Is the Segre embedding of two real varieties a real variety?

$$\newcommand{\complex}{\mathbb{C}}\newcommand{\real}{\mathbb{R}}\newcommand{\proj}{\mathbb{P}}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Seg{Seg}$$I apologize in advance for my naïve understanding of real algebraic geometry. I define a real projective variety to be a subset of $$\mathbb{P}(\real^n)$$ that is the zero locus of some finite collection of real homogeneous polynomials $$f_1,...,f_p \in \real[x_1,...,x_n]$$.

Let $$X, Y \subseteq \mathbb{P}(\real^n)$$ be real projective varieties, and let

$$\Seg: \proj(\real^n)\times \proj(\real^n) \rightarrow \proj(\real^{n}\otimes \real^n)$$

be the Segre embedding $$(v,w)\mapsto v\otimes w$$. Is it true that $$\Seg(X \times Y) \subseteq \proj(\real^n \otimes \real^n)$$ is a real projective variety?

Assuming that $$\Seg(X \times Y)$$ is a real projective variety, I have a follow-up question: Let $$\Pi\in \Hom_{\real}(\real^n \otimes \real^n)$$ be the linear map that acts as $$\Pi(v \otimes w)=\frac{1}{2}(v\otimes w + w \otimes v)$$. Is it true that $$\Pi(\Seg(X \times Y)) \subseteq \proj(\real^n \otimes \real^n)$$ is a real projective variety?

• Segre varieties are determinantal, and their defining equations have integer coefficients. Oct 6, 2022 at 10:02
• Is the question equivalent to whether the homomorphism (to the Segre variety) is surjective on real points? (In principle it makes no sense to say that a subset of the real projective space is a variety. However, it makes sense to ask whether it is Zariski closed.)
– YCor
Oct 6, 2022 at 10:06
• @FrancescoPolizzi You claim that the equations for $Seg(X\times Y)$ are easy to write down in terms of the equations for $X$ and $Y$ and the $2\times 2$ determinants?
– Ben
Oct 6, 2022 at 10:08
• @YCor Why doesn’t it make sense? Perhaps I should rephrase everything in terms of real varieties in $R^n$ that form cones. Or maybe this is true for arbitrary real varieties in $R^n$.
– Ben
Oct 6, 2022 at 10:13
• Each real algebraic subset of $\mathbb{R}^n$ has a canonical structure of affine algebraic variety over $\mathbb{R}$. Some authors use this implicitly. But for this question, to avoid ambiguity I would not say $X,Y$ are varieties, but rather algebraic sets. Oct 6, 2022 at 20:57

$$\mathrm{Seg}(X\times Y)$$ is a real projective variety since the full Segre map is an isomorphism of real algebraic varieties onto its image.
As for your second question, I think the answer is "no". If you take $$X=Y=\mathbf{P}(\mathbf R^2)$$, the composition $$\Pi\circ\mathrm{Seg}$$ is nothing but a quotient map for the action of the symmetric group $$S_2$$ on $$\mathbf{P}(\mathbf R^2)\times\mathbf{P}(\mathbf R^2)$$. The diagonal is the set of fixed points of this action. In general, if a finite group acts on a smooth real algebraic variety and if there are isotropy groups of even order, the quotient is only semi-algebraic and not real algebraic.
Here, this can be seen easily: if you denote the basis of $$\mathbf{R}^2$$ by $$X,Y$$, the image of $$\Pi\circ\mathrm{Seg}$$ is contained in the projectivization $$\mathbf P(\mathrm{Sym}^2)$$ of the symmetric square $$\mathrm{Sym}^2$$ of $$\mathbf{R}^2$$ whose basis is $$X^2,XY,Y^2$$. The image is the subset of $$\mathbf P(\mathrm{Sym}^2)$$ of nonzero real quadratic forms in $$X,Y$$ having positive discriminant, which is clearly a semi-algebraic and non-algebraic subset of $$\mathbf P(\mathrm{Sym}^2)$$.