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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 $ \equiv$ 0  (mod m), the map Fd, 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 Fd, n - 2 : Q'(n)$ \to$Q'(n) extends to a map f' : Q'$ \to$Q'. Then the map C(f') : C$ \to$C, obtained by the construction of (1.1), restricts to Fd, n : Q(n + 2)$ \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($\displaystyle \widetilde{Q}$,$\displaystyle \widetilde{C}$) $\displaystyle \;\stackrel{\partial}{\rightarrow}\;$ H2n - 1($\displaystyle \widetilde{C}$,$\displaystyle \mbox{$\bf Z$}$)  
  $\displaystyle \alpha$ $\displaystyle \uparrow$       $\displaystyle \uparrow$ $\displaystyle \gamma$  
0$\displaystyle \to$ $\displaystyle \pi_{2n}^{}$($\displaystyle \widetilde{Q}$,$\displaystyle \widetilde{C}$) $\displaystyle \to$ $\displaystyle \pi_{2n-1}^{}$($\displaystyle \widetilde{C}$) $\displaystyle \to$$\displaystyle \pi_{2n-1}^{}$($\displaystyle \widetilde{Q}$)$\displaystyle \to$ 0

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

$\displaystyle \pi_{2n}^{}$(Q, C)$\displaystyle \to$$\displaystyle \pi_{2n-1}^{}$(C) $\displaystyle \cong$ $\displaystyle \pi_{2n-1}^{}$($\displaystyle \widetilde{C}$)$\displaystyle \;\stackrel{\gamma}{\rightarrow}\;$H2n - 1($\displaystyle \widetilde{C}$)

is an isomorphism, and

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

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

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

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

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

and (Fd, n)* acts by multiplication by dk on Hn(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 $\displaystyle \;\stackrel{\mu}{\rightarrow}\;$ Sn
g $\displaystyle \downarrow$   $\displaystyle \downarrow$ g
Q(n) $\displaystyle \;\stackrel{F_{d,n}}{\rightarrow}\;$ Q(n)

where $ \mu$ has degree dk. Next, an easy computation shows that $ \widetilde{L'}$ $ \cong$ Sn + 1 $ \subset$ $ \widetilde{Q^{(n)}}$ $ \subset$ $ \widetilde{Q}$ 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 ($\displaystyle \pi_{2n-1}^{}$(Q(n))$\displaystyle \to$$\displaystyle \pi_{2n-1}^{}$(C)) = h*($\displaystyle \pi_{2n-1}^{}$(Sn)) $\displaystyle \oplus$ $\displaystyle \pi_{2n-1}^{}$(L').

The action of C(f')* on the left is compatible with this decomposition, and induces $ \mu_{*}^{}$ on $ \pi_{2n-1}^{}$(Sn), and (Fd, 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  
C(f')* acts by multiplication by dn on $ \pi_{2n-1}^{}$(C) $ \otimes$ $ \mbox{$\bf Q$}$.
With respect to the direct sum decomposition

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

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

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

$\displaystyle \varphi_{1}^{}$ $\displaystyle \in$ Hom ($\displaystyle \pi_{2n}^{}$(Q, C),$\displaystyle \pi_{2n-1}^{}$(Sn)tors)  
$\displaystyle \varphi_{2}^{}$ $\displaystyle \in$ Hom ($\displaystyle \pi_{2n}^{}$(Q, C),$\displaystyle \pi_{2n-1}^{}$(Sn + 1))  
$\displaystyle \varphi_{3}^{}$ $\displaystyle \in$ Hom ($\displaystyle \mbox{$\bf Z$}$,$\displaystyle \pi_{2n-1}^{}$(Sn)tors)  

Proof:     From the Scholium 5, the action of $ \mu_{*}^{}$ on $ \pi_{2n-1}^{}$(Sn)tors is by dk, while it is by d2k = dn on $ \pi_{2n-1}^{}$(Sn) $ \otimes$ $ \mbox{$\bf Q$}$. The action of (Fd, n)* on

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

is by dk + 1. From the Scholium, this implies that (Fd, n)* acts by dk + 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),$\displaystyle \mbox{$\bf Z$}$) $\displaystyle \;\stackrel{\cong}{\rightarrow}\;$ Hn(Q(n), L;$\displaystyle \mbox{$\bf Z$}$)
$\displaystyle \uparrow$   $\displaystyle \uparrow$ $\displaystyle \wr$
$\displaystyle \pi_{n}^{}$(Q(n)) $\displaystyle \to$ $\displaystyle \pi_{n}^{}$(Q(n), L)

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

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

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

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

where $ \rho$ = $ \rho$$\scriptstyle \prime$ $ \vee$ $ \rho$$\scriptstyle \prime$$\scriptstyle \prime$, and $ \rho$$\scriptstyle \prime$,$ \rho$$\scriptstyle \prime$$\scriptstyle \prime$ are self maps of Sn of degree dk. 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 $\displaystyle \;\stackrel{C(f')}{\rightarrow}\;$ C
$\displaystyle \psi$ $\displaystyle \downarrow$   $\displaystyle \downarrow$ $\displaystyle \psi$
Sn $\displaystyle \vee$ Sn $\displaystyle \otimes$ $\displaystyle \mbox{$\bf Q$}$ $\displaystyle \;\stackrel{\rho}{\rightarrow}\;$ Sn $\displaystyle \vee$ Sn $\displaystyle \otimes$ $\displaystyle \mbox{$\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

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

by construction. Hence the map

$\displaystyle \nu$ : $\displaystyle \pi_{2n-1}^{}$(C) $\displaystyle \otimes$ $\displaystyle \mbox{$\bf Q$}$$\displaystyle \to$H2n - 1($\displaystyle \widetilde{C}$,$\displaystyle \mbox{$\bf Z$}$) $\displaystyle \oplus$ $\displaystyle \pi_{2n-1}^{}$((Sn $\displaystyle \vee$ Sn) $\displaystyle \otimes$ $\displaystyle \mbox{$\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))

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

and $ \rho_{*}^{}$ acts by ($ \rho$$\scriptstyle \prime$×$ \rho$$\scriptstyle \prime$$\scriptstyle \prime$)* on the first summand, and by $ \rho$$\scriptstyle \prime$* = $ \rho$$\scriptstyle \prime$$\scriptstyle \prime$* on the other two summands. Since $ \rho$$\scriptstyle \prime$, $ \rho$$\scriptstyle \prime$$\scriptstyle \prime$ have degree dk, one easily computes from Scholium 5 that ($ \rho$ $ \otimes$ $ \mbox{$\bf Q$}$)* acts by d2k = dn on $ \pi_{2n-1}^{}$((Sn $ \vee$ Sn) $ \otimes$ $ \mbox{$\bf Q$}$). Finally, we have an isomorphism H2n - 1($ \widetilde{C}$,$ \mbox{$\bf Z$}$) $ \cong$ H2n - 2(C, H1(S1)), so that $ \widetilde{C(f')}_{*}^{}$ acts on H2n - 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 Fd, 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 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 $ \pi_{2n-1}^{}$(C). Hence C(f'1of'2) extends to a self map of Q. $ \Box$

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