$\newcommand{\complex}{\mathbb{C}}\newcommand{\real}{\mathbb{R}}\newcommand{\proj}{\mathbb{P}}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Seg{Seg}$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?



Is the Segre embedding of two real varieties a real variety?
$\mathrm{Seg}(X\times Y)$ 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 $\mathbf P(\mathrm{Sym}^2)$.


