# Why define the dual of an ideal lattice with "Tr" rather than inner product?

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?

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