Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 17 5 Solution Created 2026-10-03 Updated 2026-10-14
Let and be the two closed hemispheres of the sphere. Use as the clutching function between the trivial bundles on these hemispheres. The clutching construction definesThe projection to is locally a vector bundle. In a collar of the equator, extend the transition function by keeping it constant in the normal coordinate; this gives explicit vector bundle trivializations across the seam. Since the transition matrices lie in the special orthogonal group, the standard fiber orientations and Euclidean inner products agree on overlaps. Thus this is an oriented rank- vector bundle with a fiber metric.
For , choose a base point on the equator. A constant change of the negative hemisphere frame replaces by , so we may suppose . With the last standard unit vector, define the section obstruction for a clutched vector bundleThe crucial point is that a nowhere-zero section can be normalized using the fiber metric. It is then represented by continuous maps satisfyingon the equator.
If , the extension-null-homotopy criterion for a sphere extends to . Take this extension as and the constant map as . They glue to a nowhere-zero section.
Conversely, suppose such a section exists. Restricting a contraction of shows that is homotopic to a constant. The target sphere is path connected, so the constant can be moved to . Multiplying this homotopy pointwise by gives a homotopy from to . The first of these maps equals and extends over , hence is null-homotopic. So is null-homotopic too. A freely null-homotopic based sphere map represents zero in its homotopy group; any change of base point sends the zero class to zero. We have proved both directions:For , the same argument reads as the extension criterion for a two-point boundary and uses the trivial set when . For , the special orthogonal group is trivial, so the vector bundle is already a trivial vector bundle; the corresponding constant map to has zero obstruction. These cases require no higher homotopy group calculation.
If , then and the permitted vanishing of lower homotopy groups of a sphere makes . Normalize the resulting nowhere-zero section to a unit section , and setLocal orthonormal frames show that these orthogonal complements form a rank- vector bundle. The explicit vector bundle isomorphismhas inverse . Orient so that a positive frame followed by is positive in . This is splitting a trivial line from a nowhere-zero section and proves the oriented splittingFinally consider the special orthogonal sphere fibrationThe fiber over consists exactly of with . Near any unit vector, the Gram-Schmidt process continuously completes it to an oriented orthonormal frame, giving local sections of and local product fiber bundle charts. In particular has the homotopy lifting property.
Let be based and suppose . The composite is a based null-homotopic map by the given vanishing of . Lift a based null-homotopy of , beginning at , to a homotopy . The lift might initially move its base point , but always fixes . Right multiplication by corrects this: the homotopystill covers the same sphere homotopy, remains based, and starts at . At its endpoint every matrix fixes , so lies in the included . Thus every homotopy class in the target is represented by one from :This proves surjectivity of special orthogonal stabilization below the sphere dimension. For , this is the corresponding assertion about path components; the positive-rank special orthogonal groups are connected. The lifting argument also explains the zero term in the long exact sequence of homotopy groups of a fibration.