# Realizing any vector in tangent space as tangent vector

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

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 $$\gamma$$. The curve $$\gamma(t) = \sigma(x_0 + \alpha t, y_0 + \beta t)$$ is a curve that lies on the surface $$S$$, and passes through $$\sigma(x_0,y_0) \in S,$$ since $$\gamma(0) = \sigma(x_0,y_0),$$ and it has $$\alpha v_1 + \beta v_2$$ as its tangent vector at $$0$$, since by the chain rule, $$\dot\gamma(0) = (\sigma_x(x_0+\alpha t,y_0+\beta t) \alpha + \sigma_y(x_0+\alpha t, y_0+\beta t)\beta)\mid_{t=0} = \alpha\sigma_x(x_0,y_0) + \beta\sigma_y(x_0,y_0) = \alpha v_1+\beta v_2.$$ More specifically, you can take the unit sphere $$S^2$$ parametrized by $$\sigma(u,v) = (\cos u \cos v, \cos u \sin v, \sin u).$$ The partial derivatives are \begin{align} \sigma_u &= (-\sin u \cos v, -\sin u \sin v, \cos u)\\ \sigma_v &= (-\cos u \sin v, \cos u \cos v, 0). \end{align}

Consider the point $$p=(u_0,v_0) =(0,0)$$ in this parametrization, which corresponds to $$(1,0,0)$$ in Cartesian coordinates. The partial derivatives at $$p$$ simplify to \begin{align} \sigma_u &= (0, 0, 1), \quad \sigma_v = (0, 1 , 0). \end{align}

Hence $$T_pS$$ is just the $$yz$$-plane (as you would expect). As you mentioned yourself, every vector in $$T_pS$$ can be written as $$\alpha \sigma_u + \beta\sigma_v$$. For the sake of simplicity, let $$\alpha = 0, \beta = 1$$. Then your curve $$\gamma$$ becomes $$\gamma(t) = \sigma(u_0 + \alpha t, v_0 + \beta t) = \sigma(0,t) = (\cos t, \sin t, 0).$$ This curve is just a circle along the equator passing through $$(0,0,1)$$ and having $$(0,1,0) = \sigma_v = \alpha\sigma_u + \beta\sigma_v$$ as its tangent vector (i.e. it is moving east).

If we instead put $$\alpha = 1, \beta = 0$$, we get $$\gamma(t) = \sigma(u_0 + \alpha t, v_0 + \beta t) = \sigma(t,0) = (\cos t, 0, \sin t).$$ This curve is a circle along the meridian passing through $$(0,0,1)$$ and having $$(0,0,1) = \sigma_u = \alpha\sigma_u + \beta\sigma_v$$ as its tangent vector (i.e. it is moving south).