A few months ago, I read Hamilton's An isoperimetric estimate for the Ricci flow on the two-sphere, which is collected in Collected papers on Ricci flow.
For two-sphere, the isoperimetric ratio can be treated as $$ C_s^2=\inf_{\Lambda} \frac{L^2}{4\pi}(\frac{1}{A_1}+\frac{1}{A_2}) \tag{1} $$ the $L$ is the length of curve $\Lambda$, and $\Lambda$ dividing two-sphere to two parts with areas $A_1,A_2$.
Hamilton proved that the isoperimetric ratio can be attained by a curve $\overline\Lambda$. And he state that the geodesic curvature of $\overline\Lambda$ is constant $$ k=\frac{L}{2}(\frac{1}{A_1}-\frac{1}{A_2}) \tag{2} $$ by usual first variation. But he just state it without any detail or step. Recently, I read the variation of geodesic. So I want to prove (2), but I have no idea. And I fail to find reference by using curvature, isoperimetric set as key words.
So, I want a framework for proving (2). I'd like to fill in the details myself as an exercise. Thanks.
PS(2023-12-31): Assuming $\overline\Lambda:[0,a]\rightarrow S^2$ is the curve of isoperimetric ratio. $f(s,t):(-\epsilon,\epsilon)\times[0,a]\rightarrow S^2$ is the variation of $\overline\Lambda$. The variation field is $$ V(t)=\partial_s f(0,t) \tag{3} $$ For fixed $s\in (-\epsilon,\epsilon)$, we denote $f_s(t) =f(s,t)$. When $s=0$, $t$ is the arc length parameter of $f_s(t)$. And, $N(s,t)$ is the normal vector of $f_s(t)$. For keeping the area of each part unchanged, we assume $$ \int_0^a\int_0^s N(\tau,t)\cdot\partial_s f(\tau,t)d\tau dt=0 \tag{4} $$ Differentiate (4), we have $$ 0= \frac{d}{ds}\int_0^a\int_0^s N(\tau,t)\cdot\partial_s f(\tau,t)d\tau dt = \int_0^a N(s,t)\cdot\partial_s f(s,t) dt \tag{4.1} $$ Since when $s=0$, (4) always be right. Therefore, (4) is equal to $$ \int_0^a N(s,t)\cdot\partial_s f(s,t) dt=0~~~~~~\forall s\in(-\epsilon,\epsilon) \tag{4.2} $$ The length of $f_s(t)$ is $L(s)=\int_0^a |\partial_t f(s,t)| dt$. Then, $$ \begin{align} 0=\frac{dL}{ds}|_{s=0} &= \frac{d}{ds}|_{s=0} \int_0^a |\partial_t f(s,t)| dt \\ &= \int_0^a \frac{ \langle \frac{D}{ds}\frac{\partial f}{\partial t},\frac{\partial f}{\partial t} \rangle }{|\partial_t f(s,t)|} dt|_{s=0} \\ &= \int_0^a \frac{ \langle \frac{D}{dt}\frac{\partial f}{\partial s},\frac{\partial f}{\partial t} \rangle }{|\partial_t f(s,t)|} dt|_{s=0} \tag{5}\\ \text{(when $s=0$, $|\partial_t f|=1$)} ~~~~~~~ &= \int_0^a \frac{d}{dt}\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\rangle -\langle \frac{\partial f}{\partial s}, \frac{D}{dt} \frac{\partial f}{\partial t} \rangle dt|_{s=0} \\ \text{(when $s=0$, $\partial_s f=V(t)$, $f=\overline\Lambda (t)$)} ~~~~~~~ &= \langle V(t), \partial_t\overline\Lambda(t)\rangle |_0^a - \int_0^a \langle V(t),\frac{D}{dt}\frac{d\overline\Lambda}{dt}\rangle dt \end{align} $$ For any $t_1,t_2\in(0,a)$, let $$ V(t)=N(t) \delta(t-t_1) -N(t)\delta(t-t_2) \tag{6} $$ where $\delta$ is Dirac delta function. Then, (5) imply that $k(\overline\Lambda(t_1))=k(\overline\Lambda(t_2))$. By the arbitrariness of $t_1,t_2$ and parameterization of $\overline\Lambda$, we know that the geodesic curvatures of all points of $\overline\Lambda$ are equal. Then, Hamilton has show $$ \int k ds = \frac{L^2}{2}(\frac{1}{A_1}-\frac{1}{A_2}) \tag{7} $$ Therefore, $k=\frac{L}{2}(\frac{1}{A_1}-\frac{1}{A_2})$.
However, the above process is not complete, since we need to prove there is variation satisfying (4) and its variation field is (6). I try to use exponential map to show the existence, but still have not finish.
Well, on the last day of 2023, I wish all who see here happy.