Talk:Normal maps - (non)-examples (Ex)
1) First note that, for any closed manifold , there is a map giving by contracting the complement of an embedded disc . At a point , this map is bijective and has local degree and so by the fact that the degree is the sum of local degrees, i.e. where and which in this case is just .
Now specialise to the case and , and consider , an arbitrary 4-dimensional real vector bundle over . If there is a normal map then, since this is a fibrewise isomorphism we must have that . Consider the first Pontryagin classes and . With respect to some identifications and , we will show that and for some which will depend on the choice of bundle .
This would complete the proof since this would imply that but also that which is divisible by two. Note that that fact that is only a stable bundle map is dealt with here by the fact that is a stable bundle invariant and so gives that same construction for any two stably isomorphic bundles of the same dimension.
To show that for and some choice of identification , we can use a number of techniques. One quick trick is to note that the Hirzebruch Signature Theorem applies to and gives that by the fact that the first polynomial is the multiplicative sequence corresponding to the -genus is . Now and the intersection form is gives that and so . Hence with appropriate identifications. Another, more elementary, method is to show that where is the tautological complex line bundle over . This allows us to compute the total Chern class immediately which can then be used to recover the Pontryagin classes using the the highly useful formula where denotes that fibrewise replacement of with .
To show that is always even is more tricky, and is asserted (but not proven) in Milnor's paper which proves the existence of exotic -spheres. Firstly let be the sphere bundle corresponding to . Recall that, by clutching construction for bundles over spheres, there is a 1-1 correspondence between fibre bundles over with fixed fibre and structure group and . Hence oriented -bundles over are classified by . One can show explicitly that and so has universal cover and therefore since covering maps induce isomorphisms on higher homotopy groups. Fix an explicit identification of with . Therefore where labels the element of corresponding to the sphere bundle.
There is a map given by mapping to . By considering the inclusion of the fibre of the corresponding real vector bundle into the total space, one can show that this map factors as group homomorphisms where is the Grassmannian , i.e. the -dimensional linear subspaces of . Hence is linear in and and so of the form . By considering the quaternionic conjugate on the fibre, we can also show that which guarantees that for some and some identification . To finish, we can check that is the tautological bundle over by using an explicit identification of with . This shows that where is the pullback of the chosen generator of along an identification . This shows that and hence which completes the proof.
See also Obstruction classes and Pontrjagin classes (Ex).
2) For any , let . Note that the stable normal bundle is trivial since the normal bundle of the codimension one embedding is trivial. Hence any map extends to a bundle map and so it suffices to find degree maps for any .
For , a degree map is given by wrapping the circle times around itself. By suspending times we get a map which we can show still has degree by using local degrees as in the previous exercise.
3) We will show that there is a degree one normal map if and only if . Firstly note that, as above, the codimension one embedding is trivial and so is trivial for all . Hence any map extends to a normal map and it remains to show that there is a degree one map if and only if .
If , then we can get a map by contracting the complement of the standard . This is degree one using local degree as in the first exercise.
Let and suppose there is a map of cohomology rings sending the generator of to the generator of . Note that is generated by in degree one and is degree two such that for and all other products of the degree one generators zero. This can be lifted from the quotient map and the structure of the cohomology ring of coming from that of . Since , the map has a non-zero element in its kernel. Suppose, without loss of generality, that . Then and so . However and by hypothesis. This is a contradiction.
This can also be proven by using that a degree one map induces a surjection and so a surjection which gives a contradiction as before.