Is every smooth $\mathbb{R}$-variety isomorphic to an affine variety?

I sadly don't know anything about formal GAGA yet, but I am at least trying to follow my intuition as often as possible.

In differential geometry we know that we can embedd every smooth $\mathbb{R}$-manifold into some $\mathbb{R}^{m}$ (one can even make precise what this $m$ is, if one wishes). Note that for complex manifolds the analogous does not work in general.

Now morally, I would like to see a smooth manifold over $\mathbb{R}$ as a smooth $\mathbb{R}$-variety and vice versa. (although this probably fails as $\mathbb{R}$ is not algebraically closed, weird things could happen)

For example, projective space should not be affine as a variety (global sections!), but in manifolds I am told that I can embedd it into some $\mathbb{R}^m$.

Thus, I am very confused. Either, by some magic every smooth $\mathbb{R}$-variety can be embedded in $\mathbb{A}^m$ or more likely my intuition for "what an $\mathbb{R}$-manifold should correspond to in the language of algebraic geometry" (which is already very handwavy) is simply wrong. I guess it is rather the later, and I would be happy if anyone could help me clear my confusion.

(My professor of Differential Geometry said today that the fact 'every smooth vector bundle is a direct summand of a trivial vector bundle' should correspond to the fact 'every vector bundle is a locally free sheaf, thus a projective sheaf'. However, I know the 'thus' part is not true if the underlying scheme is not affine. Maybe this as an explanation of where my confusion started to derive from.)

• Every smooth manifolds can be smoothly embedded into $\mathbb R^N$, but not all real analytic manifold (closer to something algebraic) admits a real analytic embedding into $\mathbb R^N$. – user99914 Nov 12 '14 at 14:24
• I understand. This already lifted my confusion quite a bit. Thank you. – Louis Nov 12 '14 at 14:32
• If you want some GAGA-type statements you should probably restrict to compact manifolds. Also, manifolds and varieties have different topologies. Varieties are usually considered with the Zariski topology. Maybe I'm misunderstanding your terminology? – Ariyan Javanpeykar Nov 12 '14 at 15:22

It's worth distinguishing between "the $\mathbb{R}$-points of a variety defined over $\mathbb{R}$", and the variety (or scheme) itself. For instance, let $X = \mathrm{Spec} (R)$ for $R = \mathbb{R}[x,y]/(x^2 + y^2 - 1)$. This is an affine variety in $\mathbb{A}^2$ (in particular, it is not compact when you consider its $\mathbb{C}$-points. In fact $X \cong \mathbb{P}^1$ minus two complex points.) And it's important to realize that $X$ does not consist only of real points -- there are lots of maximal ideals in $R$ corresponding to complex conjugate pairs of complex points.
On the other hand, the $\mathbb{R}$-points of $X$ (usually we refer to this by the notation $X(\mathbb{R})$) are the unit circle, a compact real manifold.
The funny thing here is that the closure $\overline{X}$ in $\mathbb{P}^2$ has the same $\mathbb{R}$-points as $X$, even though $\overline{X}$ is a projective variety and $X$ is affine! So $\overline{X}$ is very different from $X$ (e.g. it can't be embedded in $\mathbb{A}^2$), but the distinction is invisible if we only think about the real points.
The moral of the story is that although $X(\mathbb{R})$ is (just) a real manifold, $X$ itself contains the information about what's going on over $\mathbb{C}$. It's probably best to think of $X$ as "a variety over $\mathbb{C}$ with the action of complex conjugation", hence consisting of both real points and complex-conjugate-pairs-of-complex-points, and $X(\mathbb{R})$ as a subset.
As far as embeddings in Euclidean space go, the Whitney embedding theorem provides smooth embeddings of real manifolds in $\mathbb{R}^n$, yes. But (a) these are smooth, not necessarily algebraic, and (b) when $\overline{X}$ is projective, there will be algebraic ways to embed its real points $\overline{X}(\mathbb{R})$ into $\mathbb{A}^n$, as in the example above, but these can't yield an embedding of the whole variety $\overline{X}$ in $\mathbb{A}^n$.