Every closed odd-dimensional manifold has Euler characteristic zero. For an orientable manifold this follows by pairing complementary Betti numbers using Poincare duality; the general case follows from the orientable double cover and Euler characteristic under a finite covering.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 16 1 b Solution Created 2026-10-03 Updated 2026-10-14
For the closed orientable surface , use the fundamental group presentationSince , define a homomorphism to by sending to and every other displayed generator to zero. All commutators map to zero, so the relation is respected, and the map is surjective. Its kernel has index . The classification of connected covering spaces supplies a connected -sheeted covering space, including the identity cover when .
Every connected finite covering space of is a closed orientable surface: compactness follows from finite degree, and the base orientation lifts through local covering charts. Lift a finite cell decomposition of the base. Every cell has lifts, so the Euler characteristic under a finite covering gives . If the covering surface has genus , this reads . ThereforeThis genus of a finite cover of a closed orientable surface depends only on the sheet number, not on the particular subgroup used to construct the cover.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 14 1 3 Solution 2026-10-07
Call this simultaneous antipodal quotient of two spheres . The simultaneous antipodal map acts freely on , so is a connected closed manifold of dimension four. Each antipodal factor has mapping degree ; their product preserves orientation. Hence is orientable.
The product is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives . Its rational first Betti number is zero, and Poincare duality gives and . Thus as well.
The universal coefficient theorem for cohomology now gives andHere because . Integral Poincare duality further gives , and .
Let generate these groups in degrees respectively. The only potentially nonzero positive-degree cup product is . But , while has no nonzero torsion subgroup, so . Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 156 1 c Solution 2026-09-28
Because the action is free, the quotient is a closed orientable surface , and is a covering of degree . The Euler characteristic under a finite covering givesor equivalentlyThe quotient inherits a hyperbolic metric, so . Hence and .
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 114 2 c Solution Created 2026-09-24 Updated 2026-09-25
A closed manifold of odd dimension has Euler characteristic zero by the Euler characteristic of an odd-dimensional closed manifold. Decomposing into and the removed closed ball along givesso .
If the antipodal boundary map extended to a fixed-point-free involution of , the quotient map would be a double covering. The formula for Euler characteristic under a finite covering would implywhich is impossible. Hence no such extension exists.
For positive , quotient by . This is a free double covering space action. Its closed manifold quotient is orientable exactly when is even, because the product mapping degree is . If , the product is simply connected and the quotient has fundamental group . For , the quotient has integral cohomology groups in degrees , in degrees , and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.