Timeline for Does every positive-definite integral lattice admit an angle-preserving homomorphism into $\Bbb Z^n$ for some $n$?
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24 events
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| Aug 6, 2019 at 16:51 | history | edited | S. Carnahan♦ | CC BY-SA 4.0 |
Fixed title (won't bother with the question body)
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| Aug 6, 2019 at 6:10 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Aug 6, 2019 at 5:25 | vote | accept | M. Winter | ||
| Aug 5, 2019 at 21:41 | comment | added | Victor Protsak | I suspect that what caused most of the confusion was omitting the word "Euclidean", as in "integral Euclidean lattice". This has been ameliorated somewhat by adding routine definitions, at the expense of readability. However, the notions of isomorphism and sublattice in the current version of the question are non-standard: it is more common to define isomorphism to be an isometry, so that, in particular, a bijective isomorphism satisfies the definition of an embedding and the image of an isomorphism ${\mathcal L}\to {\Bbb Z}^n$ is a sublattice according to the chosen definitions. | |
| Aug 5, 2019 at 19:02 | history | edited | GH from MO |
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| Aug 5, 2019 at 18:51 | answer | added | Ian Agol | timeline score: 25 | |
| Aug 5, 2019 at 15:26 | comment | added | M. Winter | @Qfwfq Thank you. Especially the second paper seems to answer my question (also, if I understood correctly). While they use a fixed $\alpha\in\sqrt{\Bbb N}$, they prove a lower bound on the dimension in which lattices exist that are not a sublattice of $\Bbb Z^n$ with this particular $\alpha$. This lower bound is unbounded in $\alpha$, which means that in any fixed dimension some $\alpha$ will work. I think that the restriction of $\alpha$ to be a square root will not be a problem. | |
| Aug 5, 2019 at 15:17 | comment | added | Qfwfq | By the way, googling a bit I found a criterion for embeddability in the standard lattice of the same rank with your $\alpha=1$ (arxiv.org/pdf/1807.05098.pdf) and something related with embeddability in higher rank standard lattice but -if I have understood well the abstract- with your constant $\alpha$ fixed (jstor.org/stable/2398341?seq=1#page_scan_tab_contents). | |
| Aug 5, 2019 at 14:41 | comment | added | M. Winter | @Qfwfq Exactly my concern. I am not aiming for saturated sublattices. But I do not know whether it makes a difference when we can choose the dimension of the embedding. | |
| Aug 5, 2019 at 14:39 | comment | added | Qfwfq | I guess one possible source of doubt for me was that considering only sublattices of the form $span \{v_1,\ldots, v_k\}\cap\mathbb{Z}^n$ you are only obtaining saturated sublattices of $\mathbb{Z}^n$ (see e.g. definition 2.10 of this doc that I found on google: arxiv.org/pdf/1702.03125.pdf) | |
| Aug 5, 2019 at 14:33 | comment | added | M. Winter | @Qfwfq Still, your example showed me that my definition of sublattice might not exactly be what I initially aimed for. In fact, I am actually more looking for an angle-preserving embedding of any integral lattice $\mathcal L$ into some $\Bbb Z^n$. That is I am looking for a group homomorphism $\phi:\mathcal L\to\Bbb Z^n$ that preserves inner products up to a factor. I will leave the question as it is for now, since I am not certain whether this distinction makes any difference. | |
| Aug 5, 2019 at 14:26 | comment | added | Qfwfq | Yep, you're right. Deleted previous comment. | |
| Aug 5, 2019 at 14:23 | comment | added | M. Winter | @Qfwfq I might err, but isn't the lattice spanned by $(1,0,0,0,0)$ and $(0,1,1,1,1)$ in $\Bbb Z^5$ isomorphic to $\Bbb Z\oplus 2\Bbb Z$? | |
| Aug 5, 2019 at 14:20 | review | Close votes | |||
| Aug 8, 2019 at 0:57 | |||||
| Aug 5, 2019 at 14:15 | comment | added | M. Winter | @YCor My edit should address your question. Also, $\Bbb Z\times \{0\}$ does not fit my definition of a sublattice of $\Bbb Z^2$. | |
| Aug 5, 2019 at 14:14 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Aug 5, 2019 at 14:09 | comment | added | M. Winter | @Qfwfq Do my edits address your questions? What I mean by isomorphic is explained immediately after I use the term "isomorphic": $\mathcal L_1$ and $\mathcal L_2$ are isomorphic if they are isomorphic as groups (via some group isomorphism $\phi:\mathcal L_1\to\mathcal L_2$), and there exists a constant $\alpha\in\Bbb R$, so that $$\langle \phi(v),\phi(w)\rangle=\alpha \langle v,w\rangle,\quad\text{for all $v,w\in\mathcal L_1$}.$$ | |
| Aug 5, 2019 at 14:09 | history | edited | paul garrett | CC BY-SA 4.0 |
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| Aug 5, 2019 at 14:07 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Aug 5, 2019 at 13:54 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Aug 5, 2019 at 13:19 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Aug 5, 2019 at 13:17 | comment | added | YCor | "Is a lattice" makes no sense, you should specify: a lattice in some subspace? in some given subspace? Do you consider $\mathbf{Z}\times\{0\}$ as sublattice of $\mathbf{Z}^2$? | |
| Aug 5, 2019 at 13:12 | history | edited | M. Winter | CC BY-SA 4.0 |
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| Aug 5, 2019 at 13:04 | history | asked | M. Winter | CC BY-SA 4.0 |