Let $S$ be a surface in $\mathbb{R}^3$.
Let $U\subseteq \mathbb{R}^2$ be open (connected) and $\sigma:U\rightarrow S$ be a homeomorphism into $\sigma(U)\subset S$, with $\sigma(U)$ open in $S$, and $\sigma$ is smooth map.
Assume that $\sigma_x,\sigma_y$ are independent for every point in $U$ (i.e. for every $p=(a,b)\in U$, the partial derivatives of $\sigma$ at $p$ are independent vectors in $\mathbb{R}^3$. [This means $\sigma$ is regular].
Now fix $(x_0,y_0)\in U$ and $v_1=\sigma_x(x_0,y_0)$, $v_2=\sigma_y(x_0,y_0)$.
It can be shown algebraically that every vector $\alpha v_1 + \beta v_2$ is tangent to surface $S$ at point $\sigma(x_0,y_0)$. Define for small $t$,
$$\gamma(t)=\sigma(x_0+\alpha t, y_0+\beta t).$$
This is a curve in $S$ and $\gamma(0)=\sigma(x_0,y_0)$; also at $t=0$ we can see that the tangent to curve (hence surface) $\gamma$ is $\alpha \sigma_x(x_0,y_0)+\beta\sigma_y(x_0,y_0)$.
I did not see this description pictorially what $\gamma(t)$ expresses? For simple example of sphere, what is this $\gamma$? Can one give some pictorial explanation of above paragraph, with simple example?
Ref. Elementary Differential Geometry – Pressley [New edition], Prop. 4.2
Best Answer
Every vector in the tangent space of a surface at a point is the velocity vector of a curve on the surface that passes through that point. The curve $\gamma$ here is chosen in order to show that every linear combination $\alpha v_1 + \beta v_2$ of the two vectors $v_1$ and $v_2$ also lies in the tangent space, since it is the velocity vector of the curve $\gamma(t)=\sigma(x_0+\alpha t, y_0 + \beta t)$. The vectors $v_1$ and $v_2$ are the vectors tangent to the coordinate lines of the parametrisation.
Here is a picture of a sphere where two coordinate lines at a point are drawn in blue. The red curve is a example of a curve $\gamma$.