Can a manifold that contains a flat part be analytic? If a manifold embedded in $\mathbb{R}^n$ has a flat part where it is locally given by the surface of a hyperplane, and the rest is curved, could this be an analytic manifold? In other words, is it possible for such a manifold to be differentiable with analytic transition maps?
 A: That is the "analytic structure" I was referring to. Let me step down 1 dimension in my example. The circle is analytic and embeddable in $\Bbb R^2$ as a square. It is the embedding that fails to be analytic (or even differentiable), not the circle itself.
To show that a manifold is analytic, you have show the existence of a covering by charts whose transition maps are all analytic. For the circle two charts are sufficient: stereographic projection from the north and south poles onto the equatorial line. That is, with $S^1$ taken as the unit circle in the plane, draw a line from $(0,1)$ through a point $p = (u,v) \in S^1$ and the coordinate $x$ where this line intersects the $x$-axis defines a chart on $S^1\setminus (0,1)$:
$$(u,v) \mapsto \frac u{1-v}=\text{sgn}(u)\sqrt{\frac{1+v}{1-v}}$$
Using $(0,-1)$ as the pole instead gives the chart
$$(u,v) \mapsto \frac u{1+v}=\text{sgn}(u)\sqrt{\frac{1-v}{1+v}}$$
The transition function between these two charts is $t \mapsto \frac 1t$.
Since this is analytic everywhere it is defined, the circle is an analytic manifold.
