Viewing complex projective space as a Grassmannian manifold

Question

Complex projective space $\mathbb{CP}^n$ carries the structure of a complex manifold of dimension $n$, hence has the underlying structure of a real manifold of dimension $2n$. It is the set of complex lines (which are planes in the real vector space sense) through the origin in $\mathbb{C}^{n+1}$ (equivalently $\mathbb{R}^{2n+2}$). With this view, there is an apparent equivalence between $\mathbb{CP}^n$ and the Grassmannian $Gr_2(\mathbb{R}^{2n+2})$, i.e. the set the two-dimensional subspaces of $\mathbb{R}^{2n+2}$. But $\dim{Gr_k(V)}=k(n-k)$ if $\dim{V}=n$, and hence $\dim(Gr_2(\mathbb{R}^{2n+2}))=2((2n+2)-2)=4n$ which is twice the real dimension of $\mathbb{CP}^n$ - what am I missing here?

Answer 1

If you identify $\mathbb{R}^{2n+2}$ with $\mathbb{C}^{n+1}$ then not every two-dimensional real plane in $\mathbb{R}^{2n+2}$ is a one-dimensional complex plane in $\mathbb{C}^{n+1}$. The identification between $\mathbb{R}^{2n+2}$ and $\mathbb{C}^{n+1}$ endows $\mathbb{R}^{2n+2}$ with a complex structure $J \colon \mathbb{R}^{2n+2} \rightarrow \mathbb{R}^{2n+2}$ and a subspace $V \leq \mathbb{R}^{2n+2}$ is a complex subspace if and only if it is $J$ invariant ($JV = V$). Thus, the complex Grassmananian is only submanifold of $\operatorname{Gr}_2(\mathbb{R}^{2n+2})$.

Thus, $\mathbb{CP}^n$ can be identified with $\{ V \in \operatorname{Gr}_2(\mathbb{R}^{2n+2}) \, | \, \dim_{\mathbb{R}} V = 2, JV = V \} \subseteq \operatorname{Gr}_2(\mathbb{R}^{2n+2})$. This even fits nicely with the dimension calculations you have provided:

The complex structure $J$ acts as a diffeomorphism of finite order on $\operatorname{Gr}_2(\mathbb{R}^{2n+2})$ and $\mathbb{CP}^n$ is charaterized as the fixed locus of the two dimensional complex subspaces. At a fixed point $V$ (a complex line), the tangent space $T_{V}(\mathbb{CP}^n) \subseteq T_{V}(\operatorname{Gr}_2(\mathbb{R}^{2n+2}))$ consists of $L \in T_{V}(\operatorname{Gr}_2(\mathbb{R}^{2n+2}))$ with $(dJ)|_{V}(L) = L$. As $T_{V}(\operatorname{Gr}_2(\mathbb{R}^{2n+2}))$ is naturally identified with $\operatorname{Hom}(V,\mathbb{R}^{2n+2}/V)$, $dJ|_{V}$ acts on $L \colon V \rightarrow \mathbb{R}^{2n+2}/V$ by $(dJ)|_{V}(L) = \hat{J} \circ L \circ J^{-1}$ where $\hat{J} \colon \mathbb{R}^{2n+2}/V \rightarrow \mathbb{R}^{2n+2}/V$ is the induced quotient map. Since $V$ is a fixed point, $J|_{V}$ is a complex structure on $V$ and $\hat{J}$ is a complex structure on $\mathbb{R}^{2n+2}/V$ and so

$$ T_{V}(\mathbb{CP}^n) = \{ L \colon V \rightarrow \mathbb{R}^{2n+2}/V \, | \, L \textrm{ is a complex map with respect to } (J,\hat{J}) \}. $$

Since $\dim V = 2$, we see that $\dim \mathbb{CP}^n = T_{V}(\mathbb{CP}^n) = 2n$ as expected.