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$\DeclareMathOperator\GU{GU}$$\DeclareMathOperator\GL{GL}$I'm a bit confused by the following situation: suppose we have an Hermitian vector space $V=K^3$ of matrix $$ J=\begin{pmatrix}& & \delta^{-1}\\ & 1 & \\ -\delta^{-1} & &\end{pmatrix}, $$ with $K=\mathbb{Q}(\delta)$ imaginary quadratic, $\delta=\sqrt{-D}$. Define $G=\GU(2,1)$ as the reductive group over $\mathbb{Q}$ of matrices in $\text{Res}_{K/\mathbb{Q}}\GL_3$ which preserve the above hermitian form, up to a similitude factor. That is, $g\in G$ by definition means $${^t}\overline{g}Jg=\mu(g)\cdot J$$ for some $\mu(g)\in \mathbb{G}_m$.

Now take $X\in V$ positive, and suppose that the orthogonal of $X$ in $V$ is anisotropic i.e., there is no vector $Y\in X^\perp\subset V$ s.t. $(Y,Y)=0$. By this assumption we know that $\GU(J\mid_{X^\perp})$ is not quasi-split over $\mathbb{Q}$. My question is: in this situation, can we still say that we have an embedding $$\GU(J\mid_{X^\perp})\hookrightarrow \GU(2,1)$$ of group schemes over $\mathbb{Q}$?

Thank you for your help.

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    $\begingroup$ Surely this is obvious? Given a unitary automorphism of $X^\perp$, just extend it to an automorphism of $V$ by defining it to be trivial on $X$. (Is it the idea of embedding a non-quasi-split space inside a quasi-split one which troubles you? If so, play with some explicit matrices until it stops troubling you.) $\endgroup$
    – David Loeffler
    Commented Dec 7, 2023 at 17:20
  • $\begingroup$ Thank you David. That's what I told myself, but yes I was a bit troubled but what you say. I'll think it over some more. $\endgroup$
    – Fra
    Commented Dec 7, 2023 at 17:24

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