Let $RP^n$ the real projective space. This is a manifold and i have take the usual charts in order to prove it. The problem is that i don't know how to define a vector field on that.
Since $RP^n$ is a set of lines that passes through the point (0,...,0) could I claim that the vector space is a map that takes every point of $RP^n$ - line and sends it to a vector $\overline {OP}$ ,where O is the origin and P is a point of the line ?
Is there a concrete example on $PR^2$?
We can visualize $RP^2$ as the plane in $\mathbb R^3$ at $z=1$.
We can write a point as $(x:y:1)$. Or more generally as $(x:y:z)$, which we consider equivalent to $(\frac xz:\frac yz:1)$ if $z\ne 0$. Consequently all points on a line through the origin are equivalent. And a point $(x:y:0)$ is a point at infinity, which is also an element of $RP^2$.
An obvious choice for a local chart is $\phi: (x:y:1)\mapsto (x,y)$.
Taking the partial derivatives gives us the representations $\frac\partial{\partial x}=(1:0:0)$ and $\frac\partial{\partial y}=(0:1:0)$. Note that these correspond exactly with the standard unit vectors in the plane $z=1$.
A vector field $X: RP^2\to TRP^2$ can now locally be written as: $$X(\mathbf x) = X(x:y:1) = \xi(\mathbf x)\frac\partial{\partial x} + \eta(\mathbf x)\frac\partial{\partial y} = (\xi(\mathbf x):\eta(\mathbf x):0)$$
