17
$\begingroup$

This question is partly motivated by a few comments here. Let me denote by $R$ the (real-closed) field of real numbers $\mathbb{R}$; everything is probably the same over an arbitrary real-closed field.

When one has a polynomial subset $V$ of $R^n$, the following two are equally sensible ways of putting a structure sheaf on $V$:

  1. One is by considering regular functions in the sense of usual scheme theory: in this case the global regular functions are $R$-polynomials in $n$ variables modulo the ideal of those polynomials vanishing on $V$. If, more precisely, we call $X$ the corresponding $R$-scheme (with all of its non-$R$-points too, which by the way are reconstructible from the set $V\subseteq R^n$) and $O_X$ its structure sheaf, then $X(R)=V\subseteq R^n$ and $O_X(X)\simeq R[x_1,\ldots, x_n]/I_X$.

  2. The other way is by declaring that a regular function is a ratio of polynomials with nonvanishing denominator. We will call such functions $R$-regular, and $R_V$ the resulting structure sheaf. We call $(V,R_V)$ an $R$-algebraic variety. This definition seems to be standard in real algebraic geometry, see e.g. Bochnak-Coste-Roy - Real algebraic geometry (Section 3.2). I think it doesn't change much if we consider the topological space $X$ of the scheme in point 1) instead, endowed with the sheaf $R_X$ that sends an open set $U$ to the rational functions on $U\subseteq X$ that are regular at each point of $U\cap X(R)$.

The resulting structure sheaves are not the same. For example, consider the real line: the function $\frac{1}{1+x^2}$ is an $R$-regular function which is not (scheme-theoretically) regular.

Likewise, one can define abstract $R$-algebraic varieties, and $R$-regular maps thereof.

The curious thing is that every projective $R$-algebraic variety is $R$-biregularly isomorphic to an affine one. Indeed, the set theoretic map (example 1.5 in Ottaviani - Real algebraic geometry. A few basics or theorem 3.4.4 in BCR) $$\mathbb{P}^n(R)\to \operatorname{Sym}^2(R^{n+1})\;\;,\quad (x_0:\ldots : x_n)\mapsto \frac{x_ix_j}{\sum_{h=1}^n x_h^2}$$ is an $R$-regular embedding. This does not correspond to an everywhere-defined morphism of schemes, as is immediately seen by looking at any component of the map in a standard affine chart of $\mathbb{P}^n$.

Are there non-(quasi-)affine abstract $R$-algebraic varieties at all?

Edit: I think the "quasi" in "quasi-affine" may be pleonastic: I haven't checked the details but a quasi-affine $R$-algebraic variety should very often be affine. Indeed, if $X=W\smallsetminus Y$, $Y\subset W \subseteq R^n$ with $W$ affine and $Y$ closed (maybe with some assumptions on $Y$), the real blowup $\operatorname{Bl}_Y W$ is closed in some $\mathbb{P}^{m}\times W$ and the latter is affine; but now the "missing" set $E$ has become a divisor: $X\simeq (\operatorname{Bl}_Y W)\smallsetminus E$, and affine minus a divisor is still affine.


The above example (the one of projective space embedding in an affine space) shows that the category $\text{$R$-Var}$ of $R$-algebraic varieties is not a full subcategory of schemes over $\operatorname{Spec}(R)$. On the other hand, I think the category $\operatorname{Sch}'_R$ of finite type separated reduced schemes over $\operatorname{Spec}(R)$ is a full subcategory of $\text{$R$-Var}$. [Edit: following the comment of Julian Rosen, we probably also want to require the schemes in $\operatorname{Sch}'_R$ to have dense $R$-points]

Are there two non-isomorphic schemes in $\operatorname{Sch}'_R$ that become isomorphic in $\text{$R$-Var}$?

Edit: even before posting, I found example 3.2.8 in BCR. There is also proposition 3.5.2 in BCR, the $R$-biregular isomorphism between the circle $x^2+y^2=1$ and $\mathbb{P}^1_R$. And between the "quadric" sphere and the "Riemann" sphere (i.e. complex projective line thought of as a real algebraic variety).


In which other ways does $\text{$R$-Var}$ deviate from $\operatorname{Sch}'_R$?

Note: I'm not asking how real algebraic geometry deviates from complex algebraic geometry (which is surely addressed in a preexisting MO question).


Edit: (added following question)

For non real-closed fields, or fields of positive characteristic, do people consider varieties in the sense of 1) or in the sense of 2)?

For example, should $1/(1+x^2)$ be a regular function on the line over $\mathbb{F}_7$? (It's a well defined function on a finite field, so there will be a polynomial realizing its values set theoretically, but should it be enough?) -- Or, should 1/(x^2-3) be a regular function on the line over $\mathbb{Q}(\sqrt{2})$?

$\endgroup$
  • 2
    $\begingroup$ For your second question, the scheme $\operatorname{Spec} R[x]/(x^2+1)$ becomes isomorphic to the empty scheme in $R$-Var. Maybe the definition of $\operatorname{Sch}'_R$ should include only those schemes whose $R$-points are Zariski-dense (this is true for every scheme arising in the construction (1)). $\endgroup$ – Julian Rosen Jun 11 at 3:34
  • $\begingroup$ Thanks. Edited accordingly. $\endgroup$ – Qfwfq Jun 11 at 12:28
3
$\begingroup$

As for your first question, concerning nonaffine R-varieties as you call them, yes, there are nonaffine R-varieties. However, they are considered pathological. Example 12.1.5 on page 301 of Bochnak-Coste-Roy, Real algebraic geometry, constructs an R-line bundle over $\mathbf R^2$ whose total space is not affine. In fact, it is not affine since it does not have any separated complexification. Note that the R-variety itself, however, is separated!

The essential point here is that the set of real points of an irreducible affine scheme over $\mathbf R$ can be reducible. In the aforementioned example, the irreducible scheme in question is the one defined by the irreducible polynomial $$p=x^2(x-1)^2+y^2\in\mathbf R[x,y,z].$$ The set of real points in $\mathbf R^3$ defined by $p$ is the disjoint union of the affine lines $$L_0=\{(0,0)\}\times\mathbf R\ \mathrm{and}\ L_1=\{(1,0)\}\times\mathbf R. $$ This is clearly a reducible subset of $\mathbf R^3$. The separated R-variety that does not have a separated complexification is the one obtained by gluing the open subsets $$ U_0=\mathbf R^3\setminus L_0\ \mathrm{and}\ U_1=\mathbf R^3\setminus L_1 $$ along the open subsets $$ U_{01}=U_0\cap U_1\subseteq U_0\ \mathrm{and}\ U_{10}=U_0\cap U_1\subseteq U_1 $$ via the regular isomorphism $$ \phi_{10}\colon U_{01}\rightarrow U_{10} $$ defined by $$ \phi_{10}(x,y,z)=(x,y,pz). $$ Note that this is indeed a regular isomorphism since the map $\phi_{01}=\phi_{10}^{-1}$ is the regular map $$ \phi_{01}\colon U_{10}\rightarrow U_{01} $$ defined by $$ \phi_{01}(x,y,z)=(x,y,\tfrac{z}{p}). $$

Now, it is easy to see that the R-variety $U$ one obtains is separated, as defined in the founding paper of the whole theory: Faisceaux algébriques cohérents by Jean-Pierre Serre. Indeed, one easily checks that the diagonal in $U\times U$ is closed. However, if one wants to construct a real scheme $X$ whose set of real points coincides with $U$, then, inevitably, $X$ will not be separated. Indeed, the polynomial $p$ defines a nonclosed point $x_0$ in any scheme-wise thickening $X_0$ of $U_0$ since $p$ has zeros in $U_0$, and similarly it defines a non closed point $x_1$ of any scheme-wise thickening $X_1$ of $U_1$. The gluing morphisms $\phi_{01}$ and $\phi_{10}$ will extend to open subsets $X_{01}$ of $X_0$ and $X_{10}$ of $X_1$, but they won't contain $x_0$ and $x_1$, respectively. This is because the polynomial $p$ vanishes at $x_0$. As a result, any scheme-wise thickening of $U$ will be nonseparated!

As for your second question, if I understand correctly, you are asking whether the functor $$ F\colon Sch_R'\rightarrow R-Var $$ defined by $F(X)=X(\mathbf R)$ is an equivalence onto a full subcategory, where $Sch_R'$ is the category of finite type separated reduced schemes over $Spec(\mathbf R)$ having dense sets of real points. This is an equivalence onto a full subcategory, its image category, if you localize $Sch_R'$ with respect to inlcusions of open subsets containing all real points: any morphism of $R$-varieties wil extend to a morphims defined on some open subset containing the real points. Uniqueness is implied by density of real points and separation.

As for your third question, I can't think of other differences between $R$-varieties and schemes over $\mathbf R$ that differ essentially from phenomena already present in the example above.

As for your final question about varieties in the sense of $R$-varieties over other fields, Serre certainly did define them in the paper I mentioned above. I'm not sure whether that has had a follow-up for other fields than real or algebraically closed fields.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the detailed answer. As for the second question, in the edit immediately following it I reported a couple of examples, that I found in Coste et al. just after having written that question, which show $Sch'_R\to R-Var$ cannot be full if you don't localize. If I get it correctly, your answer now adds the observation that localizing is the only thing left to do to obtain fullness. $\endgroup$ – Qfwfq Sep 28 at 12:50
  • $\begingroup$ As for essential surjectivity, your "non separated scheme theoretic thickening" example shows that $F$ cannot be ess. surjective (cause my $Sch_R'$ by definition only contained separated schemes). On the other hand we can't just extend $F$ from the cat. $Sch_R''$ of all possibly non separated schemes (with the rest of conditions unchanged) cause we would land outside $R-Var$. I'm wondering if the "thickification" is a functor $G:R-Var\to Sch_R$, and if we have an adjuction with the suitable extention $\tilde{F}:G(R-Var)\to R-Var$ where $G(R-Var)\subset Sch_R$ is the essential image in schemes. $\endgroup$ – Qfwfq Sep 28 at 12:51
  • $\begingroup$ Well, I think I said that the functor $F$ after localization is an equivalence onto a full subcategory. As you said, it cannot be essentially surjective. $\endgroup$ – Johannes Huisman Sep 28 at 12:54
  • 1
    $\begingroup$ Concerning a thickification functor, I think it can only be defined if at both sides you allow nonseparated objects. If you localize suitably, the functor of taking-real-points is then probably an equivalence of categories, so in particular, you would have an adjunction, yes. $\endgroup$ – Johannes Huisman Sep 28 at 13:02
  • 1
    $\begingroup$ Re your first comment (and my second one): you're right. (Btw, I wasn't making a correction, just recapping and asking an additional question) $\endgroup$ – Qfwfq Sep 28 at 15:51

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.