In the paper [LPR12], I've learned that ideal lattices are ideals in algebraic number fields. However, I can't understand why we define the dual lattice of an ideal lattice with $\operatorname{Tr}$: $$ {L}^{\vee}=\{x \in K: \operatorname{Tr}(x {L}) \subseteq \mathbb{Z}\} $$

In detail, I mean, for any algebraic number field $K$, there's an embedding that embed it into space $H$. For $K=\mathbb Q[\zeta]$, let $f\in\mathbb Q$ be the minimal polynomial of $\zeta$. Suppose $\zeta$ has $s_1$ real roots and $s_2$ pair of complex roots (and $n=s_1+2s_2$), then $$ H=\left\{\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{s_{1}} \times \mathbb{C}^{2 s_{2}}: x_{s_{1}+s_{2}+j}=\overline{x_{s_{1}+j}}, \forall j \in\left[s_{2}\right]\right\} \subseteq \mathbb{C}^{n} $$ The embedding is done by the canonical embedding $\sigma$ s.t. for any $\alpha\in K$, $\sigma(\alpha)=(\sigma_i(\alpha))_{i\in[n]}$. Moreover, $H$ can be further embedded into $\mathbb R^n$ by isomorphism $h$, by embedding the conjugate pairs $(a+bi,a-bi)$ to $(\sqrt2a,-\sqrt2b)$. (The author said this is the geometry of the ideal lattice.) Till, now seems everything goes on well.

However, the for $\alpha,\beta\in K$, which maps to $\sigma(\alpha)=v=(v_i)_{i\in[n]},\sigma(\beta)=w=(w_i)_{i\in[n]}$, the inner product of $v$ and $w$ is defined as $$ \langle v,w\rangle=\sum_{i\in [n]} v_i\overline{w_i} $$ which equals $\langle h(v),h(w)\rangle$ in $\mathbb R^n$.

However, the definition of dual lattice of an ideal lattice use $\operatorname{Tr}$ instead of such of inner product. We have $$ \operatorname{Tr}(\alpha\beta)=\sum_{i\in[n]}\sigma_i(\alpha)\sigma_i(\beta)=\sum_{i\in[n]}v_iw_i $$ which seems different from inner product.

For an typical example, I'd like to work with $K=\mathbb Q[i]$. It has two embeddings $\sigma_1(a+bi)=a+bi,\sigma_2(a+bi)=a-bi$ to $\mathbb C^2$, so the canonical embedding is $$ \sigma(a+bi)=(a+bi,a-bi) $$ Working with an ideal lattice $(1+2i)\mathbb Z+(-2+i)\mathbb Z$, and mapping the basis to $\mathbb R^2$ by $h\circ \sigma$, we have $L'=h\circ\sigma(L)=(\sqrt2,-2\sqrt2)\mathbb Z+(-2\sqrt2,-\sqrt2 )\mathbb Z$. Then we treat $L'$ as common lattice and compute its dual lattice as $(L')^\ast= (\frac{\sqrt 2}{10},-\frac{\sqrt2}5)\mathbb Z+(-\frac{\sqrt2}5,-\frac{\sqrt 2}{10})$. If we compute the dual lattice of $L$ by $\operatorname{Tr}$ definition, the dual lattice would be $$ L^\vee=(\frac1{10}-\frac15i)\mathbb Z+(-\frac15-\frac1{10}i)\mathbb Z $$ which can be embedded into $\mathbb R^2$ as $$ (h\circ\sigma)(L^\vee)=(\frac{\sqrt2}{10},\frac{\sqrt 2}5)\mathbb Z+(-\frac{\sqrt 2}5,\frac{\sqrt 2}{10})\mathbb Z $$ which is different from $(L')^\ast$, it replaces the second entry of basis vectors with their opposite number. Why this happens? Have I done something wrong? Or is their another geometric view of ideal lattice that make sense?


1 Answer 1


You haven’t made any mistake. The direct answer to your question is that there are two definitions of “dual” in the literature, one for ideals and one for lattices over the complex numbers. They are slightly inconsistent, but are identical up to complex conjugation—this is why the entries corresponding to complex parts are negated in your example.

(The paper states elsewhere in the preliminaries that $\sigma(I^\vee)$ is the complex conjugate of $\sigma(I)^*$.)

We adopted the “trace-based” definition for ideals because it is natural and commonly used in mathematics, lets us use known facts relating to the (co)different ideals, and doesn’t require the canonical embedding to define.


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