geodesics on sphere loring tu help

Question

Loring Tu in his book Differential geometry [page 104] states (in my own words) the following on geodesics on spheres:

Consider the 2 sphere of radius $a$ in $\mathbb{R}^3$ .

1. Parameterise a great circle by arc length $\gamma$.
2. $\gamma''(t)$ is perpendicular to $\gamma'(t)$
3. $\gamma''(t)$ lies in the plane of the circle (does this mean $T_{\gamma(t)}M?)$
4. $\gamma''(t)$ is perpendicular to the tangent plane at $\gamma(t)$

I'm not sure why $(4).$ holds, why is $\gamma''(t)$ in the normal space $N_{\gamma(t)}M$?

Please do not explain using curvature.

Answer 1

Note that the plane of the great circle contains the origin (i.e. the center of the sphere). $\gamma''$ is orthogonal to $\gamma'$, so because $\gamma'$ is parallel to the circumference, $\gamma''$ must be parallel to the radius of the great circle.

Now, note that the radius of the great circle is also a diameter of the sphere. So, $\gamma''$ is parallel to the radius of the sphere and hence orthogonal to its tangent plane.

Answer 2

$\gamma(t)$ lies in a plane so $\gamma'(t)$ and $\gamma''(t)$ also lie in the same plane. (If you want to be precise: planes are closed subsets, and $\gamma'(t)$ is a limit of a difference quotients with values in the plane.)

Differentiating $\langle\gamma(t),\gamma(t)\rangle=a$ and $\langle\gamma'(t),\gamma'(t)\rangle=1$ yields $\langle\gamma(t),\gamma'(t)\rangle=\langle\gamma(t'),\gamma''(t)\rangle=0$. Hence for each $t$ ,$\gamma(t)$ and $\gamma'(t)$ are an orthogonal basis of the plane and $\gamma''(t)$ is a scalar multiple of $\gamma(t)$.

But the normal space at a point $p$ of a sphere centered at the orgin consists exactly of the scalar multiples of $p$, so $\gamma''(t)$ lies in the normal space.