Let $n$ be a positive integer, and define a map $\beta : GL(n, \mathbb{C}) \rightarrow GL(2n, \mathbb{R})$ by

$$ \beta \begin{pmatrix} a^1_1 + i b^1_1 & \cdots & a^n_1 + i b^n_1\\ \vdots & & \vdots \\ a^1_n + i b^1_n & \cdots & a^n_n + i b^n_n\\ \end{pmatrix} = \begin{pmatrix} \begin{matrix} a^1_1 & -b^1_1 \\ b^1_1 & a^1_1 \\ \end{matrix} & \cdots & \begin{matrix} a^n_1 & -b^n_1 \\ b^n_1 & a^n_1 \\ \end{matrix}\\ \vdots & & \vdots \\ \begin{matrix} a^1_n & -b^1_n \\ b^1_n & a^1_n \\ \end{matrix} & \cdots & \begin{matrix} a^n_n & -b^n_n \\ b^n_n & a^n_n \\ \end{matrix}\\ \end{pmatrix} $$

My book, Lee's Smooth Manifolds(2nd Ed), on pp. 158, stats that

It is straightforward to verify that $\beta$ is an injective Lie group homomorphism whose image is a properly embedded Lie subgroup of $GL(2n, \mathbb{R})$.

In the previous page, there are two propositions as followings

Let $F:G\rightarrow H $ be a Lie group homomorphism. The kernel of $F$ is a properly embedded Lie subgroup of $G$, whose codimension is equal to the rank of $F$.

If $F: G \rightarrow H$ is an injective Lie group homomorphism, the image of $F$ has a unique smooth manifold structure such that $F(G)$ is a Lie subgroup of $H$ and $F:G \rightarrow F(G)$ is a Lie group isomorphism. (In the proof, $F(G)$ turns out just an immersed submanifold.)

I expected that the image of $\beta$ is just an immersed submanifold by the second proposition. However the author says the result of the first proposition. So I tried to prove it directly. Since the image is clearly a closed subset of $GL(2n,\mathbb{R})$, we need only to show that $\beta$ is a topological embedding. I got stuck on this point. I want to have help. Thank you.


If you consider all (not necessarily invertible) matrices, then the obvious extension of $\beta$ to a map $M_n(\mathbb{C})\to M_{2n}(\mathbb{R})$ is a linear injection, and hence a topological embedding. Since $GL(n,\mathbb{C})$ and $GL(2n,\mathbb{R})$ are just open subspaces of $M_n(\mathbb{C})$ and $M_{2n}(\mathbb{R})$, it follows that $\beta$ is also a topological embedding.

(To prove that any linear injection $i:\mathbb{R}^m\to \mathbb{R}^n$ is an embedding, note that there is an invertible linear map $T:\mathbb{R}^n\to\mathbb{R}^n$ such that $Ti:\mathbb{R}^m\to\mathbb{R}^n$ is just the standard inclusion of $\mathbb{R}^m$ into $\mathbb{R}^n$. Thus $Ti$ is an embedding, and hence so is $i$ since $T$ is a homeomorphism.)


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