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$n \times n$ matrix $A$ with complex entries is called Hermitian if $A^{*}=A,$ where $A^{*}=\bar{A}^{T}$

$$ H(\Bbb{C}):=\left\{A \in M_{2}(\Bbb{C}) \mid A^{*}=A\right\} $$ $H(\Bbb{C})$ consists of 2 by 2 the matrices $$ A=\left(\begin{array}{ll} a & b \\ \bar{b} & d \end{array}\right) $$ where $a, d \in \Bbb{R}$ and $b \in \Bbb{C}$. If $A \in H(\Bbb{C})$ then the associated binary Hermitian form is the semi quadratic map $$ Q: \mathbb{C} \times \mathbb{C} \rightarrow \Bbb{R} $$ defined by $$ Q(X, Z)=(X, Z)\left(\begin{array}{c} a & b \\ \bar{b} & d \end{array}\right)(X, Z)^{*}=a|X|^2+b X Z+\bar{b} \bar{X} Z+d|Z|^2 $$

Let $H^{+}(\mathbb{C})$ denote the set of positive definite binary quadratic hermitian forms and $\Delta(Q):=\operatorname{det}(Q)=a d-|b|^{2}$

Definition: Let $\mathcal{H}_3=\Bbb{C}\times (0,\infty)$ be the upper half space. The map $\xi: H^{+}(\mathbb{C}) \rightarrow \mathcal{H}_{3}$ defined by $$ \xi\begin{pmatrix}a & b \\ \overline{b} & d \end{pmatrix}=\frac{b}{a}+\frac{\sqrt{\Delta(Q)}}{a} \cdot j $$ is called the "zero map" for binary quadratic Hermitian forms.

My question: Why is it called zero map? I expect that $Q(X,1)=a|X|^2+bX+\overline{bX}+d$ admits $\frac{b}{a}+\frac{\sqrt{\Delta(Q)}}{a} j$ as a zero since the zero of an positive definite binary quadratic form $f(x,y)=ax^2+bxy+cy^2$ is defined as $\frac{-b+\sqrt{\Delta}}{2a}$, which is really a zero of $f(x,1)$. But when I set $X=\frac{b}{a}+\frac{\sqrt{\Delta(Q)}}{a} j$, I get $Q(X,1)=\frac{a^2+b^2+\overline{b^2}+d^2}{d}$ using the algebraic properties of quaternions. Do I have wrong expectation?

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No, it isn't a quaternionic zero of $H(X,1)=0$. For instance (and I think you can reduce to this case with a linear change of variable) consider $H(X,Z)=|X|^2+|Z|^2$. Clearly $|X|^2=-1$ has no solutions in $\mathbb{H}$.

More important is the equivariance of the $PSL(2,\mathbb{C})$-action on forms/zero sets. Take a look at Elstrodt, Grunewald, Mennicke, Groups Acting on Hyperbolic Space for more on the subject.

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