# Affine space or vector space

When I am studying about differential geometry, I have many exercise about write the equation of the tangent to curve $\gamma = r(t)$, with $r: I \to \mathbb{R}^3$, at point $r(x_0)$. That is the line through $r(x_0)$ and has direction vector $r'(x_0)$.

I confused when $r(x_0)$ and $r'(x_0)$ are elements of $\mathbb{R}^3$, why the first one is viewed as "point", while the other one is viewed as "vector".

Moreover, I wonder the space $\mathbb{R}^3$ is endowed with affine space structure or vector space structure.

As an easier example, when studying analytic geometry in high school, we write that a point $M \in \mathbb{R^3}$ is a 3-tuple $(x,y,z)$, while a "vector" $\overrightarrow{m}$ is the same with 3-tuple $(u,v,w)$. What is the different between two notations?

• Very simply because we want the line passing through the point $r(t_0)$ with direction vector $r'(t_0)$. Some people and texts do distinguish between point $(a,b,c)$ and vector $\langle a,b,c\rangle$, but the astute reader/student usually figures out how to interpret things correctly if we use the same notation. – Ted Shifrin Dec 8 '16 at 6:44

If you do things carefully, then you view your curve as having values in the three-dimensional Euclidean (or affine) space, which I denote by $E^3$ to distinguish it from the vector space. (You can see this from the fact that if you draw the situation, then you can determine the tangent without knowing where the origin lies.) But it is easy to see that via the affine structure, you can identify each tangent space $T_xE^3$ with the modelling vector space $\mathbb R^3$. The affine structure in particular allows you to form $y=x+v\in E^3$ for $x\in E^3$ and $v\in\mathbb R^3$. Using this, the identification of $\mathbb R^3$ with $T_xE^3$ is given by mapping $v$ to the derivative at $t=0$ of the curve $t\mapsto x+tv$. So for a smooth curve $r:I\to E^3$ you get $r'(t)\in T_{r(t)}E^3\cong\mathbb R^3$ (and the tangent line at $t_0$ is the unique affine line which passes through $r(t_0)$ has the same derivative there as $r$).