CC BY
2068
Probl. Anal. Issues Anal. Vol. 6(24), No. 1, 2017, pp. 68-81
DOI: 10.15393/j3.art.2017.3870
UDC 517.28, 517.54, 517.41
V. V. Starkov
STRUCTURE OF KELLER MAPPINGS, TWO-DIMENSIONAL CASE
Abstract. A Keller map is a polynomial mapping f : ^ (or Cn ^ Cn) with the Jacobian Jf = const = 0. The Jacobian conjecture was first formulated by O. N. Keller in 1939. In the modern form it supposes injectivity of a Keller map. Earlier, in the case n = 2, the author gave a complete description of Keller maps with deg f < 3. This paper is devoted to the description of Keller maps for which deg f < 4. Significant differences between these two cases are revealed, which, in particular, indicate the irregular structure of Keller maps even in the case of n = 2.
Key words: Jacobian conjecture, Keller maps
2010 Mathematical Subject Classification: 14R15
Let us denote the set of all polynomial mappings f = (f,..., fn) : Rn ^ Rn (or Cn ^ Cn) of degree m G N by Pn(m): fk is a polynomial of n variables of degree degfk < m for each k = 1,... ,n. As usual, Df is for the Jacobi matrix and Jf for the Jacobian. The Jacobian Conjecture (JC) formulated by Keller [3] in 1939 in its modern form is:
If f is a polynomial mapping and Jf = const = 0 then f is injective in Rn (Cn ).
The proof of the conjecture would allow to use it widely in a number of branches of mathematics. Beside the one given above, other equivalent formulations also exist. Many publications are devoted to this conjecture: see, e.g., [1], [2], [4], [6], [11]. In particular, in [10] the conjecture is proved for m = 2 for any n. In [5] it is checked for f G P2 (100). However, it has not been proved neither to be true nor to be false for the general case of any n. It is included in the list of "Mathematical Problems for the Next Century" [8].
©Petrozavodsk State University, 2017 IMiHiHI
A polynomial map f is called a Keller map if Jf = const = 0. It is obvious that it is enough to prove the JC for the Keller mappings with Jf = 1 and Df (0) = I (I is for the identity matrix, f (0) = 0). Therefore these properties are assumed in the sequel.
As the conjecture has neither been proved nor rejected for many years, it seems important to try to describe Keller mappings or their subclasses and then apply criteria or sufficient conditions of injectivity. Besides, Keller mappings for which the JC is true (see, e.g., [1]) are widely used in applications, which also make such study important.
In [9] Keller mappings were completely described for n = 2 and m = 3: Theorem A. [9] Let G e P(3), G(0) = 0, DG(0) = I. Then JG(x,y) = 1 if and only if G = A-1 o g o A, where g(x, y) = (U(x, y), V(x, y)),
U (x,y) = x + «2 (x + y)2 + «3(x + y)3,V (x,y) = y - «2 (x + y)2 - «3 (x + y)3,
a2 and a3 are arbitrary fixed constants, A is a linear homogeneous nonde-generate mapping.
Injectivity of such mappings G e P2 (3) follows from injectivity of Keller mappings from P2 (100) proved in [5]. Subclasses of Keller mappings where JC is true were described for any n and m in [7]. Theorem B. [7] Let g(X) = (g1,... , gn) e Pn(m), where X = (x1,..., xn), and for k = 1,..., n
gk (X) = xk + Yk [«2(xi + ... + xn )2 + «3(xi + ... + xn )3 + ... +
+«m(xi + ... + xn )m ],
n
aj, j = 1,... ,m, and Yk are arbitrary fixed constants with J2 Yk = 0.
k=i
Then g is a Keller map and g is injective.
Note that for n > 3 (but not for n = 2!) [7] gave wider compared to theorem B subclasses of Keller mappings. Also note that in the case n = 2, m = 3 the mappings g from Theorems A and B coincide. This gave birth to the conjecture (see [9]) that all Keller mappings in P2(m) have the form G = A-1 o g o A, where g(x, y) = (u(x, y), v(x, y)),
mm
u(x,y)= x ^X] ak(x + y)k, v(x,y)= y ak(x + y)k, (1) k=2 k=2
a2,..., am are arbitrary fixed constants, A is from Theorem A.
This conjecture is indirectly supported by Theorem C. With X = (x,y) e R2, consider f(X) = (u(X),v(X)), where u and v are as in (1) and
h(X) = (u(X) + w(X), v(X) + W (X)),
where W and w are homogeneous polynomials of degree (m +1) of x and y. If Jh(X) = 1, then h = A-1 o F o A, where A is a linear homogeneous nondegenerate mapping and
F(X) = (u(X) + am+1(x + y)m+1, v(X) - am+1(x + y)m+1),
for some constant am+1. The Jacobian conjecture is true for the mapping h.
This article is devoted to description of Keller mappings in P2 (4). The structure of Keller mappings in P2(m) turned out to change when m = 2, 3 changes to m = 4. P2 (4) contains new Keller mappings beside those of form (1). The following main result is proved:
Theorem 1. The class of Keller mappings f e P2 (4) with deg f = 4 contains not only mappings G of form (1) but also polynomial mappings
a2
^a,b,d(x,y) = (^1, ), where a = 0 = 6, d is any number, d = —,
= x + — (ax + by)2, 2a
$2 = y - ^[(ax + 6y)2 + x2 (26d - a2)] + 26
+x(ax+^ (26 - a)+Qb- (ax+6y)4.
Also it contains mappings Fc,s (x, y) = (F1, F2) with c = 0 = s,
O O O O A
F1 = x + cy , F2 = y - sx - 2csxy - c sy .
In particular, this theorem rejects the conjecture from [9] mentioned above about the structure of Keller mappings in P2 (m). The case turnes out to be much more complicated.
Definition 1. Let f, h e Pn(m),f (0) = h(0) = 0, Df (0) = Dg(0) = I. Mappings f and h are equivalent (denoted by f ~ h) if a linear non-degenerate mapping A exists, such that f = A-1 o h o A.
Theorem 2 below answers the question if equivalent mappings exist in classes {$a,b,d} and {Fc,s} and how many different (i.e., non-equivalent) mappings are there.
Theorem 2. Consider mappings {$abd} and {Fc s} defined in Theo-
a2
rem 1. Then for any a = 0 = b, c = 0 = s and any d = —
2b
Fc
$1,1,1 =
(x+y)2
V y - ^[(s + y)2 + ®2] - 1 x(x + y)2 - + y)4 )
$
a,b,d.
Join Theorems 1 and 2 to see that beside familiar Keller mappings of form (1) P2(4) contains at least one non-equivalent to then mapping $1 ,1 , 1.
Remark. Proofs of Theorems 1 and 2 are valid both in the real and in the complex case, as well as Theorems A and B.
Proof of Theorem 1. Let f = (f1,f2) E P2(4) be a Keller mapping, f (0) = 0, Df (0)= I, deg f = 4. Then
f1 = x + l(x,y) + l(2)(x,y)+ w(x,y),f2 = y + L(x,y) + L(2)(x,y) + W (x,y), where l and L are homogeneous polynomials of degree 2, l(2) and
L(2)
of degree 3, w and W of degree 4. Compare the highest degrees in the identity Jf = 1 to obtain Wxwy — wxWy = 0. This identity implies the equality
w = AW (2)
with some constant A.
Identities w = 0 and W = 0 can not be valid simultaneously, because deg f = 4, equality (2) holds for w = 0 and W = 0 with A = 0 (in the symmetrical case w = 0 and W = 0 the similar equality W = 5w holds). Therefore, we can assume that w = 0 and W = 0. Denote
4 4
w(x, y) = ^ akxky4-k = y4 ^ aktk = y4p(t), k=0 k=0
4
W(x, y) = £ 6kxky4-k = y4 £ 6ktk = y4q(t), t = x. (3) k=0 k=0 y
Then
Wx = y3p'(t), Wy = y3(4p(t) - tp'(t)),
Wx = y3q'(t), Wy = y3(4q(t) - tq'(t)).
Without loss of generality wx = 0 = Wx; otherwise pass to the equivalent mapping A-1 o f o A « f that satisfies these inequalities. Rewrite the equality Wxwy - wx Wy = 0 in the form
Wy _ wy 4q(t) - tq' (t) _ 4p(t) - tp' (t) q'(t) _ p' (t)
wx = wx = p'(t) w = p(ty.
Therefore, p(t) = Aq(t), A = const, i.e., w = AW.
Compare the fifth degree in equality Jf = 1 to obtain the equation
Wy(1(2) - AL(2))x + Wx(AL(2) - 1(2))y = 0.
It is obvious that this identity holds if 1(2) = AL(2). This is assumed further in this article leaving the other case /(2) = AL(2) outside. So,
ffY, _ fx + I + AL(2) + AW\ f(Xy + L + L(2) +
If A = 0, pass from the mapping f to the equivalent mapping
F = A o f o A-1 =( ~x +/2) — J \y + L + L(2) + Wy
with matrix A = ^q ( . Here I and L are polynomials of degree 2,
deg L(2) = 3, deg W = 4. Compare the fourth, third, second, and first degrees in the equality
Jf =
(Q+ lxl _ ly
(Lx + Lx2) + Wx) (1 + Ly + Ly2) + Wy)
1
to get, respectively, the equations
T Wy - ly Wx = 0, (4)
Wy + TXLy2) - lyL2 = 0, (5)
Lf + lxLy - TyLx = 0, (6)
Ly + lx = 0. (7)
Rewrite W(x,y) as
2
W = y4 l(t), l « y2 ^ Ck tk = y2 r(t)
k=0
similarly to (3). Note that l = 0; otherwise equality JF = 1 yields 0 = = Ly = Ly2 = Wy, i.e., L, L(2\ W depend only on x. Pass from F to the
equivalent mapping Fi = A-1 o F o A with matrix A = ^q and check
that F1 coincides with g from (1). So, the case l = 0 gives no new Keller
mappings beside those listed in (1).
Only one of the two possibilities holds: either lx = 0 = Wx or ly = 0 =
= Wy. Otherwise if lx = 0 then (4) and the condition l = 0 imply Wx = 0.
But then the symmetrical condition ly = 0 = Wy holds because W = 0.
So in the sequel lx = 0 = Wx can be assumed to hold. Rewrite (4) in the
Wy ly 4q 2r _ 2. ,
form ^^ = . Then — = — =>- q = ^r (t),u = const, follows in the
Wx lx q' r'
same way as in consideration of the equation Wxwy — wxWy = 0; then
W (x,y)= y4q(t) = rf2 (x,y), Wx = 2^1x1 Wy = 2^lyl. Now (5) can be rewritten as
q (2lJl — Lx2))+ IL2 =0. (8)
^ ~ a
Denote lx = ax + by; then l = -x2 + bxy + cy2 (here a, b, c are
2
constants). This together with (7) implies
L = —axy — \y2 — dx2, d = const. 2
From (6) derive Lyf2 = (lx)2 + lyLx = (ax + by)2 — (bx + 2cy)(ay + 2dx).
First consider the case b = 0, when
L(2) = ¿(ax + by)3 - ^y3 + 2bdx2y + (ab + 2cd) xy2
3b 3 \ 2 J
+ ux3,
32 u = const. Taking this last equality into account, rewrite (8) in the form
(6x + 2cy) 2^^x2 + 6xy + cy2j - a (ax + 6y)2 + 46dxy+
+ ( OT+2cd)y2—3ux2 +(ax+by)[(ax+by)2 — (bx+2cy)(ay+2dx)] =0
(bx + 2cy) 2ßf °x2 + bxy + cy2 j — 0 (ax + by)2 + 4bdxy+ 2b
(ab \ i
y + 2cdjy2 — (ax + by)(ay + 2dx) — 3ux2 + (ax + by)3 = 0. (9)
a
a
ab
y + 2cd i?/2 — (ax + by)(ay + 2dx) — 3ux2
Compare coefficients at equal powders in (9) to obtain the following equations:
x3 : ßab — 2abd — 3nu = 0, (10)
x2y : 2ßb2 + 2acß — 2a3C + 2b2 d — 4cad — 6cu = 0, (11)
3
xy2 : 6ßbe +-ab2 — 6a2c + 6bcd = 0, (12)
2
y3 : 4c2ß — 3abc + 4c2 d + b3 = 0. (13)
From (10) we determine
3u = a(ß — 2d). (14)
a3
Substitute (14) into (11) to get ß = c — d.
b3
Now (12) and (13) can be rewritten as the system of equations
ab4 + 4a3 c2 — 4a2 b2c = 0, 4a3 c3 + b6 — 3ab4 c = 0.
Substract the first equation from the second one multiplied on c to obtain
a(b4 — 4acb2 + 4a2 c2) = 0, b2(b4 — 4acb2 + 4a2 c2) = 0,
62
i.e., 62 = 2ac. So, a = 0 and c = —.
2a
Now we see that in the considered case the Jacobian JF = 1 if and only if for any d
b2 a3 c a2 /a2 \
a = 0 = b, c = —, ß = —3— d = — d, 3u = a( — — 3d). (15)
2a b 2b V 2b /
3
a3c
So
7 a 2 7 62 1/ 7\2
7 = -x + bxy+= 2a(ax+by),
6 \ 1
L = — dx2 + axy y J = — [(ax + by)2 + x2(2bd — a2)],
L(2) = ¿(ax + by)3 — ^y3 + 2bdx2y + (^ + 2cd)xy2 3b 3 2
+ ux3 =
1 / , x3 rb2 3 , 2 /ab db2\ 2 = — (ax + by)3 — y y3 + 2bdx2y + (y + — jxy2
1 , , x3 b2 3 ab 2 a3 3 / b2 2 2\
^ — (ax + by)--y--xy +—rx — dx 2bxy +--y + ax =
3b 3 2 6b V a /
+ ( üb— ad^x3 = b2
3 2 2 abxy2 dx 2 2 a d
= — x + a x y +------(ax + by) = (ax + by) x I —--I.
2b 2 a 2b a
a dN
1
d
^ = ^(l)2 = (86 - ) (ax + 6y)4.
Let us show that the polynomial mapping
F = ^a,b,d = 1 y + L + L(2) + W
is not equivalent to the mapping g from (1) for all a, 6, d (a = 0 = 6).
Assume that this is not true, so a mapping g and a matrix A from (17) exist, such that
g(AX )= A($a,b,d(X)). (16)
For the polynomial mapping f of deg f = m denote
f = £[f]k,
k-0
where [f ]k are homogeneous mappings of degree k. Let the matrix A = ( a l ) • ( XI ):= AX = ( + £ ) , detA = 0. (17)
zi := XI + yi. Then (16) implies A[$a,M(X)]4 = [g(AX)]4, i.e.,
0 \ ( 4 - ^22)(ax + by)4\
M (8b - 4^ + by)4
\8b 4a2/
H8b- 4a^)(ax+by)4 )
41
= «4Zi ( _i
a4 is the constant from (1). Then
5 = —¡3, ax + by = n(a + y)x, n = const;
this is possible only if b = 0 which contradicts the current assumption b = 0. So, the mappings §a,b,d with parameters defined in (15) are new polynomial mappings with JF = 1, not equivalent to the mappings G = = A-i o g o A from (1).
Now we need to consider the case b = 0. Then L(2) has the form
L(2) = a2x2y — {-OCy3 + 2cdxy2^ + ux3 and (8) can be written down as
2cy 2^x2 + cy2^j — 2a2xy + 2cdy2 — ax(ay + 2dx) — 3ux2 + (ax)3 = 0.
(18)
Equation a = 0 corresponds to equation (10). Keeping this in mind, rewrite (18):
2cy
2^,cy2 + 2cdy2 — 3ux2
= 0.
This equality is equivalent to one of the two conditions holding: 2c(^ + d) = 0,
i) c = 0 orii^ 3u = 0
i) Let c = 0. Then l = 0 and W = 0; this contradicts the assumptions of Theorem 1: deg f = 4. So, c = 0.
ii) In this case a = b = 0 = u, c = 0, ^ = —d; then I = cy2, L = —dx2, L(2) = —2cdxy2, W = —dc2y4 with d = 0, otherwise W = 0. Therefore,
F = Fc'd = ( X + dX2 — 2cdxy2 — dc2y4 ) , c = 0 = d'
Let us show that the mappings Fc,d and g from (1) are significantly different, i.e., g « Fc,d. To do this assume the contrary and consider the equality [A-1 o Fc,d o A]4 = [g]4 using notation (17):
( —dc2y4 ) = ( —4,4 ) (X + y)4;
This implies (7X + 5y) = n(x + y), n = const, i.e., 7 = 5 = n. Then
—dc2n4(x + y)0 = ( 7 7 ) ( —0 a4(X + y)4'
Finally dc2 = 0 and we have got a contradiction with the condition c = 0 = d. Therefore, g « Fc,d. Theorem 1 is proved. □
Proof of Theorem 2. Show that Fc,s « ^m for any c = 0 = s, i.e., a matrix A (see (17)) exists, such that A$1;1;1 (X) = Fc,s(AX).
A$i,i,i (X ) = AX+A
( 2(x + y)2 — 1
^ — 1[(x + y)2 + x2] — 1(x + y)2x — 8(x + y)4 ) Fc,s(AX ) = AX+
. c(Yx + ¿y)2
—s(ax + ßy)2 — 2cs(ax + ßy)(Yx + ¿y)2 — c2s(yx + ¿y)4
Compare the first coordinates in the equality [A$1,1,1(X)]4 = [Fc,s(AX)]4 to obtain ^ - q)(x + y)4 = 0, i.e., ^ = 0. Comparing the second coordinates yields
-1 5(x + y)4 = -c2 s(yx + 5y)4 y = 5 = 8c2 s54, 5 = 1
8 ^ " ' ' 2A3/C2
c2 s
Comparing the first coordinates of the identity [A$i,i,1(X)]2 = [Fc,s (AX)]2 obtain a = 2c52. The second coordinates give the relation
5 = 2sa2 = 2s • 4c2 54 ^ 1 = 8c2s53
we already have. The same equality appears if we consider the identity
[A$i,i,i(X)]3 = [Fc,s(AX)]3 : 5
— -(x + y)2x = — 2csa52x(x + y)2 5 = 4csa52 = 8c2s54. 2
So, for any c = 0 = s a matrix A from (17) exists and
$ = 0, y = 5 = 1_, a = 2c52 = 1
3c2 s
i.e., FCjs « $1,1,1.
To complete the proof of Theorem 2 it is enough now to show that
a2
$11 1 « $ab d for any a = 0 = b, d = — : we need to find a matrix B
2b
with det B = 0 such that $1 , 1 , 1(BX) = B$a , b, c(X). Let us show that the matrix
B =i4a^ 0 ^ ^ = /KIT
B V 25a(1 — 25) 2b5 J ' V 4a2 8
satisfies this condition. The condition of equivalence of the mappings implies 5 = 0. Evaluate
$1 ^ , 1(BX ) = BX +( ^(cl^by)2 — 8a2 54x2 1 +
+
0
—854ax(ax + by)2 — 254(ax + by)
B$a , b , d(X ) = BX + B
^ 2a(ax + by)2 1
+
y — — [(ax + by)2 + (2bd — a2)x2]
+B
x(ax+by)2( 2b— a) + (8b — 4^)(ax+by)4
1d
4
0
The condition [$1,1,1 (BX)]4 = [B$a,b,d(X)]4 is equivalent to the equality
i4 = 2Wf 1- — W 1
-254 = 2b5 ( —- - — ^ 53 = 0 .
V8b 4a V 4a2 8
Compare the first coordinates in the condition
[$1,1,l(BX )]2 = [B$0,M(X )]2 4a52
to get the obvious equality 252 = -; the second coordinates give an-
2a
other obvious equality
-252(ax + by)2 - 8a254x2 = ¿(1 - 25)(ax + by)2 - 5[(ax + by)2 +
+ (2bd - a2)x2] ^ 8a253 = (2bd - a2). To complete the proof, we need now only to check the identity
[$i,i,i(BX )]3 = [B$a,M(X )]3 for the second coordinates:
a d \ , , db 1
a/ ' " 4a2 8'
(a d\
—--)x(ax + by)2 ^^ 53
2b a /
a2
Therefore, for any a = 0 = b, d = —, c = 0 = s the relation
~ $1,1,1 ~ Fc,s holds. Theorem 2 is proved. □ The next statement declares existence of equivalent Keller mappings g of form (1): a subclass of P2 (4) of different (pairwise non-equivalent) mappings of form (1) is constructed.
Proposition. Any mapping g = ga2,a3,a4 £ P2(4) of form (1) with
a2 = 0 = defines a constant s = 422 such that ga2,a3,a4 ~ g1,1,s-
a2
Proof. Choose arbitrary constants a2 =0, a3 = 0, and a4 in the mapping g = ga2,a3,a4 £ P2(4). Let us prove that a constant s and a matrix A from (17) exist, such that
ga2 ,a3 ,«4
(AX) = Ag1,1,s (X),
i.e.,
ga2 ,«3,«4
(AX) = AX + («2 [(a + Y )x + (ß + ¿)y]2 +
+a3[(a + y)x + (% + 5)y]3 + a4[(a + y)x + (% + 5)y]4 ) ( ^
= Ag^(X) = AX + [(x + y)2 + (x + y)3 + s(x + y)4] ( a — % Compare coefficients in this equality to get the folowing equations: a — % = — (y — 5) ^^ a + y = % + 5,
a2 (a + y)2 = a — %, a3 (a + y)3 = (a — % ), a4 (a + y)4 = s(a — %).
Denote A = a + y = % + 5 and rewrite these equations in the form a2A2 = a — %' a3A3 = a — %, a4A4 = s(a — %). Then the values of A and s can be determined:
a4A4 a4 A4 a4 A2
a4a2
A = «2/«3 = 0, s =-- = —=-= —.
a — $ a2 A2 a2 a§
The elements of the matrix A are constructed as follows: choose any a and y = A — a, $ = a — a2A2, 5 = A — $ = A — a + a2A2. Check that det A = 0:
det A = a5 — $y = a(A — a + a2 A2) — (a — a2 A2)(A — a) = A3a2 = 0. The proposition is proved. □
Remark. Note that in the Proposition the constant s significantly depends on the chosen a2, a3, a4.
Acknowledgment. The work is supported by the Russian Science Foundation under grant 17-11-01229 and performed in Petrozavodsk State University.
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Received May 24, 2017. In revised form, June 8, 2017. Accepted June 8, 2017. Published online June 16, 2017.
Petrozavodsk State University
33, Lenina pr., Petrozavodsk 185910, Russia
E-mail: vstar@petrsu.ru
