm{/}y$: $y\cosh y-\sinh y$ vanishes at $0$ and has derivative $y\sinh y>0$, so $y\cosh y>\sinh y$ and $\coth y=\cosh y\mathrm{/}\sinh y>1\mathrm{/}y$. Also $\cos x\le 1$ and $\cosh y\ge 1$, with the latter inequality strict for $y>0$.

The model Jacobi field is strictly shorter at larger curvature. [F1, F2] For fixed $t>0$ put $m_k(t):={\int }_0^{t}x{\mathrm{sn}}_k(x)dx\mathrm{.}$ Since ${\mathrm{sn}}_k>0$ on $(0,D_k)$, where $D_k:=\pi \mathrm{/}\sqrt{k}$ for $k>0$ and $D_k:=+\mathrm{\infty }$ for $k\le 0$, one has $m_k(t)>0$ for $0<t<D_k$. From the explicit formulas, for $k\mathrm{\ne }0$, ${\mathrm{\partial }}_k{\mathrm{sn}}_k(t)=\frac{t{\mathrm{cs}}_k(t)-{\mathrm{sn}}_k(t)}{2k}=-\frac{m_k(t)}{2},$ the first equality following by differentiating the displayed quotient $\sin (\sqrt{k}t)\mathrm{/}\sqrt{k}$ (respectively $\sinh (\sqrt{-k}t)\mathrm{/}\sqrt{-k}$) in $k$, and the second from $t{\mathrm{cs}}_k-{\mathrm{sn}}_k=(t{\mathrm{cs}}_k-{\mathrm{sn}}_k)(0)+{\int }_0^t(-kx{\mathrm{sn}}_k(x))dx=-k{m}_k(t)$, which uses $({\mathrm{cs}}_k{)}^{\mathrm{\prime }}=-k{\mathrm{sn}}_k$ [F2]. At $k=0$, [F11] gives ${\mathrm{sn}}_k(t)=t-k{t}^3\mathrm{/}6+o(k)$ from either side, so ${\mathrm{\partial }}_k{\mathrm{sn}}_k(t){\mathrm{\mid }}_{k=0}=-t^3\mathrm{/}6=-m_0(t)\mathrm{/}2$. The derivative formula extends continuously across $k=0$ since the explicit model functions converge uniformly on the finite interval $[0,t]$. Hence ${\mathrm{\partial }}_k{\mathrm{sn}}_k(t)<0$: for fixed $t>0$ the function $k\mapsto {\mathrm{sn}}_k(t)$ is continuously differentiable and strictly decreasing on $(-\mathrm{\infty },{\pi }^2\mathrm{/}{t}^2)$. In particular ${\mathrm{sn}}_k(t)<t$ for $0<k<{\pi }^2\mathrm{/}{t}^2$, ${\mathrm{sn}}_0(t)=t$ and ${\mathrm{sn}}_k(t)>t$ for $k<0$; by [F5] the same comparison is the strict sign of the Rauch comparison between the space forms of curvatures $k_1<k_2$: for matched initial derivatives of the same norm $\mathrm{\mid }E\mathrm{\mid }>0$, one has $\mathrm{\mid }{J}_{k_2}(t)\mathrm{\mid }={\mathrm{sn}}_{k_2}(t)\mathrm{\mid }E\mathrm{\mid }<{\mathrm{sn}}_{k_1}(t)\mathrm{\mid }E\mathrm{\mid }=\mathrm{\mid }{J}_{k_1}(t)\mathrm{\mid }$. If $E=0$, both fields are identically zero.

The model cotangent is strictly smaller at larger curvature. [F1, F3] From the explicit formulas, ${\mathrm{ct}}_0(t)=1\mathrm{/}t$, ${\mathrm{ct}}_k(t)=\sqrt{k}\cot (\sqrt{k}t)$ for $k>0$ and ${\mathrm{ct}}_k(t)=\sqrt{-k}\coth (\sqrt{-k}t)$ for $k<0$, and ${\mathrm{ct}}_k(t)\to 1\mathrm{/}t$ as $k\to 0$ in either sign. For $k>0$, with $x=\sqrt{k}t\in (0,\pi )$, $\frac{d}{dk}{\mathrm{ct}}_k(t)=\frac{\sin x\cos x-x}{2\sqrt{k}{\sin }^{2}x}<0,$ because $\sin x\cos x\le \sin x<x$ for $x>0$ by step 1.1 (for $x\le \pi \mathrm{/}2$ use $\sin x<x$ and $\cos x\le 1$; for $x>\pi \mathrm{/}2$, $\sin x\cos x<0<x$). For $k<0$, with $y=\sqrt{-k}t>0$, $\frac{d}{dk}{\mathrm{ct}}_k(t)=-\frac{\sinh y\cosh y-y}{2\sqrt{-k}{\sinh }^{2}y}<0,$ because $y<\sinh y\cosh y$ by step 1.1 applied to $\sinh (2y)>2y$. Hence for fixed $t>0$ the function $k\mapsto {\mathrm{ct}}_k(t)$ is strictly decreasing on $(-\mathrm{\infty },{\pi }^2\mathrm{/}{t}^2)$, and ${\mathrm{ct}}_k(t)<1\mathrm{/}t$ for $k>0$, $=1\mathrm{/}t$ for $k=0$, $>1\mathrm{/}t$ for $k<0$. By [F8] this is exactly the strict sign of the Hessian and Laplacian comparisons between the model of curvature $k$ and the flat model: the curvature bound $K\ge k^{\mathrm{\prime }}$ holds in $M_k^n$ whenever $k\ge k^{\mathrm{\prime }}$ [F6, F7], and the attained model value ${\mathrm{ct}}_k$ is strictly below ${\mathrm{ct}}_{k^{\mathrm{\prime }}}$ when $k>k^{\mathrm{\prime }}$.

Equality in the constant-$k$ model. The five comparisons are equalities in $M_k^n$ and $M_k^2$:

• Jacobi: by [F5] the matched normal Jacobi field of the model is $J(t)={\mathrm{sn}}_k(t)P_{t}E$, so the Rauch bound is attained at every $t$ of the domain.
• Hessian and Laplacian: by [F8], $\mathrm{Hess}r$ and ${\mathrm{\Delta }}_{g}r$ equal the comparison expressions of [F6] and [F7] exactly, and the two curvature hypotheses hold with equality.
• Volume: by [F9], ${\mathrm{vol}}_g(B(p,r))=V_k^{\star }(r)$, so the Bishop–Gromov ratio is identically $1$.
• Toponogov: by [F4] a comparison triangle in $M_k^2$ is unique up to isometry, so the comparison triangle of an admissible model triangle is congruent to the triangle itself and every model angle equals its comparison angle; in the hinge form the model opposite side is by definition $c_k(a,b,\theta )$ [F10], so both comparisons hold with equality. In each case the bounding quantity is exactly the model quantity, so no strictness remains inside a fixed model; strictness can only appear between models of different curvature, which is what steps 1.2, 1.3, 2.1 and 5.1 compute.

The primitives $m$, $G$ and $\rho$. [F1] For the rest of the computation fix $k$ and write $m(t):=m_k(t)={\int }_0^{t}x{\mathrm{sn}}_k(x)dx$ on the positive domain, $h(t):={\mathrm{sn}}_k(t{)}^2$, $G(t):={\mathrm{sn}}_k(t)m(t)=h(t)\rho (t)$ with $\rho (t):=m(t)\mathrm{/}{\mathrm{sn}}_k(t)$, and ${\tan }_k(t):={\mathrm{sn}}_k(t)\mathrm{/}{\mathrm{cs}}_k(t)$. Then $m(0)=0$, $m^{\mathrm{\prime }}=t{\mathrm{sn}}_k(t)$, $m$ is odd and $m>0$ on $(0,D_k)$; $h$ is even and strictly increasing on $(0,D_k\mathrm{/}2)$ (for $k>0$ its restriction is ${\sin }^2(\sqrt{k}t)\mathrm{/}k$, for $k\le 0$ it is $t^2$ or ${\sinh }^2(\sqrt{-k}t)\mathrm{/}(-k)$), with $h(t)>0$ for $0<t<D_k$; $G$ is even and $G>0$ on $(0,D_k)$; and on $(0,D_k\mathrm{/}2)$ one has ${\mathrm{cs}}_k>0$, hence ${\tan }_k>0$.

Half-angle form of the model cosine law. [F3, F4] Let $(a,b,c)$ be admissible with angle $\theta \in (0,\pi )$ opposite $c$ as in [F4], and put $s=(a+b)\mathrm{/}2$, $u=(a-b)\mathrm{/}2$, $v=c\mathrm{/}2$. Then ${\sin }^2\frac{\theta }{2}=\frac{h(v)-h(u)}{{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)},{\cos }^2\frac{\theta }{2}=\frac{h(s)-h(v)}{{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)},$ and consequently the two identities $h(s)-h(u)={\mathrm{sn}}_k(a){\mathrm{sn}}_k(b),h(v)={\cos }^2\frac{\theta }{2}h(u)+{\sin }^2\frac{\theta }{2}h(s)\mathrm{.}$ For $k\mathrm{\ne }0$ this is a computation from the cosine law of [F4]: the addition formulas of [F3] give ${\mathrm{cs}}_k(a\pm b)={\mathrm{cs}}_k(a){\mathrm{cs}}_k(b)\mp k{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)$, so $1-\cos \theta =\frac{{\mathrm{cs}}_k(a-b)-{\mathrm{cs}}_k(c)}{k{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)},1+\cos \theta =\frac{{\mathrm{cs}}_k(c)-{\mathrm{cs}}_k(a+b)}{k{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)},$ and ${\mathrm{cs}}_k(2x)=1-2k{\mathrm{sn}}_k(x{)}^2$ converts the right hand sides into the two displayed expressions, whose sum is $1$ and gives the third identity; the fourth follows by substituting the third into the first. For $k=0$ the same identities are the elementary Euclidean computation: ${\sin }^2(\theta \mathrm{/}2)=(v^2-u^2)\mathrm{/}(ab)$, ${\cos }^2(\theta \mathrm{/}2)=(s^2-v^2)\mathrm{/}(ab)$, $s^2-u^2=ab$ and $v^2={\cos }^2(\theta \mathrm{/}2)u^2+{\sin }^2(\theta \mathrm{/}2)s^2$ follow from $c^2=a^2+b^2-2ab\cos \theta$. Also, the map $\theta \mapsto c_k(a,b,\theta )$ is strictly increasing on $(0,\pi )$: the first identity writes $h(c_k(a,b,\theta )\mathrm{/}2)=h(u)+{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b){\sin }^2(\theta \mathrm{/}2)$ whose right-hand side varies strictly increasingly in $\theta$. For $k>0$ the model opposite side is less than $D_k$, so its half lies in $(0,D_k\mathrm{/}2)$, where $h$ is strictly increasing; for $k\le 0$ the same holds for every positive half-side. Hence $c_k(a,b,\theta )$ and $\theta$ determine each other strictly increasingly.

The saturated model volume is strictly smaller at larger curvature. [F1, F9, step 1.2] For $k_1<k_2$, put $D_i:=D_{k_i}$, with $D_i=+\mathrm{\infty }$ when $k_i\le 0$. Then $D_2\le D_1$, and step 1.2 gives $A_{k_2}(t)<A_{k_1}(t)$ for $0<t<D_2$. For every $r>0$, $V_{k_i}^{\star }(r)={\int }_0^{\min \{r,D_i\}}{A}_{k_i}(t)dt\mathrm{.}$ The $k_1$ integral covers at least $(0,\min \{r,D_2\})$, on which its density is strictly larger. Hence $V_{k_2}^{\star }(r)<V_{k_1}^{\star }(r)(r>0)\mathrm{.}$ Thus the larger-curvature model has strictly smaller saturated balls, the strict form of the flat-versus-curved Bishop–Gromov comparison.

The ratio $\rho$ is strictly increasing. In the three cases, using the explicit forms of ${\mathrm{sn}}_k$ and the identity $m(t)=({\mathrm{sn}}_k(t)-t{\mathrm{cs}}_k(t))\mathrm{/}k$ for $k\mathrm{\ne }0$ (both sides vanish at $0$ and have derivative $t{\mathrm{sn}}_k(t)$):

• $k>0$: with $x=\sqrt{k}t$, $\rho (t)=(1-x\cot x)\mathrm{/}k$ and ${\rho }^{\mathrm{\prime }}(t)=(x-\sin x\cos x)\mathrm{/}(\sqrt{k}{\sin }^{2}x)>0$ by step 1.1;
• $k=0$: $\rho (t)=t^2\mathrm{/}3$, so ${\rho }^{\mathrm{\prime }}(t)=2t\mathrm{/}3>0$;
• $k<0$: with $y=\sqrt{-k}t$, $\rho (t)=(y\coth y-1)\mathrm{/}(-k)$ and ${\rho }^{\mathrm{\prime }}(t)=(\sinh y\cosh y-y)\mathrm{/}(\sqrt{-k}{\sinh }^{2}y)>0$ by step 1.1. Thus $\rho$ is strictly increasing on $(0,D_k)$. No monotonicity of $G=h\rho$ on the whole positive domain is asserted.

The addition identity for $G$. [F2, F3, step 1.5] For all $a,b>0$, with $s=(a+b)\mathrm{/}2$ and $u=(a-b)\mathrm{/}2$, ${\mathrm{sn}}_k(u)m(u)+\frac{1}{2}[m(a){\mathrm{sn}}_k(b)+{\mathrm{sn}}_k(a)m(b)]={\mathrm{sn}}_k(s)m(s)\mathrm{.}$ For $k\mathrm{\ne }0$, multiply the difference of the two sides by $k$ and use $m(t)=({\mathrm{sn}}_k(t)-t{\mathrm{cs}}_k(t))\mathrm{/}k$ together with the product-to-sum consequences of [F3]: ${\mathrm{sn}}_k(x){\mathrm{cs}}_k(x)={\mathrm{sn}}_k(2x)\mathrm{/}2$, ${\mathrm{sn}}_k(x){\mathrm{sn}}_k(y)=({\mathrm{cs}}_k(x-y)-{\mathrm{cs}}_k(x+y))\mathrm{/}(2k)$, ${\mathrm{cs}}_k(x){\mathrm{cs}}_k(y)=({\mathrm{cs}}_k(x-y)+{\mathrm{cs}}_k(x+y))\mathrm{/}2$, ${\mathrm{sn}}_k(x){\mathrm{cs}}_k(y)=({\mathrm{sn}}_k(x+y)+{\mathrm{sn}}_k(x-y))\mathrm{/}2$; then the difference equals $[h(u)-h(s)+{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)]+[-\frac{u}{2}{\mathrm{sn}}_k(2u)-\frac{1}{2}(a{\mathrm{cs}}_k(a){\mathrm{sn}}_k(b)+b{\mathrm{sn}}_k(a){\mathrm{cs}}_k(b))+\frac{s}{2}{\mathrm{sn}}_k(2s)],$ whose first bracket is zero by the addition formula for ${\mathrm{cs}}_k(2\cdot )$ and whose second bracket is zero because $a{\mathrm{cs}}_k(a){\mathrm{sn}}_k(b)+b{\mathrm{sn}}_k(a){\mathrm{cs}}_k(b)=s{\mathrm{sn}}_k(2s)-u{\mathrm{sn}}_k(2u)$; hence the difference is zero. For $k=0$ the identity reads $u^4\mathrm{/}3+(a^{3}b+a{b}^3)\mathrm{/}6=s^4\mathrm{/}3$ and follows from $s^4-u^4=(s-u)(s+u)(s^2+u^2)=ab(a^2+b^2)\mathrm{/}2$.

The conjugate function $\mathrm{\Gamma }$ is strictly convex. [F2, step 2.2, step 1.5] Let $\mathrm{\Gamma }$ be defined on the set of values of $h$ by $\mathrm{\Gamma }(h(t)):=G(t)$ for $t\in (0,D_k\mathrm{/}2)$; this is well defined because $h$ is a strictly increasing bijection from $(0,D_k\mathrm{/}2)$ onto its image. The formulas $G(t)\to 0$ and $h(t)\to 0$ as $t\downarrow 0$ extend $\mathrm{\Gamma }$ continuously by $\mathrm{\Gamma }(0)=0$; when $k>0$, the explicit model formulas likewise extend it continuously to $h(D_k\mathrm{/}2)$. Then $\frac{d\mathrm{\Gamma }}{dh}=\frac{\rho (t)}{2}+\frac{t}{2}{\tan }_k(t),\frac{d^2\mathrm{\Gamma }}{d{h}^2}=\frac{{\rho }^{\mathrm{\prime }}(t)+{\tan }_k(t)+t\mathrm{/}{\mathrm{cs}}_k(t{)}^2}{4{\mathrm{sn}}_k(t){\mathrm{cs}}_k(t)}>0$ at $h=h(t)$, $t\in (0,D_k\mathrm{/}2)$. Indeed, $dh\mathrm{/}dt=2{\mathrm{sn}}_k{\mathrm{cs}}_k$ and $dG\mathrm{/}dt={\mathrm{cs}}_{k}m+{\mathrm{sn}}_k{m}^{\mathrm{\prime }}={\mathrm{sn}}_k({\mathrm{cs}}_k\rho +t{\mathrm{sn}}_k)$, so $\frac{d\mathrm{\Gamma }}{dh}=(dG\mathrm{/}dt)\mathrm{/}(dh\mathrm{/}dt)=\rho \mathrm{/}2+(t\mathrm{/}2){\tan }_k$; differentiating the last expression with respect to $t$ and dividing by $2{\mathrm{sn}}_k{\mathrm{cs}}_k$ gives the displayed second derivative, whose numerator is positive because ${\rho }^{\mathrm{\prime }}>0$ (step 2.2), ${\tan }_k>0$, $t>0$ and ${\mathrm{sn}}_k,{\mathrm{cs}}_k>0$ on $(0,D_k\mathrm{/}2)$, while $({\tan }_k{)}^{\mathrm{\prime }}=(1+k{\tan }_k^2)={\mathrm{cs}}_k^{-2}$ by the Wronskian identity [F2]. Hence $\mathrm{\Gamma }$ is strictly convex on the interior and, by its continuous endpoint extension, on every closed subinterval of the extended domain.

The half-angle numerator is positive. [step 1.5, step 1.6, step 2.2, step 2.3, step 3.1] Let $(a,b,c)$ be admissible, $\theta \in (0,\pi )$ its angle at the vertex between the sides $a$ and $b$, and set $M_1:={\cos }^2\frac{\theta }{2}G(u)+{\sin }^2\frac{\theta }{2}G(s)-G(v),s=\frac{a+b}{2},u=\frac{a-b}{2},v=\frac{c}{2}\mathrm{.}$ Then $M_1>0$. To see this, put $t_s:=s$ if $k\le 0$ or $s\le D_k\mathrm{/}2$, and $t_s:=D_k-s$ if $k>0$ and $s>D_k\mathrm{/}2$. Then $t_s\in (0,D_k\mathrm{/}2]$ and $h(t_s)=h(s)$. Also $\mathrm{\mid }u\mathrm{\mid }<v<t_s$: when $t_s=s$ this follows from the strict triangle inequality, and when $s>D_k\mathrm{/}2$ it follows from $a+b+c<2D_k$. Since $\rho$ is strictly increasing and $h(s)=h(t_s)>0$, $G(t_s)\le G(s)$, strictly if $t_s<s$. Thus $\mathrm{\mid }u\mathrm{\mid }<v<t_s\le D_k\mathrm{/}2$, and $h(u),h(v),h(t_s)$ lie in the increasing range of $h$; the fourth identity of step 1.6 makes $h(v)$ a strict convex combination of $h(u)$ and $h(t_s)$ with weights ${\cos }^2(\theta \mathrm{/}2)$ and ${\sin }^2(\theta \mathrm{/}2)$, both in $(0,1)$. Strict convexity of $\mathrm{\Gamma }$ (step 3.1) on the closed interval between $h(u)$ and $h(t_s)$ gives $G(v)=\mathrm{\Gamma }(h(v))<{\cos }^2\frac{\theta }{2}\mathrm{\Gamma }(h(u))+{\sin }^2\frac{\theta }{2}\mathrm{\Gamma }(h(t_s))={\cos }^2\frac{\theta }{2}G(u)+{\sin }^2\frac{\theta }{2}G(t_s)\le {\cos }^2\frac{\theta }{2}G(u)+{\sin }^2\frac{\theta }{2}G(s),$ so $M_1>0$; here $G(u)=G(\mathrm{\mid }u\mathrm{\mid })$ because $G$ is even. This is the only place where the saturation $s>D_k\mathrm{/}2$ of the positive-curvature model is handled.

The model angle is strictly increasing in the curvature. [F1, F4, step 1.2, step 1.6, step 4.1] Fix an admissible side datum $(a,b,c)$ and let $\theta (k)$ be the model angle opposite $c$ between the sides $a$ and $b$ in $M_k^2$, as $k$ ranges over the open admissible interval $(-\mathrm{\infty },k_{\max })$ (where $k_{\max }:=\min \{(\pi \mathrm{/}\max \{a,b,c\}{)}^2,(2\pi \mathrm{/}(a+b+c){)}^2\}$); on this interval the datum stays nondegenerate and $\theta (k)\in (0,\pi )$. Then ${\theta }^{\mathrm{\prime }}(k)=\frac{2M_1(k)}{{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)\sin \theta (k)}>0,$ where $M_1(k)$ is the positive quantity of step 4.1 computed with the model functions at $k$. Indeed, differentiating the fourth identity of step 1.6 in $k$ gives $-G(v)=\frac{d}{dk}\left[{\cos }^2\frac{\theta }{2}\right](h(u)-h(s))-{\cos }^2\frac{\theta }{2}G(u)-{\sin }^2\frac{\theta }{2}G(s),$ because at each fixed argument $t$, ${\mathrm{\partial }}_{k}h(t)=2{\mathrm{sn}}_k(t){\mathrm{\partial }}_k{\mathrm{sn}}_k(t)=-G(t)$ by step 1.2; since $\frac{d}{dk}{\cos }^2(\theta \mathrm{/}2)=-\frac{1}{2}\sin \theta {\theta }^{\mathrm{\prime }}$ and $h(u)-h(s)=-{\mathrm{sn}}_k(a){\mathrm{sn}}_k(b)$ by step 1.6, this rearranges to the displayed formula. Its right-hand side is positive because $M_1(k)>0$, ${\mathrm{sn}}_k(a),{\mathrm{sn}}_k(b)>0$ (the datum is admissible at $k$) and $\sin \theta (k)>0$. Therefore $\theta$ is strictly increasing on the admissible interval: for $k_1<k_2$ with the datum admissible at both, ${\alpha }_{k_2}(a,b,c)=\theta (k_2)>\theta (k_1)={\alpha }_{k_1}(a,b,c)$. For fixed $\theta \in (0,\pi )$, if $c_{k_2}(a,b,\theta )\ge c_{k_1}(a,b,\theta )$, strict increase of the angle at fixed side lengths in $k$ and of the model side at fixed $k$ in $\theta$ would give an angle greater than $\theta$ at $k_2$ for its own side length, a contradiction. Thus $c_{k_2}(a,b,\theta )<c_{k_1}(a,b,\theta )$.

The five strict signs and their directions. Assemble the computations. For $k_1<k_2$ with the relevant data admissible at both curvatures:

• Rauch: for matched initial derivatives with common norm $\mathrm{\mid }E\mathrm{\mid }>0$, the model Jacobi field at $k_2$ is strictly shorter than at $k_1$ at every positive time in their common domain (step 1.2); in particular, against the flat model, positive curvature gives strictly shorter fields and negative curvature strictly longer fields, while inside a fixed model the bound is an equality (step 1.4). This is consistent with [F5]: in $M_{k_2}^n$ one has $\sec =k_2\ge k_1$, so Rauch's inequality $\mathrm{\mid }{J}_{k_2}\mathrm{\mid }\le \mathrm{\mid }{J}_{k_1}\mathrm{\mid }$ is strict for $E\mathrm{\ne }0$; for $E=0$ both sides are zero.
• Hessian and Laplacian: ${\mathrm{ct}}_{k_2}(t)<{\mathrm{ct}}_{k_1}(t)$ for $t>0$ in the common domain (step 1.3), so the space-form values ${\mathrm{ct}}_k$ of [F8] give strict inequalities relative to the comparison values of [F6], [F7]: the flat-versus-positive sign is ${\mathrm{ct}}_k(t)<1\mathrm{/}t$ and the flat-versus-negative sign is ${\mathrm{ct}}_k(t)>1\mathrm{/}t$. The directions match $\mathrm{Hess}r\le {\mathrm{ct}}_{k}g$ and ${\mathrm{\Delta }}_{g}r\le (n-1){\mathrm{ct}}_k$ under $K\ge k$, $\mathrm{Ric}\ge (n-1)kg$.
• Volume: $V_{k_2}^{\star }(r)<V_{k_1}^{\star }(r)$ for every $r>0$ (step 2.1): the model of larger curvature has strictly smaller balls, matching the Bishop–Gromov upper bound, with equality inside a fixed model (step 1.4).
• Toponogov: ${\alpha }_{k_2}>{\alpha }_{k_1}$ and $c_{k_2}(a,b,\theta )<c_{k_1}(a,b,\theta )$ for nondegenerate data (step 5.1): the model of larger curvature has strictly fatter triangles and strictly shorter opposite sides, matching the hinge inequality $c\le c_k(a,b,\theta )$ and the angle inequality $\alpha \ge \bar{\alpha }$ under $K\ge k$ [F10], with equality inside a fixed model (step 1.4). All equalities inside a fixed model are those of step 1.4. The example is a direction diagnostic only: every statement above is about the explicitly displayed model functions, no item of either page lists this example among its dependencies, and no later proof uses it.