A quotient space of complex projective space Let $\mathbb{C}P^n$ be the $n$-dimensional complex projective space and denote $[z_0:\dots:z_n]$ its points. If we glue $[z_0:\dots:z_n]$ and $[\overline{z_0}:\dots:\overline{z_n}]$ for any $[z_0:\dots:z_n]\in\mathbb{C}P^n$, where $\overline{z}$ denotes the complex conjugation of $z$, then we obtain a quotient space $\overline{\mathbb{C}P^n}:=\mathbb{C}P^n/[z_0:\dots:z_n]\sim[\overline{z_0}:\dots:\overline{z_n}]$.
Now I am interested in these $\overline{\mathbb{C}P^n}$'s, but I know neither if these $\overline{\mathbb{C}P^n}$'s have a formal name nor where I can find their properties.
I'd appretiate if you can provide some relavant references.
 A: This doesn't help with $\mathbb{CP}^n$, but see Atiyah and Berndt https://arxiv.org/abs/math/0206135 for a discussion of the quotient space for the other projective planes.
A: For $n=1$ the quotient space is $D^2$.
For $n=2$ the quotient spaces is $S^4$ , that was proved by Arnold, Kuiper and Massey (searching "Arnold-Kuiper-Massey Theorem" should yield the relevant references).
For higher dimensions note that the quotient space has an open set homeomorphic to $\mathbb{C}^{n}/\tau$ where $\tau$ is conjugation. Note that there will be an $\mathbb{RP}^{n-1}$ cone singularity along the fixed point set. I am not sure if there is somewhere where these orbifolds are studied in detail, maybe try searching papers which cite the ones mentioned above.
A: This answer gives information about the cohomology of $\overline{\mathbb CP^n}$. Perhaps someone will recognize this as the cohomology of a familiar space.
The conjugation is an action of $\Sigma_2$ on $\mathbb CP^n$. We are interested in the orbit space of this action. Recall that the fixed point of this action is homeomorphic to $\mathbb RP^\infty$.
The quotient space $\mathbb CP^n/\mathbb RP^n$ has a free action of $\Sigma_2$ (away from the basepoint). The fact that for free action strict quotient is equivalent to homotopy quotient implies that there is a homotopy pushout square
$$
\begin{array}{ccc}
\mathbb RP^n \times \mathbb RP^\infty & \to & \mathbb RP^n \\
\downarrow & & \downarrow \\
\mathbb CP^n \times_{\Sigma_2} E\Sigma_2 & \to & \overline{\mathbb CP^n} 
\end{array}
$$
It is perhaps instructive to consider what happens when $n=\infty$. One can realize the limit diagram as a diagram of classifying spaces:
$$
\begin{array}{ccc}
B\Sigma_2\times B\Sigma_2 & \to & B\Sigma_2 \\
\downarrow & & \downarrow \\
BO(2) & \to & \overline{\mathbb CP^\infty} 
\end{array}
$$
From this square one can, in principle, calculate the cohomology of $\overline{\mathbb CP^n}$ using the Meier-Vietoris exact sequence. If one localizes away from the prime $2$, the answer is pretty simple. In this case $\mathbb RP^\infty\simeq *$, and we obtain an equivalence away from the prime $2$:
$$
\mathbb CP^n \times_{\Sigma_2} E\Sigma_2  \xrightarrow{\simeq}  \overline{\mathbb CP^n}.
$$
If $\Lambda$ is a ring where $2$ is invertible, then $H^*(BO(2);\Lambda)\cong \Lambda[p_1]$, where $p_1$ is a class in dimension $4$ (the first Pontryagin class). The cohomology of $\mathbb CP^n \times_{\Sigma_2} E\Sigma_2$ is isomorphic to the truncation $\Lambda[p_1]/_{(p_1^{\left\lfloor \frac{n}{2}\right\rfloor+1})}$. Notice that for $n=1$ you get the cohomology of a point, and for $n=2$ you get the cohomology of a sphere, as expected.
Cohomology with mod 2 coefficients is more complicated/interesting. In the limit when $n=\infty$ you get the following diagram in cohomology
$$
\begin{array}{ccc}
H^*(\overline{\mathbb CP^\infty};\mathbb Z/2) & \to & \mathbb Z/2[w_1, w_2] \\
\downarrow & & \downarrow \\
\mathbb Z/2[x] & \to &\mathbb Z/2[x, y]
\end{array}
$$
Where the homomorphism on the right side sends $w_1$ to $x+y$ and $w_2$ to $xy$. The Poincare series of $\tilde H^*(\overline{\mathbb CP^\infty};\mathbb Z/2)$ comes out to be $\frac{t^4}{(1-t)(1-t^2)}$.
For finite $n$, we have to take a truncation of this diagram. If I am not mistaken, the diagram in cohomology comes out to be the following
$$
\begin{array}{ccc}
H^*(\overline{\mathbb CP^n};\mathbb Z/2) & \to & \mathbb Z/2[w_1, w_2]/_{(w_2^{n+1})} \\
\downarrow & & \downarrow \\
\mathbb Z/2[x]/_{(x^{n+1})} & \to &\mathbb Z/2[x, y]/_{(x^{n+1})}
\end{array}
$$
According to my calculations, the Poincare series of $\tilde H^*(\overline{\mathbb CP^n};\mathbb Z/2)$
turns out to be $$t^4\frac{(1-t^{n-1})(1-t^n)}{(1-t)(1-t^2)}.$$
More explicitly, this equals to
$$t^4(1+t+t^2+ \cdots +t^{n-2})(1+t^2+t^4+\cdots +t^{n-2})$$ if $n$ is even, and $$t^4(1+t^2+t^4+\cdots + t^{n-3})(1+t+t^2+\cdots +t^{n-1})$$ if $n$ is odd. Once again, this seems to give the right answer for $n=1, 2$, so I hope this is a good sign.
