I'm following Neukirch's algebraic number theory. The situation is as follows:
Let $K$ be a number field of degree $n$. Then $n = r + 2s$, where $r$ is the number of real embeddings $\rho : K \to \Bbb C$ (i.e. those embeddings $\rho$ such that $\rho(K) \subseteq \Bbb R$) and $s$ is the number of conjugate pairs of complex embeddings $\sigma : K \to \Bbb C$. Let $\mathcal{R}$ denote the set of real embeddings and $\mathcal{C}$ the set of complex embeddings. We fix a choice of complex embedding $\sigma \in \mathcal{C}$ so that we may distinguish between a complex embedding and its conjugate. Also, denote by $\bar{\cdot}$ complex conjugation.
Let $K_\Bbb C := \prod_{\tau \in \mathcal{R} \cup \mathcal{C}} \Bbb C$ and
$$K_\Bbb R = \lbrace (z_\tau) \in K_\Bbb C : z_\rho \in \Bbb R,\ z_{\overline{\sigma}} = \overline{z}_\sigma\rbrace.$$
There is an isomorphism $\psi : K_\Bbb R \to \prod_{\tau \in \mathcal{R}\cup \mathcal{C}}\Bbb R = \Bbb R^{r + 2s}$ such that $(z_{\tau}) \mapsto (x_\tau)$ with
$$x_\tau := \begin{cases} z_\tau,\ &\text{if}\ \tau \in \mathcal{R},\\ \operatorname{Re}(z_\tau),\ &\text{if}\ \tau = \sigma\in \mathcal{C},\\ \operatorname{Im}(z_\tau),\ &\text{if}\ \tau = \overline{\sigma}\in \mathcal{C}.\end{cases}$$
The $\Bbb C$-vector space $K_\Bbb C$ is equipped with a hermitian scalar product
$$\langle x, y \rangle_{c} = \sum_{\tau} x_\tau \overline{y}_\tau,$$
called the canonical metric.
The claim is that $\psi$ induces a scalar product
$$\langle z_\tau, z_\tau^\prime\rangle_M = \sum_{\tau} \alpha_\tau z_\tau z_\tau^\prime$$
with $\alpha_\tau = 1$ if $\tau \in \mathcal{R}$ and $\alpha_\tau = 2$ if $\tau \in \mathcal{C}$.
When I attempt to show that this is actually the case I run into a problem: I would expect that $\langle \psi(z_\tau), \psi(z_\tau^\prime)\rangle_c = \langle z_\tau, z_\tau^\prime\rangle_M$, but when I perform the calculation the following happens:
$$\langle \psi(z_\tau), \psi(z_\tau^\prime)\rangle_c = \sum_\tau \psi(z_\tau)\overline{\psi(z_\tau^\prime)} = \sum_{\rho \in \mathcal{R}} \psi(z_\rho)\overline{\psi(z_\rho^\prime)} + \sum_{\sigma \in \mathcal{C}} \psi(z_\sigma)\overline{\psi(z_\sigma^\prime)} + \sum_{\overline{\sigma} \in \mathcal{C}}\psi(z_\overline{\sigma})\overline{\psi(z_\overline{\sigma}^\prime)}$$
and applying $\psi$ we get
$$\sum_{\rho \in \mathcal{R}} x_\rho x_\rho^\prime + \sum_{\sigma \in \mathcal{C}} x_\sigma x_\sigma^\prime + \sum_{\overline{\sigma} \in \mathcal{C}} x_\overline{\sigma}x_\overline{\sigma}^\prime = \sum_{\rho \in \mathcal{R}} x_\rho x_\rho^\prime + \sum_{\sigma \in \mathcal{C}} x_\sigma x_\sigma^\prime + x_\overline{\sigma}x_\overline{\sigma}^\prime$$
and in the last equality I would expect a factor of $2$ to appear in the second sum on the right of the equality (for $\alpha_\tau$'s definition to be justified).
Can I somehow collapse this sum using the given definitions such that a factor of $2$ appears (i.e. some correction term arising from an undercounting of the complex embeddings or something)? Also, I'm not sure if splitting the sum $\sum_\tau \psi(z_\tau)\overline{\psi(z_\tau^\prime)}$ into three sums based on $\tau$'s membership is justified; looks like I'm also double counting the elements of $\mathcal{C}$ !