Homogeneous quasi-translations in dimension 5

We give a proof in modern language of the following result by Paul Gordan and Max Nöther: a homogeneous quasi-translation in dimension 5 without linear invariants would be linearly conjugate to another such quasi-translation x+H\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x + H$$\end{document}, for which H5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H_5$$\end{document} is algebraically independent over C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb C}$$\end{document} of H1,H2,H3,H4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H_1, H_2, H_3, H_4$$\end{document}. Just like Gordan and Nöther, we apply this result to classify all homogeneous polynomials h in 5 indeterminates, for which the Hessian determinant is zero. Others claim to have reproved ‘the result of Gordan and Nöther in P4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb P}^4$$\end{document}’ as well, but their proofs have gaps, which can be fixed by using the above result about homogeneous quasi-translations. Furthermore, some of the proofs assume that h is irreducible, which Gordan and Nöther did not. We derive some other properties which H would have. One of them is that degH≥15\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\deg H \ge 15$$\end{document}, for which we give a proof which is less computational than another proof of it by Dayan Liu. Furthermore, we show that the Zariski closure of the image of H would be an irreducible component of V(H), and prove that every other irreducible component of V(H) would be a 3-dimensional linear subspace of C5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb C}^5$$\end{document} which contains the fifth standard basis unit vector.


Introduction
Throughout this paper, we will write x for an n-tuple (x 1 , x 2 , . . . , x n ) of variables, where n is a positive integer. We write J F for the Jacobian matrix of a polynomial map F = (F 1 , F 2 , . . . , F m ) with respect to x, where m is another positive integer, i.e.
We write H f for the Hessian matrix of a polynomial f with respect to x, i.e.
In addition, we can write exp(D) and exp(t D) for the automorphisms corresponding to the maps x + H and x + t H respectively. But in order to make the article more readable for readers that are not familiar with derivations, we will omit the terminology of derivations further in this article. Gordan and Nöther (1876) studied (homogeneous) quasi-translations to obtain results about (homogeneous) polynomials h with det Hh = 0. One such a result is the classification of homogeneous polynomials in 5 indeterminates for which the Hessian determinant is zero. This classification has been reproved in Franchetta (1954) and Garbagnati and Repetto (2009), but only for the case where h is an irreducible polynomial. In Russo (2016, Ch. 7), the proof of Garbagnati and Repetto (2009) is extended to the case where h is a square-free polynomial. With an easy argument, which the reader may find, one can extend these results to the case where h is a power of such a polynomial. But then, you still do not have all polynomials h.
However, Russo (2016), told me that by way of de Bondt (2008, Th. 2.2) one can reduce the general case to the case where h is square-free. This is indeed true, because of the following.

Proposition 1.2 Let h ∈ C[x]
and leth be the square-free part of h.
Proof (i) This is a special case of de Bondt (2008, Th. 2.2).
(ii) Suppose thath ∈ A, and let f be an arbitrary factor of h over C [x]. It suffices to show that f ∈ A.
The connection between quasi-translations and polynomial Hessians with determinant zero, which comes from Gordan and Nöther (1876), is given at the beginning of Sect. 4. This connection is used in Garbagnati and Repetto (2009) and Russo (2016, Ch. 7) as well, and appears as Garbagnati and Repetto (2009, p. 33) and Russo (2016, Lem. 7.3.7) respectively. Garbagnati and Repetto (2009) and Russo (2016, Ch. 7) contain classifications in dimensions less than 5 as well, but with the same limitations as above on the factorization of h. These limitations are not present in Lossen (2004), which follows the approach of Gordan and Nöther (1876) in proving the classifications in dimensions less than 5.
In Watanabe (2014), it is claimed that rk J H = 3 if x + H is a quasi-translation in dimension n = 5, but this is not true. Hence the proof in Watanabe (2014) of the classification of homogeneous polynomials in 5 indeterminates, for which the Hessian determinant is zero, has a gap. The paper (Franchetta 1954) has an error and hence a gap on the same point. This gap can be fixed by proving that rk J H = 3 indeed, if x + H is associated to a polynomial for which the Hessian determinant is zero, which can be done by way of the results on linear invariants of quasi-translations, as given in Gordan and Nöther (1876) and this paper: see Remark 4.7 at the end of Sect. 4. Garbagnati and Repetto (2009) and Russo (2016, Ch. 7) on one hand, and de Bondt (2009, Th. 5.3.7) on the other hand, treat the case where rk J H = 3 incorrectly as well. But both incorrect treatments are only on subcases which do not overlap, so Russo (2016, Ch. 7) and de Bondt (2009, Th. 5.3.7) fix each other's errors. The error in Garbagnati and Repetto (2009) and Russo (2016, Ch. 7) can be repaired by way of Theorem 4.6, which comes from Gordan and Nöther (1876). The error in de Bondt (2009, Th. 5.3.7) can be repaired by way of Lemma 4.4, which gives a simpler argument than that in Gordan and Nöther (1876).
It is easy to show that for any homogeneous polynomial map H such that rk J H = 1, x + H has n − 1 independent linear invariants. Gordan and Nöther (1876) proved that any homogeneous quasi-translation x + H such that rk J H = 2 has at least 2 independent linear invariants. In their study of homogeneous quasitranslations x + H in dimension n = 5 with rk J H = 3, Gordan and Nöther (1876) distinguished two cases, namely "Fall a)" and "Fall b)", of which "Fall a)" had two subcases, which we indicate by (a1) and (a2).
The quasi-translations of subcase (a1) in Gordan and Nöther (1876) are the homogeneous quasi-translations x + H in dimension 5 with Jacobian rank three, for which the Zariski closure of the image of H is a 3-dimensional linear subspace of C 5 . The quasi-translations of case (b) in Gordan and Nöther (1876) are the homogeneous quasi-translations in dimension 5 with Jacobian rank three, which are linearly conjugate to another such quasi-translation x + H , for which H 5 is algebraically independent over C of H 1 , H 2 , H 3 , H 4 , but for which the Zariski closure of the image of H is not a 3-dimensional linear subspace of C 5 .
The quasi-translations of subcase (a2) in Gordan and Nöther (1876) are categorized by a somewhat technical property, which is the existence of p (1) and p (2) as in (iii) of Theorem 3.8. Let us just say for now that they are the homogeneous quasi-translations in dimension 5 with Jacobian rank three, which do not belong to case (b) or subcase (a1) in Gordan and Nöther (1876). As a consequence of Theorem 3.8, we deduce in Corollary 3.10 that quasi-translations of case (a2) in Gordan and Nöther (1876) have at least one linear invariant, by showing that the linear span of the image of H is 4-dimensional. Having reasoned about these three cases, one can wonder whether they actually exist.

Example 1.3
The following three H 's are chosen in such a way, that x + H with n = 5 is a quasi-translation which belongs to the above-described case (a1), (a2), and (b), respectively.
The quasi-translations for (a1) and (a2) were found by using techniques of de Bondt (2006, §2). The quasi-translations for (b) was found by applying Propositions 2.4 and 2.5, on the quasi-translation x + H with n = 4 and H = (1, x 4 , x 2 4 , 0). An unsolved question is whether a homogeneous quasi-translation in dimension 5 always has a linear invariant or not. We reprove the following results obtained in Gordan and Nöther (1876) in modern language: a homogeneous quasi-translation in dimension 5 without a linear invariant can only belong to case (b) in Gordan and Nöther (1876). Furthermore, we give a somewhat less computational proof of the result in Liu (2011) that a homogeneous quasi-translation in dimension 5 without a linear invariant must have degree 15 at least.
In dimension 6 and up, homogeneous quasi-translations do not need to have linear invariants, see (de Bondt 2006 Th. 2.1). If we substitute x 5 = 1 in the quasi-translations of cases (a2) and (b) in Example 1.3 and remove the last component, we get nonhomogeneous quasi-translations in dimension 4 without linear invariants.
The rest of the paper is organized as follows. In the next section, we show some basic concepts about quasi-translations.
In Sect. 3, we prove some geometric results about homogeneous quasi-translations x +H for which rk J H ≤ (n+1)/2. As a consequence, we deduce that a homogeneous quasi-translation in dimension 5 without linear invariants can only belong to case (b) in Gordan and Nöther (1876).
In Sect. 4, we apply the result that a homogeneous quasi-translation in dimension 5 without a linear invariant can only belong to case (b) in Gordan and Nöther (1876), to classify all homogeneous polynomials in 5 indeterminates for which the Hessian determinant vanishes.
In Sect. 5, we study homogeneous quasi-translations in dimension 5 that belong to case (b) in Gordan and Nöther (1876), with the purpose of getting properties of possible homogeneous quasi-translations in dimension 5 without linear invariants. One of these properties is that the degree of such a quasi-translation is at least 15.
In Sect. 6, we prove some geometric results about quasi-translations which gives us the following result about quasi-translations which belong to case (b) in Gordan and Nöther (1876): the Zariski closure of the image of H is an irreducible component of V (H ), which contains a linear 1-dimensional subspace L of C 5 , such that every other irreducible component of V (H ) is a 3-dimensional linear subspace of C 5 which contains L. Here, V (H ) is the set of common zeroes of H 1 , H 2 , . . . , H n .

Some basics about quasi-translations
In Proposition 2.2 below, we will show that quasi-translations are also characterized by H (x + t H) = H and by that J H · H is the zero vector. We need the following lemma to prove Proposition 2.2.

Lemma 2.1 Assume that x + H is a polynomial map and f
in case one of the following assumptions is satisfied. ( for all m ∈ N. This is only possible if (2.1) holds.
(2) By the chain rule and J H · H = (0 1 , 0 2 , . . . , 0 n ), we get where I n is the unit matrix of size n. Since J f · H = 0, it follows from the above that Suppose that t divides the right hand side of (2.2) exactly r < ∞ times. Then t divides f (x + t H) − f (x) more than r times. Hence t divides the left hand side of (2.2) more than r times as well, which is a contradiction. So both sides of (2.2) are zero. Since the right hand side of (2.2) is zero, we get (2.1).

Proposition 2.2
Let H : C n → C n be a polynomial map. Then the following properties are equivalent: Furthermore, if any of (1), (2) and (3) is satisfied, then The middle hand side of (2.3) gives the left hand side by substituting t = 1 and the right hand side by taking the coefficient of t 1 . Lemma 2.1 gives the converse implications by way of (1) and (3). Hence (2.3) follows as soon as we have the equivalence of (1), (2) and (3). By taking the Jacobian of (2), we get (J H )| x=x+t H · (I n + tJ H ) = J H , which gives (2.4) after substituting t = −t. Therefore, it remains to show that (1), (2) and (3) are equivalent.

Proposition 2.3 Assume x + g H is a quasi-translation over
is nonzero. Then x + H is a quasi-translation over C as well. Furthermore, the invariants of x + H and x + g H are the same. If additionally H is homogeneous of positive degree, then rk J g H = rk J H.
Proof By (1) ⇒ (2) of Proposition 2.2, we see that g(x +tg H )· H i (x +tg H ) = g · H i . We can substitute t = g −1 t in it, to obtain that on account of (2.3), and by substituting t = g we see that f is an invariant of x + g H. The converse follows in a similar manner by substituting t = g −1 .
Suppose that H is homogeneous of positive degree. From Proposition 1.2.9 of either van den Essen (2000) or de Bondt (2009), we deduce that in order to prove that rk J g H = rk J H , it suffices to show that trdeg C C(g H) = trdeg C C(H ). For that purpose, we show that for any R ∈ C[y], both R(g H) and R(H ) are zero if one of them is. Proposition 2.4 Assume x + H is a quasi-translation in dimension n over C, and F is an invertible polynomial map in dimension n over C with inverse G. Then is a quasi-translation as well, if and only if deg t G i (x +t H) ≤ 1 for all i. In particular, if T is an invertible matrix of size n over C, we have that is a quasi-translation as well.
Proof Assume first that deg t G i (x + t H) ≤ 1 for all i. Then we can write By substituting t = 1 on both sides, we obtain that G •(x + H )• F = x + G (1) (F) and substituting t = −1 tells us that its inverse G Since H (x + m H) = H on account of (1) ⇒ (2) of Proposition 2.2, we obtain Hence deg t G(x + t H) ≤ 1, as desired.
Proposition 2.5 gives a tool to obtain homogeneous quasi-translations over C from arbitrary quasi-translations x + H over C. Hence we can obtain results about arbitrary quasi-translations by studying homogeneous ones.

Proposition 2.5 Assume x + H is a quasi-translation over C in dimension n, and
is a homogeneous quasi-translation over C in dimension n + 1.

Proof
Denote Using that JH n+1 is the zero row, we see that it suffices to show that This is indeed the case, because the chain rule tells us that Proposition 2.6 below connects quasi-translations with homogeneity.

Proposition 2.6 Assume H is a homogeneous polynomial map over C. Then the assertions
(1) J H 2 is the zero matrix, n be the map which multiplies each term in any of the n components by its own degree. Then one can verify that In order to prove (2) ⇒ (3), assume that (2) holds. By looking at the coefficient of To show the second claim of (3), assume that rk J H > 1. Write H = gH , where g ∈ C[x], such that gcd{H 1 ,H 2 , . . . ,H n } = 1. Since rk J H > 1, we have degH ≥ 1. Furthermore, V (H ) cannot be written as a zero set of a single polynomial. Since C[x] is a unique factorization domain, we see that dim V (H ) ≤ n − 2.
Using Proposition 2.3, Proposition 1.2.9 of either van den Essen (2000) or de Bondt (2009), and the above obtainedH (H ) = 0 and dim V (H ) ≤ n − 2, in that order, we deduce that which gives the second claim of (3).

The image of the map H of quasi-translations x + H
We prove several results about quasi-translations with geometrical arguments. Some of these results have been claimed by Gordan and Nöther (1876). For the last two sections, we need several parts of Corollary 3.10 in this section.
Since the results may essentially be useful for non-homogeneous quasi-translations as well, it does not seem to be a good idea to work with projective varieties. But we will need the completeness of complex projective space in some manner. The lemma below gives us an affine version of that.
Lemma 3.1 LetZ ⊆ C m+kn be closed with respect to the Euclidian topology. Assume that for every point ofZ , the projection onto its last kn coordinates gives a point of C kn with complex norm √ k. LetX be the image of the projection ofZ onto its first m coordinates.
Suppose that there is an irreducible variety X ⊆ C m and a Zariksi open set U of X , such that U ⊆X ⊆ X . ThenX = X.
Proof Since the set of points in C kn whose complex norm is √ k form a compact space, the projection ofZ ontoX is closed with respect to the Euclidean topology. Hencẽ X is closed in the Euclidean topology. SoX contains the Euclidean closure of U in X . On account of O' Meara et al. (2011, Th. 7.5.1), the Euclidean closure of U in X is the same as the Zariksi closure of U in X , which is X . Hence X ⊆X . SoX = X indeed.
Notice that reverting to Euclidean topology is not only because the complex inner product cannot be expressed as a polynomial, but also because the Zariski topology of a product is not the corresponding product topology.
We also need a weak form of the projective fiber dimension theorem in some manner. Lemma 3.3 below is an affine version of that. But first, we need another lemma.
To prove the last claim, suppose that H has no constant part. Then 0 ∈ V (H ). From a weak version of the affine fiber dimension theorem (or from Lemma 3.3 below, applied on the map (H, Lemma 3.3 Suppose that H : C n → C n is a polynomial map and p ∈ C n , such that the linear span C p of p contains infinitely many points of the image of H . Then there exists an irreducible component X of H −1 (C p) such that H (X ) has infinitely many points, and the dimension of any such X is larger than n − rk J H.
contains infinitely many points and Y has finitely many irreducible components, there is an irreducible component Since H (X ) has infinitely many points of C p, it follows that C p is contained in the Zariski closure of H (X ).
Lemma 3.4 Assume x +H is a homogeneous quasi-translation over C. Suppose that p and q are independent and contained in the image of H . Then there exists an algebraic set X of dimension at least n − 2(rk J H − 1), such that H (c + t p) = H (c + tq) = 0 for all c ∈ X.
Proof On account of Lemma 3.3, there exist irreducible algebraic sets X p and X q of dimension at least n +1−rk J H , such that H (X p ) and H (X q ) contain infinitely many points of C p and Cq respectively. The set By a similar argument with q instead of p, we see that H (c + t p) = H (c) = H (c+tq) is dependent of both p and q for every c ∈ X p ∩ X q . Due to the homogeneity of H , 0 ∈ X p ∩ X q . Hence it follows from Hartshorne (1977, Ch. I, Prop. 7.1) that the dimension of X p ∩ X q is at least n − 2(rk J H − 1). So X = X p ∩ X q suffices. Proof V (H ) contains only finitely many (n − 2)-dimensional linear subspaces of C n because dim V (H ) ≤ n − 2. Furthermore, the Zariski closure of the image of H is irreducible on account of Lemma 3.2. From those two facts, we can deduce that it suffices to show that every nonzero p in the image of H is contained in an (n − 2)-dimensional linear subspace of C n which is contained in V (H ).
So take any nonzero p in the image of H . Take q independent of p such that q is the image of H as well. From Lemma 3.4, it follows that there exists an algebraic set X of Take c ∈ X • , such that c is independent of p and q if n = 5. Then the linear span of c, p and q has dimension at least max{2, n − 2}. Since H (c + t p) = 0, the linear span L of c and p is contained in V (H ). Since c ∈ L ⊆ V (H ) and c ∈ X • , it follows from the irreducibility of L that L ⊆ X .
In as similar manner, it follows that for everyc ∈ L ∩ X • , hence for allc ∈ L, the linear span ofc and q is contained in V (H ). So the linear span of L and q is contained in V (H ). This linear span has dimension at least max{2, n−2}. Since dim V (H ) ≤ n−2, it follows that n ≥ 4 and that p is contained in an (n − 2)-dimensional linear subspace of C n which is contained in V (H ).
Theorem 3.6 (Gordan and Nöther) Assume x+H is a homogeneous quasi-translation over C, such that deg H ≥ 1.
(i) If rk J H ≤ 1, then the image of H is a line through the origin and x + H has n − 1 independent linear invariants.
(iii) If rk J H = 2, then x + H has at least two independent linear invariants.
Proof For the moment, we prove (iii) only for the case where n ≤ 5, because we do not need the case where n ≥ 6 in this paper. To prove the general case of (iii), one can replace the use of Lemma 3.5 by that of the more general Corollary 6.5 in the last section.
Let W be the Zariski closure of the image of H . From Lemma 3.2, it follows that W is irreducible and that dim W = rk J H .
Since H is homogeneous and dim W = rk J H = 1, it follows from the irreducibility of W that the image of H can only be a line through the origin. Hence there are n − 1 independent linear forms l 1 , l 2 , . . . , l n−1 which vanish on the image of H . So l 1 , l 2 , . . . , l n−1 are invariants of x + H . (ii) Assume that gcd{H 1 , H 2 , . . . , H n } = 1. Since deg H ≥ 1, it follows from (i) that rk J H ≥ 2. From (2) ⇒ (3) of Proposition 2.6, it follows that rk J H ≤ n − 2, but its proof tells us that even rk , such that gcd{H 1 ,H 2 , . . . ,H n } = 1. Since rk J H = 2 > 1, we have degH ≥ 1.
On account of Proposition 2.3, rk JH = rk J H = 2. Furthermore, 2 ≤ dim V (H ) ≤ n − 2 on account of (ii), so n ≥ 4. From Lemma 3.5, it follows that the linear span of the image ofH is contained in V (H ). Since dim V (H ) ≤ n −2, the linear span of the image ofH has dimension at most n − 2 as well. Hence there are at least two independent linear forms l 1 and l 2 which vanish on the image ofH . Thus l i (H ) = 0 and l i (H ) = g · 0 = 0 for both i ≤ 2. So l 1 and l 2 are invariants of x + H .
Definition 3.7 Let H be a polynomial map. We define a GN-plane of H as a 2dimensional linear subspace of C n which is contained in V (H ).  Take c ∈ X nonzero. Since H is homogeneous, we deduce by substituting t = t −1 that H (tc + p) = H (tc + q) = 0. In the general case, consider the sets By applying proper substitutions in t, we see that the imageX of the projection ofZ onto its first 2n coordinates is equal to that of Z . SinceX contains an open subset of X := W × W , it follows from Lemma 3.1 thatX = X , which gives (i). (ii) Suppose that there exists a p ∈ W for which there are only finitely many GNplanes L p p. Let Y be the set of q ∈ W for which there are infinitely many GN-planes L q q. It is clear that (ii) holds if Y = {0}, so assume that there exist a q ∈ Y which is nonzero. Take P := {c ∈ V (H ) | H (c + t p) = 0} and Q := {c ∈ V (H ) | H (c + tq) = 0}. Since H is homogeneous, we see that both P and Q are unions of GN-planes. Furthermore, dim P = 2 and dim Q ≥ 3 because of the cardinality assumptions on the GN-planes in P and Q. Let L be a generic linear subspace of dimension n−2 of C n , so that dim(L ∩ P) = 0. Then L ∩ P = {0} ⊆ L ∩ Q and on account of Hartshorne (1977, Ch. I, Prop. 7 By applying proper substitutions in t, we see that the imageX of the projection ofZ onto its first n coordinates is equal to that of Z . Furthermore q is contained inX , but p is not. Since q ∈ Y \ {0} was arbitrary, we see that Y ⊆X . If Y would contain an open subset of W , then Lemma 3.1 tells us thatX = W , which contradicts that p is not contained inX . So Y does not contain an open subset of W , and W \ Y is not contained in a proper closed subset of W indeed. (iii) We can simplify (iii) by changing both the quantization set of q and the quantization order, to get the following.
(iii ) Suppose that p (1) , p (2) , . . . , p (k) ∈ W , such that p (i) is contained in only finitely many GN-planes of H for each i. Then for each q ∈ W which contains only finitely many GN-planes of H , there exist a GN-plane L q of H and GN-planes L p (1) p (1) , L p (2) p (2) , …, L p (k) p (k) of H , such that L q and L p (i) intersect nontrivially for each i.
The case where k = 1 of this simplification follows from (i). The case where k ≥ 2 of this simplification follows from the case where k = 1 of the unsimplified (iii) with p (1) = q, which may be assumed by induction on k. So it remains to deduce (iii) from its simplification. For that purpose, define Y as We can write Y as a union of algebraic sets of the form for each i, and rk q c (1) c (2) · · · c (k) ≤ 2 (3.1) where L p (i) p (i) is a GN-plane of H for each i. This union is finite by assumption. Let f be the projection of C n+2kn onto its first n coordinates. From the simplified version of (iii), it follows that the image of f | Y contains all q ∈ W which contains only finitely many GN-planes of H . On account of (ii), the image of f | Y is not contained in a proper algebraic subset of W . Hence there exists an irreducible component Z of Y such that the image of f | Z is not contained in a proper algebraic subset of W . From Mumford (1999, §1.8, Th. 3), it follows that the image of f | Z contains an open subset of W . Since Y is a finite union of algebraic subsets of the form (3.1) and Z is irreducible, we deduce that Z is contained in an algebraic subset of the form (3.1). Takẽ By applying proper substitutions in t and y 1 , y 2 , . . . , y k , we see that the imagẽ X of f |Z is the same as that of f | Z , soX contains an open subset of W . From Lemma 3.1, it follows thatX = W . SinceX is the image of the restriction of f on an algebraic subset of the form (3.1), the unsimplified (iii) follows. Definition 3.9 Let X be any subset of C n . We say that a ∈ C n is an apex of X if (1 − λ)c + λa ∈ X for all λ ∈ C and all c ∈ X .
We say that a p ∈ C n is a projective apex of X if p = 0 and c + λp ∈ X for all λ ∈ C and all c ∈ X .
If X is the Zariski closure of the image of a map H , then we say that a and p as above are an image apex of H and a projective image apex of H respectively.

apex projective apex
One may convince oneself that a projective apex is in fact an apex on the projective horizon.
If X is a zero set of homogeneous polynomials, e.g. because X is the Zariski closure of the image of a homogeneous map, then 0 is an apex of X . If 0 is an apex of X , then a projective image apex is the same as a nonzero apex. In that case, we will parenthesize the word projective.
Proof From Lemma 3.2, it follows that W is irreducible and that rk J H = dim W .
(1) ⇒ (2) Assume that dim V (H ) = rk J H ≤ 3 and that (2) does not hold. Then there exists a nonzero p ∈ W which contains infinitely many GN-planes of H that are contained in W . Suppose that W is the zero set of g 1 , g 2 , . . . , g m and let Then Y has an irreducible component Z which contains infinitely many GN-planes of H . Hence dim Z ≥ 3. Since Z ⊆ Y ⊆ W and dim W = rk J H ≤ 3, it follows from the irreducibility of Z and W that Z = W . So p is a nonzero (projective) apex of W and (1) does not hold. So if p is contained in infinitely many GN-planes of H , then (2) cannot hold. Hence assume that p is contained in only finitely many GN-planes of H . Since (3) does not hold, there exists a nonzero c ∈ V (H ) which shares a GN-plane of H with every q ∈ W . Inductively, we can choose p (i) in the interior of W outside L p (1) , L p (2) , . . . , L p (i−1) for i = 1, 2, 3, . . ., such that c ∈ L p (i) for each i. As we have seen above, L p (i) ⊆ W for each i, so c is a counterexample to the claim of (2).
(3) ⇒ (4) Assume that (3) is satisfied. From (ii) of Theorem 3.8, it follows that there exist a p (1) ∈ W and a p (2) ∈ W as in (iii) of Theorem 3.8. Take L p (1) and L p (2) as in (iii) of Theorem 3.8. Since there is no nonzero c ∈ V (H ) which shares a GN-plane of H with every q ∈ W , W cannot be equal to any linear span. Hence it suffices to show that W is contained in the linear span of L p (1) and L p (2) . In the case where L p (1) ∩ L p (2) = {0}, this follows directly from (iii) of Theorem 3.8, so assume that there exist a nonzero c ∈ L p (1) ∩ L p (2) . Let From (3), it follows that Y is a proper algebraic subset of W . Since W irreducible and contained in the union of Y and the linear span of L p (1) and L p (2) , W is contained in the linear span of L p (1) and L p (2) . (4) ⇒ (5) Assume that (4) is satisfied. If rk J H ≤ 1, then W is a line through the origin on account of (i) of Theorem 3.6, which gives (5). So assume that rk J H ≥ 2. Then W is properly contained in a 4-dimensional linear subspace of C n and hence rk J H = dim W < 4. The case where k = 2 of (iii) of Theorem 3.8 is obtained on Gordan and Nöther (1876, p. 566), and is used on the same page to prove the case where n = 5 and rk J H = 3 of Corollary 3.10. Gordan and Nöther (1876) classified all homogeneous polynomials with singular Hessians in dimension 5 as follows.

Theorem 4.1 (Gordan and Nöther) Assume h ∈ C[x] is a homogeneous polynomial in dimension n = 5. If det Hh = 0 and h is not a polynomial in n − 1 = 4 linear forms in C[x], then there exists an invertible matrix T over C such that h(T x) is of the form
where f and a 1 , a 2 , a 3 are polynomials over C in their arguments.
The proof that is given below uses results about homogeneous quasi-translations in dimension five and follows the approach of Gordan and Nöther more or less.
The following connection exists between singular Hessians and quasi-translations.
be the map which multiplies each term by its own degree in y. Then one can verify that E y R = y t ∇ R, and that E y R is a linear combination of the homogeneous parts R * of R. So J h · H = (y t ∇ R) y=∇h = (E y R) y=∇h = 0. Hence h(x + t H) = h on account of (2.3) in Proposition 2.2.
In order to prove Theorem 4.1, we need the classification of all homogeneous polynomials with singular Hessians in dimensions less than 5, which is as in Theorem 4.3 below. Our proof of Theorem 4.3 is somewhat different from that by Gordan and Nöther. A proof of Theorem 4.3 which is based on that by Gordan and Nöther can be found in Lossen (2004).

Theorem 4.3 (Gordan and Nöther) Assume h ∈ C[x]
is a homogeneous polynomial in dimension n ≤ 4. If det Hh = 0, then the components of ∇h are linearly dependent over C.
Proof Suppose that the components of ∇h are linearly independent over C. Then deg ∇h ≥ 1 because det Hh = 0. Let H = (∇ R)(∇h) as in Proposition 4.2, such that R has minimum degree. Then H is a nonzero quasi-translation and deg H ≥ 1 because deg R ≥ 2 and deg ∇h ≥ 1. Furthermore, H is homogeneous because R and ∇h are homogeneous. From (2) ⇒ (3) of Proposition 2.6, it follows that r := rk J H ≤ max{n − 2, 1} ≤ 2. Using (i) and (iii) of Theorem 3.6, we can deduce that x + H has n − r < n linear invariants.
Since n − r < n, there exists a nonzero p ∈ C n which is a zero of all these n − r linear invariants. From Proposition 1.2.9 of either van den Essen (2000) or de Bondt (2009), it follows that trdeg C (H ) = r . Hence the n − r linear invariants of . Consequently, p is a projective image apex of H . From Lemma 4.4 below, it follows that J h · p = 0, so the components of ∇h are linearly dependent over C indeed.

Lemma 4.4 Let h ∈ C[x]
and R ∈ C[y], such that R * (∇h) = 0 for every homogeneous part R * of R. Then Proof From Proposition 4.2, it follows that h(x + t H) = h. By taking the Jacobian on both sides, we obtain Suppose that p is a projective image apex of H . Take T ∈ GL n (C) such that the last column of T equals p. Then e n is a projective image apex ofH := T −1 H . SoH n is algebraically independent ofH 1 ,H 2 , . . . ,H n−1 . Hence trdeg C C(H ) = trdeg C C (H 1 ,H 1 , . . . ,H n−1 ) + 1. From proposition 1.2.9 of either van den Essen (2000) or de Bondt (2009), it follows that the last row of JH is independent of the rows above it. But Theorem 4.1 is formulated as de Bondt and van de Essen (2004, Th. 3.6). The starting point of the proof of de Bondt and van de Essen (2004, Th. 3.6) is de Bondt and van de Essen (2004 Th. 2.1 iii)), which is not accompanied by a proof and comes down the following.

Theorem 4.5 (Gordan and Nöther) Assume h ∈ C[x]
is a homogeneous polynomial in dimension n = 5. Suppose that R(∇h) = 0, such that R has minimum degree. Then R can be expressed as a polynomial in three linear forms over y.
Proof Notice that R is homogeneous because h is homogeneous and R has minimum degree. We distinguish two cases.

• R cannot be expressed as a polynomial in four linear forms over y.
Then the components of ∇ R are linearly independent over C. Since R has minimum degree, the components of H := (∇ R)(∇h) are linearly independent over C as well. Write H = gH , where g ∈ C[x], such that gcd{H 1 ,H 2 , . . . ,H n } = 1. Since the components of H and hence alsoH are linearly independent over C, we have degH ≥ 1. On account of Proposition 2.3, rk JH = rk J H . Since the components ofH are linearly independent over C, it follows from Theorem 3.6 that 3 ≤ rk JH ≤ dim V Since H is homogeneous, we can substitute t = g −1 t to deduce that p is a projective image apex of H as well. From Lemma 4.4, it subsequently follows that J h · p = 0. Hence the components of ∇h are linearly dependent over C. Since R has minimum degree, we conclude that deg R = 1, so R is a linear form in C[y]. Contradiction. • R can be expressed as a polynomial in four linear forms over y.
Then there is an i ≤ 5 such that y i is not a linear combination of these four linear forms. Say that i = 5. Then R is of the formR(y 1 + c 1 y 5 , y 2 + c 2 y 5 , y 3 + c 3 y 5 , y 4 + c 4 y 5 ), where c i ∈ C for each i. Furthermore,R ∈ C[y 1 , y 2 , y 3 , y 4 ] is homogeneous andR(∇h) =R(∇ĥ) = 0, wherẽ Sinceh is homogeneous, say of degree d, it follows thath = x d 5ĥ (x −1 5 x) and that ∇h and x d−1 5 (∇ĥ)(x −1 5 x) agree on the first 4 components. From this, we can deduce thatR, as a homogeneous polynomial in C[y 1 , y 2 , y 3 , y 4 ] such that R(∇ĥ) = 0, has minimum degree as well. From Theorem 4.6 below, we obtain thatR can be expressed as a polynomial in three linear forms in C[y 1 , y 2 , y 3 , y 4 ].
Hence R can be expressed as a polynomial in three linear forms in C[y].

Theorem 4.6 Let n = 4 and h ∈ C[x], not necessarily homogeneous. Suppose that R ∈ C[y] is homogeneous, such that R(∇h) = 0. If R has minimum degree, then R can be expressed as a polynomial in three linear forms in C[y].
Proof Suppose that R has minimum degree. Leth be the leading homogeneous part of h, and define H := (∇ R)(∇h). From Proposition 4.2, it follows that h(x + t H) = h. By taking the leading coefficient with respect to t, we deduce thath(H ) = 0.
Sinceh is homogeneous and R(∇h) = 0, it follows from Theorem 4.3 that the components of ∇h are linearly dependent over C, say that L(∇h) = 0 for some linear form L ∈ C[y]. Assume first that rk Hh = 3. Then the relations between the components of ∇h form a prime ideal of height one, which is a principal ideal because C[y] is a unique factorization domain. Since L is irreducible, (L) must be that principal ideal, and L | R because R(∇h) = 0. Since R has minimum degree, R is irreducible, so R is linear.
Assume next that rk Hh ≤ 2. Since there exists a linear relation between the components of ∇h, there exists a T ∈ GL n (C) such that the last component of T t ∇h is zero. Hence the last component of Since H(h(T x)) = T t (Hh)| x=T x T , we see that rk H(h(T x)) ≤ 2. It follows from Theorem 4.3 again thath(T x) can be expressed as a polynomial in two linear forms. Henceh =h(T (T −1 x)) can be expressed as a polynomial in two linear forms as well.
Sinceh is homogeneous in addition,h decomposes into linear factors, and one of these factors is already a relation between H 1 , H 2 , H 3 , H 4  Remark 4.7 The proof of the first case in the proof of Theorem 4.5 is different from that given in Gordan and Nöther (1876, p. 568), where the second claim of Lemma 4.4 is obtained by way of differentiation on the inverse of H . Since the inverse of H is not a map, the above proof of this first case seems much easier. The proof of this first case as given in de Bondt (2009, Th. 5.3.7) is incorrect.
The proof of the second case in the proof of Theorem 4.5 comes from Gordan and Nöther (1876, p. 567). This seems a little odd, because Lemma 4.6 is about not necessarily homogeneous polynomials, which Gordan and Nöther did not consider in Gordan and Nöther (1876). But in spite of that, the proof of Lemma 4.6 comes from Gordan and Nöther (1876, p. 567) indeed.
On Gordan and Nöther (1876, p. 567), Gordan and Nöther additionally prove that rk J H ≤ 2, as follows. They assume that H 1 = H 2 = 0 on account of Theorem 4.5 and Proposition 2.4, and show the first claim of Lemma 4.4 that J h · J H = 0, to conclude that either h ∈ C[x 1 , x 2 ] or that the rows of J (H 3 , H 4 , H 5 for all i in the first case, so that the row space of J H is generated by J ( ∂ ∂ x 1 h) and J ( ∂ ∂ x 2 h). Unlike Gordan and Nöther, we do not need to show that rk J H ≤ 2 here, because for the techniques in de Bondt and van den Essen (2004), linear dependences between the components of H are the only thing that matters. But the result of Gordan and Nöther can be used to fix the gap in Watanabe (2014), which is caused by the incorrect (Watanabe 2014, Lm. 5.2), and a gap on the same point in Franchetta (1954).

Homogeneous 5-dimensional quasi-translations of 'Fall b)'
In this section, we study homogeneous quasi-translations in dimension 5 which corresponds to 'Fall b)' in Gordan and Nöther (1876, §8). In Corollary 5.2, we will show that homogeneous quasi-translations in dimension 5 which are not of this type always have a linear invariant.
Theorem 5.1 Assume x + H is a homogeneous quasi-translation in dimension 5 over C, such that gcd{H 1 , H 2 , H 3 , H 4 , H 5 } = 1 and H 5 is algebraically independent over C of H 1 , H 2 , H 3 , H 4 . If x + H does not have two independent linear invariants, then rk J H = dim V (H ) = 3 and the following holds.

(iv) There exists an invariant a ∈ C[x] of degree at most 1 of x + H , such that every invariant of x + H which can be expressed as a polynomial in four linear forms in C[x] is contained in C[a]. (v) If H has no linear invariants at all, then g ∈ C.
Proof From (ii) of Theorem 3.6, it follows that 2 ≤ rk J H ≤ dim V (H ) ≤ 3. Assume that x + H does not have two independent linear invariants. From (iii) of Theorem 3.6, it follows that rk J H = 2, so rk J H = dim V (H ) = 3.
(i) Since H 5 is algebraically independent over C of H 1 , H 2 , H 3 , H whence deg t g(x + t H) = 0 and g is an invariant of x + H . Similarly, any linear form in p and q that divides H i is an invariant of x + H as well. If there is at most one independent linear form in p and q that divides H i for any i ≤ 4 such that H i = 0, then deg g = deg H and x + H has three independent linear invariants, which is a contradiction. Hence there are two independent linear forms in p and q that are invariants of x + H . Since p and q are in turn linear forms in these invariants, p and q are invariants of x + H themselves.
Since g is an invariant of x + H , it follows from (2.3) in Proposition 2.2 that J g · H = 0. Hence Now ∂ ∂ x 5 g = 0 contradicts the assumption that gcd{g, H 5 } | gcd{H 1 , (iii) Let r be the degree with respect to x 5 of (H 1 , H 2 , H 3 , H 4 ). If the degree with respect to x 5 of H 5 is larger than r +1, then deg x 5 J H 5 · H = deg x 5 ( ∂ ∂ x 5 H 5 )· H 5 > 2r + 1, which contradicts (3) ⇒ (1) of Proposition 2.2. Take forH i all terms of degree r with respect to x 5 of H i if i ≤ 4, and forH 5 all terms of degree r + 1 with respect to x 5 of H 5 . Then the part of degree 2r with respect to x 5 of J (H 1 , H 2 , H 3 , H 4 ) · H equals J (H 1 ,H 2 ,H 3 ,H 4 ) ·H and the part of degree 2r + 1 of J H 5 · H equals JH 5 ·H . Since J H i · H = 0 for all i on account of (1) ⇒ (3) of Proposition 2.2, we have JH ·H = 0. On account of (3) ⇒ (2) of Proposition 2.2, deg tH 5 (x + tH ) = 0. Since x 5 |H 5 , deg t (x + tH ) 5 = 0 as well. HenceH 5 = 0 and deg x 5 H 5 ≤ r = deg x 5 (H 1 ,  H 2 , H 3 , H 4 ). By taking leading parts with respect to x 5 , we see that for homogeneous and hence any R ∈ C[y 1 , y 2 , y 3 , y 4 ], R (H 1 , H 2 , H 3 , H 4 ) x 4 ], we deduce that x +H can be regarded as a quasi-translation in dimension four (over its first four coordinates). By (i) and (iii) of Theorem 3.6, there are two independent linear forms l 1 and l 2 in Suppose that the leading parts of p and q with respect to x 5 are independent and of the same degree with respect to x 5 . Since (H 1 ,H 2 ,H 3 ,H 4 ) is the leading part of (H 1 , H 2 , H 3 , H 4 ) with respect to x 5 , it follows that (H 1 ,H 2 ,H 3 ,H 4 ) = h(p,q), wherep andq are the leading parts of p and q respectively with respect to x 5 . By assumption,p andq are independent, so we can deduce from l 1 (H ) = l 2 (H ) = 0 that l 1 (h) = l 2 (h) = 0 and hence also l 1 (H ) = l 2 (H ) = 0. Contradiction, thus the leading parts of p and q with respect to x 5 are dependent or have different degrees with respect to x 5 , as desired. (iv) Take for a the linear invariant of x + H , if it has any, and take a = 1 otherwise.
Let f be a non-constant invariant of x + H which can be expressed in four linear forms. We distinguish two cases. • On account of (iii) above, we can obtain that deg x 5 p < deg x 5 q, namely by replacing p and q by linear combinations of p and q, and adapting h accordingly. If we replace H by T −1 H (T x) and ( f, p, q) by ( f (T x), p(T x), q(T x)) for some T ∈ GL 5 (C) such that the last column of T is equal to the fifth unit vector, the form of H does not change and neither do deg x 5 f, deg x 5 p and deg x 5 q. By choosing T appropriate, we can obtain −∞ ≤ deg y 2 h 1 < deg y 2 h 2 < deg y 2 h 3 < deg y 2 h 4 . On account of (2.3) in Proposition 2.2, J f · H = 0. By looking at the leading coefficient with respect to x 5 in J f · H , we can successively deduce that Hence f is a polynomial over C in the invariant x 1 of x + H , and f was a polynomial over C in the invariant x 5 ] and last column of T is equal to the fifth unit vector. Just as above, we replace H by T −1 H (T x) and ( f, p, q) by ( f (T x), p(T x), q(T x)). So we may assume that f ∈ C[x 1 , x 2 , x 3 , x 5 ]. From (2.3), it follows that J f · H = 0 and that any homogeneous part of f is an invariant of x + H as well, so we may assume that f is homogeneous. Since x + H has at most one linear invariant, we can use techniques in the proof of (i) of Theorem 3.6 to show that rk J (H 1 , H 2 , H 3 ) = 2. Hence the ideal b := (R ∈ C[y 1 , y 2 , y 3 ] | R(H 1 , H 2 , H 3 ) = 0) has height 1, and since C[y] is a unique factorization domain, b is principal. Say that R is a generator of b. By looking at the leading homogeneous part of f (x + H ) = f , we see that f (H ) = 0. Since H 5 is algebraically independent of H 1 , H 2 , H 3 , we deduce that R(x 1 , x 2 , x 3 ) | f . From (2.3), it follows f (x + t H) = f , from which we can deduce that every factor of f is an invariant of where a is as in (iv where h is homogeneous of degree at least 3 and ( p, q) is homogeneous of degree at least 4.
such that h and ( p, q) are homogeneous. Furthermore, it follows from (iii) and (iv) of Theorem 5.1 that we may assume that deg x 5 q > deg x 5 p and deg x 5 p > 0 respectively. On account of (2.3) in Proposition 2.2, q(x +t H) = q(x), and looking at the leading coefficient with respect to t gives q(H ) = 0. Since e 5 is a projective apex of H , we even have q (H 1 , H 2 , H 3 , H 4 , H 5 + t) = 0. Hence deg x 5 q ≤ deg q − 1 and in case of equality, looking at the leading coefficient with respect to t in q(x 1 , x 2 , x 3 , x 4 , t) gives a linear form l 1 such that l 1 (H 1 , H 2 , H 3 , H 4 ), which contradicts that x + H has no linear invariants. Thus deg x 5 q ≤ deg q − 2. If we combine this with the conclusion of the previous paragraph, then we obtain If deg h < 3, then there exists a linear form l 2 ∈ C[x 1 , x 2 , x 3 , x 4 ] such that l 2 (h)= 0 and hence also l 2 (H 1 , H 2 , H 3 , H 4 ) = 0, which contradicts that x + H has no linear invariants. Hence deg h ≥ 3.
The following theorem has been proved in Liu (2011) as well. The proof that is given below is somewhat less computational than that in Liu (2011 Proof Just like in the proof of Corollary 5.2, we may assume that H is as in (5.1) such that h is homogeneous of degree at least 3 and ( p, q) is homogeneous such that (5.2) is satisfied. If deg q ≥ 5, then deg H ≥ 5 deg h ≥ 15 indeed, because ( p, q) is homogeneous. Hence assume that deg q ≤ 4. We shall derive a contradiction.
(i) From (iv) of Theorem 5.1, it follows that deg x 4 p ≥ 1 and deg x 4 q ≥ 1. From (5.2), we deduce that deg x 5 p = 1, deg x 5 q = 2 and deg q = 4. Assume without loss of generality that deg Let r be the leading coefficient with respect to x 5 of q. On account of (2.3) in Proposition 2.2, J q · H = 0. By looking at the leading coefficient with respect to x 5 of J q · H = 0, we deduce from (iii) of Theorem 5.1 that r ∈ C[x 1 , x 2 , x 3 ]. Since q(H 1 , H 2 , H 3 , H 4 , t) = 0, r (H 1 , H 2 , H 3 ) = 0 as well. By looking at the leading coefficient with respect to x 5 in r (H 1 , H 2 , H 3 ), we see that the coefficients of x 2 3 and x 2 x 3 of r are zero. Hence r is of the form r = (λ 1 x 1 +λ 2 x 2 +λ 3 x 3 )x 1 −λ 2 4 x 2 2 , where λ i ∈ C for all i. Since r is irreducible, we have λ 3 λ 4 = 0. f . On account of (iii) of Theorem 5.1, deg If we change a factor x i in a product into t H i , the degree with respect to x 5 of that product will increase deg Having to do such a change v times, starting with a term u ∈ C[x], we deduce from deg x 5 u = deg x 4 ,x 5 u − deg x 4 u for terms u ∈ C[x] that for any term and hence any polynomial u ∈ C[x], the coefficient of t v of u(x + t H) has degree at most with respect to x 5 . Since b(u) is affinely linear in deg x 4 ,x 5 u as a function on terms u ∈ C[x], the part of degree b( f ) with respect to x 5 of the coefficient of = deg x 5f that the degree with respect to Hence Since p and q are invariants of x + H , and deg x 5 p = 1 and deg x 5 q = 2, we have deg x 4 ,x 5 p = 1 and deg x 4 ,x 5 q = 2. (ii) LetH i be the part of H i that has degree 2 deg h − 1 with respect to (x 4 , x 5 ), for i = 1, 2, 3, andH j the part of H j that has degree 2 deg h with respect to (x 4 , x 5 ), for j = 4, 5. HenceH 5 = b −1 aH 4 is linearly dependent over C ofH 4 . Assume next thatH 3 = 0. Letq be the leading and quadratic part of q with respect to (x 4 , x 5 ). Thenq |H 4 , so deg tq (x + tH ) ≤ deg tH 4 (x + tH ) = 0 on account of (3) ⇒ (2) of Proposition 2.2. Since λ 3 = 0 and the leading term with But the right hand side has degree deg y 2 h k with respect to x 5 on account of (5.3). Contradiction, so deg q ≥ 5.

The kernel of the map H of quasi-translations x + H
In the beginning of the proof of Theorem 5.1, we have shown that for quasitranslations x + H which belong to case (b) in Gordan and Nöther (1876), dim V (H ) = rk J H = 3. Hence the Zariski closure of the image of H is an irreducible component of V (H ) for such quasi-translations. Corollary 6.4 in this section subsequently gives us several results about quasi-translations which belong to case (b) in Gordan and Nöther (1876), among which a result about such quasi-translations without linear invariants.
First we prove some geometric results about quasi-translations to obtain Theorem 6.3. Next, we use Theorem 6.3 to prove Corollary 6.4. At last, we use Theorem 6.3 to prove Corollary 6.5, which gives us the case where n ≥ 6 of (iii) of Theorem 3.6. Lemma 6.1 Assume x + H is a quasi-translation in dimension n over C. Let X ⊆ C n be an irreducible variety such that H | X is not the zero map, so that the Zariski closure Y of the image of H | X is nonzero.
Then for each c ∈ X , there exists a nonzero p ∈ Y such that g(c + t p) = g(c), for every invariant g of x + H , where t is a new indeterminate.
Proof Let G be the set of invariants of x + H . We first prove this lemma for all c in a nonempty open subset of X . The generic property of c that we assume is that H (c) = 0. Since H | X is not the zero map, we are considering a nonempty open subset of X indeed. From (2.3) in Proposition 2.2, it follows that g(x + t H) = g(x) for every invariant g of x + H . Hence g(c + t p) = g(c) for every g ∈ G, if we take p = H (c) = 0.
In the general case, consider the sets Z := {(c, p, b) ∈ X × (C n ) 2 | g(c + t p) = g(c) for every g ∈ G and b t p = 1} andZ := {(c, p, b) ∈ Z | b is the complex conjugate of p} By applying proper substitutions in t, we see that the imageX of the projection ofZ onto its first n coordinates is equal to that of Z . SinceX contains an open subset of X , it follows from Lemma 3.1 thatX = X , which gives the desired result.
Lemma 6.2 Assume x + H is a quasi-translation in dimension n over C. Let W be the Zariski closure of the image of H . Then for any linear subspace L of C n , the assertions (1) dim L > dim V (H ); (2) every irreducible component of H −1 (L) has dimension greater than dim V (H ); (3) for each c ∈ V (H ), there exists a nonzero p ∈ L ∩ W such that H (c + t p) = 0; satisfy (1) ⇒ (2) ⇒ (3).

Then the Zariski closure W of the image of H is contained in V (H ). Furthermore, every irreducible component X of V (H ) which is not equal to W is a 3-dimensional
linear subspace of C n for which dim(X ∩ W ) = 2.
If W has a nonzero (projective) apex p and V (H ) has a component X which does not contain p, then W is contained in the 4-dimensional linear subspace of C n which is spanned by X and p.
Proof Using (2) ⇒ (3) of Proposition 2.6 and Lemma 3.2, we deduce that W is irreducible and that W ⊆ V (H ). Let X be an irreducible component of V (H ) which is not equal to W . Since X = W and dim V (H ) ≤ 3, we have dim(X ∩ W ) ≤ 2. From gcd{H 1 , H 2 , . . . , H n } = 1, we deduce that dim(X ∩ W ) ≤ 2 ≤ n − dim V (H ). On account of Theorem 6.3, X ∩ W has an irreducible component Z of dimension n − dim V (H ) = 2 = dim(X ∩ W ), such that c + q ∈ X for all c ∈ X and all q in the linear span of Z .
Notice that dim X ≤ dim V (H ) ≤ 3. Suppose that dim X ≤ 2. Then X ⊆ W because X is irreducible and dim(X ∩ W ) = 2. Since W is irreducible, this contradicts the fact that X is an irreducible component of V (H ) which is not equal to W . Thus dim X = 3. Let r be the dimension of the linear span of Z . If r ≥ 3, then X contains the linear span of r independent q ∈ Z , whence X is equal to the linear span of r = 3 independent q ∈ Z . If r ≤ 2, then r = 2 because dim Z = 2, and X is the linear span of two independent q ∈ Z , and any c ∈ X \ Z .
Suppose that W has a nonzero (projective) apex p and V (H ) has a component X which does not contain p. Since dim(X ∩W ) = 2, there are infinitely many GN-planes spanned by p and a nonzero q ∈ X ∩ W . Any proper algebraic subset of W can only have finitely many GN-planes, because W is irreducible and dim W = 3. Hence the set of infinitely many GN-planes spanned by p and a nonzero q ∈ X ∩ W is dense in W . It follows that W is contained in the linear span of X and p. Corollary 6.5 Assume x + H is a homogeneous quasi-translation over C, such that rk J H + dim V (H ) ≤ n. Then H (c + p) = 0 for all c ∈ V (H ) and all p in the linear span of the image of H . In particular, x + H has at least rk J H linear invariants.
Proof The case where deg H ≤ 0 is easy, so assume that deg H ≥ 1. Let W be the Zariski closure of the image of H and X be an irreducible component of V (H ). From Lemma 3.2, it follows that W is irreducible and that dim(X ∩ W ) ≤ dim W = rk J H ≤ n − dim V (H ). Using Theorem 6.3, we subsequently deduce that X ∩ W has an irreducible component Z of dimension n − dim V (H ), such that c + p ∈ X for all c ∈ X and all p in the linear span of Z .
If W X , then by the irreducibility of W , dim Z ≤ dim(X ∩ W ) < dim W = rk J H ≤ n − dim V (H ), which contradicts dim Z = n − dim V (H ). Hence W ⊆ X , and by irreducibility of W once again, the only irreducible component of X ∩ Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.