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 quotient space of complex projective space
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.


