We know [$H^d(\mathbb{RP}^5,\mathbb{Z}\_2)=\mathbb{Z}\_2$](https://topospaces.subwiki.org/wiki/Cohomology_of_real_projective_space#Coefficients_in_a_module_over_a_2-divisible_ring). So there are two classes of $\mathbb{Z}\_2$ generators, trivial and nontrivial, for $d=0,1,2,3,4,5$.


Wha are the Poincaré dual $(5-d)$-dimensional manifolds of the generators PD($a^d$) of $a^d\in H^d(\mathbb{RP}^5,\mathbb{Z}\_2)$, for $d=0,1,2,3,4,5$?


\begin{array}{|c|c|}
\hline
\text{PD}(a^d)& a^d =0 \in H^d(\mathbb{RP}^5,\mathbb{Z}\_2) &a^d \neq 0 \in H^d(\mathbb{RP}^5,\mathbb{Z}\_2)\\ \hline
d=0 &?& \mathbb{RP}^5?\\ 
\hline
d=1 &? &\mathbb{RP}^4?\\ 
\hline
d=2 &?& \mathbb{RP}^3?\\ 
\hline
d=3 &?&\mathbb{RP}^2?\\ 
\hline
d=4 &?&\mathbb{RP}^1?\\ 
\hline
d=5 &?&\mathbb{RP}^0=\text{a point}?\\
\hline
\end{array}


* It is said that if $a^d=0$, a Poincaré dual for $a\in H^d(M;\mathbb{Z}/2)$ is any embedded closed $(5-d)$-manifold which bounds, such as a small $S^{5-d}$ around a point, and this is orientable. Is this a correct generator for the trivial class $a^d =0 \in H^d(\mathbb{RP}^5,\mathbb{Z}\_2)$?



Poincaré dual of the generators of $H^d(\mathbb{RP}^5,\mathbb{Z}_2)$
Yes, all of them are true.$\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$


First, let's show $\RP^4\subset\RP^5$ is Poincaré dual to $a\in H^1(\RP^5;\Z/2)$. In this case only, there's a nice
geometric shortcut: $a$ determines a principal $\Z/2$-bundle $P\to\RP^5$ with $w\_1(P) = a$, unique up to
isomorphism. If $a\ne 0$, a codimension-1 submanifold $M\subset\RP^5$ is a Poincaré dual of $a$ iff $P$ is trivial
over $\RP^5\setminus M$.


The double cover $P = S^5\twoheadrightarrow\RP^5$ is the unique nontrivial double cover of $\RP^5$, and $a$ is the
unique nonzero element of $H^1(\RP^5;\Z/2)$, so $w\_1(P) = a$. $\RP^4$ embeds in $\RP^5$ as the equatorial
$S^4\subset S^5$ followed by taking the quotient by the antipodal map. Therefore, on $\RP^5\setminus\RP^4$, $P$ is
trivial: it's the double cover of the northern and southern hemispheres onto a ball.




---


To get at higher powers of $a$, one can use the fact that cup product is Poincaré dual to the intersection pairing.
That is, take two copies of $\RP^4\subset\RP^5$ and make them transverse to each other. Then, their intersection is
a Poincaré dual to $a^2$. However, thinking about intersections in $\RP^n$ can be confusing, so let's employ
another trick I like: instead of working on $\RP^n$, work on $S^n$, but only do things that are invariant under the
antipodal map.


Thus, instead of $\RP^4\subset\RP^5$, we're looking at the equatorial $S^4\subset S^5$; explicitly, under the
standard embedding $S^5\hookrightarrow\mathbb R^6$ as the unit sphere, this is the submanifold $\{x\_1 = x\_2 = x\_3 = x\_4 = x\_5 = 0\}\cap S^5$.
Another $S^4$ inside $S^5$ which is preserved by the antipodal map is $\{x\_2 = x\_3 = x\_4 = x\_5 = x\_6 = 0\}\cap S^5$, and
in the quotient this produces an $\RP^4$ homologous to the original one. Their intersection in $S^5$ is $\{x\_2 =
x\_3 = x\_4 = x\_5 = 0\}\cap S^5$, which is an $S^3\subset S^5$; taking the quotient by $\Z/2$, this is the usual
$\RP^3\subset\RP^5$, and therefore $\RP^3$ is a Poincaré dual to $a^2$.


Continuing in this way, you can show that $\RP^2$ is a Poincaré dual to $a^3$, $\RP^1$ is a Poincaré dual to
$a^4$, and $a^5$ is Poincaré dual to a point.




---


The remaining statements are all sort of tautological. A zero class in cohomology is Poincaré dual to a zero class
in homology, which is represented by a submanifold of the correct dimension which bounds. The Poincaré dual $N$ to
$1\in H^0(\RP^5;\Z/2)$ represents the mod 2 fundamental class in that the map $N\to\RP^5$ sends
$$[N]\mapsto[\RP^5]\in H\_5(\RP^5;\Z/2),$$
and $N = \RP^5$ and the identity map works.


