How do I find a smooth map $\operatorname{Gr}_{\mathbb{C}}(k, n) \to \operatorname{Gr}(2k, 2n)$?

Question

I am trying to find a smooth bijective map from the complex Grassmannian of $k$-dimensional subspaces of $\mathbb{C}^n$ to the Grassmannian of $2k$-dimensional subspaces of $\mathbb{R}^{2n}$, but I do not know how to construct it.

I think it is related to the

$$a+bi\mapsto\begin{bmatrix}a & -b\\b&a\end{bmatrix}$$

map but I am not sure how to transform it in the case of Grassmannians.

Answer 1

If $\varphi:G_{\mathbb{C}}(k,n)\to G(2k,2n)$ is a smooth bijective map (of real manifolds), then for any $k$-complex linear subspace $X$ of $\mathbb{C}^n$ there exists a $(2k)$-real linear subspace $Y$ of $\mathbb{R}^{2n}$ such that $\varphi(X)=Y$ and $Y$ is $J$-invariant, where $J$ is an $\mathbb{R}$-linear automorphism of $\mathbb{R}^{2n}$ such that $J^2=-Id_{\mathbb{R}^{2n}}$.

But there exist $(2k)$-real linear subspaces $Z$ of $\mathbb{R}^{2n}$ such that $Z$'s are not $J$-invariant!, and therefore $\varphi$ can not be bijective.

Remark: I do not need of the smoothness of $\varphi$!

Answer 2

Lemma: If $f : M \to N$ is a smooth bijection, then $\dim M = \dim N$.

Proof: By Sard's Theorem, there is $p \in M$ such that $df_p$ is surjective and hence $\dim M \geq \dim N$. By Proposition $4.1$ of Lee's Introduction to Smooth Manifolds (second edition), there is an open neighbourhood $U$ of $p$ such that $f|_U$ is a submersion. The Rank Theorem (Theorem $4.12$ in Lee's book) applied to $f|_U$ shows that there are coordinates about $p$ and $f(p)$ so that the map $f$ takes the form $(x^1, \dots, x^n, x^{n+1}, \dots, x^m) \mapsto (x_1, \dots, x^n)$ where $m = \dim M$ and $n = \dim N$. As the map $f$ is injective, we see that $m = n$.

Note that

\begin{align*} \dim_{\mathbb{R}}\operatorname{Gr}_{\mathbb{C}}(k, n) &= 2\dim_{\mathbb{C}}\operatorname{Gr}_{\mathbb{C}}(k, n) = 2k(n-k)\\ \dim_{\mathbb{R}}\operatorname{Gr}(2k, 2n) &= 2k(2n - 2k) = 4k(n-k). \end{align*}

For $k = 0, n$, both $\operatorname{Gr}_{\mathbb{C}}(k, n)$ and $\operatorname{Gr}(2k, 2n)$ are points and hence there is a smooth bijection between them. If $0 < k < n$ however, then $\dim \operatorname{Gr}_{\mathbb{C}}(k, n) < \dim \operatorname{Gr}(2k, 2n)$, so there is no smooth bijection $\operatorname{Gr}_{\mathbb{C}}(k, n) \to \operatorname{Gr}(2k, 2n)$.

However, there is a natural map $\operatorname{Gr}_{\mathbb{C}}(k, n) \to \operatorname{Gr}(2k, 2n)$ given by sending a complex $k$-plane to its underlying real $2k$-plane. This is a smooth embedding, but it is not a bijection (unless $k = 0$ or $n$) because it is not surjective: there are many real $2k$-planes which are not invariant under the complex structure on $\mathbb{R}^{2n}$ (induced by the identification $\mathbb{C}^n \cong \mathbb{R}^{2n}$), and hence can not be viewed as complex $k$-planes in $\mathbb{C}^n$. In fact, by Sard's Theorem, the image has Lebesgue measure zero.