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Conjecture 1 (Jacobian Conjecture). If F : k n → k n is a polynomial map such that det J (F ) ∈ k \ {0}, then F is a polynomial automorphism, that is, there exists a polynomial map G : k n → k n satisfying F (G) = X.

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VOL. 78 1998 NO. 1

ON THE FORMAL INVERSE OF POLYNOMIAL ENDOMORPHISMS

BY

PIOTR O S S O W S K I (TORU ´ N)

Let k be a field of characteristic 0. We begin by recalling some facts about the Jacobian Conjecture. We denote by J (F ) the Jacobian matrix of a polynomial map F .

Conjecture 1 (Jacobian Conjecture). If F : k n → k n is a polynomial map such that det J (F ) ∈ k \ {0}, then F is a polynomial automorphism, that is, there exists a polynomial map G : k n → k n satisfying F (G) = X.

Yagzhev [9] and Bass, Connell and Wright [1] showed that, if the Jacobian Conjecture is true for all n ≥ 2 and all polynomial maps of the form F = X − H with H homogeneous of degree 3, then it is true for all polynomial maps.

For the Jacobian matrix of a polynomial map F the hypothesis det J (F ) ∈ k \ {0} is equivalent to the nilpotence of J (H).

Let G = (G 1 , . . . , G n ) with G i ∈ k[[X 1 , . . . , X n ]] be the formal inverse of F = X − H, that is, F (G) = X. It is obvious that F is an automorphism if and only if G 1 , . . . , G n are polynomials.

Since in X − H all the non-zero homogeneous components have odd degree, G has the same property. Let G i = P

d≥0 G (d) i , where each G (d) i is homogeneous of degree 2d + 1 and i = 1, . . . , n. Several formulas for G (d) i are known. In those given by Bass, Connell and Wright [1] and Dru˙zkowski and Rusek [2], the components G (d) i are expressed as Q-linear combinations of polynomials indexed by rooted trees. Our aim is to prove that the polyno- mials, corresponding in the above mentioned expansions to the same rooted tree, differ by a rational factor depending on the structure of the rooted tree.

1. Rooted trees. If T is a non-directed tree, then V (T ) denotes the set of its vertices, and the set of its edges is a symmetric subset E(V ) ⊆ V (T ) × V (T ). A tree T with a distinguished vertex rt T ∈ V (T ), a root , is called a rooted tree. By induction we define the sets V j (T ) of vertices of 1991 Mathematics Subject Classification: Primary 13B25; Secondary 13B10, 14E09, 05C05.

Key words and phrases: polynomial automorphisms, Jacobian Conjecture, rooted trees.

[97]

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height j. Let V 0 (T ) = {rt T }. For j > 0 let v ∈ V j (T ) iff for some w ∈ V j−1 (T ) there exists an edge (w, v) ∈ E(T ) and v 6∈ V i (T ) for all i < j. Moreover, let ht(T ) = max{j : V j (T ) 6= ∅}.

If v ∈ V j (T ), then let

v + = {w ∈ V j+1 (T ) : (w, v) ∈ E(T )}.

By a leaf of a rooted tree T we mean a vertex v ∈ V (T ) such that v + = ∅.

The rooted trees form a category; a morphism T → T 0 is a map f : V (T ) → V (T 0 ) such that f (rt T ) = rt T

0

and (f × f )(E(T )) ⊆ E(T 0 ). If T is a rooted tree, then Aut(T ) denotes the group of all automorphisms of T and α(T ) = |Aut(T )|. (|X| is the cardinality of the set X.)

For a rooted tree T and a vertex v ∈ V (T ) we define a rooted tree T v to be a subtree of T such that rt T

v

= v and w ∈ V (T v ) if v belongs to a path from w to the root.

Let T be a rooted tree and

rt + T = {v 11 , . . . , v 1m

1

, . . . , v s1 , . . . , v sm

s

}.

Moreover, let {T v

11

, . . . , T v

1m1

}, . . . , {T v

s1

, . . . , T v

sms

} be the isomorphism classes of the rooted trees T v

ij

. It is easy to see that

α(T ) =

s

Y

j=1

(α(T v

j1

) m

j

· m j !) (1)

(cf. [6]).

In this note we assume that there exists an empty rooted tree ∅ with V (∅) = ∅ and E(∅) = ∅.

2. Bass–Connell–Wright formal inverse expansion. Let H = (H 1 , . . . , H n ), where H 1 , . . . , H n ∈ k[X 1 , . . . , X n ] are homogeneous of de- gree 3. Let n = {1, . . . , n}. For i ∈ n, a rooted tree T and a function f : V (T ) → n such that f (rt T ) = i, in [1] there are defined polynomials

P T ,f = Y

v∈V (T )

 Y

w∈v

+

D f (w)  H f (v) 

and

σ i (T ) = X

f :V (T )→n f (rt

T

)=i

P T ,f .

In [1, Ch. III, 5.(4)] it is shown that if T contains a vertex such that

|v + | > 3, then σ i (T ) = 0. We denote by T 0 d a fixed set of representatives

of the isomorphism classes of rooted trees with d vertices and with |v + | ≤ 3

for each v ∈ V (T ). Note that T 0 0 = {∅}. Using these observations, we can

quote the following theorem.

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Theorem 2 (Bass–Connell–Wright [1]). Let F = X − H : k n → k n , where H is homogeneous of degree 3 and the matrix J (H) is nilpotent. Then G (0) i = X i and

G (d) i = X

T ∈ T

0d

1

α(T ) σ i (T ) for d ≥ 1.

Theorem 2 suggests the following definition: σ i (∅) = X i for i ∈ n.

In the sequel we use the below description of the numbers α(T ).

Definition 3. For a rooted tree T and a vertex v ∈ V (T ) let α(v, T ) =

s

Y

j=1

m j !

where m 1 , . . . , m s are the cardinalities of the isomorphism classes of the rooted trees from {T w : w ∈ v + }. Note that α(v, T ) = 1 for each leaf v.

Remark. One can rewrite the formula (1) in the form α(T ) = α(rt T , T ) Y

v∈rt

+T

α(T v ).

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Lemma 4. If T is a rooted tree, then α(T ) = Y

v∈V (T )

α(v, T ).

P r o o f. Use (2) and induction with respect to the height of T .

3. Dru ˙zkowski–Rusek formal inverse. In [2] we can find another de- scription of the formal inverse. We suppose that F = X − H, where H is homogeneous of degree 3. It is well known that there exists a unique 3-linear symmetric polynomial map e H : k n × k n × k n → k n such that H(X, X, X) = H(X). e

Theorem 5 (Dru˙zkowski–Rusek [2]). If G = P

d≥0 G (d) is the formal inverse of F = X − H, then G (0) = X and

G (d) = X

p+q+r=d−1

H(G e (p) , G (q) , G (r) ) for d ≥ 1.

For small indices we have:

G (0) = X,

G (1) = e H(X, X, X),

G (2) = 3 e H(X, X, e H(X, X, X)), (3)

G (3) = 9 e H(X, X, e H(X, X, e H(X, X, X)))

+ 3 e H(X, e H(X, X, X), e H(X, X, X)).

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We shall see that each G (d) is a linear combination of polynomial maps cor- responding to rooted trees.

Definition 6. For any rooted tree T ∈ T 0 d with d ≥ 1 we define P(T ) to be a multiset (i.e., a set with repeated elements; see [7]) containing repre- sentatives of the isomorphism classes of the rooted trees T v for v ∈ rt + T and 3 − |rt + T | empty trees. Thus the multiset P(T ) has exactly 3 elements.

Example 7. (Always the lowest vertex is the root.) P( r) = {∅, ∅, ∅},

P  r

r r J J

r 

= { r, r r , ∅},

P( r r

) = {r, ∅, ∅},

P  r % %

r r e e

r r  = { r, r, r r }.

Definition 8. For a rooted tree T ∈ T 0 d we define, by induction on d ≥ 0, a polynomial homogeneous map τ (T ) : k n → k n of degree 2d + 1 as follows:

τ (∅) = X (the identity map),

τ (T ) = e H(τ (T 1 ), τ (T 2 ), τ (T 3 )) for d ≥ 1 and P(T ) = {T 1 , T 2 , T 3 }.

Now, let us describe the coefficients in linear combinations like (3).

Definition 9. For a rooted tree T and a vertex v ∈ V (T ) we define β(v, T ) to be the number of different sequences of elements of the multiset P(T v ).

Lemma 10. If d ≥ 0, then G (d) = X

T ∈ T

0d

β(T )τ (T ), where β(T ) = Y

v∈V (T )

β(v, T ).

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P r o o f. We prove this lemma by induction on d.

For d = 0 the equality (4) is obvious. Note that β(∅) = 1.

Let d > 0. Then G (d) = X

p+q+r=d−1

H(G e (p) , G (q) , G (r) )

= X

p+q+r=d−1

H e  X

T

1

∈ T

0p

β(T 1 )τ (T 1 ), X

T

2

∈ T

0q

β(T 2 )τ (T 2 ), X

T

3

∈ T

0r

β(T 3 )τ (T 3 ) 

= X

p+q+r=d−1

X

T

1

∈ T

0p

X

T

2

∈ T

0q

X

T

3

∈ T

0r

β(T 1 )β(T 2 )β(T 3 ) · e H(τ (T 1 ), τ (T 2 ), τ (T 3 )).

All maps of the form e H(τ (T 1 ), τ (T 2 ), τ (T 3 )) are homogeneous of degree 2p +

1 + 2q + 1 + 2r + 1 = 2(p + q + r) + 3 = 2d + 1. Collecting summands with

the same map τ (T ), for T ∈ T 0 d , we see that the coefficient of τ (T ) is equal

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to β(rt T , T ) · β(T 1 )β(T 2 )β(T 3 ) = β(T ), where P(T ) = {T 1 , T 2 , T 3 }.

4. Main theorem. We are going to compare the expressions for G (d) given in the previous subsections.

Definition 11. For a rooted tree T ∈ T 0 d and a vertex v ∈ V (T ) we define numbers %(v, T ) and %(T ) as

%(v, T ) = α(v, T )β(v, T ) and %(T ) = Y

v∈V (T )

%(v, T ).

In particular, %(∅) = 1.

Corollary. If T ∈ T 0 d , then %(T ) = α(T )β(T ).

Lemma 12. If T ∈ T 0 d and v ∈ V (T ), then

%(v, T ) = 3!

(3 − |v + |)! =

1 for |v + | = 0, 3 for |v + | = 1, 6 for |v + | ∈ {2, 3}.

P r o o f. It is sufficient to collect the numbers α(v, T ), β(v, T ) and %(v, T )

= α(v, T )β(v, T ) in a table. In the second column we assume that the rooted trees T 1 , T 2 , T 3 , ∅ are all distinct.

|v + | P (T

v

) α(v, T ) β(v, T ) %(v, T )

0 {∅, ∅, ∅} 1 1 1

1 {T 1 , ∅, ∅} 1 3 3

2 {T 1 , T 1 , ∅} 2 3 6

{T 1 , T 2 , ∅} 1 6 6

3 {T 1 , T 1 , T 1 } 6 1 6

{T 1 , T 1 , T 2 } 2 3 6

{T 1 , T 2 , T 3 } 1 6 6

Now, compare the first and last columns. The last column is obviously equal to 3!/(3 − |v + |)!.

Corollary. For T ∈ T 0 d we have

%(T ) = 2 |{v∈V (T ) : |v

+

|≥2}| · 3 |V (T )\Leaf(T )| , where Leaf(T ) is the set of all leaves of the rooted tree T .

In the proof of Theorem 13 we make use of the following polarization formula:

H e i (U, V, W ) = 1 3!

n

X

p,q,r=1

U p V q W r3 H i

∂X p ∂X q ∂X r

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(see [5, p. 251]). We also recall Euler’s formula: if F (X) is homogeneous of degree p, then

n

X

p=1

X p

∂F (X)

∂X p = p · F (X).

We are now in a position to formulate and prove the main theorem of our paper.

Theorem 13. If i ∈ n and T ∈ T 0 d for d ≥ 0, then σ i (T ) = %(T )τ i (T ),

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where τ (T ) = (τ 1 (T ), . . . , τ n (T )).

P r o o f. We argue by induction on the number of vertices of T . The case d = 0 is obvious:

σ i (∅) = X i = 1 · X i = %(∅)τ i (∅).

Suppose now that T ∈ T 0 d (d ≥ 1) and (6) is true for all rooted trees with less than d vertices. Let P(T ) = {T 1 , . . . , T s , ∅, . . . , ∅}, where 0 ≤ s ≤ 3 and T 1 , . . . , T s are non-empty. For i ∈ n, using a “tree surgery” (see [1], [8]), we can write

σ i (T ) = σ i

 r e % e % T 1 . . . T s 

=

n

X

j

1

,...,j

s

=1

σ j

1

(T 1 ) . . . σ j

s

(T s ) ∂ s H i

∂X j

1

. . . ∂X j

s

.

Let us apply Euler’s formula 3 − s times and let T j = ∅ for j = s + 1, . . . , 3:

σ i (T ) = 1 (3 − s)!

n

X

j

1

,j

2

,j

3

=1

σ j

1

(T 1 ) . . . σ j

s

(T s )X j

s+1

. . . X j

3

3 H i

∂X j

1

∂X j

2

∂X j

3

= 1

(3 − s)!

n

X

j

1

,j

2

,j

3

=1

σ j

1

(T 1 )σ j

2

(T 2 )σ j

3

(T 3 ) ∂ 3 H i

∂X j

1

∂X j

2

∂X j

3

.

Hence by assumption, σ i (T ) = %(T 1 )%(T 2 )%(T 3 )

(3 − s)!

n

X

j

1

,j

2

,j

3

=1

τ j

1

(T 1 )τ j

2

(T 2 )τ j

3

(T 3 ) ∂ 3 H i

∂X j

1

∂X j

2

∂X j

3

,

and by (5),

σ i (T ) = 3! · %(T 1 )%(T 2 )%(T 3 )

(3 − s)! H e i (τ (T 1 ), τ (T 2 ), τ (T 3 )).

Finally, we apply Lemma 12, Definition 8 and Definition 11 to get

σ i (T ) = %(rt T , T )%(T 1 )%(T 2 )%(T 3 )τ i (T ) = %(T )τ i (T ),

and the proof is complete.

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Corollary. If T ∈ T 0 d for d ≥ 0, then σ(T ) = %(T )τ (T ), where σ(T ) = (σ 1 (T ), . . . , σ n (T )).

5. Remarks. Theorem 13 and Lemma 10 give us an alternative proof of Theorem 2. Indeed,

G (d) = X

T ∈ T

0d

β(T )τ (T ) = X

T ∈ T

0d

1

α(T ) %(T )τ (T ) = X

T ∈ T

0d

1

α(T ) σ(T ).

This proof looks simpler than the original one in [1].

It is well known that a polynomial map F = X − H : k n → k n with H homogeneous of degree 3 and J (H) nilpotent has a polynomial inverse iff G (d) = 0 for deg G (d) = 2d + 1 > 3 n−1 . Bass, Connell and Wright [1]

conjectures that not only G (d) = 0 but also σ(T ) = 0 in the case T ∈ T 0 d

and 2d + 1 > 3 n−1 . A counterexample was given in [4]. On the other hand, Gorni and Zampieri [3] showed that there is a polynomial automorphism of the form X − H as above such that for any n there exists a rooted tree T with τ (T ) 6= 0 and with the number of vertices of T greater than n. In both papers, the counterexample is the same polynomial map:

F = (X 1 + X 4 (X 1 X 3 + X 2 X 4 ), X 2 − X 3 (X 1 X 3 + X 2 X 4 ), X 3 + X 4 3 , X 4 ), given by van den Essen for other reasons. In view of Theorem 13, τ (T ) 6= 0 iff σ(T ) 6= 0, and problems solved in [3] and [4] are equivalent. Moreover, we can exhibit rooted trees T for which τ (T ) 6= 0 (Gorni and Zampieri have not done it).

If

T 0 = r % % r e e r r

∈ T 0 4 , T s = J J r r

T s−1

∈ T 0 2s+4 for s ≥ 1, then (see [4]) σ(T s ) 6= 0 for all s ≥ 0 and therefore the polynomial maps

τ (T 0 ) = e H( e H(X, X, X), e H(X, X, X), e H(X, X, X)), τ (T s ) = e H(X, e H(X, X, X), τ (T s−1 )) for s ≥ 1 are non-zero.

REFERENCES

[1] H. B a s s, E. H. C o n n e l l and D. W r i g h t, The Jacobian conjecture: Reduction of degree and formal expansion of the inverse, Bull. Amer. Math. Soc. (N.S.) 7 (1982), 287–330.

[2] L. M. D r u ˙z k o w s k i and K. R u s e k, The formal inverse and the Jacobian con-

jecture, Ann. Polon. Math. 46 (1985), 85–90.

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[3] G. G o r n i and G. Z a m p i e r i, Yagzhev polynomial mappings: on the structure of the Taylor expansion of their local inverse, Ann. Polon. Math. 64 (1996), 285–290.

[4] P. O s s o w s k i, A counterexample to a conjecture of Bass, Connell and Wright , Colloq. Math. 77 (1998), 315–320.

[5] V. L. P o p o v and E. B. V i n b e r g, Invariant theory , in: Algebraic Geometry IV, Encyclopaedia Math. Sci. 55, Springer, 1994, 123–278.

[6] C. E. P r a e g e r and P. S c h u l t z, On the automorphisms of rooted trees with height distributions, in: Combinatorial Mathematics X (Adelaide, 1982), Lec- ture Notes in Math. 1036, Springer, 1983, 319–334.

[7] R. P. S t a n l e y, Enumerative Combinatorics, Vol. 1, Cambridge Stud. Adv.

Math. 49, Cambridge Univ. Press, 1997.

[8] D. W r i g h t, Formal inverse expansion and the Jacobian conjecture, J. Pure Appl. Algebra 48 (1987), 199–219.

[9] A. V. Y a g z h e v, On Keller’s problem, Sibirsk. Mat. Zh. 21 (1980), no. 5, 141–

150 (in Russian).

Faculty of Mathematics and Informatics Nicholas Copernicus University

Chopina 12/18 87-100 Toru´ n, Poland

E-mail: ossowski@mat.uni.torun.pl

Received 30 January 1998;

revised 16 March 1998

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