4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.