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Let $G$ be a finite group and let $X_g$ be variables indexed by $G$. Consider the complex algebraic set defined by \begin{align} X_e &= 0\\ X_g &= X_{g^{-1}}\;\;\text{ for all }g\in G,\\ X_g &= X_{g'}\;\;\text{ whenever $g$ and $g'$ are conjugate in $G$},\tag{1}\\ \sum_{h\in G}X_hX_{h^{-1}g} &= X_g-X_g^2\;\;\text{ for all }g\in G\setminus\{e\}\tag{2}. \end{align} Note that condition (1) is vacuous if $G$ is abelian. It seems to be that every complex solution of this system is actually real. I have verified this for all groups of order $\le15$, all non-abelian groups of order $\le23$, and the non-solvable groups $\text{Alt}_5$ and $\text{Sym}_5$. So I expect that this holds in general and that there should be a simple reason which I fail to see. (Something like the slick argument that eigenvalues of real symmetric matrices are real.)

Question: Does anyone know this or some related result, or can tell a simple reason for that?

Remark 1: This empirical observation came up when playing with Conjecture 6.1 in this preprint by Deninger, Grundhöfer and Kramer.

Remark 2: Summing the equations (2) for $g\in G\setminus\{e\}$ shows that each complex solution $x_g$ satisfies $\sum_{g\in G}x_g=0$ or $1$. In the latter case $y_g$ with $y_g=x_g-1/(|G|-1)$ for $g\ne e$ and $y_e=0$ is a solution of the system too. So without loss of generality we may add the condition $\sum_{g\in G}X_g=0$.

Remark 3: If one drops condition (1), then $12$ is the smallest order of a group for which the system has a non-real solution.

Remark 4 (regarding Benjamin Steinberg's comment): Indeed, it is tempting to write a solution $x_g$ as a sum $x_g=\sum a_{\chi}\cdot\chi(g)$, where $\chi$ runs through the irreducible characters of $G$, and rewrite the conditions on $x_g$ in terms of the coefficients $a_{\chi}$. However, for instance if $G$ is cyclic, then the conditions (and the question of reality) on the $a_{\chi}$'s happen to be exactly the same as those on the $x_g$'s.

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    $\begingroup$ The algebraic set in question (call it $Y$) is defined over $\mathbb{Q}$. If the property is true, and $y\in Y(\mathbb{C})$ is any point, then $\sigma(y)$ is real for every endomorphism $\sigma$ of $\mathbb{C}$. This means that $y\in Y(F)$ where $F\subset \overline{\mathbb{Q}}$ is the field of totally real numbers. We conclude that $Y$ is finite and consists of totally real points. $\endgroup$ Jun 28, 2022 at 16:29
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    $\begingroup$ If you require this to be a class function then It is a linear combination of characters. Can you not exploit that characters take conjugate values on inverses? $\endgroup$ Jun 28, 2022 at 19:28
  • $\begingroup$ @BenjaminSteinberg Indeed I had tried that. But for instance if $G$ is cyclic one is back to the original question. The convolution becomes a square, and the square a convolution. $\endgroup$ Jun 28, 2022 at 20:01

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