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 $\equiv$ 0  (mod m), the map F_{d, n} : Q^{(n + 2)}$\to$Q^{(n + 2)} extends to a continuous map f : Q$\to$Q (of degree dⁿ).

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

	H2n($$\widetilde{Q}$$,$$\widetilde{C}$$)	$$\;\stackrel{\partial}{\rightarrow}\;$$	H2n - 1($$\widetilde{C}$$,$$\mbox{$\bf Z$}$$)
	$$\alpha$$ $$\uparrow$$		$$\uparrow$$ $$\gamma$$
0$$\to$$	$$\pi_{2n}^{}$$($$\widetilde{Q}$$,$$\widetilde{C}$$)	$$\to$$	$$\pi_{2n-1}^{}$$($$\widetilde{C}$$)	$$\to$$$$\pi_{2n-1}^{}$$($$\widetilde{Q}$$)$$\to$$ 0

where the vertical maps are Hurewicz maps. Since $\partial$,$\alpha$ are isomorphisms, the composite

$$\pi_{2n}^{}$$(Q, C)$$\to$$$$\pi_{2n-1}^{}$$(C) $$\cong$$ $$\pi_{2n-1}^{}$$($$\widetilde{C}$$)$$\;\stackrel{\gamma}{\rightarrow}\;$$H_{2n - 1}($$\widetilde{C}$$)

is an isomorphism, and

$$\pi_{2n-1}^{}$$(C) $$\cong$$ $$\pi_{2n}^{}$$(Q, C) $$\oplus$$ $$\pi_{2n-1}^{}$$(Q) $$\cong$$ $$\pi_{2n}^{}$$($$\widetilde{Q}$$,$$\widetilde{C}$$) $$\oplus$$ $$\pi_{2n-1}^{}$$($$\widetilde{Q}$$).

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

0$$\to$$$$\pi_{i}^{}$$(Sⁿ)$$\to$$$$\pi_{i}^{}$$($$\widetilde{Q}$$)$$\to$$$$\pi_{i}^{}$$(S^{n + 1})$$\to$$0,

where the splitting is obtained from a homotopy section of (*). Since Q⁽ⁿ⁾ = Q^{(n + 1)}, the map $\pi_{i}^{}$(Q⁽ⁿ⁾)$\to$$\pi_{i}^{}$(Q) is an isomorphism for i = n, and a surjection for i = n + 1. In particular, the homotopy section of (*), and the inclusion of the fibre Sⁿ $\subset$ $\widetilde{Q}$ of (*), factor through Q⁽ⁿ⁾. Hence $\pi_{2n-1}^{}$(Q⁽ⁿ⁾)$\to$$\pi_{2n-1}^{}$(Q) is surjective. Thus

$$\pi_{2n-1}^{}$$(C) $$\cong$$ $$\pi_{2n}^{}$$(Q, C) $$\oplus$$ im ($$\pi_{2n-1}^{}$$(Q⁽ⁿ⁾)).

We can refine this a little. Since $\pi_{n}^{}$(Q⁽ⁿ⁾) $\cong$ $\pi_{n}^{}$(Q) $\cong$ $\mbox{$\bf Z$}$, there is a map $\widetilde{g}$ : Sⁿ$\to$$\widetilde{Q^{(n)}}$, inducing a map g : Sⁿ$\to$Q⁽ⁿ⁾, such that $\widetilde{g}$, g represent generators of $\pi_{n}^{}$, and $\widetilde{g}$ is homotopic to the inclusion of the fibre of (*); further,

$$\pi_{n}^{}$$($$\widetilde{Q^{(n)}}$$) $$\cong$$ H_n($$\widetilde{Q^{(n)}}$$,$$\mbox{$\bf Z$}$$) $$\hookrightarrow$$ H_n(Q⁽ⁿ⁾,$$\mbox{$\bf Z$}$$),

and (F_{d, n})_* acts by multiplication by d^k on H_n(Q⁽ⁿ⁾,$\mbox{$\bf Z$}$). Then there is an inclusion h : $\pi_{2n-1}^{}$(Sⁿ) $\hookrightarrow$ $\pi_{2n-1}^{}$(C) induced by g and a homotopy commutative diagram

Sn	$$\;\stackrel{\mu}{\rightarrow}\;$$	Sn
g $$\downarrow$$		$$\downarrow$$ g
Q(n)	$$\;\stackrel{F_{d,n}}{\rightarrow}\;$$	Q(n)

where $\mu$ has degree d^k. Next, an easy computation shows that $\widetilde{L'}$ $\cong$ S^{n + 1} $\subset$ $\widetilde{Q^{(n)}}$ $\subset$ $\widetilde{Q}$ maps isomorphically onto S^{n + 1} in the fibration (*). Further F_{d, n} restricts to a self map of L' of degree d^k. Thus we have a decomposition

im ($$\pi_{2n-1}^{}$$(Q⁽ⁿ⁾)$$\to$$$$\pi_{2n-1}^{}$$(C)) = h_*($$\pi_{2n-1}^{}$$(Sⁿ)) $$\oplus$$ $$\pi_{2n-1}^{}$$(L').

The action of C(f')_* on the left is compatible with this decomposition, and induces $\mu_{*}^{}$ on $\pi_{2n-1}^{}$(Sⁿ), and (F_{d, n})_* on $\pi_{2n-1}^{}$(L'). Note that since n is even, $\pi_{2n-1}^{}$(Sⁿ) $\cong$ $\mbox{$\bf Z$}$ $\oplus$ $\pi_{2n-1}^{}$(Sⁿ)_tors (where the subscript ``tors'' denotes the torsion subgroup).

Lemma 8  

(xiii)

C(f')_* acts by multiplication by dⁿ on $\pi_{2n-1}^{}$(C) $\otimes$ $\mbox{$\bf Q$}$.

(xiv)

With respect to the direct sum decomposition

$$\pi_{2n-1}^{}$$(C) = $$\pi_{2n}^{}$$(Q, C) $$\oplus$$ ($$\mbox{$\bf Z$}$$ $$\oplus$$ $$\pi_{2n-1}^{}$$(Sⁿ)_tors) $$\oplus$$ $$\pi_{2n-1}^{}$$(L'),

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

$$\left(\vphantom{ \begin{array}{cccc} d^n & 0 & 0 & 0 \\ 0 & ... ... &\varphi_3 & d^k & 0 \\ \varphi_2 & 0 & 0 & d^{k+1} \end{array} }\right.$$$$\begin{array}{cccc} d^n & 0 & 0 & 0 \\ 0 & d^n & 0 & 0 \\ \varphi_1 &\varphi_3 & d^k & 0 \\ \varphi_2 & 0 & 0 & d^{k+1} \end{array}$$$$\left.\vphantom{ \begin{array}{cccc} d^n & 0 & 0 & 0 \\ 0 & ... ... &\varphi_3 & d^k & 0 \\ \varphi_2 & 0 & 0 & d^{k+1} \end{array} }\right)$$

where

$$\varphi_{1}^{} \in \pi_{2n}^{} \pi_{2n-1}^{}$$

$$\varphi_{2}^{} \in \pi_{2n}^{} \pi_{2n-1}^{}$$

$$\varphi_{3}^{} \in \mbox{$\bf Z$} \pi_{2n-1}^{}$$

Proof:     From the Scholium 5, the action of $\mu_{*}^{}$ on $\pi_{2n-1}^{}$(Sⁿ)_tors is by d^k, while it is by d^2k = dⁿ on $\pi_{2n-1}^{}$(Sⁿ) $\otimes$ $\mbox{$\bf Q$}$. The action of (F_{d, n})_* on

$$\pi_{n+1}^{}$$($$\widetilde{L'}$$) $$\cong$$ H_{n + 1}($$\widetilde{L'}$$,$$\mbox{$\bf Z$}$$) $$\cong$$ H_n(L', H₁(S¹))

is by d^{k + 1}. From the Scholium, this implies that (F_{d, n})_* acts by d^{k + 1} on $\pi_{2n-1}^{}$(L'). Hence (ii) follows, once we prove (i). Now Q⁽ⁿ⁾ = L' $\cup$ L'' where L' $\cap$ L'' = L $\cong$ $\mbox{$\bf P$}$^{k - 1}. Since $\pi_{n}^{}$(L) is finite, we see that $\pi_{n}^{}$(Q⁽ⁿ⁾)$\to$$\pi_{n}^{}$(Q⁽ⁿ⁾, L) is injective. We have a diagram, whose vertical arrows are Hurewicz maps,

Hn(Q(n),$$\mbox{$\bf Z$}$$)	$$\;\stackrel{\cong}{\rightarrow}\;$$	Hn(Q(n), L;$$\mbox{$\bf Z$}$$)
$$\uparrow$$		$$\uparrow$$ $$\wr$$
$$\pi_{n}^{}$$(Q(n))	$$\to$$	$$\pi_{n}^{}$$(Q(n), L)

so that the Hurewicz map on $\pi_{n}^{}$(Q⁽ⁿ⁾) is injective. Consider the quotient map

Q⁽ⁿ⁾ = L' $$\cup$$ L''$$\to$$(L' $$\cup$$ L'')/L $$\cong$$ Sⁿ $$\vee$$ Sⁿ.

This induces an isomorphism on H_n, and hence an injection on $\pi_{n}^{}$. We have a diagram

Q(n)	$$\;\stackrel{F_{d,n}}{\longrightarrow}\;$$	Q(n)
$$\downarrow$$		$$\downarrow$$
Sn $$\vee$$ Sn	$$\;\stackrel{\rho}{\longrightarrow}\;$$	Sn $$\vee$$ Sn

where $\rho$ = $\rho$^$\prime$ $\vee$ $\rho$^{$\prime$$\prime$}, and $\rho$^$\prime$,$\rho$^{$\prime$$\prime$} are self maps of Sⁿ of degree d^k. Let (Sⁿ $\vee$ Sⁿ) $\otimes$ $\mbox{$\bf Q$}$ be the space obtained from Sⁿ $\vee$ Sⁿ by localising at $\mbox{$\bf Q$}$. The map Q⁽ⁿ⁾$\to$(Sⁿ $\vee$ Sⁿ) $\otimes$ $\mbox{$\bf Q$}$ extends to a map $\psi$ : C$\to$(Sⁿ $\vee$ Sⁿ) $\otimes$ $\mbox{$\bf Q$}$, since $\pi_{i}^{}$(Sⁿ $\vee$ Sⁿ) $\otimes$ $\mbox{$\bf Q$}$ = 0 for n < i < 2n - 1. Further, the diagram

C	$$\;\stackrel{C(f')}{\rightarrow}\;$$	C
$$\psi$$ $$\downarrow$$		$$\downarrow$$ $$\psi$$
Sn $$\vee$$ Sn $$\otimes$$ $$\mbox{$\bf Q$}$$	$$\;\stackrel{\rho}{\rightarrow}\;$$	Sn $$\vee$$ Sn $$\otimes$$ $$\mbox{$\bf Q$}$$

commutes upto homotopy, since there are no obstructions to extending the constant homotopy on Q⁽ⁿ⁾. The map $\psi_{*}^{}$ : $\pi_{2n-1}^{}$(C) $\otimes$ $\mbox{$\bf Q$}$$\to$$\pi_{2n-1}^{}$((Sⁿ $\vee$ Sⁿ) $\otimes$ $\mbox{$\bf Q$}$) is injective on the summand

($$\pi_{2n-1}^{}$$(Sⁿ) $$\oplus$$ $$\pi_{2n-1}^{}$$(L')) $$\otimes$$ $$\mbox{$\bf Q$}$$ = $$\pi_{2n-1}^{}$$(Sⁿ) $$\otimes$$ $$\mbox{$\bf Q$}$$

by construction. Hence the map

$$\nu$$ : $$\pi_{2n-1}^{}$$(C) $$\otimes$$ $$\mbox{$\bf Q$}$$$$\to$$H_{2n - 1}($$\widetilde{C}$$,$$\mbox{$\bf Z$}$$) $$\oplus$$ $$\pi_{2n-1}^{}$$((Sⁿ $$\vee$$ Sⁿ) $$\otimes$$ $$\mbox{$\bf Q$}$$)

is injective. The action of C(f')_* $\otimes$ $\mbox{$\bf Q$}$ is obtained by restricting the action of $\widetilde{C(f')}_{*}^{}$ $\oplus$ ($\rho$ $\otimes$ $\mbox{$\bf Q$}$)_* to the image of $\nu$. We have an isomorphism (see [W] XI (1.6), (1.7))

$$\pi_{2n-1}^{}$$(Sⁿ $$\vee$$ Sⁿ) $$\cong$$ $$\pi_{2n}^{}$$(Sⁿ×Sⁿ) $$\oplus$$ $$\pi_{2n-1}^{}$$(Sⁿ) $$\oplus$$ $$\pi_{2n-1}^{}$$(Sⁿ),

and $\rho_{*}^{}$ acts by ($\rho$^$\prime$×$\rho$^{$\prime$$\prime$})_* on the first summand, and by $\rho$^$\prime$_* = $\rho$^{$\prime$$\prime$}_* on the other two summands. Since $\rho$^$\prime$, $\rho$^{$\prime$$\prime$} have degree d^k, one easily computes from Scholium 5 that ($\rho$ $\otimes$ $\mbox{$\bf Q$}$)_* acts by d^2k = dⁿ on $\pi_{2n-1}^{}$((Sⁿ $\vee$ Sⁿ) $\otimes$ $\mbox{$\bf Q$}$). Finally, we have an isomorphism H_{2n - 1}($\widetilde{C}$,$\mbox{$\bf Z$}$) $\cong$ H_{2n - 2}(C, H₁(S¹)), so that $\widetilde{C(f')}_{*}^{}$ acts on H_{2n - 1}($\widetilde{C}$,$\mbox{$\bf Z$}$) by dⁿ. This completes the proof of (i). $\Box$
We now easily complete the proof of the Theorem. Assume by induction that, for all d $\equiv$ 0  (mod m'), the map F_{d, n - 2} : Q'⁽ⁿ⁾$\to$Q'⁽ⁿ⁾ extends to a map f' : Q'$\to$Q'. Let m = (m'N)², where N annihilates $\pi_{2n-1}^{}$(C)_tors. If d $\equiv$ 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 F_{d1, n - 2}, F_{d2, n - 2} respectively. Then C(f'₁of'₂) = C(f'₁)oC(f'₂) is an extension of F_{d, n}, and one readily computes from the above lemma that it acts by multiplication by dⁿ on $\pi_{2n-1}^{}$(C). Hence C(f'₁of'₂) extends to a self map of Q. $\Box$