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I asked the same question on math.stackexchange recently (https://math.stackexchange.com/questions/2134978/is-it-possible-to-orbifold-torus-td-into-a-sphere-sd-using-mathbbz-2), but it didn't receive much attention there, so I decided to move the question to this forum.

Consider a torus $T^d$ constructed as a hypercube $[-1,1]^d$ with identified opposite faces. Then additionally identify pairs points $\boldsymbol{r}$ and $-\boldsymbol{r}$ within this cube. For $d>1$, what is the result of this orbifold quotient? Specifically, is the underlying space homeomorphic to sphere $S^d$ or not?

For $d=2$ it is easy to visualize this procedure simply by taking half of a square $[-1,1]^2$ and by properly "gluing" the edges. It turns out that indeed $T^2/\mathbb{Z}_2\cong S^2(2,2,2,2)$, which has underlying space $S^2$. I am wondering whether such a statement (or a similar one) also generalizes to $d\geq 3$.

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    $\begingroup$ Are you sure it's true for d=1? That seems to me homeomorphic to the interval, not the circle. By the way, this question seems very related mathoverflow.net/questions/213001 $\endgroup$
    – j.c.
    Commented Feb 23, 2017 at 14:13
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    $\begingroup$ It's not going to be homeomorphic to any manifold. The origin will be locally a cone over a projective space, and this is a disc only in dimension 2. $\endgroup$
    – Bruno Martelli
    Commented Feb 23, 2017 at 14:28
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    $\begingroup$ Do you have $2^{2d}$ fixed points, right? My impression is that for $d$ even this is homeomorphic to the quotient of a $d$-dimensional complex torus by the Kummer involution $(-1)$, whose quotient is not a sphere for $d >2$. For instance, when $d=4$ the quotient is homeomorphic to the quotient of a $2$-dimensional complex torus by the action of $(-1)$, and the quotient is a Kummer surface, namely a complex $K3$ surface with sixteen conical singularities, which is obviously not homeomorphic to $S^4$ (for instance, because the Euler number is $8$). $\endgroup$
    – Francesco Polizzi
    Commented Feb 23, 2017 at 14:29
  • $\begingroup$ @j.c. You are perfectly right, I screwed up the $d=1$ case... so there is perhaps nothing to generalize to higher dimensions. $\endgroup$
    – Tomáš Bzdušek
    Commented Feb 23, 2017 at 14:41
  • $\begingroup$ @FrancescoPolizzi: I think it's only $2^d$ stable points. E.g. for $d=2$ these are $[0,0]$, $[0,1]$, $[1,0]$ and $[1,1]$. $\endgroup$
    – Tomáš Bzdušek
    Commented Feb 23, 2017 at 14:43

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Given the specific quotient described in the question, the answer is no. Under this quotient there is a fixed point at the origin and the neighborhood of this point quotients to a cone over $RP^{d-1}$. For $d=2$, $RP^1=S^1$ so the cone has underlying space a disk, however for $d>2$, the underlying cone will not be homeomorphic a $d$-ball and so it is not even a manifold (a necessary condition to be $S^d$).

For $d\geq 3$, even if we allow more interesting fixed points sets (say knots and links), we still cannot obtain $S^d$ as the 2-fold quotient (or in fact cyclic) of $T^d$. This is more commonly said in reverse that $T^d$ is not the cyclic-branched cover over $S^d$, which follows from:

Hirsch, U., & Neumann, W. D. (1975). On cyclic branched coverings of spheres. Mathematische Annalen, 215(3), 289-291.

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