Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 16 / 3

Let $m=\dim X$ be odd. Use mod-two Poincare duality, which applies even when the closed manifold is nonorientable. It gives $b_k^{(2)}=b_{m-k}^{(2)}$, where $b_k^{(2)}={\dim }_{{\mathbb{F}}_2}{H}_k(X;{\mathbb{F}}_2)$. The Euler characteristic can be computed with any field of coefficients: for a finite cell model, the alternating sum of chain dimensions equals the alternating sum of homology dimensions, and the former counts cells independently of the field.

Consequently $\chi (X)={\sum }_{k=0}^m(-1{)}^k{b}_k^{(2)}$. Pair each term with index $m-k$. The paired Betti numbers are equal and their signs opposite because $m$ is odd; there is no unpaired middle index. ThereforeThis argument applies componentwise if the manifold is disconnected and does not require an orientation.

Write the alternating bilinear form as $B$. Alternation gives $B(v,w)=-B(w,v)$ by expanding $B(v+w,v+w)=0$. If $V\ne 0$, choose $v\ne 0$. Nondegeneracy supplies $w$ with $B(v,w)\ne 0$; rescale $w$ so that $B(v,w)=1$. The vectors are independent, and the restriction to $E=\mathrm{span}\{v,w\}$ has matrix $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, whose determinant is one.

Every $z\in V$ has a unique decomposition into a vector in $E$ and a vector orthogonal to $E$: subtract $B(z,w)v-B(z,v)w$ to obtain the latter. Thus $V=E\oplus E^{\perp }$. If a vector in $E^{\perp }$ annihilates $E^{\perp }$ under $B$, it also annihilates $E$, and hence all of $V$; nondegeneracy makes it zero. The restricted alternating bilinear form on $E^{\perp }$ is therefore again nondegenerate. Induction splits off two dimensions at each step and ends at the zero space. HenceThe proof works in characteristic two as well: alternation, rather than merely skew symmetry, is the needed hypothesis.

Write $\dim X=4r+2=2m$, with $m=2r+1$ odd. Rational Poincare duality gives $b_k=b_{2m-k}$. Pairing every term outside the middle yieldsThe Poincare duality pairing on $H^m(X;\mathbb{Q})$ is $B(\alpha ,\gamma )=⟨\alpha ⌣\gamma ,[X]⟩$. It is nondegenerate, and graded commutativity of the cup product gives $B(\alpha ,\gamma )=-B(\gamma ,\alpha )$ because $m$ is odd. In particular $B(\alpha ,\alpha )=0$ over $\mathbb{Q}$. Part (b) therefore makes $b_m$ even, soFor the failure without orientability, ${\mathbb{RP}}^2$ is a closed nonorientable surface with one cell in each dimension zero, one and two; hence $\chi =1-1+1=1$. More generally, ${\mathbb{RP}}^{4r+2}$ is nonorientable and has Euler characteristic one. Thus the orientation hypothesis in the Euler characteristic parity in dimensions congruent to two modulo four is essential.