Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 16 5 Solution Created 2026-10-03 Updated 2026-10-14
Over the Grassmannian , the complex tautological bundle on a Grassmannian isIts fibre at is the k-plane itself. A neighbourhood of a fixed plane consists of graphs of linear maps ; gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle .
Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates , and a subordinate partition of unity with support contained in . Defineextending each component by zero off its support. This is continuous and satisfies , so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is . Thus they define a map , and identifies with . This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.
We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff ; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices . Compactness gives a subdivision withRestriction of to the image at is injective: if it killed , then , impossible by the strict norm bound unless . Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.
Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings and of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.
Write . For the assumed fiber bundle with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor this gives group isomorphisms. For , the relevant pointed-set part and connectedness of give , corresponding to the single component . Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle . Since , its exact sequence yields . The given identification and the degree calculation therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.
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.
For an oriented rank- bundle over , with , normalize its clutching function to be based at the identity. A nowhere-zero section exists precisely when the displayed homotopy class vanishes. If it vanishes, extend across one hemisphere and use the constant section on the other. Conversely normalize any nowhere-zero section using the fiber metric. Its two hemisphere maps satisfy on the boundary. The first hemisphere makes homotopic to a constant; the second makes null-homotopic. Thus is null-homotopic. If , the lower homotopy group of the sphere vanishes and a section always exists. Rank one is already trivial because is trivial.