Cell decomposition for $\mathbb{C}P^n$ that has $\mathbb{R}P^n$ as a subcomplex?

Real projective space $\mathbb{R}P^n$ embeds in a natural way in complex projective space $\mathbb{C}P^n$. (Using standard projective coordinates on $\mathbb{C}P^n$, $\mathbb{R}P^n$ is the subspace consisting of points that have a representative in which all coordinates are real.)

I know (very standard) cell decompositions for $\mathbb{C}P^n$ and $\mathbb{R}P^n$ that realize them as CW-complexes: $\mathbb{C}P^n$ has one cell in every even dimension while $\mathbb{R}P^n$ has one cell in every dimension. So in this standard CW-complex structure on $\mathbb{C}P^n$, $\mathbb{R}P^n$ does not occur as a subcomplex.

What I would like to know is this:

Is there a cell decomposition of $\mathbb{C}P^n$ that reveals $\mathbb{R}P^n$ as a subcomplex? What is it?

[An addition to the bounty notice. In case it makes a difference I will consider a non-trivial but non-general case as well, say $n=3$ or some such, for bounty. A general answer (affirmative or negative) is obviously preferred. Equally obviously, I'm not speaking for Ben. Have fun, JL]

I believe the following works. Consider a cell decomposition made of subsets of $\mathbb{CP}^n$ of the form

$$\{ (x_1+s_1 y_1 i : x_2+s_2 y_2 i : \ldots : x_k+s_k y_k i : 1 : 0 : \ldots : 0) \mid x_i \in \mathbb{R}, y_i \in \mathbb{R}_{>0} \},$$

where $s_i \in \{-1,0,+1\}$. We label such a cell simply by $s_1 s_2 s_3 \ldots s_k$ (if $k=0$ we denote it by $\varnothing$).

The dimension of each cell is $k + \sum_{j=1}^k |s_j|$ (because each $s_j$ is associated with one degree of freedom $x_j$ if it is zero and with two $(x_j, y_j)$ if it is nonzero). It is easy to see that the boundary of the cell $s_1 s_2 s_3 \ldots s_k$ consists of cells where either one or more of the $s_j$ are removed, or some nonzero $s_j$ are replaced by $0$.

The real projective space $\mathbb{RP}^n$ is then explicitly realized as a subcomplex with cells $\varnothing, 0, 00, 000,$ etc.

Examples:

• The cell decomposition for $\mathbb{CP}^1$ (which is the Riemann sphere) consists of two hemispheres $+$ and $-$ separated by an "equator" $0$ with a marked point $\varnothing$.

• The cell decomposition for $\mathbb{CP}^2$ consists of the following set of 13 cells:

$$\{ \varnothing, 0, +, -, 00, +0, -0, 0+, 0-, ++, +-, -+, -- \}.$$

The number of $D$-dimensional cells with $D = 0, 1, 2, 3, 4$ is 1, 1, 3, 4 and 4 respectively. To check that this makes sense, we can for example compute the Euler characteristic: $\chi = 1-1+3-4+4=3$, as expected (recall that $\chi(\mathbb{CP}^n)=n+1$).

• The cell decomposition for $\mathbb{CP}^3$ has 40 cells in total:

$$\{ \varnothing, 0, +, -, 00, +0, -0, 0+, 0-, ++, +-, -+, --, 000, +00, 0+0, 00+, -00, 0-0, 00-, ++0, +0+, 0++, +-0, +0-, 0+-, -0+, -0+, 0-+, --0, -0-, 0--, +++, -++, +-+, ++-, --+, -+-, +--, --- \}.$$

The Euler characteristic is $\chi = 1-1+3-5+10-12+8=4$, which again checks out.