CLASSIFICATION OF DEGREE 2 POLYNOMIAL AUTOMORPHISMS OF C 3

For the family of degree at most 2 polynomial self-maps ofC3 with nowhere vanishing Jacobian determinant, we give the following classification: for any such map f , it is affinely conjugate to one of the following maps: (i) An affine automorphism; (ii) An elementary polynomial autormorphism E(x, y, z) = (P (y, z) + ax,Q(z) + by, cz + d), where P and Q are polynomials with max{deg(P ),deg(Q)} = 2 and abc 6= 0. (iii)  H1(x, y, z) = (P (x, z) + ay,Q(z) + x, cz + d) H2(x, y, z) = (P (y, z) + ax,Q(y) + bz, y) H3(x, y, z) = (P (x, z) + ay,Q(x) + z, x) H4(x, y, z) = (P (x, y) + az,Q(y) + x, y) H5(x, y, z) = (P (x, y) + az,Q(x) + by, x) where P and Q are polynomials with max{deg(P ),deg(Q)} = 2 and abc 6= 0.


Introduction
In this note, we will give a classification theorem for the family of degree at most 2 polynomial self-maps of C 3 with nowhere vanishing Jacobian determinant.Note that any polynomial automorphism has a nowhere vanishing Jacobian determinant.Our Theorem 2.1 implies that any degree at most 2 polynomial self-map of C 3 with nowhere vanishing Jacobian determinant is a polynomial automorphism.
Let G be the group of polynomial automorphisms of C 2 .Let A be the group of affine automorphisms of C 2 and let E be the group of elementary polynomial automorphisms of C 2 such that each e ∈ E is of the form: e(x, y) = (ax + P (y), by + c) where P is a polynomial and a, b = 0. Note that E is the group of all polynomial automorphisms that carry each line of the form y = constant to a line of the form y = constant .Then Jung's Theorem [J] asserts that G is generated by A and E. Applying Jung's Theorem, Friedland and Milnor [FM] classified the polynomial automorphisms of C 2 : Any polynomial automorphism of C 2 is affinely conjugate to one of the following types of maps: (i) an affine automorphism; (ii) an elementary polynomial automorphism; (iii) A finite composition of generalized Hénon mappings.Each generalized Hénon mapping is of the form h(x, y) = (P (x) − ay, x) where p is a polynomial of x of degree at least 2 and a = 0.
It seems to be difficult to extend Jung's Theorem to C n for n ≥ 3.So we cannot follow Friedland and Milnor's proof to classify polynomial automorphisms in higher dimensions.But if we restrict to polynomials of degree at most 2 in C 2 , it is not necessary to apply Jung's Theorem for the classification, see [HO] for a proof.In this paper, we give the classification of degree at most 2 polynomial self-maps of C 3 with nowhere vanishing Jacobian determinant up to affine conjugation.The organization of this paper is as follows: in Section 2, we give the statement of our classification Theorem 2.1 and we also include some remarks on the dynamical differences between the various classes in our theorem.In Section 3, the proof of Theorem 2.1 is given and in Section 4, we briefly give some discussions of some basic dynamical properties of these maps.

The statement of main theorem and some remarks
Theorem 2.1.If f : C 3 → C 3 is a degree at most 2 polynomial selfmap with nowhere vanishing Jacobian determinant, then f is affinely conjugate to one of the following maps: (1) An affine automorphism; (2) An elementary polynomial automorphism where P is a polynomial of y, z of degree at most 2, Q is a polynomial of z of degree at most 2 and abc = 0. Note that it maps every hyperplane z = k to a hyperplane z = k and maps every line where P and Q are polynomials with max{deg(P ), deg(Q)} = 2 and abc = 0.
First of all the elementary maps and the class H 1 distinguish themselves from the other classes by the fact that the maps fix a hypersurface and the orbits of all points outside this hypersurface converge to it.Hence the dynamics reduces to two dimensions.In the case of H 1 , the maps reduces to a Hénon map on the fixed hypersurface z = α.In the case of the elementary maps, the orbits in the fixed hypersurface z = α converge to the fixed curve y = β on which the maps are automorphisms.In fact both the elementary maps and the maps of class H 1 are semidirect products over a mapping A(z) = cz + d, i.e. there is a function Hence we need only to find dynamical differences between the classes H 2 , H 3 , H 4 , H 5 .
First we can observe that it is natural to consider the maps H 2 , H 3 together as opposed to the maps H 4 , H 5 .There is a dynamical difference in the asymptotic dynamics.For the maps H 2 , H 3 , the orbits generically converge to one point at infinity.For example, for the map then this is the point [A: B: 0: 0] at infinity in projective coordinates.On the other hand, for the maps H 4 , H 5 the generic orbit converges to a complex line at infinity.
It remains to distinguish dynamically the maps H 2 and H 3 as well as to distinguish the maps H 4 and H 5 .
Comparing the maps H 2 and H 3 , we observe that the map H 2 is a Hénon map in the last two coordinates, (y, x) → (Q(y) + bz, y).In other words such a map F is a semi-direct product over a mapping h(y, z) ∈ Aut(C 2 ), i.e. there is an analytic function (y, z) → A y,z ∈ Aut(C) such that F (x, y, z) = (A y,z (x), h(y, z)).This sets H 2 apart from H 3 .
Comparing the maps H 4 and H 5 we consider again their behaviour at infinity.We see that there is a P 1 at infinity which is mapped to itself.For H 4 this map is a second degree polynomial, while for H 5 this map is rational of degree 2, i.e. has a more complicated dynamics.

The Proof of Theorem 2.1
Let G be the family of degree at most 2 polynomial self-map of C 3 with nowhere vanishing Jacobian determinant.For any f ∈ G, we can write f in the following form: Because the degree of f is at most 2, the Jacobian matrix of f is as follows: Since the determinant of the Jacobian matrix f (x, y, z) is a nonzero constant, all coefficients of the polynomial det(f ) must be zero except the constant.In particular, the coefficients of x 3 , y 3 and z 3 must be zero, i.e., ( 1) It is easy to see that both b 1 and a 2 are the coefficient of xy in f 1 (x, y, z), this implies that b 1 = a 2 .By looking at the coefficients of xy, yz, and xz in f = (f 1 , f 2 , f 3 ), we obtain the following table: By using the above table we can write our function f in the following form: ( Let's introduce the following trivial lemma which is useful in our proof of Theorem 2.1.Lemma 3.1.Let F = (f 1 , . . ., f n ) be a polynomial self-map of C n of degree at most 2, with nowhere vanishing Jacobian.Let g and h be any affine automorphisms of C n .We denote the degree 2 homogeneous part of F by (φ 1 , . . ., φ n ) and the degree 2 homogeneous part of g • F • h by (ψ 1 , . . ., ψ n ).Then the following statements are equivalent: Proof: Clear.Remark 3.2.For the map F in above lemma, if we want to prove that n j=1 α j φ j ≡ 0, we can simplify J F , the Jacobian matrix of F , by composing constant invertible matrices in both sides of J F .Note that e., we have to use the new variables for the Jacobian matrix of F , but this doesn't matter because J h is a constant matrix and therefore we may keep the original notation as the new variables.
Proof: If for all constants α, β and γ with |α| + |β| + |γ| = 0, αφ 1 + βφ 2 + γφ 3 ≡ 0, then by Lemma 3.1 and Lemma 3.4 we may assume that In this case, the Jacobian matrix of f is as follows: Since the Jacobian determinant is a nonzero constant, then det This is a contradiction.
Proof of the Theorem 2.1: If deg(f ) = 1, then it is easy to see that f is an affine automorphism.
If deg(f ) = 2, then by Lemma 3.5 we can assume that there exist constants k 1 and k 2 such that where φ j are degree 2 homogeneous polynomial and L j (x, y, z) are degree 1 polynomial and written as L j (x, y, z) := α j x + β j y + γ j z + ρ j .In particular, We will classify L 6 into the following 3 cases: where φ j are degree 2 homogeneous polynomials and L j (x, y, z) are degree 1 polynomials and written as L j (x, y, z) := α j x + β j y + γ j z + ρ j .
Then the Jacobian matrix of F 2 has the following form: Then since det(F 2 (x, y, z)) is a nonzero constant, Then for any fixed x, we may consider as a degree 2 polynomial self-map of C 2 with nowhere vanishing Jacobian determinant.
We can write and det J Fx (y, z) = a nonzero constant independent on x.
(1) Since b 7 y 2 + c 7 z 2 + d 7 yz and b 8 y 2 + c 8 z 2 + d 8 yz are proportional, we may compose an invertible constant matrix, say M 1 , to the left of the matrix of J Fx (y, z) to kill the y, z terms in the first row or second row.WLOG, we may assume that we killed the y, z terms in the first row.Now our Jacobian matrix becomes Note that e 7 , k 1 , f 7 , k 2 are different from the original values, for simplicity we still use the same notation.The following discussion will follow the same rule.
(3) If e 7 = 0, f 7 = 0, then we may compose an invertible constant matrix to the right of the matrix M 1 J Fx to kill f 7 .Now the matrix becomes Since the determinant of this matrix is a nonzero constant, we have c 8 = d 8 = f 8 = 0.If e 8 = 0, we will use the fact e 7 = 0 to kill e 8 .So now the matrix ix The determinant of this matrix is e 7 k 4 x − 2b 8 yk 2 + k 1 k 4 − k 2 k 3 which is a nonzero constant.This implies that e 7 k 4 = 0, b 8 k 2 = 0. e 7 = 0 implies that k 4 = 0. k 2 cannot be 0 otherwise the determinant is 0, so b 8 = 0. Now our matrix is pretty simple: It is easy to see that our claim is true for the map F x with above matrix as its Jacobian matrix.By Lemma 3.1 and Remark 3.2, we proved our claim.
(4) If one of {e 7 , f 7 } is 0 and the other is different from 0, then go to the case (3).
If α 8 = 0, then β 8 = 0.In this case, following a similar discussion as above, we may rewrite our map F 3 as the following form: (11) where p(x, z) is a degree at most 2 polynomial of x and z.
Therefore, in the case (i) the map F 2 is affinely conjugate to one of the maps H 2 , H 3 , H 4 , H 5 .
Case (ii): α 6 = 0 and β 6 = 0.In this case, we construct the following affine map g 5 : where φ j are degree 2 homogeneous polynomials and L j (x, y, z) are degree 1 polynomials and written as L j (x, y, z) := α j x + β j y + γ j z + ρ j .
Then following the same arguments in the proof of case (i), we can prove that F 5 is conjugate to one of the following maps: Remark 4.2.For the map H 1 (x, y, z) = (P (x, z)+ay, x+Q(z), cz+d), if P (x, z) doesn't depend on x, then H 2 1 is a special case of our elementary map E. So the discussion of its dynamics goes to the study of elementary maps.If c = 1, d = 0 and P (x, z) = kx 2 + • • • with k = 0, then this map is essentially an Hénon map of C 2 for any fixed z.Therefore in general we believe that H 1 has interesting dynamics Remark 4.3.For the map H 2 (x, y, z) = (P y, z) + ax, z + Q(y), y), if a = 1 and P (y, z) = 0, then the map H 2 is essentially an Hénon map of C 2 for any fixed x.So if Q is a degree 2 polynomial, we believe that the map H 2 has interesting dynamics.Otherwise we have some trivial examples like this: H 2 (x, y, z) = (y 2 − x 2 + x, y, z).It is easy to see that H 2 2 is an identity map.
Remark 4.4.For H 3 (x, y, z) = (P (x, z) + ay, z + Q(x), x), we have the following facts about its fixed points: (ii) If P (x, x) + aQ(x) = 0 and (ii.a)The degree of the polynomial P (x, x) + aQ(x) ≤ 1, then it is easy to see that the Fix(H 3 ) could be an empty set, one point or an entire curve; (ii.b)The degree of the polynomial P (x, x) + aQ(x) = 2, then it is easy to see that the H 3 has exactly 2 isolated fixed points counted with multiplicity.We believe that the H 1 in this case has rich dynamics.
Example 4.5.There are some interesting examples of H 3 in the case of P (x, x) + aQ(x) = 0 and a = 1 with H 3 3 is an identity map.For example, H 3 (x, y, z) = (z 2 + y, z − x 2 , x).Remark 4.6.Follow the same discussion as in Remark 3.6, we know that H 4 (x, y, z) = (P (x, y) + az, x + Q(y), y) has 2 isolated fixed points counted multiplicity if P (y − Q(y), y) + Q(y) + (a − 1)y is a degree 2 polynomial of y.We also believe that H 4 has rich dynamics in this case.
Remark 4.7.For the map H 5 (x, y, z) = (P (x, y) + az, by + Q(x), x), if b = 1 and Q(x) = 0, then the map H 5 is essentially an Hénon map of C 2 for any fixed y.If P and Q are degree 2 polynomials, we believe that the map H 5 has interesting dynamics and different from the dynamics of Hénon map in C 2 .
There are also some uninteresting examples like this: H 3 (x, y, z) = (2xz, 2y − x 2 , x).It is easy to check this map has only one periodic point which is the fixed point at the origin.
Remark 4.8.For every map of H j and E, the degree of its inverse polynomial could be 3 or 4 if the deg(P ) = deg(Q) = 2.But it must be 2 if either deg(P ) ≤ 1 or deg(Q) ≤ 1.
Remark 4.9.The detailed discussion of the dynamical properties of our maps H j and E will appear in our forthcoming papers.