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?
