Say i have the smooth manifold $ M=\Re^2 $, and a smooth curve $\gamma:\Re \to M$ with $\gamma(t)=(t,t)$.
Can i draw this curve to the manifold without the use of any chart?
Maybe it sounds a silly question, but i ask it because if $\Re^2$ is only a smooth manifold why should its points be located at certain places as if it were a vector space? I don't think it has enough structure to have lets say (2,0) at the right of (1,0)
Thanks
On
Looking at the comment in José's answer, I think what your confusion is.
Let $\phi\colon U =\mathbb{R}^2 \to M=\mathbb{R}^2$ be a chart of $\mathbb{R}^2$. You really have to distinguish the domain $U=\mathbb{R}^2$ and the manifold $M=\mathbb{R}^2$, even though they are the same set. In this example they are the same set, but in general there not.
Example. The chart $\phi \colon \mathbb{R}^2\to\mathbb{R}^2: (r,\theta)\mapsto(r\cos \theta,r\sin\theta)$. In the manifold $\mathbb{R}^2$ your curve will be the straight line with slope $1$, the first bissectrice. But in the domain $U$, the $(r,\theta)$ coordinate-space, the associated coordinate curve will look like a horizontal line ($\theta=\frac{\pi}{4}$).
A similar thing holds for all coordinate charts: your curve $\gamma$ will always be the same line in the manifold, but the associated coordinate curve $\phi^{-1}\circ\gamma$ in the domain of the chart, depends on the choice of coordinates.
But $\mathbb{R}^2$ is a set, independently of the manifold structure that we decide to endow it with. And on the specific set, we have the usual Cartesian coordinates to represent its points. So, in particular, yes, $(2,0)$ is to the right of $(1,0)$ (with respect to these coordinates). Besides, what have “left” and “right” to do with charts?