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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!

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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 $\Lambda.$

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    $\begingroup$ This formula for the dual works if the lattice is full rank; otherwise, one needs to restrict it to the span of the lattice. $\endgroup$ Commented Aug 27, 2020 at 12:16

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