basic question about closed curve From Ted Shifrin's Diff Geo notes, p. 25, He said: What curve $\Gamma$ in the unit sphere can be the Gauss map of some closed space curve $\alpha$? Since $\alpha(s)=\alpha(0)+ \int_{0}^{s}T(u)du$, it should be that $\int_{0}^{L}T(u)du=0$, where $L$ is the length of $\Gamma$. Then for any (unit) vector $A$, we have $0=A \cdot \int_{0}^{L}T(u)du=\int_{0}^{L}A \cdot T(u)du$ and so the average value of $T \cdot A$ must be 0. In particular, the tangent indicatrix must cross the great circle with normal vector $A$. Here is my basic question: I can't see why the tangent indicatrix must cross the great circle with normal vector $A$. Is it deduced from the equation $T \cdot A=0$? If so, why should it be? And also, why is $A \cdot x \geq 0$ the equation of hemisphere? I guess it is because of same reason for $T \cdot A=0$, but I don't have an idea. Thank you for your answer in advance.

If $A$ is a unit vector it also describes a point on the unit sphere. The tangent space at this point consists of the vectors $v \in \mathbb{R}^3$ that satisfy $v\cdot A=0$.

The equation $A\cdot x \ge 0$ describes a half space of $\mathbb{R}^3$. If you know that $x$ is on the sphere, than this is exactly a hemisphere.