Next: 3 The even dimensional
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Previous: 1.2 Computation of Homotopy
We begin by recalling the statement of Theorem 1.
Theorem 1
Let
Q be a smooth quadric hypersurface in
^{n + 1}, where
n = 2
k + 1.
Then for any positive integer
d 0 (
mod 2
^{k}) there exist
continuous maps
f :
^{n}Q, 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 2^{k  1}), the induction hypothesis gives us
a map
f' : ^{n  2}Q' satisfying
f'^{*}((1)) = (d').
Since ^{n  1} is the projective cone over ^{n  2}, Construction 1.1
yields a map
C(f') : ^{n  1}C.
The obstruction to extending this map to a map ^{n}Q is a class
O(C(f')) H^{2n}(^{n},(Q)). By Lemma 4, the group
(Q) has exponent 4. Let be the composite
where is the restriction of the map
: ^{n}^{n} preserving
the cell structure, and given in suitable homogeneous coordinates by
(
z_{0} :
^{ ... } :
z_{n})
(
z_{0}^{2} :
^{ ... } :
z_{n}^{2}).
The obstruction to extending to a map ^{n}Q is
O(
) =
(
O(
C(
f')))
H^{2n}(
^{n},
(
Q)).
But clearly = 0 on this cohomology group.
Let
f : ^{n}Q be an extension of . We have a
commutative diagram of integral cohomology groups
H^{2}(Q) 

H^{2}(^{n}) 



H^{2}(Q') 

H^{2}(^{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 = 2
k + 1.
Then there exists a continuous map
f :
QQ of degree
d^{n} whenever
 (viii)

d 0 (mod 2^{k}), or
 (ix)

d = e^{2n  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
F_{d, n} : Q^{(n + 2)}Q^{(n + 2)} extends to a map f : QQ.
By induction, we may assume
that
F_{d, 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 = e^{2n  3} for some e > 0.
We have constructed (see (1.1)) a map
C(f') : CC which satisfies
C(f') = F_{d, n}. By computing C(f')_{*} on
(C) we will show that the obstruction to extending
the fourfold composite C(f')^{4} to a self map of Q vanishes.
There is a filtration F on
(C) given as follows.
Take
F_{0} = (C); F_{1} is the kernel of the composite
where is the Hurewicz map; and finally
F_{2} = im ((Q^{(n)})(C)).
Clearly C(f')_{*} is compatible with this filtration and induces a
map
gr ^{F}C(f')_{*} on
gr ^{F}(C). This map may be
computed as follows.
Lemma 6
 (x)

F_{2}, F_{1}/F_{2} are vector spaces over
/2.
 (xi)
 The natural composite map
(
Q,
C)
(
C)
F_{0}/
F_{1} H_{2n  1}(
,
)
is an isomorphism, giving a direct sum decomposition
 (xii)

gr ^{F}C(f')_{*} is multiplication by d^{n}.
Proof:
We have a commutative diagram with exact bottom row

H_{2n}(,) 

H_{2n  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
F_{1} (). The self map
of
is of degree d^{n} and this gives (iii) for F_{0}/F_{1}.
Since Q^{(n)} = L is a linear projective subspace of ^{n + 1},
is an S^{n} in ; it is easy to check that,
in the fibration (*), this maps
isomorphically to a great sphere
S S^{n + 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 S^{n + 1} and let U be its
inverse image in . We have the following
Sublemma 7
Let
i :
S^{n} be a fibre of (*) lying over a point of
D^{}.
A unit tangent vector field
v on
S S^{n}
gives a map
v :
S^{n} 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
S^{n + 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
F_{2} im (
(
)
(
)) =
im (
(
S^{n})
(
)),
where S^{n}Q is the inclusion of the fibre of (*). Hence
F_{2} is a vector space over
/2, and by Scholium 5
the action of
gr ^{F}C(f')_{*} on it is by
d^{k + 1} d^{n} (mod 2). Further, we obtain an isomorphism
F_{1}/
F_{2} (
)/
im (
(
))
_{2}(
(
S^{n + 1},
D^{})),
so that F_{1}/F_{2} is a
/2vector space.
Let
g : (D^{n + 1}, S^{n})(,) be the
generator of
(,) (Q^{(n + 1)}, L) .
We have a diagram, commutative
upto homotopy,
(D^{n + 1}, S^{n}) 

(,) 



(D^{n + 1}, S^{n}) 

(,) 
where is a map of degree d^{k + 1}. From the
Scholium 5
we see that
()_{*} induces multiplication by
d^{k + 1} on
(D^{n + 1}, S^{n}).
By the sublemma we have isomorphisms
(
,
)
(
,
)
(
S^{n + 1},
D^{}),
so that the composite
: (D^{n + 1}, S^{n})(S^{n + 1}, D^{}) of
g and the natural map
(,)(S^{n + 1}, D^{}) is also a generator
for
(S^{n + 1}, D^{}). By the Freudenthal suspension
theorem, is an isomorphism on
. From the
diagram

H_{2n}(,;) 

H_{2n  1}(,;) 






0 
(,) 

(,) 
(,) 0 
where , are isomorphisms,we see that
im (
(
D^{n + 1},
S^{n})
(
,
))
= ker(
(
,
)
H_{2n  1}(
,
;
)).
In particular, F_{1}/F_{2} is contained in this image; thus
C(f')_{*} acts by multiplication by
d^{k + 1} d^{n} (mod 2) on
F_{1}/F_{2},and this completes the proof of (iii).
From this lemma we see that we have
Hom ((Q, C), F_{1}) = F_{1} and
Hom (F_{1}/F_{2}, F_{2}) End (F_{1}) such that, for all pairs
(a, b) in
(C) = (Q, C) F_{1}
we have the equation
C(
f')
_{*}(
a,
b) = (
d^{n}a,
d^{n}b +
(
b) +
(
a)).
Since both
F_{1}, F_{1}/F_{2} are of exponent 2, it follows that the
fourfold composite of C(f') satisfies
C(f')^{4}_{*}(a, 0) = (d^{4n}a, 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
20021121