Timeline for Lattice question
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2017 at 18:30 | comment | added | Francisco Santos | Requiring the fundamental unit not to have lattice points other than its vertices excludes two-dimensional counter-examples (since the only lattice polygons with that property are a unit parallelogram and half of it) and it also excludes counter-examples constructed using a Minkowski sum of segments as a fundamental unit. The latter was convenient for counter-examples since Minkowski sums have few different edge vectors. But I would expect the statement to still fail in this restrict version. | |
| Feb 4, 2017 at 18:30 | comment | added | Francisco Santos | Excluding interior points in the "fundamental unit" is certainly not enough: a counter-example with interior points can be converted into one without by Cartesian product (of both $P$ and the fundamental unit) with a segment in one extra dimension. I guess you mean that the fundamental unit does not have any lattice point apart of its vertices. | |
| Feb 4, 2017 at 15:18 | comment | added | Alex | @FranciscoSantos I think I understand why your counterexample works and what I misdefined. Let's start with a polytope $P$. And consider copies of $P$ such that all the created vertices form a lattice. Now $P$ will be a fundamental unit for this created lattice. That is not to say that there are no other polytopes which copied can lead to the same lattice. There are in fact lots of such choices. In your counterexample you try to argue using lattice points inside the fundamental unit. But in the definition I wanted, the fundamental unit has no interior lattice points.Apologies for the confusion | |
| Feb 4, 2017 at 15:11 | answer | added | Alex | timeline score: 0 | |
| Feb 2, 2017 at 18:23 | comment | added | Alex | @FranciscoSantos In the example you gave in the answer, notice that $(0, 1)$ is a side of the fundamental unit. In particular you can use it to immediately connect $(2, i)$ with $(2, j)$. | |
| Feb 2, 2017 at 15:48 | answer | added | Francisco Santos | timeline score: 1 | |
| Feb 2, 2017 at 15:26 | comment | added | Francisco Santos | No. My "fundamental unit" is the Minkowski sum of the five vectors I state. All its edges (as vectors) are parallel to one of the five defining vectors. (Less important, there was a mistake in my comment. I intended $P$ to be $[0,2]^3$ so that it is indeed a lattice polytope since I assumed that is one requirement you pose. So its lattice points are the eight vertices $\{0,2\}^3$ of the cube plus the six centers of the facets of the cube). | |
| Feb 2, 2017 at 11:14 | comment | added | Alex | @FranciscoSantos The lattice points in $P$ are $z = (0, 0, 0)$ and $z_{i, j, k, l} = (-1)^k\varepsilon_i + (-1)^l\varepsilon_j$ where $1 \leq i < j \leq 3$, $\varepsilon_i$ are the standard basis vectors in $\mathbb{R}$^3 and $k, l \in \{0, 1\}$. The fundamental unit your considering has (among others) sides which (considered as vectors) are equal to $k_1 = (0, 1, -1), k_2 = (1, -1, 0), k_3 =(2, 0, 0), k_4 = (0, 2, 0), k_5 = (0, 0, 2)$. But then lattice connectivity of $P$ is immediate. | |
| Feb 1, 2017 at 23:00 | comment | added | Francisco Santos | I still have problems with your definitions, but maybe the following is a counter-example to your question: Let $P$ be the cube $[-1,1]^3$, with respect to the lattice $L=\{(x,y,z)\in \Z^3 : x+y+z \in 2\Z\}$. Take as "fundamental unit" the (Minkowski sum of) the vectors $(2,0,0)$, $(0,2,0)$, (0,0,2)$, $(3,1,0)$ and $(0,3,1)$. | |
| Feb 1, 2017 at 14:53 | comment | added | Alex | @TMM I'm predefining what movement is. So movement, or adjacency, is predefined. This extra data that tells you where you can move is what I call fundamental unit. | |
| Feb 1, 2017 at 14:52 | comment | added | TMM | But why can/can't you move to $(x+1, y+1)$ or $(x+4, y-17)$? What defines when another point is "close enough"? If the lattice is generated by $(1, 0)$ and $(0.01, 1)$ instead, would anything change in terms of your "adjacency"? You really need to find a proper definition for your vague concepts for this to make any sense. | |
| Feb 1, 2017 at 14:47 | comment | added | Alex | @TMM Consider $\mathbb{Z}^2$ in $\mathbb{R}^2$. This is of course a lattice. Now, I would like to consider paths on this lattice. I need to define movement. I have choices in doing so. I can either say that from a point $(x, y)$ I can move in any of the points $(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)$ or, for instance, I can say that I can also move diagonally into $(x + 1, y + 1), (x - 1, y - 1)$ (on top of the previous list). The first case is the regular "math page" lattice or a rectangular lattice, while the second case is a triangular lattice. | |
| Feb 1, 2017 at 13:52 | comment | added | TMM | "However there are of courses lattices that have the same set of points L but the directions and repeating pattern distinct. How do we morally define them as "different lattices". - This in particular does not make sense - they are not different lattices. | |
| Feb 1, 2017 at 13:50 | comment | added | TMM | Lattices are well defined concepts, and your "definitions" do not seem to match this. There also does not exist a "standard lattice" per se, nor can you define a lattice as a matrix. And if two lattices contain exactly the same sets of points, then they are the same lattice - you are just talking about different bases for the same lattice. | |
| Feb 1, 2017 at 12:33 | history | edited | Alex | CC BY-SA 3.0 |
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| Feb 1, 2017 at 9:51 | comment | added | Francisco Santos | You should clearly define what you mean by "fundamental unit". Do you mean "set of non-zero shortest vectors in the lattice"? | |
| Jan 27, 2017 at 21:29 | comment | added | Alex | @mukhujje Yes and No. Yes in the sense that if there is an automorphism of $\mathbb{R}^n$ that maps one fundamental unit to the other then of course the statement holds for one lattice iff it holds for the other. And No in the sense that there is no choice for a fundamental unit to be a hexagon for instance if the lattice is spanned by the two vectors $v_1 = (1, 0)$ and $v_2 = (0, 1)$. So whatever holds for this lattice can't be transported to a hexagonal lattice for instance. | |
| Jan 27, 2017 at 18:10 | comment | added | Alex | @Fedor Petrov In the case of the triangular lattice in the second example, the fundamental unit is... well... just a triangle, that is, the points $e^{\frac{i\pi}{3}}$ and $(1, 0)$ are connected by a segment. The lattice $\mathcal{L}$ is indeed $\mathbb{Z}v_1 \oplus \mathbb{Z}v_2$ but there are choices for the fundamental unit. I chose the triangle with sides $v_1, v_2$ and $v_1 - v_2$, but of course I could've taken the paralelogram with sides $v_1$ and $v_2$. So the fundamental unit tells you what you're allowed "to draw" | |
| Jan 27, 2017 at 13:26 | comment | added | mukhujje | If you can prove for the standard lattice then can't you change basis to reduce to the standard lattice argument? | |
| Jan 27, 2017 at 12:01 | comment | added | Fedor Petrov | What do you mean by a fundamental unit? I supposed this is a parallelepiped, not a triangle. | |
| Jan 27, 2017 at 11:22 | comment | added | Alex | @FedorPetrov I made some corrections and provided a couple of examples. I am trying to prove some polytope is lattice path connected but I couldn't find any results in the literature so I'm trying to come up with all sorts of criterions when this might hold true. | |
| Jan 27, 2017 at 11:20 | history | edited | Alex | CC BY-SA 3.0 |
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| Jan 26, 2017 at 15:37 | comment | added | Fedor Petrov | I am confused. How is this possible? Would you give some example on the plane? | |
| Jan 26, 2017 at 12:04 | history | edited | Alex | CC BY-SA 3.0 |
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| Jan 26, 2017 at 11:31 | history | asked | Alex | CC BY-SA 3.0 |