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Assume that $M$ is a manifold.

Is there an embedding of $M$ in some $\mathbb{R}^{n}$ such that the image of $M$ in $\mathbb{R}^{n}$ is invariant under each reflection $(x_{1},x_{2},\ldots x_{i},\ldots,x_{n}) \mapsto (x_{1},x_{2},\ldots , -x_{i},\ldots,x_{n})$, for all $i\in \{1,2,\ldots ,n\}$?

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    $\begingroup$ Why can't you simply embed $M$ into $\mathbb{R}^m$ in any way possible and then embed the $\mathbb{R}^m$ symmetrically into $\mathbb{R}^{2m}$? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Mar 27, 2016 at 14:52
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    $\begingroup$ Sorry, I misread the question. I was thinking of only one reflection. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Mar 27, 2016 at 17:34
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    $\begingroup$ This would imply first of all that $(\mathbb Z/2\mathbb Z)^n$ acts on $M$ in a nontrivial way. More precisely, if the $i$-th of the $n$ generators acts trivially, then the embedding factors through $\mathbb R^{n-1}$, where the $i$-th coordinate is missing. If a generator acts nontrivially, it has to reverse orientation. If $M$ was compact and orientable, this has consequences for the cohomology ring (e.g., $M$ cannot be an even-dimensional Kähler manifold), and all the Pontryagin numbers would have to vanish. On the other hand, for orientable compact surfaces, such embeddings obviously exist. $\endgroup$
    – Sebastian Goette
    Commented Mar 27, 2016 at 17:34
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    $\begingroup$ $S^2\subset\mathbb R^3$ is easy. To embed a compact orientable surface of genus $g>0$ into $\mathbb R^3$, assume the centers of its "holes" are at $(g-1,0,0),(g-3,0,0),\dots, (1-g,0,0)$, and the circles in the $xy$-plane of radius $1/2$ around these centers are contained in $M$. Now, build the rest of $M$ as symmetrically around these circles as possible. $\endgroup$
    – Sebastian Goette
    Commented Mar 27, 2016 at 22:20
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    $\begingroup$ @TomChurch I was a bit too quick. If $\omega$ is the Kähler form, then $\gamma_i^*\omega=\pm\omega$ is impossible because then $\gamma_i^*\omega^{2k}=\omega^{2k}$, but $\gamma_i$ reverses orientation. But $\gamma_i^*\omega$ is a Kähler form, too, so we have at most hyperkähler holonomy (or maybe a symmetric space?). So I was wrong: at least $T^{4k}$ admits such an embedding. However, $K3$ does not because its signature is nonzero (which is a Pontryagin number by Hirzebruch's theorem). So what about higher dimensional hyperkähler manifolds and Kähler symmetric spaces? $\endgroup$
    – Sebastian Goette
    Commented Mar 28, 2016 at 15:18

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No. There exist closed manifolds which do not admit any compact group actions, in particular, no $Z_2$ actions. For example, Shultz showed in "Group actions on hypertoral manifolds. II." that in dimensions $\ge 4$ every oriented cobordism class contains such a manifold.

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