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Relation to @HenriCohen's answer https://mathoverflow.net/a/481802

edited Nov 5, 2024 at 2:49

$\DeclareMathOperator\Tr{Tr}$I think that it means that we regard $F \otimes_{\mathbb Q} \mathbb C$ as a 2-dimensional vector space over $\mathbb C$ by action on the second factor. Then $\Tr((a + b\sqrt D) \otimes z)$ equals $z\Tr(a + b\sqrt D) = 2a z$, by definition.

If we define $F \otimes_{\mathbb Q} \mathbb C \to \mathbb C^2$ by $v \otimes z \mapsto (v z, \overline v z)$, where $\overline\cdot$ is the Galois conjugation of $F/\mathbb Q$, then we may transport the natural $\mathbb C$-linear action of $F$ on the source to such an action on the target. Specifically, $(z_1, z_2) \in \mathbb C^2$ is the image of $$\frac1 2(1 \otimes z_1 + \sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(1 \otimes z_2 - \sqrt D \otimes z_2\sqrt D^{-1}),$$ so that $u(z_1, z_2)$ is the image of $$\frac1 2(u \otimes z_1 + u\sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(u \otimes z_2 - u\sqrt D \otimes z_2\sqrt D^{-1}),$$ whose trace in the sense above is $$\frac1 2(z_1 + z_2)\Tr(u) + \frac1{2\sqrt D}(z_1 - z_2)\Tr(u\sqrt D).$$ If $u$ equals $a + b\sqrt D$, then this last equals $$ \frac1 2(z_1 + z_2)(2a) + \frac1{2\sqrt D}(z_1 - z_2)(2b D) = %(z_1 + z_2)a + (z_1 - z_2)b\sqrt D = (a + b\sqrt D)z_1 + (a - b\sqrt D)z_2. $$$$ \frac1 2(z_1 + z_2)(2a) + \frac1{2\sqrt D}(z_1 - z_2)(2b D) = %(z_1 + z_2)a + (z_1 - z_2)b\sqrt D = (a + b\sqrt D)z_1 + (a - b\sqrt D)z_2. %\tag{$*$}\label{481799_star} $$ This agrees with the formula in @HenriCohen's answer, since, in the notation there, we can number the real embeddings so that $u^{(1)} = a + b\sqrt D$ and $u^{(2)} = a - b\sqrt D$. It is off by a factor of $\sqrt D$ from the explicit expression in the reference. I did not find the formula there on a quick scan, so cannot explain the discrepancy, but it could come simply from a different identification of $F \otimes_{\mathbb Q} \mathbb C$ with $\mathbb C^2$.

If we define $F \otimes_{\mathbb Q} \mathbb C \to \mathbb C^2$ by $v \otimes z \mapsto (v z, \overline v z)$, where $\overline\cdot$ is the Galois conjugation of $F/\mathbb Q$, then we may transport the natural $\mathbb C$-linear action of $F$ on the source to such an action on the target. Specifically, $(z_1, z_2) \in \mathbb C^2$ is the image of $$\frac1 2(1 \otimes z_1 + \sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(1 \otimes z_2 - \sqrt D \otimes z_2\sqrt D^{-1}),$$ so that $u(z_1, z_2)$ is the image of $$\frac1 2(u \otimes z_1 + u\sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(u \otimes z_2 - u\sqrt D \otimes z_2\sqrt D^{-1}),$$ whose trace in the sense above is $$\frac1 2(z_1 + z_2)\Tr(u) + \frac1{2\sqrt D}(z_1 - z_2)\Tr(u\sqrt D).$$ If $u$ equals $a + b\sqrt D$, then this last equals $$ \frac1 2(z_1 + z_2)(2a) + \frac1{2\sqrt D}(z_1 - z_2)(2b D) = %(z_1 + z_2)a + (z_1 - z_2)b\sqrt D = (a + b\sqrt D)z_1 + (a - b\sqrt D)z_2. $$ This is off by a factor of $\sqrt D$ from the explicit expression in the reference. I did not find the formula there on a quick scan, so cannot explain the discrepancy, but it could come simply from a different identification of $F \otimes_{\mathbb Q} \mathbb C$ with $\mathbb C^2$.

If we define $F \otimes_{\mathbb Q} \mathbb C \to \mathbb C^2$ by $v \otimes z \mapsto (v z, \overline v z)$, where $\overline\cdot$ is the Galois conjugation of $F/\mathbb Q$, then we may transport the natural $\mathbb C$-linear action of $F$ on the source to such an action on the target. Specifically, $(z_1, z_2) \in \mathbb C^2$ is the image of $$\frac1 2(1 \otimes z_1 + \sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(1 \otimes z_2 - \sqrt D \otimes z_2\sqrt D^{-1}),$$ so that $u(z_1, z_2)$ is the image of $$\frac1 2(u \otimes z_1 + u\sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(u \otimes z_2 - u\sqrt D \otimes z_2\sqrt D^{-1}),$$ whose trace in the sense above is $$\frac1 2(z_1 + z_2)\Tr(u) + \frac1{2\sqrt D}(z_1 - z_2)\Tr(u\sqrt D).$$ If $u$ equals $a + b\sqrt D$, then this last equals $$ \frac1 2(z_1 + z_2)(2a) + \frac1{2\sqrt D}(z_1 - z_2)(2b D) = %(z_1 + z_2)a + (z_1 - z_2)b\sqrt D = (a + b\sqrt D)z_1 + (a - b\sqrt D)z_2. %\tag{$*$}\label{481799_star} $$ This agrees with the formula in @HenriCohen's answer, since, in the notation there, we can number the real embeddings so that $u^{(1)} = a + b\sqrt D$ and $u^{(2)} = a - b\sqrt D$. It is off by a factor of $\sqrt D$ from the explicit expression in the reference. I did not find the formula there on a quick scan, so cannot explain the discrepancy, but it could come simply from a different identification of $F \otimes_{\mathbb Q} \mathbb C$ with $\mathbb C^2$.

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answered Nov 4, 2024 at 21:52

LSpice

If we define $F \otimes_{\mathbb Q} \mathbb C \to \mathbb C^2$ by $v \otimes z \mapsto (v z, \overline v z)$, where $\overline\cdot$ is the Galois conjugation of $F/\mathbb Q$, then we may transport the natural $\mathbb C$-linear action of $F$ on the source to such an action on the target. Specifically, $(z_1, z_2) \in \mathbb C^2$ is the image of $$\frac1 2(1 \otimes z_1 + \sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(1 \otimes z_2 - \sqrt D \otimes z_2\sqrt D^{-1}),$$ so that $u(z_1, z_2)$ is the image of $$\frac1 2(u \otimes z_1 + u\sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(u \otimes z_2 - u\sqrt D \otimes z_2\sqrt D^{-1}),$$ whose trace in the sense above is $$\frac1 2(z_1 + z_2)\Tr(u) + \frac1{2\sqrt D}(z_1 - z_2)\Tr(u\sqrt D).$$ If $u$ equals $a + b\sqrt D$, then this last equals $$ \frac1 2(z_1 + z_2)(2a) + \frac1{2\sqrt D}(z_1 - z_2)(2b D) = %(z_1 + z_2)a + (z_1 - z_2)b\sqrt D = (a + b\sqrt D)z_1 + (a - b\sqrt D)z_2. $$ This is off by a factor of $\sqrt D$ from the explicit expression in the reference. I did not find the formula there on a quick scan, so cannot explain the discrepancy, but it could come simply from a different identification of $F \otimes_{\mathbb Q} \mathbb C$ with $\mathbb C^2$.