Indecomposable elements in a lattice 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?
 A: Let $I$ be the set of indecomposable elements in $L$.


*

*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.

*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 $b \in A$. $\square$


*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.
A: 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).
