3
$\begingroup$

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.)

$\endgroup$
  • 3
    $\begingroup$ 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 $\mathbb R^N$. $\endgroup$ – user99914 Nov 12 '14 at 14:24
  • $\begingroup$ I understand. This already lifted my confusion quite a bit. Thank you. $\endgroup$ – Louis Nov 12 '14 at 14:32
  • $\begingroup$ 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? $\endgroup$ – Ariyan Javanpeykar Nov 12 '14 at 15:22
6
$\begingroup$

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 $\mathbb{A}^n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.