Let $K/k$ be a finite separable extension. If necessary, we can assume $[K : k] = 2$. Let $H$ be a $K$-closed subgroup of $\operatorname{GL}_n$, and let $\tilde{H} = \operatorname{Res}_{K/k}H$. Since $\tilde{H}$ is a linear algebraic group over $k$, it should embed into some $\operatorname{GL}_m$. I was wondering if there is a nice way to do this in general if, say, we fix a basis of $K/k$.

Of course since restriction of scalars respects closed subgroup immersions, it suffices to do this for $H = \operatorname{GL}_n$.

Example: $K = \mathbb C, k = \mathbb R$, $H = \operatorname{GL}_{1,\mathbb C}$. If $Z$ is the center of $\operatorname{GL}_{2,\mathbb R}$, we can realize $\tilde{H}$ inside $\operatorname{GL}_{2,\mathbb R}$ as the product $Z.\operatorname{SO}_2$, since a nonzero complex number $re^{i\theta}, e^{i\theta} = a+bi$ identifies with

$$\begin{pmatrix} r \\ & r \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$

(Non)-example: Again $K = \mathbb C, k = \mathbb R$, $H = \operatorname{GL}_{n,\mathbb C}$. Then $\tilde{H}$ is $\mathbb R$-isomorphic to a Levi subgroup of the unitary group $U(n,n)$. But this isn't quite what I want, because $U(n,n)$ is an outer form of $\operatorname{GL}_{2n}$. In other words, this embedding $\tilde{H} \rightarrow \operatorname{GL}_{2n}$ is not defined over $\mathbb R$.

(Non)-example 2: If $K/k$ is quadratic with nontrivial automorphism $\sigma$, then we have an embedding $\operatorname{GL}_n(K) \rightarrow \operatorname{GL}_{2n}(k)$ on points by

$$g \mapsto \begin{pmatrix} g \\ & \sigma(g) \end{pmatrix}$$

This embedding respects the Galois action. However, it is not a morphism of group schemes.

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    $\begingroup$ Yes; $\operatorname{Res}_{K/k}(H)$ is realised inside $\operatorname{Res}_{K/k}(\operatorname{GL}_N)$, which embeds inside $\operatorname{GL}_{[K : k]N}$. The latter embedding is obtained by viewing an $N$-dimensional $K$-vector space as an $[K : k]N$-dimensional $k$-vector space. $\endgroup$ – LSpice Mar 30 at 21:50
  • $\begingroup$ Thanks, that is a really nice way to think about it $\endgroup$ – D_S Mar 31 at 17:27

Thanks to LSpice for answering. In the quadratic case, we have $K = k(\sqrt{d})$ for $d \in k$. A basis $e_1, ... , e_n$ of an $n$-dimensional $K$-vector space has basis $e_1, ... , e_n, \sqrt{d} e_1, ... , \sqrt{d}e_n$ over $k$. For $g = (\alpha_{ij}) \in \operatorname{GL}_n(K)$ with $\alpha_{ij} = a_{ij} + \sqrt{d}b_{ij}$, we have

$$g.e_k = a_{1k}e_1 + \cdots + a_{nk}e_n + b_{1k} \sqrt{d}e_1 + \cdots + b_{nk} \sqrt{d}e_n$$

$$g.\sqrt{d}e_k = b_{1k}d e_1 + \cdots + b_{nk}d e_n + a_{1k} \sqrt{d}e_1 + \cdots + a_{nk}\sqrt{d}e_n$$

Writing $g$ uniquely as $A + \sqrt{d}B$ for $A, B \in \operatorname{Mat}_n(k)$, the map

$$g \mapsto \begin{pmatrix} A & dB \\ B & A \end{pmatrix}$$ embeds $\operatorname{GL}_n(K)$ into $\operatorname{GL}_{2n}(k)$. It's easy to check that the corresponding subgroup variety of $\operatorname{GL}_{2n}$ is defined over $k$ and isomorphic to $\operatorname{Res}_{K/k}\operatorname{GL}_n$.


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