# Conjugate action on a complex quasi-projective variety

Based on the joint work with Matteo Gallet The diffeomorphism type of small hyperplane arrangement is combinatorially determined, I would like to characterize those line arrangements with up to 10 lines that have combinatorially-determined topology of their complement manifold.

In particular, I need some help in order to to better understand the conjugate action on the complex realization space of a matroid.

$\textbf{Set up.}$ Let $f_{1},\ldots,f_{m}\in\mathbb{R}[t_{1},\ldots,t_{d}]$ be polynomials with real coefficients and let $X$ be the complex variety defined by $X=\{P\in\mathbb{C}^{d}\mid f_{i}(P)=0, 1\leq i\leq m\}$. Let $\sim$ be the equivalence relation defined on $X$ by $P\sim Q\Longleftrightarrow P=Q\text{ or }P=\overline{Q}$ (here $\overline{Q}$ is the point obtained conjugating componentwise the entries of $Q$). Finally, let us consider the quotient $X/\sim$.

$\textbf{Question.}$ Is that possible to explicit describe the quotient $X/\sim$, perhaps in terms of a certain family of equations?

• You probably already know this (and it might not be very useful): Let $\mathcal{O}(X)$ be the coordinate ring of $X$, so that $\mathcal{O}(X) = \mathbb C[t_1,\ldots,t_d]/(f_1,\ldots, f_m)$. If $\sigma$ is an algebraic involution of $X$, then $X/<\sigma>$ is an affine variety and its coordinate ring is the ring of $\sigma$-invariants in $\mathbb C[X]$. One can compute the latter ring explicitly. However, the involution defined by complex conjugation on $X$ is not algebraic... Nov 16 '16 at 11:32

I do think Ariyan's approach is useful. Since conjugation is an $\mathbb{R}$-homomorphism, one has to regard $X$ as a variety in $\mathbb{R}^{2 d}$. But over $\mathbb{R}$, the map of varieties given by the invariant ring will no longer be surjective.
First look at the case $X = \mathbb{R}^{2 d}$. Then a quotient is given by the surjective map $$\mathbb{R}^{2 d} \to \mathbb{R}^d \times (\mathbb{R}_{\ge 0})^d, \ (x_1,\ldots, x_d,y_1, \ldots, y_d) \mapsto (x_1,\ldots, x_d,y_1^2, \ldots, y_d^2),$$ so it is given by inequalities rather than equalities. To get the quotient for the general case, restrict the above map to $X$ and determine the image of the restriction: the image is the desired quotient $X/\sim$.
Explicitly, one needs to express the polynomials $f_i$ as polynomials in the real parts $x_i$ and the imaginary parts $y_i$ and then intersect the ideal generated by these polynomials with the subalgebra $\mathbb{R}[x_1,\ldots,x_d,y_1^2,\ldots,y_d^2]$. This can be computed with Gröbner bases. I don't think there's a general formula for the resulting ideal. If the intersection is generated by polynomials $g_i$, then then $X/\sim$ is given as a subset of $\mathbb{R}^{2 d}$ by these equations together with the inequalities saying that the last~$d$ components must be non-negative.