Next: Bibliography Up: Continuous Self Maps of Previous: 2 The odd dimensional

# 3 The even dimensional case

In this section we prove the following theorem

Theorem 2   Let Q be a smooth quadric hypersurface of dimension n = 2k. There is an integer m such that for any positive d 0  (mod m), the map Fd, n : Q(n + 2)Q(n + 2) extends to a continuous map f : QQ (of degree dn).

Proof:     Suppose that for some d > 0, the map Fd, n - 2 : Q'(n)Q'(n) extends to a map f' : Q'Q'. Then the map C(f') : CC, obtained by the construction of (1.1), restricts to Fd, n : Q(n + 2)Q(n + 2). We compute below the obstruction to extending C(f') to a map f : QQ. As in the odd dimensional case, we begin by observing that we have the diagram with exact bottom row

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

where the vertical maps are Hurewicz maps. Since , are isomorphisms, the composite

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

is an isomorphism, and

(C) (Q, C) (Q) (,) ().

From Lemma 4, we have a split exact sequence for each i

0(Sn)()(Sn + 1)0,

where the splitting is obtained from a homotopy section of (*). Since Q(n) = Q(n + 1), the map (Q(n))(Q) is an isomorphism for i = n, and a surjection for i = n + 1. In particular, the homotopy section of (*), and the inclusion of the fibre Sn of (*), factor through Q(n). Hence (Q(n))(Q) is surjective. Thus

(C) (Q, C) im ((Q(n))).

We can refine this a little. Since (Q(n)) (Q) , there is a map : Sn, inducing a map g : SnQ(n), such that , g represent generators of , and is homotopic to the inclusion of the fibre of (*); further,

() Hn(,) Hn(Q(n),),

and (Fd, n)* acts by multiplication by dk on Hn(Q(n),). Then there is an inclusion h : (Sn) (C) induced by g and a homotopy commutative diagram

 Sn Sn g g Q(n) Q(n)

where has degree dk. Next, an easy computation shows that Sn + 1 maps isomorphically onto Sn + 1 in the fibration (*). Further Fd, n restricts to a self map of L' of degree dk. Thus we have a decomposition

im ((Q(n))(C)) = h*((Sn)) (L').

The action of C(f')* on the left is compatible with this decomposition, and induces on (Sn), and (Fd, n)* on (L'). Note that since n is even, (Sn) (Sn)tors (where the subscript tors'' denotes the torsion subgroup).

Lemma 8
(xiii)
C(f')* acts by multiplication by dn on (C) .
(xiv)
With respect to the direct sum decomposition

(C) = (Q, C) ( (Sn)tors) (L'),

C(f')* has a matrix of the form

where
 Hom ((Q, C),(Sn)tors) Hom ((Q, C),(Sn + 1)) Hom (,(Sn)tors)

Proof:     From the Scholium 5, the action of on (Sn)tors is by dk, while it is by d2k = dn on (Sn) . The action of (Fd, n)* on

() Hn + 1(,) Hn(L', H1(S1))

is by dk + 1. From the Scholium, this implies that (Fd, n)* acts by dk + 1 on (L'). Hence (ii) follows, once we prove (i). Now Q(n) = L' L'' where L' L'' = L k - 1. Since (L) is finite, we see that (Q(n))(Q(n), L) is injective. We have a diagram, whose vertical arrows are Hurewicz maps,

 Hn(Q(n),) Hn(Q(n), L;) (Q(n)) (Q(n), L)

so that the Hurewicz map on (Q(n)) is injective. Consider the quotient map

Q(n) = L' L''(L' L'')/L Sn Sn.

This induces an isomorphism on Hn, and hence an injection on . We have a diagram

 Q(n) Q(n) Sn Sn Sn Sn

where = , and , are self maps of Sn of degree dk. Let (Sn Sn) be the space obtained from Sn Sn by localising at . The map Q(n)(Sn Sn) extends to a map : C(Sn Sn) , since (Sn Sn) = 0 for n < i < 2n - 1. Further, the diagram

 C C Sn Sn Sn Sn

commutes upto homotopy, since there are no obstructions to extending the constant homotopy on Q(n). The map : (C) ((Sn Sn) ) is injective on the summand

((Sn) (L')) = (Sn)

by construction. Hence the map

: (C) H2n - 1(,) ((Sn Sn) )

is injective. The action of C(f')* is obtained by restricting the action of ( )* to the image of . We have an isomorphism (see [W] XI (1.6), (1.7))

(Sn Sn) (Sn×Sn) (Sn) (Sn),

and acts by (×)* on the first summand, and by * = * on the other two summands. Since , have degree dk, one easily computes from Scholium 5 that ( )* acts by d2k = dn on ((Sn Sn) ). Finally, we have an isomorphism H2n - 1(,) H2n - 2(C, H1(S1)), so that acts on H2n - 1(,) by dn. This completes the proof of (i).
We now easily complete the proof of the Theorem. Assume by induction that, for all d 0  (mod m'), the map Fd, n - 2 : Q'(n)Q'(n) extends to a map f' : Q'Q'. Let m = (m'N)2, where N annihilates (C)tors. If d 0  (mod m), then d = d1d2 where d1 = em'N and d2 = m'N for some integer e. We then have self maps f'1, f'2 extending Fd1, n - 2, Fd2, n - 2 respectively. Then C(f'1of'2) = C(f'1)oC(f'2) is an extension of Fd, n, and one readily computes from the above lemma that it acts by multiplication by dn on (C). Hence C(f'1of'2) extends to a self map of Q.

Next: Bibliography Up: Continuous Self Maps of Previous: 2 The odd dimensional
Kapil Hari Paranjape 2002-11-21