Is the space of all linear complex structures of $\mathbb{R}^{2n}$ an embedded submanifold of $GL_{2n}(\mathbb{R})$?

Question

Denote $G = GL_{2n}(\mathbb{R})$, and let $F : G \to G$ be the map $F(X)=X^2$. Let $\mathcal{J} = F^{-1}(-\operatorname{Id})$, the space of all linear complex structures of $\mathbb{R}^{2n}$. Is $\mathcal{J}$ an embedded submanifold of $GL_{2n}(\mathbb{R})$?

$G$ acts on itself from the left by conjugation $A \overset{B\cdot}{\mapsto} BAB^{-1}$, and under this action $\mathcal{J}$ is the orbit of the standard complex structure of $\mathbb{R}^{2n}$: $$J_0 = \begin{pmatrix} 0 & -\operatorname{Id} \\ \operatorname{Id} & 0 \end{pmatrix}$$

The stabilizer of $J_0$ is a closed Lie subgroup which can be identified with $H = GL_n(\mathbb{C})$. Therefore, the quotient $G/H$ has a smooth structure, and the orbit map $A \mapsto A J_0 A^{-1}$ transcends to a smooth, injective, $G$-equivariant immersion $\iota : G/H \to G$ whose image is $\mathcal{J}$. But, the action of $G$ is not proper, as the stabilizer of $J_0$ is not compact. So, we cannot deduce that $\iota$ is a proper map, and therfore an embedding.

Is there any other way to show this is an embedding? Or is it wrong?

Another way I tried to approach this is by using the fact that $F : G \to G$ is also $G$-equivariant, and therefore has a constant rank on every orbit. But, this doesn't give me a constant rank in an open neighborhood of $\mathcal{J}$, so I cannot argue it's a level set of a map with a constant rank.

Answer 1

I will need a lemma from the theory of transformation groups, due to Arens:

Lemma. Suppose that $G$ is a Lie group (in fact, one needs less), $G\times X\to X$ is a continuous action on a completely metrizable topological space, with a closed orbit $Gx\subset X$ and $G_x$ is the stabilizer of $x$ in $G$. Then the orbit map $$ f: Z=G/G_x\to Gx=Y $$ is a homeomorphism (with $Y\subset X$ equipped with the subspace topology). For completeness, here is a proof:

First of all, as a closed subset of a completely metrizable space, $Y$ has the Baire property (since the restriction of a complete metric from $X$ to $Y$ is again complete). Observe also that $Z=G/G_x$ is a locally compact space. Moreover, for every subset $V\subset Z$ with nonempty interior, there are countably many elements $g_i\in G$ such that the $g_i$-translates of $V$ cover $Z$. (I am assuming Lie groups to be 2nd countable.)

I claim that for every $z\in G/G_x$ there exist a neighborhood basis consisting of compact neighborhoods whose images have nonempty interior in $Y$. Suppose not. Then there exists a compact $K\subset Z$ with nonempty interior such that the closed subset $f(K)\subset Y$ has empty interior. A countable union of the translates, $g_iK, i\in {\mathbb N}$, equals $Z$. By the $G$-equivariance of $f$, we have $$ \bigcup_{i\in {\mathbb N}} f(g_i K)= Y. $$ But each $f(g_i K)= g_i f(K)$ also has empty interior in $Y$. Thus, $Y$ is a union of countably many subsets with empty interiors contradicting the Baire Property.

Thus, $f(K)$ has nonempty interior for every compact $K\subset Z$ such that $int(K)\ne \emptyset$. Taking the open set $U=f^{-1}( int f(K))$ we see that every point in $Z$ has a neighborhhood basis consisting of open sets whose images under $f$ are open. Hence, $f$ is an open map and, thus, is a homeomorphism. qed

Now, back to your problem: You have an injective immersion $\iota: G/H\to GL_{2n}({\mathbb R})$, where $H$ is the stabilizer of $J_0$. The image of $\iota$ is closed, as the preimage of a point under a continuous map. By Arens' Lemma, $\iota$ is a homeomorphism to its image, hence, an embedding. Thus, its image is an embedded submanifold.

See also my answer here.

Answer 2

In this case, it's not difficult to show that the inverse map $\iota^{-1} : \mathcal{J} \to G / H$ is continuous.

Let $J \in \mathcal{J}$, and suppose $J = AJ_0A^{-1}$. Observe that $$ JAe_i = AJ_0e_i = Ae_{i + n} $$ for $i = 1, \dots, n$, where $e_i$ is the $i$th standard basis vector of $\mathbb{R}^{2n}$. This shows that the columns of $A$ are of the form $$ A = (v_1, \dots, v_n, Jv_1, \dots, Jv_n). $$ Now, consider the subset $U \subseteq \mathcal{J}$ consisting of all $J' \in \mathcal{J}$ such that the matrix $A_{J'}$ with columns $$ A_{J'} = (v_1, \dots, v_n, J'v_1, \dots, J'v_n) $$ is invertible. Then $U$ is an open neighborhood of $J$ since it is the preimage of $\mathbb{R} \setminus \{0\}$ under the map $$ J' \mapsto \det A_{J'}. $$ Observe that $$ J'A_{J'}e_i = A_{J'}e_{i + n} = A_{J'}J_0e_i, \qquad J'A_{J'}e_{i + n} = (J')^2A_{J'}e_i = -A_{J'}e_i = A_{J'}J_0e_{i + n} $$ for $i = 1, \dots, n$. This proves $J' = A_{J'}J_0A_{J'}^{-1}$ for all $J' \in U$, i.e., $$ \iota([A_{J'}]) = J'. $$ Hence, we have $$ \iota^{-1}(J') = [A_{J'}] $$ for all $J' \in U$. This shows that $\iota^{-1}$ is continuous over $U$, so we are done.