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

  • $\begingroup$ Segre varieties are determinantal, and their defining equations have integer coefficients. $\endgroup$ Oct 6, 2022 at 10:02
  • $\begingroup$ 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.) $\endgroup$
    – YCor
    Oct 6, 2022 at 10:06
  • $\begingroup$ @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? $\endgroup$
    – Ben
    Oct 6, 2022 at 10:08
  • $\begingroup$ @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 $R^n$. $\endgroup$
    – Ben
    Oct 6, 2022 at 10:13
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 6, 2022 at 20:57

1 Answer 1


$\mathrm{Seg}(X\times Y)$ 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 $\mathbf P(\mathrm{Sym}^2)$.


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