Is $\int_\gamma k_g ds =0$ for any shortest simple closed curve $\gamma$ dividing $S^2$ into two equal parts?

Question

Consider a smooth Riemannian manifold $(S^2,g)$, where $S^2$ is the topological sphere. $\gamma$ is the shortest simple closed curve which dividing $S^2$ to two parts and the two parts have equal areas (as shown in the figure below, where $A_1=A_2$). Then, whether there is $$ \int_\gamma k_g ds =0 $$ where $k_g$ is the geodesic curvature. And how to show it?

PS(2023-7-24): In Kurt G.'s answer, I feel there is a way to keep the equator is the shortest curve.

When $f^2\le \sin^2\theta$ (not always equal), then the shortest curve maybe not the equator.

But when $f^2\ge \sin^2\theta$, the shortest curve should be the equator.

The $f(\frac{\pi}{2})=1$ is not keypoint, it just be used to keep the equators of $(S^2,g)$ and unit sphere are same.

In fact, I think the $\int_\gamma k_g ds =0$ should be wrong in general. Although I didn't quite get the Deane's comment, but I get an idea. Consider a surface liking dumbbell (picture below). At beginning, $A_1=A_2$, suppose at beginning, $k_g\equiv0$ on the black curve. Then, we enlarge $A_2$ a little. So, the curve must move right. When $A_2-A_1$ is enough small, the black curve doesn't rotate without translating, since after rotate, the curve elongate more than after translating. Therefore, we can structure a pipe (the middle part of dumbbell) which only at a meridian $k_g\equiv 0$. So, when it move right, there is not $\int k_g ds =0$.

Besides, I even suspect that $k_g$ is constant maybe wrong. Since we can strcture some special crease on the pipe such that after moving right, the $k_g$ of curve is not constant. But, about this, I am very unclear.

Answer 1

As I understand it now the question is quite interesting and I think it is generally not true that $\int_\gamma k_g\,ds=0\,.$ By the example below the geodesic curvature $k_g$ is not identically zero on a curve that divides the surface into two parts of equal area:

What we know:

On the unit sphere $S^2$ in polar coordinates, $\theta$ polar angle, $\varphi$ azimuthal angle, the standard metric is $$ g=\begin{pmatrix}1&0\\0&\sin^2\theta\end{pmatrix}\,. $$ The equator $\{\theta=\pi/2\}$ is a geodesic because is is a great circle and it divides the sphere into two equal parts whose area is $2\pi$ each.

Modification:

We equip the sphere with the metric $$ h=\begin{pmatrix}1&0\\0&f^2(\theta)\end{pmatrix} $$ where $f$ is a function satisfying the conditions $$\tag{1} f(0)=f(\pi)=0\,,\quad f'(0)=-f'(\pi)=1\,,\quad f^{(n)}(0)=f^{(n)}(\pi)=0\,,\; \forall n\text{ even, } $$ of Ch. 1, section 3.4 of [1]. (As pointed out by mollyerin we need this to have a smooth metric on $S^2\,.$)

Let's also assume that $f>0$ on $(0,\pi)\,.$

The non zero Christoffel symbols are \begin{align} %\Gamma_{\varphi\theta\varphi}&=\Gamma_{\phi\theta\varphi}=f(\theta)f'(\theta)\,, %&\Gamma_{\theta\varphi\varphi}&=-f(\theta)f'(\theta)\,,\\ {\Gamma^\varphi}_{\theta\varphi}&={\Gamma^\varphi}_{\theta\varphi}=\frac{f'(\theta)}{f(\theta)}\,, &{\Gamma^\theta}_{\varphi\varphi}&=-f(\theta)f'(\theta)\,. \end{align} The geodesic equations $\ddot\theta + \Gamma^\theta_{\varphi\varphi}\dot\varphi\dot\varphi=0=\ddot\varphi+2\,\Gamma^\varphi_{\theta\varphi}\dot\theta\dot\varphi$ are then \begin{align} 0&=\ddot \theta-f(\theta)f'(\theta)\,\dot\varphi^2\,,& 0&=\ddot\varphi+\frac{2f'(\theta)\,\dot\theta\,\dot\varphi}{f(\theta)}\,. \end{align} The first equation shows directly that a latitude $\{\theta=\text{const}\}$ with $\theta\not\in\{0,\pi\}$ is a geodesic only when $f'(\theta)=0\,.$

The area of the part of the sphere that lies between $\theta_1$ and $\theta_2$ is $$ S(\theta_1,\theta_2)=\int_0^{2\pi}\int_{\theta_1}^{\theta_2}f(\theta)\,d\theta\,d\varphi=2\pi\int_{\theta_1}^{\theta_2}f(\theta)\,d\theta\,. $$ The latitude $\theta^*$ that divides the sphere into two parts of equal area is defined by $$ 2S(0,\theta^*)=S(0,\pi)\,. $$ I believe there exist many functions $f$ that satisfy the requirements (1) where $f'(\theta^*)$ is not zero. Asymmetric perturbations of the classic $f(\theta)=\sin\theta$ should probably work.

Each such case is an example where the curve consisting of the latitude dividing the sphere into two equal parts is not a geodesic and therefore does not have geodesic curvature $k_g$ zero.

[1] P. Petersen, Riemannian Geometry.