I realize this question Poincare dual PD[$A$] of a manifold $M$ and $A\in H^1(M,\mathbb{Z}_k)$ has something to do with the Steenrod problem.
I want to ask a more specific question.
Let $X \in H^1(S^5/\mathbb{Z}_k, \mathbb{Z}_k)=\mathbb{Z}_k$.
Let $p_1(M)$ is the 1st Pontryagin class of $M=S^5/\mathbb{Z}_k$.
How to evaluate $$ \int_{M=S^5/\mathbb{Z}_k} X p_1(M)=? $$ The integration of $X p_1(M)$ over $M=S^5/\mathbb{Z}_k : = L(k;1,1,1)$ defined as Lens space https://en.wikipedia.org/wiki/Lens_space#Definition?
The relation to a previous question is that if we know PD($X$), then we may evaluate $p_1$(PD($X$)). But this may not give what we need for this
We can take two examples, one is solved already as my trial, the other is my question for you:
- $k=2$, we have $M=S^5/\mathbb{Z}_2=\mathbb{RP}^5$, since $p_1(M)=w_2(M)^2 \mod 2$,
$w_2(M)=\binom{6}{2}X^2=X^2 \mod 2$,
$X p_1(M)=X w_2(M)^2=X^5 \mod 2$
So $\int_{M=S^5/\mathbb{Z}_2} X p_1(M)=1$, which we get the answer.
- $k=3$, what is the answer on $S^5/\mathbb{Z}_3 : = L(3;1,1,1)$?
Thank you!
Consider a $(2n+1)$ dimensional lens space $L=L_k(l_1,l_2,\ldots, l_n)$. Let $\Bbb Z_k$ act on $S^{2n+1}\times \Bbb C$ by $(u,z)\mapsto(A{\bf{u}},\zeta z)$, where $\zeta=e^{i2\pi/k}$ and $A=\operatorname{diag}(\zeta^{l_0},\ldots,\zeta^{l_n})$ (or in terms of real entries, $A$ is a block diagonal matrix, where each block represents a rotation by $2l_i\pi/k$ radians). Passing to the orbit space, we have a complex line bundle $\eta$ over the lens space. If the action is generated by $(u,z)\mapsto(A {\bf u},\zeta^j z)$, then we obtain the complex line bundle $\eta^{\otimes j}$ (the $j$-fold tensor product of $\eta$). By Gysin sequence, the bundle $\eta$ is non-trivial, with $c_1(\eta)$ generating $H^2(L;\Bbb Z)$.
Consider the bundle $TS^{2n+1}\oplus\varepsilon\cong \varepsilon^{2n+2}$, where $\varepsilon$ is the trivial (real) line bundle. Let $\Bbb Z_k$ act trivially on the trivial summand and by the differential on $TS^{2n+1}$ summand. For the trivial $(2n+2)$-plane bundle, we define the action on $S^{2n+1}\times \Bbb R^{2n+1}$ to be generated by $({\bf x},{\bf u})\mapsto (A{\bf x},A{\bf u})$, where $A$ is realized as a real block diagonal matrix as mentioned before. By direct computation, we see that the group actions commute with the bundle isomorphism, i.e., let $E(TS^{2n+1}\oplus \epsilon^{2n+2})\to S^{2n+1}\times \Bbb R^{2n+2}$ be the bundle isomorphism defined as $({\bf{x}},({\bf{v}},t))\mapsto({\bf x},{{\bf{v}}+t\bf{x}})$, then if we apply the group action followed by bundle isomorphism, we end up getting $$({\bf{x}},({\bf{v}},t))\mapsto (A{\bf{x}},(A{\bf{v}},t))\mapsto (A{\bf{x}},{A\bf{v}}+t{A\bf x})=(A{\bf{x}},A({\bf{v}}+t{\bf x}))$$ Swapping the order of these operations gives, $$({\bf{x}},({\bf{v}},t))\mapsto ({\bf{x}},{\bf{v}}+t{\bf x})\mapsto (A{\bf x},A({\bf v}+t{\bf x}))$$ Passing to the orbit space, we get the bundle isomorphism $TL\oplus\varepsilon\cong TS^{2n+1}/\Bbb Z_k\oplus \varepsilon\cong\varepsilon^{2n+2}/\Bbb Z_k$. If we identify $\Bbb R^{2n+2}$ with $\Bbb C^{n+1}$, then $\varepsilon^{2n+2}/\Bbb Z_k\cong (\eta^{\otimes l_0}\oplus\eta^{\otimes l_1}\oplus\ldots\oplus\eta^{\otimes l_n})_{\Bbb R}$. Note that this is an isomorphism of the underlying real vector bundles.
Using $TL\oplus\varepsilon\cong(\eta^{\otimes l_0}\oplus\eta^{\otimes l_1}\oplus\ldots\oplus\eta^{\otimes l_n})_{\Bbb R}$, we can compute the Pontrjagin class as $$p(L)=\prod_{i=0}^n(1+l_i^2c_1(\eta)^2)$$ Let $\tilde{c_1}$ denote the mod $k$ reduction of $c_1(\eta)$, then for $5$-dimensional lens spaces, this simplifies to $$\tilde{p}(L)=1+(l_0^2+l_1^2+l_2^2)\tilde{c_1}^2\pmod{k}$$
Let $[L]$ be the fundamental class in $H^5(L;\Bbb Z)$, and let $[L]_k$ be its image under the coefficient reduction. Then, $$\langle X\tilde{p_1}(L),[L]_k\rangle=(l_0^2+l_1^2+l_2^2)\langle X\tilde{c_1}^2,[L]_k\rangle$$ Without further computation, we conclude that $\langle X\tilde{p_1}(L_3(1,1,1)),[L_3(1,1,1)]_3\rangle=0\in\Bbb Z_3$ for any $X\in H^1(L_3(1,1,1),\Bbb Z_3)$.
In general, we need to compute the intersection product $\langle X\tilde{c_1}^2,[L]_k\rangle$. According to this answer, this is viable if we can express $\tilde{c_1}$ in terms of a particular generator (which I don't know how to do).
Edit:
I found a class of examples where the intersection product can be computed directly. Suppose $k=3,4,6$, then there are precisely two choices of generators for each $H^i$. Choose $x\in H^1$ such that $x\beta(x)^2=[L]_k$, where $\beta$ is the Bockstein homomorphism, then we have $\tilde{c_1}=\pm\beta(x)$ and hence $\langle x\tilde{c_1}^2,[L]_k\rangle=1$. Therefore, if $X=ax$ for some $a\in\Bbb Z$, then $$\langle Xp_1(L_k(l_0,l_1,l_2)),[L]_k\rangle\equiv a(l_0^2+l_1^2+l_2^2)\pmod{k}$$