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.