# Why does a complex linear normalization of a real algebraic surface inherit a real structure?

Could you recommend any references to (some of) the following very basic assertions in algebraic geometry? (It seems unreasonable to reprove them in a research paper.)

(1) Let a surface $$X$$ in $$\mathbb{P}^n$$ be the solution set of a system of polynomial equations with real coefficients. Assume that $$X$$ has a smooth linear normalization $$\bar X$$ in $$\mathbb{P}^N$$. Then the complex conjugation on $$\mathbb{P}^n$$ lifts to an antiholomorphic involution on the linear normalization $$\bar X$$ (i.e., a normalization map $$\bar X\to X$$ commutes with the involutions on $$\bar X$$ and $$X$$).

EDIT3. The following examples show that this is not completely trivial:

Ex1. Even if $$\bar X$$ is a solution set of a system of polynomial equations with real coefficients as well, a complex normalization map $$\bar X \to X$$ may not be given by real polynomials, and complex conjugation in $$\mathbb{P}^N$$ may not be the required antiholomorphic involution on $$\bar X$$. E.g., the cone $$\bar X=\{x^2-y^2-z^2=0\}$$ in $$\mathbb{P}^3$$ is a normalization of the cone $$X=\{x^2+y^2-z^2=0\}$$ in $$\mathbb{P}^3$$ but the map $$\bar X\to X$$, $$(x,y,z)\mapsto (x,iy,z)$$ is not given by polynomials with real coefficients. And the map is indeed a complex linear normalization map because the latter is only defined up to composition with a complex isomorphism.

Ex2. The lift of a fixed point may not be a fixed point. E.g., the curve $$y^2=x^2(x-1)$$ in $$\mathbb{P}^2$$ has linear normalization $$(1:t:t^2:t^3)$$ in $$\mathbb{P}^3$$. A normalization map is $$x(t)=t^2+1, y(t)=t^3+t$$ (or more accurately $$(x_0:x_1:x_2:x_3)\mapsto (x_0:x_0+x_2:x_1+x_3)$$). Then the real point $$(x,y)=(0,0)$$ of the curve is covered by two distinct complex conjugate points $$(1:\pm i:-1:\mp i)$$ of the linear normalization.

Assertion (1) might seem a tautology, but it has highly nontrivial consequences: e.g., (1)-(3) together with [1, Theorems 5-7 and Proposition 1] imply that a (nonruled) surface in $$\mathbb{P}^n$$ with a 2-dimensional set of real points parametrized by complex polynomials of degree $$2$$ has a parametrization by real polynomials of degree $$2$$ as well. And analogous assertion for higher degree polynomials does not remain true [1, Remark 4] although there is still a nice map $$\mathbb{C}P^2\to X$$, just not a linear mormalization anymore.

(2) (The closure of) the Veronese surface $$(1:u:v:u^2:uv:v^2)$$ in $$\mathbb{P}^5$$ is biregular to $$\mathbb{P}^2$$.

(3) The only antiholomorphic involution of $$\mathbb{P}^2$$ up to projective automorphism is the complex conjugation.

And also a question:

(4) Is (the closure of) the ruled surface $$(1:u:v:u^2:uv)$$ in $$\mathbb{P}^4$$ smooth? Which 'standard' surface is it isomorphic to? What are the antiholomorphic involutions on (a desingularization of) the surface?

Ideally, a reference to a particular published theorem, which can be just applied as is' by a nonspecialist, is requested. Notice that (2)-(3) are mentioned in wikipedia and mathoverflow, but without a proof or a reference.

One should remark that there are some references where a version of (1) is stated even without the assumption that $$\bar X$$ is smooth but these versions cannot be correct just because an antiholomorphic involution' is undefined for a nonsmooth surface (https://en.wikipedia.org/wiki/Complex_manifold). EDIT: In fact a generalization to locally ringed spaces is used there without mentioning that; thanks to Angelo for clarification.

[1] J. Schicho, The multiple conical surfaces // Contrib. Algeb. Geom. 42:1 (2001), 71-87.

• I am a bit confused by what you're asking. Are you asking why the normalization of a variety defined over $\mathbb{R}$ is defined over $\mathbb{R}$? Nov 18, 2019 at 21:38
• @DenisNardin Asking about assertion (1) precisely. A reference to a precise statement of the result you mention ("normalization of a variety defined over R is defined over R") would be helpful. Maybe it implies (1), cannot see without a precise statement (does or does not it say that the projection $\bar X\to X$ is real as well?). Nov 19, 2019 at 7:09
• I think there's some kind of language barrier, but yes the variety $\tilde{X}$ is defined by a set of equations with real coefficients in $\mathbb{P}^N$ for some $N$ and the projection $\tilde{X}\to X$ is determined by polynomials with real coefficients. I'm not sure where it is spelled out, but it is immediate from the standard construction of normalization for schemes (see for example section 4.1.2 in Liu's Algebraic Geometry and Arithmetic Curves, in particular proposition 4.1.27 there). Nov 19, 2019 at 7:21
• I'm a bit confused by what your counterexample is supposed to show. The real plane projective curve $X$ with equation $y^2=x^2(x-1)$ has normalization $\mathbb{P}^1\to X$ given by $t\mapsto(1+t^2,t+t^3)$ (projectivizing all this, of course): I agree that the real singular point $(0,0)$ on $X$ does not come from a real point on $\mathbb{P}^1$, but the normalization can still be given by real polynomials. Nov 24, 2019 at 9:52
• @MikhailSkopenkov I am confused. You seem to be considering the normalization as given by a given embedding in some projective space. This is not at all how I intend the term: for me the normalization is an abstract algebraic variety, and all I'm saying is that there exists some embedding in $\mathbb{P}^n$ given by real coefficients such that the projection $\tilde{X}\to X$ is given by real polynomials (this is because the map $\tilde{X}\to X$ is finite and so projective, moreover it is a morphism of varieties over $\mathbb{R}$ by definition!). I suspect some miscommunication is present. Nov 24, 2019 at 10:06

The notion of anti-analytic involution is perfectly well defined for general analytic spaces: it is an involution of locally ringed spaces, that is antilinear with respect to complex scalars. Any projective variety $$X \subseteq \mathbb P^n$$ defined by equations with real coefficients has an antilinear involution, obtained by restricting the antilinear involution on $$\mathbb P^n$$.

If you use the language of schemes, very suitable for this type of questions, the point is the following. If $$X$$ is a projective scheme over $$\mathbb R$$, the base change $$X_{\mathbb C} := \mathop{\rm Spec}{\mathbb C} \times_{\mathop{\rm Spec}{\mathbb R}}X$$ has an involution $$X_{\mathbb C} \simeq X_{\mathbb C}$$, coming from conjugation. The normalization of a projective scheme over $$\mathbb R$$ is also a projective scheme over $$\mathbb R$$, and it stays normal after base changing to $$\mathbb C$$.

But you can avoid thinking about schemes defined over $$\mathbb R$$. Let $$X \subseteq \mathbb P^n$$ be a projective variety defined by equations with real coefficients, $$\nu\colon \overline X \to X$$ the normalization map, $$\tau\colon X \simeq X$$ the anti-analytic involution. The normalization has a universal property: if $$f\colon Y \to X$$ is an analytic, or anti-analytic, map from a (reduced) variety $$Y$$, which does not map any irreducible component of $$Y$$ into the non-normal locus of $$X$$, then $$f$$ lifts uniquely to an analytic, or anti-analytic, map $$Y \to \overline X$$. This allows you to lift the involution $$\tau$$ to an anti-analytic involution of $$\overline X$$.

[Added later]: maybe the scheme-theoretic proof is not so bad, it boils down to a simple fact in commutative algebra. First of all, the existence of the anti-analytic involution is local in the Zariski topology: since the involution is unique, if it exists, local involutions can be glued together. So, we can assume that $$X$$ is an algebraic subvariety of $$\mathbb C^n$$, so it corresponds to a finitely generated $$\mathbb R$$-algebra $$A$$. Let $$\overline A$$ be the normalization of $$A$$; then my claim is that $$\overline X$$ corresponds to the algebra $$A\otimes_{\mathbb R} \mathbb C$$. More concretely, if $$\overline A = \mathbb R[x_1, \dots, x_m]/(f_1, \dots, f_r)$$ (the normalization of a finitely algebra over a field is also finitely generated), then $$\overline A\otimes_{\mathbb R} \mathbb C = \mathbb C[x_1, \dots, x_m]/(f_1, \dots, f_r)$$. This defines the anti-holomorphic involution: it comes from conjugation in $$\mathbb C$$.

To check the fact above, call $$K$$ the field of fractions of $$A$$; then the field of rational functions of $$A\otimes_{\mathbb R} \mathbb C$$ is $$K\otimes_{\mathbb R} \mathbb C$$. This boils down to the fact that an element $$f + i g$$ (here $$f$$ and $$g$$ are in $$K$$) is integral over $$A \otimes_{\mathbb R} \mathbb C$$, or, equivalently, over $$A$$, if and only if $$f$$ and $$g$$ are. One direction is standard: if $$f$$ and $$g$$ are integral, so is $$f + ig$$, because $$i$$ clearly is integral. In the other direction, if $$f + ig$$ is integral over $$A$$, so is its conjugate $$f + ig$$, and so $$f = ((f+ig)+(f-ig))/2$$ is integral, and analogously for $$g$$.

• Thank you very much for a clear explanation, especially for the last down-to-Earth part avoiding schemes. Notice that the question is primarily about a reference, not a proof (a reference to mathoverflow is not reliable enough for a research paper). Could you recommend a reference to the mentioned universal property, if you have time? So far looked up Liu's Algebraic Geometry and Arithmetic Curves - but Definition 1.19 there asserts universality only for dominant morphisms, not for `involutions of locally ringed spaces, that are antilinear with respect to complex scalars'. Nov 20, 2019 at 9:13
• (cont'd) And $\tau:X\to X$ is not a morphism and cannot be anti-analytic in the usual sense, once $X$ is not smooth. Thank you also for the explanation what is meant by anti-analytic involution in the books; this comment is really helpful (because the authors seem to find this too obvious to mention:). Finally, a bit offtopic remark: imho generalization to locally ringed spaces does not prove (1) but just moves the issue to a deeper level - the required properties of the spaces seem to be harder to prove than (1) itself. But let me apologize for too much muttering, and thank you again! Nov 20, 2019 at 9:21
• I added a different, more algebraic proof. It's very standard commutative algebra, not something I would put in a paper. Nov 20, 2019 at 17:28
• Thank you very much again! Personally, would include this type of argument into a paper, if there is no reference available, especially if the rest of the paper is far from algebraic geometry and commutative algebra. Just to make sure: does your argument imply in particular that the projection $\bar X\to X$ is given by linear polynomials with real coefficients? Nov 21, 2019 at 4:24
• Yes, sure, the projection is also defined by real polynomial maps. Nov 21, 2019 at 7:18

Because of the lack of references, the detailed elementary proofs of (2)-(3) and a version of (1) have been written down in arXiv:2002.01355.