In this paper ( http://cseweb.ucsd.edu/classes/wi10/cse206a/lec2.pdf ), it says that for two sub-lattices $L_1$ and $L_2$ of $\mathbb{Z}^n$, it is easy to see that: $$ \widehat{L_1\cap L_2}=\widehat{L}_1+\widehat{L}_2$$ where $\widehat{L}$ is the dual of lattice $L$ and $``+"$ stands for the linear combinations of lattices.

Well, the relation $\widehat{L}_1+\widehat{L}_2\subset \widehat{L_1\cap L_2}$ is easy, but I can not verify that : $ \widehat{L_1\cap L_2}\subset\widehat{L}_1+\widehat{L}_2$. How can I check it? That is not a homework, I just want to profoundly understand all parts of the lecture note.



1 Answer 1


Let us use the notations given in http://cseweb.ucsd.edu/classes/wi10/cse206a/lec2.pdf. Let $B$ and $B'$ be two bases and $D$ and $D'$ be dual bases of the $B$ and $B'$ respectively, i.e., $D^TB = I, D'^TB' = I$. We want to show that $\widehat{\mathcal{L}(B) \cap \mathcal{L}(B')} = \mathcal{L}(D|D')$ which is equivalent to $\mathcal{L}(B) \cap \mathcal{L}(B') = \widehat{\mathcal{L}(D|D')}$. For the equality to be true, it is necessary that $span(B) = span(B')$. So, we will consider $B, B' \in \mathbb{Z}^{n \times m}$.

Let us show that $\mathcal{L}(B) \cap \mathcal{L}(B') \subseteq \widehat{\mathcal{L}(D|D')}$. Let $v \in \mathcal{L}(B) \cap \mathcal{L}(B')$, i.e., $v = Bz = B'z'$ where $z, z' \in \mathbb{Z}^m$. For any $w \in \mathcal{L}(D|D')$, i.e., $w = Dy + D'y'$ where $y, y' \in \mathbb{Z}^m$, we have $$<v, w>\ =\ <Bz, Dy> + <B'z', D'y'>\ \in \ \mathbb{Z}$$ which implies $v \in \widehat{\mathcal{L}(D|D')}$. Since, this is true for every $v \in \mathcal{L}(B) \cap \mathcal{L}(B')$, we have $\mathcal{L}(B) \cap \mathcal{L}(B') \subseteq \widehat{\mathcal{L}(D|D')}$.

Let us now show that $\widehat{\mathcal{L}(D|D')} \subseteq \mathcal{L}(B) \cap \mathcal{L}(B')$. From the definition of dual lattice, we have $span(D) = span(B)$ and $span(D') = span(B')$ which implies $span(D) = span(D') = span(D|D')$. So, for any $v \in \widehat{\mathcal{L}(D|D')}$, we can write $v = Bc = B'c'$ where $c, c' \in \mathbb{Q}^m$. If $c \notin \mathbb{Z}^m$, then there exist an $i$ such that $c_i \notin \mathbb{Z}$. This implies that for $w = De_i \in \mathcal{L}(D|D')$, $$<v, w> = <Bc,De_i> = c_i \notin \mathbb{Z}$$ But, this is contrary to the assumption that $v \in \widehat{\mathcal{L}(D|D')}$. Hence, $c \in \mathbb{Z}^m$ and using similar argument, we have $c' \in \mathbb{Z}^m$. Therefore, $\widehat{\mathcal{L}(D|D')} \subseteq \mathcal{L}(B) \cap \mathcal{L}(B')$.


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