Polynomials Maps and Even Dimensional Spheres


                             Francisco-Javier TURIEL

                Geometr'ia y Topolog'ia, Facultad de Ciencias Ap. 59,

                               29080 M'alaga, Spain

                            email: turiel@agt.cie.uma.es



    Abstract. We construct, for every even dimensional sphere Sn, n   2, and ev*
 *ery odd

integer k, a homogeneous polynomial map f : Sn ! Sn of Brouwer degree k and alg*
 *ebraic

degree 2 | k | -1.

                           ____________-

    A polynomial map from X   Rm  to Y   Rr is the restriction to X of a polyno*
 *mial

map F : Rm ! Rr such that F (X)   Y . When each component of F is homogeneous of

degree k, we will say that the polynomial map from X   Rm to Y   Rr is homogene*
 *ous of

degree k. As usual Sn is the sphere on Rn+1 defined by the equation x21+ ... + *
 *x2n+1= 1, in

short || x ||2= 1, whereas Snr, 1   r   n, will be the differentiable manifold,*
 * diffeomorphic

to Sn , defined by the equation (x21+ ... + x2r)2 + x2r+1+ ... + x2n+1= 1. In t*
 *his work we

show:



    Theorem 1. Suppose n even and   2. Let k be an integer. Then:

(a) If k is odd, there exists a homogeneous polynomial map from Sn to Sn of Bro*
 *uwer

degree k and algebraic degree 2 | k | -1.

(b) If k is even there exists, for each 2   2r   n, a polynomial map from Sn2rt*
 *o Sn of

Brouwer degree k.



    Representing elements of ssn(Sn) by polynomial maps is an old question [1] *
 *which

was affirmatively solved by Wood, in 1968, provided that n is odd (theorem 1 of*
 * [3],

see[4] as well for the complex sphere). Nevertheless, as far as I know, this pr*
 *oblem is still


                                        1

open for n even; our theorem settles it when the Brouwer degree is odd.  In bot*
 *h cases

the polynomial maps constructed are homogeneous; therefore the problem of repre*
 *senting

elements of ssn(Sn) by homogeneous polynomial maps is solved now, since only ze*
 *ro and

the odd topological degrees may be represented in this way when n is even [2].

    The proof of theorem 1 of [3] makes use of a natural polynomial map of topo*
 *logical

degree 2 (lemmas 4 and 5). Nothing similar is known for n even; however the pol*
 *ynomial

map x 2 R3 ! (x21- x22, 2x1x2, x3) 2 R3 send S22into S2 with topological degree*
 * 2. Part

(b) of our theorem generalizes this map.

    For proving theorem 1 we start constructing a family of real polynomials in*
 * one vari-
               X`
able. Let '` =    ajtj be the Taylor expansion of ' = (1 - t)-1=2, at zero, up *
 *to order
               j=0

`; that is to say aj = (2j_-_1)(2j_-_3)_._.2.j1..j!Since the radius of converge*
 *nce of the power

       X1
series    ajtj is 1 and each aj > 0 , we have a0 = 1   '`   (1 - t)-1=2, t 2 [0*
 *, 1), whence
       j=0
(t - 1)'2`(t) + 1   0 and '`(t)   1 when t   0 (both inequalities are obvious i*
 *f t   1).
                                                    1X
    On the other hand if we set ' = t`+1R + '` then    tj = (1 - t)-1 = '2 = t`*
 *+1R"+ '2`
                                                    j=0
                                   X`
on (-1, 1); Therefore '2`= t`+1S +    tj where S is a polynomial in t. It follo*
 *ws, from
                                   j=0
that, the existence of a polynomial ~` of degree ` such that (t - 1)'2`(t) + 1 *
 *= t`+1~`.



    Lemma 1. For every ` one has (t - 1)'2 + 1 = t`+1~` where ~` is a polynomia*
 *l of

degree `. Moreover ~`(t)   0 and '(t) > 0 for each t 2 R if ` is even, and for *
 *any t   0 if

` is odd.



    Proof. It will suffice to show that ~`(t)   0 and '`(t) > 0 if `   2 is eve*
 *n and t < 0.

First we will prove, by induction on `, the existence of a ffi` > 0 such that '*
 *` is strictly

decreasing on (-1, -1 + ffi`).  Note that '` = a`t` + a`-1t`-1 + '`-2 = a((2` -*
 * 1)t` +

2`t`-1) + '`-2 where a > 0.

    By induction hypothesis or because '0 = 1, the polynomial '`-2 is decreasin*
 *g on


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(-1, -1 + ffi`-2), or on R if ` = 2.  But the derivative ((2` - 1)t` + 2`t`-1)0*
 * = ((2` -

1)`t`-1 + 2`(` - 1)t`-2) < 0 on (-1, -1], so '` is strictly decreasing on some *
 *interval

(-1, -1 + ffi`).

    We show now that '`(t) > (1-t)-1=2 > 0 if t < 0. As (1-t)-1=2 is strictly i*
 *ncreasing,
                                                                               *
 *X1
it is enough to prove our result on (-1, 0). On this interval lim`!1 {'`(t)} = *
 *   ajtj =
                                                                               *
 *j=0
                            X1
(1 - t)-1=2. But the series    ajtj is alternating and the sequence {aj | t |j}*
 *j2N, whose
                            j=0
limit is zero, strictly decreasing; then '`(t) > (1 - t)-1=2 > 0 for ` even.

    Finally, if '`(t) > (1 - t)-1=2 > 0 for any t < 0, a straightforward calcul*
 *ation shows

that (t - 1)'2`(t) + 1 < 0, whence ~`(t)   0 since t`+1 < 0.


    Recall that any polynomial ~ which do not takes negative values has even de*
 *gree and

can be write ~ = ~21+ ~22, where ~1 and ~2 are polynomials of degree   half of *
 *degree of

~. Therefore by setting k = ` + 1, ff = '`, ~` = ~, fi1 = ~1 and fi2 = ~2 one h*
 *as:



    Corollary 1. For any odd natural number k there exist three polynomials ff,*
 * fi1, fi2,

the first one of degree k - 1 and the other two with degree   k_-_1_2, such tha*
 *t ff(t) > 0 and

(1 - t)ff2(t) + tk(fi21(t) + fi22(t)) = 1 anywhere.



    Let us proof part (a) of theorem 1. Since topological degrees  1 may be rep*
 *resented

by linear maps, we can assume k   1. On C x Rn-1 = Rn+1, endowed with coordinat*
 *es

(z, y) = (z, y1, ..., yn-1) for which Sn = {(z, y); | z |2 +y21+ ... + y2n-1= 1*
 *}, we define


                   F (z, y) = ((fi1(| z |2) + ifi2(| z |2))zk, ff(| z |2)y)


where ff, fi1 and fi2 are as in corollary 1. Then F (Sn)   Sn.

    Set S1 = {(z, 0); | z |2= 1}   Sn. As ff(t) > 0 for each t 2 R, F -1(S1) = *
 *S1 and F

preserves the orientation transversely to S1. Hence the maps F|S1and F|Sn have *
 *the same

topological degree, that is to say k.

    By construction all the monomials of F have odd degree   2k-1. Multiplying *
 *everyone


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of them by a suitable power of | z |2 +y21+ ... + y2n-1the map F becomes homoge*
 *neous of

algebraic degree 2k - 1, whereas F|Sn do not change.

    For proving (b), first we set "~`(t) = ~`(t2) and "'`(t) = '`(t2). By lemma*
 * 1 we have

(t2 - 1)'"2`(t) + 1 = t2`+2"~`(t), "'`(t) > 0 and "~`(t)   0 for any t 2 R. Thi*
 *s allows us to

find out, for every natural number "k  1, three polynomials "ff, "fi1, "fi2such*
 * that "ff(t) > 0

and (1 - t2)"ff2(t) + t2"k(f"i21(t) + "fi22(t)) = 1 anywhere.

    Consider on Rn+1 = R2rx Rn-2r+1 coordinates (x, y) = (x1, ..., x2r, y1, ...*
 *, yn-2r+1).

Let f : R2r ! R2r be a homogeneous polynomial map of algebraic degree 2"k, send*
 *ing

S2r-1 into S2r-1 with topological degree k =  2"k, which always exists (see [3]*
 *) and

J : R2r ! R2r the isomorphism given by Jx = (-x2, x1, ..., -x2r, x2r-1), that i*
 *s to say the

canonical complex structure of R2r. One defines (if "k= 0 just consider a const*
 *ant map):


               F (z, y) = (f"i1(k x k2)f(x) + "fi2(k x k2)Jf(x), "ff(k x k2)y)


    Then F (Sn2r)   Sn and the same argument as in part (a), applied to S2r-1 =*
 * {(x, 0); k

x k2= 1}   Sn2r, shows that the topological degree of F : Sn2r! Sn equals k.



    References


    1.  Baum, P.F.:  Quadratics maps and stable homotopy groups of spheres, Ill*
 *inois

J.Math. 11 (1967), 586-595.

    2. Golasi'nski, M. and G'omez Ruiz. F.: Polynomial and regular maps into Gr*
 *assman-

nians, K-Theory 26 (2002), 51-68.

    3.  Wood, R.:  Polynomial maps from spheres to spheres, Invent.  Math.  5 (*
 *1968),

163-168.

    4.  Wood, R.:  Polynomial maps of affine quadrics, Bull.  London Math.  Soc*
 *.  25

(1993), 491-497.


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