# If $H \subset \operatorname{GL}(n)$, can we realize $\operatorname{Res}_{K/k} H$ inside $\operatorname{GL}([K : k]n)$?

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.

• 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. – LSpice Mar 30 at 21:50
• Thanks, that is a really nice way to think about it – 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 $$\operatorname{Res}_{K/k}\operatorname{GL}_n$$.