# Dual of a complex lattice

We know that for a real full-ranked lattice $$\Lambda$$, with real square matrix $$\mathbf{B}$$, the dual lattice $$\Lambda^{\vee}$$ has matrix $$(\mathbf{B}^{-1})^T$$.

Now If we have a complex lattice with $$N \times N$$ complex matrix $$\mathbf{C}$$, what would be the matrix of its dual lattice? Would it be the same? Or we need to put complex conjugation?

Thank you!

• I'm not sure about your formula: cseweb.ucsd.edu/classes/wi12/cse206A-a/LecDual.pdf Commented Aug 26, 2020 at 8:18
• The formula works in the case that $B$ is full-rank, as $D = B(B^t B)^{-1} = BB^{-1} B^{-t} = B^{-t}$ Commented Sep 2, 2020 at 22:50

Let $$G=\{a+ib:a,b\in \mathbb{Z}\}$$ be the Gaussian integers (something similar can be done with the Eisentstein integers defined via third complex roots of unity as well).
$$G$$ is the ring of integers of the complex field $$\mathbb{C}$$. With the inner product of $$x,y \in \mathbb{C}^n$$ defined via the hermitian inner product $$x\cdot \overline{y}=x_1 \overline{y_1}+\cdots+x_n\overline{y_n},$$ the dual lattice of a lattice $$\Lambda \subset \mathbb{C}^n$$ is defined as $$\Lambda^{\ast}=\{x \in \mathbb{C}^n: x \cdot \overline{u} \in G~for~all~ u \in \Lambda\}.$$ The lattice $$\Lambda$$ itself is of course the set of all vectors $$\xi M,$$ where $$\xi=(\xi_1,\ldots,\xi_n) \in G^n$$ is arbitrary and $$M$$ is a complex $$n\times m$$ generating matrix for $$\Lambda.$$