Does $GL(n,\mathbb{C})$ inject into $GL^+(2n, \mathbb{R})$ for all $n$?

Question

I know that $GL(n, \mathbb{C}) \hookrightarrow GL^+(2n,\mathbb{R})$ for n = 1 and 2. I just did these with pen and paper (and mathematica). But is this statement true in general? What is a proof of this? Thanks for the help.

Oh, how about since as Lie groups, $GL(n,\mathbb{C}) \subset GL(2n, \mathbb{R})$ and $GL(n,\mathbb{C})$ is connected but $GL(2n, \mathbb{R})$ has two connected components, one for positive determinant and one for negative determinant? And the identity has positive determinant, so it must lie in that component.

Is there an algebraic proof though?

Answer 1

The claim is: If $V$ is an $n$-dimensional complex vector space with underlying $2n$-dimensional real vector space $W$, then the canonical group monomorphism $\mathrm{GL}(V) \to \mathrm{GL}(W)$ lands inside $\mathrm{GL}^+(W)=\{f \in \mathrm{GL}(W) : \det(f)>0\}$. The purpose of this abstract reformulation is that we may use operations on vector spaces in order to simplify the problem: If $V'$ is another finite-dimensional complex vector space with underlying real vector space $W'$, the diagram $$\begin{array}{c} \mathrm{GL}(V) \times \mathrm{GL}(V') & \rightarrow & \mathrm{GL}(W) \times \mathrm{GL}(W') \\ \downarrow && \downarrow \\ \mathrm{GL}(V \oplus V') & \rightarrow & \mathrm{GL}(W \oplus W') \end{array}$$ commutes, and the image of $\mathrm{GL}^+(W) \times \mathrm{GL}^+(W')$ is contained in $\mathrm{GL}^+(W \oplus W')$. Therefore, if some element in $\mathrm{GL}(V \oplus V')$ lies in the image of $\mathrm{GL}(V) \times \mathrm{GL}(V')$, it suffices to consider the components. Combining this with the fact that $\mathrm{GL}(V)$ is generated by elementary matrices (after chosing a basis of $V$), we may reduce the whole problem to the following three types of matrices:

• the $1 \times 1$-matrices $(\lambda)$,
• the $2 \times 2$-matrices $\begin{pmatrix} 1 & 0 \\ \lambda & 1 \end{pmatrix}$,
• and the $2 \times 2$-matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

Write $\lambda=a+ib$ with $(a,b) \in \mathbb{R}^2 \setminus \{(0,0)\}$. Then, the complex $1 \times 1$-matrix $(\lambda)$ becomes the real $2 \times 2$-matrix $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$, which has determinant $a^2+b^2 > 0$. The complex $2 \times 2$-matrix $\begin{pmatrix} 1 & 0 \\ \lambda & 1 \end{pmatrix}$ becomes the real $4 \times 4$-matrix $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ a & -b & 1 & 0 \\ b & a & 0 & 1 \end{pmatrix}$, which has determinant $1$. Finally, the complex $2 \times 2$-matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ becomes the real $4 \times 4$-matrix $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$, which has determinant $1$.