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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 2k - 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'))
H2n(
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
(
z0 :
... :
zn)

(
z02 :
... :
zn2).
The obstruction to extending
to a map
n
Q is
O(

) =

(
O(
C(
f')))
H2n(
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
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 = 2
k + 1.
Then there exists a continuous map
f :
Q
Q 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 : Q
Q.
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') : C
C 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
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
- (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,
d
v(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)
d
v(
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 Sn
Q 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
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