Timeline for A symmetric embedding of manifolds
Current License: CC BY-SA 3.0
19 events
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| Mar 29, 2016 at 8:11 | comment | added | Sebastian Goette | @TomChurch Having thought about hyperkähler manifolds: if there is an $S^2$ worth of Kähler forms, then $\gamma_i^*$ cannot act linearly on those without fixing one of them or mapping it to its antipode. So, $M$ cannot be an irreducible, nonflat hyperkähler manifold. | |
| Mar 29, 2016 at 7:20 | comment | added | Ali Taghavi | @SebastianGoette thanks again for your very helpful comments. | |
| Mar 28, 2016 at 18:51 | vote | accept | Ali Taghavi | ||
| Mar 28, 2016 at 15:18 | comment | added | Sebastian Goette | @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? | |
| Mar 28, 2016 at 14:35 | comment | added | Tom Church | @SebastianGoette: do you mean that M cannot be a $4k$-dimensional Kähler manifold? | |
| Mar 28, 2016 at 2:40 | answer | added | Vitali Kapovitch | timeline score: 20 | |
| Mar 27, 2016 at 22:20 | comment | added | Sebastian Goette | $S^2\subset\mathbb R^3$ 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. | |
| Mar 27, 2016 at 22:16 | comment | added | Sebastian Goette | Dear Ali, I forgot to assume that $M$ is connected. Otherwise, you could embed a connected component to $(0,\infty)^n$ and add $2^n-1$ mirror images. If $M$ is connected, then each generator $\gamma_i$ must have a fixed point $x$ (i.e., $x\in M$ with $x_i=0$). If $\gamma_i$ acts trivially on $T_xM$, it acts trivially on $M$. Otherwise, the one-dimensional $(-1)$-eigenspace of $\gamma_i$ will lie inside $T_xM$. In this case, $\gamma_i$ reverses orientation as claimed. | |
| Mar 27, 2016 at 20:35 | comment | added | Ali Taghavi | In the previous comment i mean;.....(a closed compact manifold with fixed point property....). | |
| Mar 27, 2016 at 20:30 | history | edited | Ali Taghavi |
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| Mar 27, 2016 at 20:25 | comment | added | Ali Taghavi | @SebastianGoette thank you for your very interesting comment. is it obvious that the restriction of a reflection to M is either identity or orientation reversing?Moreover may you elaborate about orientable compact surface?Moreover, i guess that a compact manifold with fixed point property, does not admit such symmetric embedding. do you think that my guess is true? | |
| Mar 27, 2016 at 17:34 | comment | added | Sebastian Goette | 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. | |
| Mar 27, 2016 at 17:34 | comment | added | José Figueroa-O'Farrill | Sorry, I misread the question. I was thinking of only one reflection. | |
| Mar 27, 2016 at 16:26 | comment | added | Ali Taghavi | @JoséFigueroa-O'Farrill the natural embedding of $\mathbb{R}^{m}$ into $\mathbb{R}^{2m}$ is a symmetric embedding, but obviously it does not solve the main question of this post.yes?, | |
| Mar 27, 2016 at 14:52 | comment | added | José Figueroa-O'Farrill | Why can't you simply embed $M$ into $\mathbb{R}^m$ in any way possible and then embed the $\mathbb{R}^m$ symmetrically into $\mathbb{R}^{2m}$? | |
| Mar 27, 2016 at 13:10 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Mar 27, 2016 at 12:50 | history | edited | Ali Taghavi |
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| Mar 27, 2016 at 12:35 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Mar 27, 2016 at 12:27 | history | asked | Ali Taghavi | CC BY-SA 3.0 |