= Solution
Work first with coefficients $\mathbb F_2$. For a rank-$r$ real <vector bundle> $E\to B$ over a <CW complex>, choose a fibre metric and its <disk bundle> $D(E)$ and <sphere bundle> $S(E)$. The mod-two <Thom class> is the unique class $U\in H^r(D(E),S(E);\mathbb F_2)$ restricting to the nonzero generator on every fibre pair. The <Thom isomorphism theorem> asserts that
$$
T:H^{j-r}(B;\mathbb F_2)\xrightarrow{\ \cong\ }H^j(D(E),S(E);\mathbb F_2),\qquad a\longmapsto\pi^*a\cup U.
$$
No orientation of the <vector bundle> is needed with these coefficients.
Let $j:H^*(D(E),S(E))\to H^*(D(E))$ forget the relative condition, and let $s:B\to D(E)$ be the zero section. Set $e_2(E)=s^*j(U)$, the top <Stiefel–Whitney class> $w_r(E)$. Since $D(E)$ retracts onto the zero section, $j(U)=\pi^*e_2(E)$. Thus the relative-to-absolute map corresponds under the <Thom isomorphism theorem> to multiplication by $e_2(E)$. Substituting these identifications into the <long exact sequence> in <relative cohomology> of $(D(E),S(E))$ gives the <unoriented Gysin sequence>
$$
\cdots\longrightarrow H^{j-r}(B;\mathbb F_2)\xrightarrow{\cup e_2(E)}H^j(B;\mathbb F_2)\xrightarrow{p^*}H^j(S(E);\mathbb F_2)\longrightarrow H^{j-r+1}(B;\mathbb F_2)\xrightarrow{\cup e_2(E)}H^{j+1}(B;\mathbb F_2)\longrightarrow\cdots,
$$
where $p$ is the <sphere bundle> projection.
Apply this to the <real tautological line bundle> $\gamma_n\to\mathbb{RP}^n$. Its <sphere bundle> is $S^n$, with projection the antipodal double cover, and put $t=e_2(\gamma_n)\in H^1(\mathbb{RP}^n;\mathbb F_2)$. Suppose $n\ge1$. The map on $H^0$ from the connected base to the connected sphere is an <isomorphism>, so exactness shows that multiplication by $t$ is injective from $H^0$ to $H^1$. Since both groups are one-dimensional, $t$ is a generator.
For $1<j<n$, the adjacent sphere <cohomology groups> in the <Gysin sequence> vanish, so multiplication by $t$ is an <isomorphism> from degree $j-1$ to degree $j$. If $n>1$, the top portion is
$$
0\longrightarrow H^{n-1}(\mathbb{RP}^n;\mathbb F_2)\xrightarrow{\cup t}H^n(\mathbb{RP}^n;\mathbb F_2)\longrightarrow H^n(S^n;\mathbb F_2)\longrightarrow H^n(\mathbb{RP}^n;\mathbb F_2)\longrightarrow0.
$$
The last nonzero arrow is surjective between one-dimensional groups, hence an <isomorphism>; the preceding arrow is zero, so multiplication by $t$ is again an <isomorphism>. For $n=1$ the earlier $H^0$ argument already gives the top multiplication. Thus $1,t,\ldots,t^n$ are the nonzero generators in their respective degrees, while $t^{n+1}=0$ by dimension. We obtain
$$
\boxed{H^*(\mathbb{RP}^n;\mathbb F_2)\cong\mathbb F_2[t]/(t^{n+1}),\qquad |t|=1.}
$$
For $n=0$ this says simply $H^*(\mathbb{RP}^0;\mathbb F_2)=\mathbb F_2$, with $t=0$. This computes the <mod-two cohomology ring of real projective space>, including its multiplication rather than only its additive groups.
Now use integral coefficients and an oriented rank-$n$ <vector bundle>. Its orientation selects an integral <Thom class> $U$. Define its <Euler class> by
$$
\boxed{e(E)=s^*j(U)\in H^n(B;\mathbb Z).}
$$
Again $j(U)=\pi^*e(E)$. The <cup product> of two relative classes in $H^*(D(E),S(E))$ agrees with the mixed relative/absolute product after forgetting the relative condition on either factor. Consequently
$$
\boxed{U\cup U=U\cup j(U)=U\cup\pi^*e(E).}
$$
This is the <cup square of a Thom class>. If $n$ is odd, <graded commutativity of the cup product> gives $U\cup U=-U\cup U$, hence $2U^2=0$. Since $U\cup\pi^*(2e(E))$ is, up to the graded sign, the <Thom isomorphism> image of $2e(E)$, its vanishing implies
$$
\boxed{2e(E)=0\qquad(n\text{ odd}).}
$$
Thus the <Euler class of an oriented odd-rank vector bundle is two-torsion>; the conclusion is integral and does not require the base <cohomology> to be torsion-free.
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