= Solution
An $R$-orientation of a rank-$d$ real <vector bundle> $E\to B$ is a locally coherent choice of generator of $H^d(E_b,E_b\setminus\{0\};R)$ in every fibre. Equivalently, it is represented by a <Thom class> $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on each fibre pair. If $e(E)\in H^d(B;R)$ is the <Euler class>, the <Gysin sequence> of its unit sphere bundle contains
$$
\cdots\to H^{q-d}(B;R)\xrightarrow{\smile e(E)}H^q(B;R)
\to H^q(S(E);R)\to H^{q-d+1}(B;R)\xrightarrow{\smile e(E)}H^{q+1}(B;R)\to\cdots.
$$
The diagonal quotient defining $L(p)$ is the <three-dimensional lens space as a circle bundle> $S^1\to L(p)\to S^2$ with Euler class $p$ times a generator. With integral coefficients, the only nontrivial Euler-class map is multiplication by $p:H^0(S^2;\mathbb Z)\to H^2(S^2;\mathbb Z)$. Exactness gives
$$
\boxed{H^j(L(p);\mathbb Z)\cong
\begin{cases}
\mathbb Z,&j=0,3,\\
\mathbb Z/p,&j=2,\\
0,&\text{otherwise}.
\end{cases}}
$$
Modulo $p$, the Euler class vanishes, so the same <Gysin sequence> gives
$$
\boxed{H^j(L(p);\mathbb F_p)\cong\mathbb F_p\quad(0\leq j\leq3),}
$$
with all other groups zero.
For the coefficient sequence $0\to\mathbb Z\xrightarrow{p}\mathbb Z\to\mathbb F_p\to0$, the <long exact sequence from a coefficient sequence> contains
$$
0=H^1(L(p);\mathbb Z)\to H^1(L(p);\mathbb F_p)
\xrightarrow{\delta}H^2(L(p);\mathbb Z)
\xrightarrow{p}H^2(L(p);\mathbb Z).
$$
The last map is zero and both middle groups have order $p$, so $\delta$ is an isomorphism. Reduction $H^2(L(p);\mathbb Z)\to H^2(L(p);\mathbb F_p)$ is likewise an isomorphism. Their composite is the <Bockstein isomorphism for a three-dimensional lens space>
$$
\boxed{\beta:H^1(L(p);\mathbb F_p)\xrightarrow{\sim}H^2(L(p);\mathbb F_p).}
$$
If $a'=na$ is another generator, linearity of the <Bockstein homomorphism> and bilinearity of the <cup product> give
$$
\boxed{t(a')=n^2t(a).}
$$
Moreover $t(a)\ne0$: $\beta(a)$ is nonzero, and the <Poincare duality> pairing $H^1\times H^2\to H^3\cong\mathbb F_p$ is nondegenerate. If $h:L(p)\to L(p)$ is an orientation-reversing <homotopy equivalence> and $h^*a=na$, naturality gives
$$
n^2t(a)=t(h^*a)
=\langle h^*(a\smile\beta a),[L(p)]\rangle
=\langle a\smile\beta a,h_*[L(p)]\rangle
=-t(a).
$$
Cancelling the nonzero $t(a)$ yields $n^2\equiv-1\pmod p$. Thus \b[$-1$ must be a <quadratic residue> modulo $p$].
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