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Let $L$ be an discrete lattice in $\mathbb R^n$. We say that a nonzero $a\in L$ is indecomposable if and only if $a$ cannot be written as $a=b+c$ with $b,c$ nonzero and $b^T c>0$.

I was initially trying to prove that the indecomposable elements generate the Voronoi cell (also called Dirichlet domain) $V=\{x\in\mathbb R^n:|x|<|x-v| \mbox{ for all } 0\ne v\in L\}$, in the sense that if we define $H_v=\{x\in\mathbb R^n:|x|<|x-v|\}$ then $V=\cap H_a$ where the intersection runs over the indecomposable elements.

Now, I have managed to show the above by establishing that $u^Tv\ge 0$ implies $H_u\cap H_v\subset H_{u+v}$. Further I wish to show that this intersection is minimal in the sense that we cannot remove any indecomposable element and still get $V(0)$. Also it is the unique minimal such set. How is that possible?

What I am possibly thinking of is to prove that if $a,b$ are both indecomposable and distinct then we cannot have $H_a\subset H_b$. But how to prove that? I am not getting an intuition of what is an indecomposable vector.

Update: I think the last two paragraphs on Pg 57 of these notes contain the answer. But I am unable to understand them almost entirely. Can someone explain?

  • Though I am not familiar with indecomposable elements, it seems that your definitions have several issues: Firstly, what is to prevent me from writing any element $a$ as $2a + (-a)$, so that $(-a)^T(2a) = -2|a|^2$, which would imply that no non-zero $a$ is indecomposable? (And even 0 can be written $e_1 + (-e_1)$, so even that is not indecomposable.) Second, your definition of $H_v$ is just a half space, which means that any intersection of many such $H_v$'s is convex, and thus cannot be a lattice. – Yly May 30 '20 at 18:03
  • What you say is perfectly true. I took the definition from Algebraic graph theory by Chris Godsil (Pg 330) but strangely cannot find it anywhere on the internet. The book is not currently accessible to me so I can't check it. –  May 30 '20 at 19:08
  • I just checked the book you refer to, and your definitions are incongruent with those of the book. The inequality in the definition of indecomposable should be the other way around, and the intersection of half spaces yields the Voronoi cell of the origin, not the lattice (which also addresses your question). It is probably also true that the indecomposable elements (properly defined) generate the lattice, but "generate" here doesn't mean what you define it to mean, and has nothing to do with the half spaces $H_v$. – Yly May 30 '20 at 20:48
  • Thanks for pointing out the errors which I have now corrected. I think the errors were typos and my original question still stands. –  May 31 '20 at 02:08

2 Answers2

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Let $I$ be the set of indecomposable elements in $L$.

  1. Hopefully you're aware that while the condition

    $u^{\top} v \geqslant 0$ implies $H_u \cap H_v \subseteq H_{u + v}$

    is essential in proving that $V(0) = \bigcap \limits_{a \in I} H_a$, it is far from being sufficient on its own.

  2. You're trying to prove that

    • $I$ is a minimal set satisfying $V(0) = \bigcap \limits_{a \in I} H_a$, i.e. for any $b \in I$ we have that $V(0) \subsetneq \bigcap \limits_{a \in I \setminus \{ b \}} H_a$;
    • $I$ is a unique such minimal set.

    For this it suffices to prove a stronger statement:

    $(*) \quad$ If $A \subseteq L$ is a subset satisfying $V(0) = \bigcap \limits_{a \in A} H_a$, then $I \subseteq A$.

    Be advised: I'm assuming that the inequality $b^{\top} c > 0$ in the definition of indecomposability should be non-strict (i.e. $b^{\top} c \geqslant 0$), as otherwise the statement from the first bullet is false - $I$ need not be minimal. An easy counterexample is $\mathbb{Z}^2 \subseteq \mathbb{R}^2$, where $V(0)$ is generated (by means of intersecting the $H_a$'s) by just four elements: $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$, but $(1, 1)$ is also indecomposable.

    It remains to prove $(*)$. First note that $0 \notin A$ as $H_0 = \varnothing$. Now fix any $b \in I$. Clearly $\frac{1}{2} b \notin V(0)$ as $\frac{1}{2}b \notin H_b$, so there is $a \in A$ such that $\frac{1}{2}b \notin H_a$, i.e. $\| \frac{1}{2} b \| \geqslant \| \frac{1}{2} b - a \|$ or (after easy transformations) $\left< a, b-a \right> \geqslant 0$. But then we may write $b = a + (b-a)$, so by the indecomposability of $b$ one of the summands must be zero, which implies $b = a$ and therefore $b \in A$. $\square$

  3. A proof of $V(0) = \bigcap \limits_{a \in I} H_a$, requested in the comment. I will use the following lemma:

    If $A \subseteq L$ is a nonempty subset, then there is $b \in L$ such that $\| b \| = \min \limits_{a \in L} \| a \|$.

    Proof: assume for contradiction that such $b$ does not exist. Then there is a sequence $(a_n)$ of elements of $A$ such that $\| a_{n+1} \| < \| a_n \|$ for each $n$. Such a sequence must be bounded, so it has a limit point. As $L$ is a subgroup, it follows that $L$ contains points arbitrarily close to $0$, which contradicts the assumption that $L$ is a lattice, so the lemma has been proved.

    Now assume for contradiction that $\bigcap \limits_{a \in I} H_a \neq V(0)$, which means that $\bigcap \limits_{a \in I} H_a \not \subseteq H_b$ for some $b \in L \setminus \{ 0 \}$. By the lemma, we can assume that $b$ has the smallest distance to $0$ of all elements of $L \setminus \{ 0 \}$ with that property, i.e. $\bigcap \limits_{a \in I} H_a \subseteq H_c$ for every $c \in L \setminus \{ 0 \}$ with $\| c \| < \| b \|$.

    Now clearly $b \notin I$, so write $b = c+d$ where $c, d \neq 0$ and $\left< c, d \right> \geqslant 0$. Then $\| b \|^2 = \| c \|^2 + \| d \|^2 + 2 \left< c, d \right> \geqslant \| c \|^2 + \| d \|^2$, thus $\| c \|, \| d \| < \| b \|$. By the choice of $b$ we have that $\bigcap \limits_{a \in I} H_a \subseteq H_c$ and $\bigcap \limits_{a \in I} H_a \subseteq H_d$. But $H_c \cap H_d \subseteq H_{c+d} = H_b$, which is a contradiction.

Adayah
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  • Thanks, can you also please add the proof for $V(0)=\cap H_a$ where $a$ runs over the indecomposable elements. I thought I had it by replacing $H_a$ for each non-indecomposable element $a=b+c$ by $H_b\cap H_c$, but I later found that this proof is incorrect/incomplete as $b$ may not be indecomposable and its removal may make $a$ reappear. –  May 31 '20 at 14:29
  • I also think it should be $b^Tc\ge 0$ but I have no longer got access to the book (Algebraic Graph Theory By Chris Godsil, Gordon F. Royle; Pg 330) and google books only shows me a fragment which says (not clearly) that $(b,c)>0$. –  May 31 '20 at 14:32
  • @Shahab I added the proof. Intuitively, it expands your attempt by repeated decomposition of the element until all the summands become indecomposable. This can not go on forever, because in every decomposition the components become closer to $0$, but the subgroup is discrete. – Adayah May 31 '20 at 15:29
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In the following, I assume without proof that the number of indecomposable elements is finite. Let $a$ be indecomposable. Then for any other indecomposable element $b$, we have: $(a-b)^Tb<0$, that is $a^Tb<b^Tb$. Since there are only a finite number of such $b$s, there exists $\varepsilon>0$ such that $x=(\dfrac{1}{2}+\varepsilon)a$ satisfies $x^Tb<\dfrac{1}{2}b^Tb$ for every indecomposable element $b \neq \pm a$. That is: $x \in H_b$ for $b \neq a$ and $x \notin H_a$. Therefore, not including $H_a$ would retain this element $x$ in $V$ (which should not have been the case).

Aravind
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