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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SL{SL}$By Will Sawin's answer to Moduli Spaces of Higher Dimensional Complex Tori the moduli space of complex $d$-tori is $X = \GL_{d}(\mathbb{C})\backslash{\GL_{2d}(\mathbb{R})}/{\GL_{2d}(\mathbb{Z})}$. What would happen if complex structures which are anti-biholomorphic equivalent were also identified? If the resulting space is $Y$ would the canonical map $Y \to X$ be a $2$-sheeted covering? Is there a reference for such moduli spaces?

I don't think so because if $d = 1$ it would be
$$X \cong \mathbb{R}^*{\SO(2)}\backslash{\GL_2(\mathbb{R})}/{\GL_2(\mathbb{Z})} \cong \SO(2)\backslash{\SL_2(\mathbb{R})}/{\SL_2(\mathbb{Z})} \cong \mathbb{H}/{\SL_2(\mathbb{Z})}\cong \mathbb{R^2}$$ and $$Y \cong \mathbb{R}^*O(2)\backslash{\GL_2(\mathbb{R})}/{\GL_2(\mathbb{Z})} \cong (\SO(2)\backslash{\SL_2(\mathbb{R})})/{\GL_2(\mathbb{Z})}\cong \mathbb{H}/{\GL_2(\mathbb{Z})} \cong \mathbb{R}^2_+ = \{(x,y)\in \mathbb{R}^2 \ | \ y \geq 0\}$$ so $X \to Y$ cannot be a $2$-sheeted covering?

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    $\begingroup$ You can probably realize $Y$ as an orbifold with $X$ its 2-sheeted covering in orbifold sense. But as you note, it will not be so as topological spaces. $Y$ will not be a manifold, it will have a "boundary" corresponding to tori isomorphic to their conjugates. $\endgroup$
    – Wojowu
    Commented Sep 30 at 12:43
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    $\begingroup$ It should be the rectangular and the "isosceles tori", the latter being ones with a lattice basis of the form $\{1,\frac{1}{2}+ib\}$ for some $b\in\mathbb R$. However your homeomorphism should be correct, since the isomorphism with $\mathbb R^2\cong\mathbb C$ corresponds to taking $j$-invariant, and the invariant tori are the ones with real $j$-invariant. $\endgroup$
    – Wojowu
    Commented Sep 30 at 15:27
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    $\begingroup$ The left and right actions of negative matrices don't coincide, so you cannot just arbitrarily forget them on the right just because they act on the left too. $\endgroup$
    – Wojowu
    Commented Oct 1 at 11:14
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    $\begingroup$ @Wojowu Ok, then my first thought was correct, so the matrix $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ acts non trivially on $\mathbb{R}^*O(2)\backslash GL_2(\mathbb{R}) \cong SO(2)\backslash SL_2(\mathbb{R})$ and translates into $z \mapsto -\bar z$ on $\mathbb{H}$ which folds the upper half plane at the imaginary axis, which gives $\mathbb{R}^*O(2)\backslash GL_2(\mathbb{R})/GL_2(\mathbb{Z}) \cong \mathbb{C}/\sim (z \mapsto \bar z) \cong \mathbb{R}^2_+$? $\endgroup$
    – psl2Z
    Commented Oct 1 at 11:45
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    $\begingroup$ (My comment crossed with yours. It looks like you're figuring it out!) The "covering" map is the quotient map from the cusped orbifold to the triangle. This is a covering of orbifolds, but not of topological spaces. $\endgroup$
    – HJRW
    Commented Oct 1 at 11:47

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For $d$ odd these two components are disconnected, because $I$ and $-I$ induce opposite orientation. For $d$ even, you have an involution, which takes a lattice to a complex conjugate lattice. This involution takes a lattice $\tau_1, \dotsc, \tau_{2d}$ to a complex conjugate lattice $\bar\tau_1, \dotsc, \bar\tau_{2d}$, and has many fixed points.

$\DeclareMathOperator\GL{GL}$However, be very careful about this: "By this https://mathoverflow.net/q/126329 the moduli space of complex $d$-tori is $X = \GL_{d}(\mathbb{C})\backslash{\GL_{2d}(\mathbb{R})}/{\GL_{2d}(\mathbb{Z})}$": for $d\geq 2$, this double quotient is very non-Hausdorff, because $\GL_{2d}(\mathbb{Z})$ acts on $\GL_{d}(\mathbb{C})\backslash{\GL_{2d}(\mathbb{R})}$ with dense orbits, as follows from Ratner theory. It's not very good idea to consider this quotient as an orbifold, because it has codiscrete topology: all open sets are either empty or everything. Detailed explanation is found in Ergodic complex structures on hyperkahler manifolds.

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  • $\begingroup$ Thank you for the reference. It looks interesting. What do you mean by "these two components"? The two sheets of the "covering"? And what do you mean by "orientation"? I think if we had real dimensions then $-I$ would induce the opposite orientation if and only if $d$ was odd? Further, then it is not interesting/ useless to study the moduli space for $d \geq 2$ topologically, i.e. only the $1$-dimensional case gives an interesting moduli space? $\endgroup$
    – psl2Z
    Commented Oct 4 at 23:05
  • $\begingroup$ In particular all moduli spaces for $d \geq 2$ would simply be homeomorphic to $(\mathbb{R},\tau)$ with the trivial topology $\tau$? $\endgroup$
    – psl2Z
    Commented Oct 4 at 23:12
  • $\begingroup$ Complex structure on a vector space defines the orientation. Two components are complex structures with the opposite orientation. The correspondence $I \mapsto -I$ identifies these two components when dimension is odd, and produces an involution on the Teichmuller space of complex structures when it is even. $\endgroup$
    – Misha Verbitsky
    Commented Oct 11 at 21:14
  • $\begingroup$ Ok. I was first confused and thought that $I$ was the indentity. But $I$ is the complex structure. In the case that $d$ is even, doesn't $I \mapsto -I$ produce the identity on the Teichmüller space and the moduli spaces of complex and of complex/anticomplex structures are thus equal in the case $d$ even? In the case $d$ odd $I \mapsto -I$ produces an involution on the Teichmüller space? For example in case $d = 1$ it gives $\cong \mathbb{C}$ for the moduli space complex structures and $\cong \{\text{Im}(z) \geq 0\}$ for the moduli space of complex/anticomplex structures? $\endgroup$
    – psl2Z
    Commented Oct 18 at 13:44
  • $\begingroup$ Since for $d=1$ it is $\mathbb{R}^*SO(2)\backslash GL_2(\mathbb{R})/GL_2(\mathbb{Z}) \cong \mathbb{C}$ and $\mathbb{R}^*O(2)\backslash GL_2(\mathbb{R})/GL_2(\mathbb{Z}) \cong \{\text{Im}(z) \geq 0\}$. $\endgroup$
    – psl2Z
    Commented Oct 18 at 13:48

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