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# 2 The odd dimensional case

We begin by recalling the statement of Theorem 1.

Theorem 1   Let Q be a smooth quadric hypersurface in n + 1, where n = 2k + 1. Then for any positive integer d 0  (mod 2k) there exist continuous maps f : nQ, where f*((1)) = (d ).

Proof:     This is obvious for n = 1. We may assume, by induction, that the theorem holds for quadrics of dimension n - 2. n has a cell decomposition where the skeleta are linear projective subspaces of smaller dimension. By Scholium 1 we have a cell decomposition of Q whose n - 1 skeleton is the projective cone C over a smooth quadric Q' of dimension n - 2. For any d' 0  (mod 2k - 1), the induction hypothesis gives us a map f' : n - 2Q' satisfying f'*((1)) = (d'). Since n - 1 is the projective cone over n - 2, Construction 1.1 yields a map C(f') : n - 1C. The obstruction to extending this map to a map nQ is a class O(C(f')) H2n(n,(Q)). By Lemma 4, the group (Q) has exponent 4. Let be the composite

n - 1n - 1C,

where is the restriction of the map : nn preserving the cell structure, and given in suitable homogeneous coordinates by

(z0 : ... : zn) (z02 : ... : zn2).

The obstruction to extending to a map nQ is

O() = (O(C(f'))) H2n(n,(Q)).

But clearly = 0 on this cohomology group. Let f : nQ be an extension of . We have a commutative diagram of integral cohomology groups

 H2(Q) H2(n) H2(Q') H2(n - 2)

where f'' is the composite of f' with the restriction of . Since f''* is multiplication by 2d', so is f*.
We now prove the following refinement of Theorem 2, in the odd dimensional case. 1 1'

Theorem 2   Let Q n + 1 be a smooth quadric hypersurface with n = 2k + 1. Then there exists a continuous map f : QQ of degree dn whenever
(viii)
d 0  (mod 2k), or
(ix)
d = e2n - 1, for some integer e.

22 Proof:     Clearly (i) follows from Theorem 1. We prove (ii) by showing that for the chosen integers d the map Fd, n : Q(n + 2)Q(n + 2) extends to a map f : QQ. By induction, we may assume that Fd, n - 2 : Q'(n)Q'(n) extends to a map f' : Q'Q', for Q' a smooth quadric hypersurface of dimension n - 2, and d = e2n - 3 for some e > 0. We have constructed (see (1.1)) a map C(f') : CC which satisfies C(f') = Fd, n. By computing C(f')* on (C) we will show that the obstruction to extending the four-fold composite C(f')4 to a self map of Q vanishes. There is a filtration F on (C) given as follows. Take F0 = (C); F1 is the kernel of the composite

(C)()H2n - 1(,),

where is the Hurewicz map; and finally F2 = im ((Q(n))(C)). Clearly C(f')* is compatible with this filtration and induces a map gr FC(f')* on gr F(C). This map may be computed as follows.

Lemma 6
(x)
F2, F1/F2 are vector spaces over /2.
(xi)
The natural composite map

(Q, C)(C)F0/F1 H2n - 1(,)

is an isomorphism, giving a direct sum decomposition

(C) F1 (Q, C).

(xii)
gr FC(f')* is multiplication by dn.

Proof:     We have a commutative diagram with exact bottom row

 H2n(,) H2n - 1(,) 0 (,) () () 0

The boundary homomorphism is an isomorphism by the long exact sequence of homology for the pair (,), and is an isomorphism by the relative Hurewicz theorem. This proves (ii) and gives an isomorphism F1 (). The self map of is of degree dn and this gives (iii) for F0/F1. Since Q(n) = L is a linear projective subspace of n + 1, is an Sn in ; it is easy to check that, in the fibration (*), this maps isomorphically to a great sphere S Sn + 1. In fact is a section of the sphere bundle of the tangent bundle of S which is contained in by Scholium 3. Let D- be a hemisphere capping S in Sn + 1 and let U be its inverse image in . We have the following

Sublemma 7   Let i : Sn be a fibre of (*) lying over a point of D-. A unit tangent vector field v on S Sn gives a map v : Sn which is homotopic within U to the inclusion i.

Proof:     Let p be the point of D- orthogonal to S (i.e. the pole''). We have a map : S×[0,]D- given by

(x, t) sin(t) . p + cos(t) . x.

For all (x, t) S×[0,] let n(x, t) be the tangent vector at the point (x, t) given by sin(t) . x - cos(t) . p. Then, dv(x) is orthogonal to n(x, t) in the tangent space of Sn + 1 at (x, t) so that we get a map H : S×[0,] given by the formula

(x, t) dv(x) + sin(t) . n(x, t).

Clearly H(x, 0) = v(x) and H(x, t) = x considered as a tangent vector at p.
Thus we have isomorphisms

F2 im (()()) = im ((Sn)()),

where SnQ is the inclusion of the fibre of (*). Hence F2 is a vector space over /2, and by Scholium 5 the action of gr FC(f')* on it is by dk + 1 dn  (mod 2). Further, we obtain an isomorphism

F1/F2 ()/im (()) 2((Sn + 1, D-)),

so that F1/F2 is a /2-vector space. Let g : (Dn + 1, Sn)(,) be the generator of (,) (Q(n + 1), L) . We have a diagram, commutative upto homotopy,

 (Dn + 1, Sn) (,) (Dn + 1, Sn) (,)

where is a map of degree dk + 1. From the Scholium 5 we see that ()* induces multiplication by dk + 1 on (Dn + 1, Sn). By the sublemma we have isomorphisms

(,)(,)(Sn + 1, D-),

so that the composite : (Dn + 1, Sn)(Sn + 1, D-) of g and the natural map (,)(Sn + 1, D-) is also a generator for (Sn + 1, D-). By the Freudenthal suspension theorem, is an isomorphism on . From the diagram

 H2n(,;) H2n - 1(,;) 0 (,) (,) (,) 0

where , are isomorphisms,we see that

im ((Dn + 1, Sn)(,))
= ker((,)H2n - 1(,;)).

In particular, F1/F2 is contained in this image; thus C(f')* acts by multiplication by dk + 1 dn  (mod 2) on F1/F2,and this completes the proof of (iii).
From this lemma we see that we have Hom ((Q, C), F1) = F1 and Hom (F1/F2, F2) End (F1) such that, for all pairs (a, b) in (C) = (Q, C) F1 we have the equation

C(f')*(a, b) = (dna, dnb + (b) + (a)).

Since both F1, F1/F2 are of exponent 2, it follows that the four-fold composite of C(f') satisfies

C(f')4*(a, 0) = (d4na, 0),

and this proves the theorem.

Next: 3 The even dimensional Up: Continuous Self Maps of Previous: 1.2 Computation of Homotopy
Kapil Hari Paranjape 2002-11-21