Introduction to the paper of Gordan-Noether in Math. Annalen bd 10, 1876 Ueber die algebraischen Formen, deren Hesse sche Determinante identisch

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1 Introduction to the paper of Gordan-Noether in Math. Annalen bd 10, 1876 Ueber die algebraischen Formen, deren Hesse sche Determinante identisch verschwindet Homogeneous polynomials whose Hessian determinant identically vanishes by Junzo Watanabe Department of Mathematics, Tokai University September 11, 2012, Workshop in Hawaii 1

2 1 Why are we interested ( in homogeneous polynomials with zero Hessian? f such that 2 ) f x k x = 0 l R = K[x 1, x 2,, x n ] A polynomial ring. A = R/I = A d i=0 A graded Artinian Gorenstein algebra. F R; degree F = d such that I = Ann R (F ) := {p(x 1,, x n ) R p( 1,, n )F = 0}, 2

3 Remark I contains no linear forms. F is a form in n variables properly. (No variable can be eliminated by a linear change of variables.) Definition A has the strong Lefschetz property if L = ξ 1 x ξ n x n, (ξ i K) such that L d 2i : A i A d i, is bijective for i = 0, 1, 2,. 3

4 Proposition Assume that F is a form in n variables properly. FCAE: 1. L d 2 : A 1 A d 1 is not a bijection for any linear form L. 2. The Hessian determinant of F is identically zero. This is why we are interested in homogeneous polynomials with zero Hessian. 4

5 2 A history of Hessian In 1851, 1856 Otto Hesse ( ) wrote two papers in Crelle s Journal and proved that if the Hessian determinant identically vanishes, then a variable can be eliminated by means of a linear transformation of the variables. Hesse s claim is not true in general. In fact his proof was weird and the validity of the proof was doubted from the beginning. On the other hand it must have been easy to see that Hesse s claim is true for binary forms as well as quadrics. 5

6 1875 Moritz Pasch proved that Hesse s claim is true for ternary and quaternary cubics. 6

7 1876 P. Gordan- M. Noether, Math. Annalen bd. 10 reached a correct statement. Gordan-Noether s results (among other things): 1. If the Hessian determinant identically vanishes, a variable can be eliminated by means of a birational transformation. Moreover they proved: 2. Hesse s claim is true if the number of variables is at most four. And 3. In C[x 1,, x 5 ], they determined all homogeneous forms with zero Hessian. 7

8 It seems that all these results and the method to prove them had been forgotten completely. It is because, in my opinion, they do not have applications and do not relate to other things. Now we have the necessity to understand their paper and to rewrite their proof from the view point of contemporary algebra Professor Hiroshi Yamada ( ) wrote a paper in which he tried to give proofs for these facts. He did not quite succeed, and the paper was not published. I call 8

9 the unfinished paper Yamada s Notes J. Watanabe gave a 15-minuet talk at Northeastern University: AMS Regional meeting Let A = d i=0 A i be a Gorenstein ring. ( ) L d 2 : A 1 A d 1 is not a bijection The form corresponding to A (in Macaulay s inverse system) has zero Hessian. And I said that Gordan-Noether s result could be used. D. Eisenbud commented: Are you sure of their results? 9

10 2000 A. Geramita suggested to me that I should write about the result in his preprint series. So I wrote a paper in Queen s Papers in Pure and Appl. Math., Vol. 119, pp What I wrote is : L d 2 : A 1 A n 1 is not bijective L F has zero Hessian 10

11 2003 T. Harima, J. Migliore, U. Nagel, and J. Watanabe wrote a paper in J. algebra, quoted Gordan-Neother s results. We had to say we have not confirmed them. I knew the existence of the following paper only two weeks ago Cristoph Lossen: When does the Hessian determinant vanish identically? Bull Braz Math Soc, New Series 35(1),

12 I knew the existence of the following paper only a week ago Alice Garbagnati and Flavia Repetto: A geometric approach to Gordan-Noether s and Franchetta s contribution of a question posed by Hesse ArXiv Math v1[math.AG]

13 T. Maeno and J. Watanabe (Illinois J. Math.) defined higher Hessians, and proved that: A zero-dimensional graded Gorenstein algebra A has the strong Lefschetz property All higher Hessians of F do not vanish identically, where F is the algebraic form corresponding to A. {Lefschets elements} = d/2 j=0 {j-th Hessian 0} We are writing a preprint On the theory of 13

14 Gordan-Noether, where we give proof for most of their results. Probably there are new things that are not contained in the above cited papers. We could not have written it without Yamada s Notes. 14

15 In Part I of this lecture I want to give an outline of proof for the first statement of Gordan-Noether s results. Theorem If Hessian determinant is identically zero, then a variable can be eliminated by means of a birational transformation. To understand the paper of Gordan-Noether, it is helpful to know some examples: 15

16 f = Example 1 (x i x j ) 1 i<j n has zero Hessian, but R/I has the strong Lefscehtz property. This is not a contradiction, because a variable can be eliminated by means of a linear transformation. y 1 = x 1 x 2, y 2 = x 2 x 3,, y n 1 = x n 1 x n. y n = x x n. The partials has the relation: f f n = 0. Ann R f is generated by the elementary symmetric functions. 16

17 Example 2 f = u 2 x + uvy + v 2 z is the simplest example of a homogeneous form with zero Hessian which does not reduce to a form in fewer variables. The Gorenstein algebra that corresponds to it is the trivial extension of K[u, v]/(u, v) 3 by the canonical module. It does not have the SLP. Replace u u, v v, x x v 2 z/u 2, y y + vz/y, z 0, then ( ) f = u 2 x ( v2 u 2)z + uv (y + ( vu ) )z 17

18 In other words, z has disappeared and f has been reduced to u 2 x + uvy. It seems that Hesse did not known this example. It is clear that Gordan and Noether knew this was the simplest counter example to Hesse s claim. 18

19 Geometric meaning of u 2 x + uvy + v 2 z H. Nasu showed me: V (u 2 x + uvy + v 2 z) P 4 is the generic projection of the image of the Segre embedding P 1 P 2 P 5. 19

20 Example 3 Let m 1,, m 10 be the 10 monomials of degree 3 in K[u, v, w]. Let x i be 10 new variables and let f = m 1 x 1 + m 2 x m 10 x 10. Then the Gorenstein algebra that corresponds to it is the trivial extension of K[u, v, w]/(u, v, z) 4 by the canonical module. Clearly it does not have the SLP, because the Hilbert function is This is also an example of non-unimovdal Gorenstein 20 Hilbert series.

21 (As I said, ) u 2 x + uvy + v 2 z is the simplest counter example to Hesse s claim. u 2 x + uvy + v 2 z + w 3 is another such example in SIX variables. (u 2 x + uvy + v 2 z)w is another such example in SIX variables. 21

22 In the first part I am going to give a proof only for (A), but we should be thinking of (B) and (C) also. THEOREM (Gordan-Noether, 1876) (A) If the Hessian determinant identically vanishes, then a variable can be eliminated by means of a birational transformation of the variables. (B) Hesse s claim is true if n 4. (C) n = 5, all forms with zero Hessian can be determined. 22

23 Gordan-Noether discovered that a form with zero Hessian satisfies a linear partial differential equation which has striking properties. It helps to know: There exists one particular partial diff equation which yields more partial diff equations. So in fact a form with zero Hessian satisfies a system of partial diff equations. For (A) and (B) first one is enough. For (C) more diff equation are necessary. There are at least TWO topics: 1. How does the diff equation arise from a form with zero 23

24 Hessian? 2. What are the solutions of the system of partial diff. equations. In other words, how and what are the polynomials with zero Hessian? Let me start with a partial differential equation without saying anything about Hessians. 24

25 3 Partial differential equations In the polynomial ring R = K[x 1, x 2,..., x n ], we consider the partial differential equation: h 1 (x) x 1 f + h 2 (x) x 2 f + + h n (x) x n f = 0, where h i = h i (x) R are polynomials. Put h = (h 1,, h n ). The set of solutions in R is denoted by Sol(h; R). It is a subring of R. We will always assume that h i are homogeneous of the same degree. Even in this case Sol(h; R) may not be finitely generated. So we do not treat it generally, but we consider certain special cases. 25

26 Consider the case h i (x) are constants. So h i (x) = a i K, a = (a 1,, a n ) 0. Then the differential equation a 1 is essentially f + a 2 f + + a n f = 0 (1) x 1 x 2 x n x n f = 0. Then in this case Sol(a; R) = K[x 1,, x n 1 ]. We want to describe it without making a linear change of variables. The above observation shows that the set of solutions of (1) is a subring of R generated by (n 1) linear forms. 26

27 In fact the algebra Sol(a; R) can be described as follows: Sol(a; R) = K[ ij 1 i < j n], where If a n 0, then ij = a i a j x i x j. 1n = a 1 x n a n x 1 2n = a 2 x n a n x 2. n 1,n = a n 1 x n a n x n 1 is a basis of Sol(a; R) Any ij is a linear combination 27

28 of in and jn. ij = 1 a n (a j in a i jn ). 1 i < j n 1. In short, we may regard A n as a union of lines and Sol(a; R) is the set of functions which take the same values on the lines. 28

29 Theorem Suppose that f(x) Sol(a; R). Then 1. Sol(a; R) = K[{ ij 1 i < j n}]. (In any case this is a polynomial ring over K in n 1 variables.) 2. If we assume that a n 0, then Sol(a; R) is generated by 1n, 2n, n 1,n as an algebra over K. 3. f(x) = f(x + ta). t K, where K is any extension field. 4. If f is not a constant, f(a) = 0 (Moreover, if deg f 1, then x f(a) = 0 j.) j 29

30 We consider a system of linear differential equations: a (1) 1 x f + a (1) 1 2 x f + + a (1) 2 n x f = 0 n a (2) 1 x f + a (2) 1 2 x f + + a (2) (2) 2 n x f = 0 n We may assume that the matrix (a (i) j ) i=1,2;j=1,,n has rank two. Thus the system is essentially x n 1 f = Then the set of solutions in R is: x n f = 0. Sol(a (1), a (2) ; R) = K[x 1, x 2,, x n 2 ]. If we want to describe the subring without making a 30

31 linear change of variables, then we can say Sol(a (1), a (2) ; R) = K[ ijk 1 i < j < k n], where If a (1) 1 a (1) 2 a (2) 1 a (2) 2 ijk = a (1) i a (1) j a (1) k a (2) i a (2) j a (2) k x i x j x k. 0, for example, we can choose { 12λ λ = 3, 4,, n} as a minimal set of generators for the subalgebra Now we will treat an arbitrary number of equations. 31

32 Consider the system of linear equations: a (1) 1 x f + a (1) 1 2 x f + + a (1) 2 n x f = 0 n a (2) 1 x f + a (2) 1 2 x f + + a (2) 2 n x f = 0 n. a (r) 1 x f + a (r) 1 2 x f + + a (r) 2 n x f = 0 n (The rank of the coefficient matrix is r.) The set of solutions is: Sol(a (1),, a (r) ; R) = K[ j1 j 2 j r 1 j 1 j 2 j r+1 n], (3) 32

33 where a (1) j a (1) 1 j a (1) 2 j r+1 a (2) j a (2) 1 j a (2) 2 j r+1 j1 j 2 j r j r+1 =.... a (r) j a (r) 1 j a (r) 2 j r+1 x j1 x j2 x jr+1 In any case we can find n r linearly independent elements, such that they generate the subring Sol(a (1),, a (r) ; R) 33

34 In the above argument all coefficients are constants. We would have the same result if we treat the coefficients as indeterminates INDEPNDENT of x 1,, x n. u (1) 1 x f + u (1) 1 2 x f + + u (1) 2 n x f = 0 n u (2) 1 x f + u (2) 1 2 x f + + u (2) 2 n x f = 0 n (4). u (r) 1 x f + u (r) 1 2 x f + + u (r) 2 n x f = 0 n Then we can describe the set of solutions in the ring R = K[{u (i) j }][x]. To give a strict proof we need the theory of determinantal ideal. 34

35 We want to introduce notation. If L K n such that L = a (1),, a (r), we write Sol(L; R) for Sol(a (1),, a (r) ; R). If a = (a 1 : : a n ) P n 1, then the value f f a a n x 1 x n is determined up to a scalar multiple. So it makes sense to speak of solutions of the differential equation. If L P n 1 is a linear subspace, we use the same notation Sol(L; R) = Sol(a (1),, a (r) ; R). 35

36 This is the end of this section. We will define a self-vanishing system. Now we consider h 1 F + + h n F = 0, x 1 x n where h j = h j (x). 36

37 4 Self-vanishing system and some properties Caution: In this context system is a vector. Definition Suppose that h = (h 1,, h n ) is polynomial vector. (h j = h j (x) K[x 1,..., x n ]). Then h is a self-vanishing system (SVS) if h j Sol(h; R) j. In other words h is an SVS if h j ( j) is a solution of the differential equation: h 1 F + + h n F = 0 x 1 x n 37

38 Example 1 A constant vector h = (a 1, a 2,..., a n ) K n is obviously a self-vanishing system. Example 2 Let h j K[x] be homogeneous polynomials (of the same degree). Suppose that h = (h 1,..., h n ) satisfy the following conditions. 1. h 1 = = h r = 0, for some integer r; 1 r < n. 2. The polynomials h r+1,..., h n are functions only in x 1,..., x r. Then h is a self-vanishing system of forms. 38

39 This should be called Gordan-Noether type. See how it is if r = 1, 2 or n 1, n 2. As you will see later, an SVS arises from a form with zero Hessian. I wish I knew if there were other types of SVS if they come from forms with zero Hessian. If SVS which come from Hessian were all Gordan-Noether type, then you could determine all forms with zero Hessian. 39

40 Self-vanishing system behaves like a constant vector. Theorem Suppose that h = (h 1 (x),, h n (x)) is an SVS. FCAE. 1. f(x) = f(x + th). t K, where K is any extension field. 2. f(x) Sol(h; R). In particular ij = h i h j x i x j Sol(h; R) Proof. Let y = (y 1,, y n ) be a set of variables. Define 40

41 the operator D yx by D yx = y y n. x 1 x n Then we have f(x + yt) = f(x) + 1 1! D yxf(x)t d! Dd yx f(x)td. Define the operator D hx by D hx = h h n. x 1 x n If h is self-vanishing, then we have f(x + ht) = f(x) + 1 1! D hxf(x)t d! Dd hx f(x)td. QED. 41

42 Corollary Suppose that h = (h 1 (x),, h n (x)) is a selfvanishing system. Let f(x) Sol(h; R)(= the set of solutions). Then 1. f(h 1,, h n ) = In particular h j (h 1,, h n ) = 0 j. 3. f(x) = f(s 1, s 2,, s n 1, 0), where s i (x) = x i h i(x) h n (x) x n, (1 i n). (Assumed that h n 0.) 4. f(x)g(x) Sol(h; R) f(x), g(x) Sol(h; R). 42

43 Proof. 1. Look at the d times polarization of f. 3. In the expression f(x) = f(x + th) substitute t = x n h n. Then, since ith component of (x + th) is x i h i h n x n, (x + th) = (s 1, s 2,, s n 1, 0). 4. This follows from f(x) = f(x + th) f Sol(h; R). QED. 43

44 5 Forms with zero Hessian I want to show that if f is a form with zero Hessian, then it satisfies a partial differential equation whose coefficients are an SVS. 44

45 Let f K[x 1,..., x n ] be a homogeneous polynomial. Let f j = f x. Assume that the Hessian 2 f j x k x = 0. l Then there exits a vector (h 1,, h n ) such that ( 2 ) f (h 1,, h n ) = 0. (5) x k x l (We will assume that GCD(h 1,, h n ) = 1. ) ( 2 ) x f 1 (h 1,, h n ). = 0. x k x l This shows (h 1,, h n ) f 1. f n x n = 0. 45

46 Hence f(x) Sol(h; R). We already knew that, for each k, f 1k (h 1,, h n ). = 0. f nk (Just look at the kth column of (5).) Thus f k (x) Sol(h; R). We have to prove that h j (x) Sol(h; R) j. This can be done by showing that there exists a poly- 46

47 nomial c(x) such that c(x)h j (x) is a polynomial of f 1 (x),, f n (x). In fact suppose that c(x)h j (x) is a polynomial in f 1,, f n. Then c(x)h j (x) K[f 1,, f n ] Sol(h; R). Hence h j (x) Sol(h; R). It remains to prove that c(x)h j (x) is a polynomial in f 1,, f n for some c(x). 47

48 Another way to obtain the vector (h 1,, h n ). Note that the Hessian 2 f x k x is the Jacobian of the l partials f 1,, f n of f. So there is an algebraic relation among the partials; g(y 1,, y n ) K[y 1,, y n ] such that g(f 1,, f n ) = 0. g is homogeneous. So we have g g y y n = (deg)g y 1 y n (Choose g so that the degree is the smallest.) 48

49 In this formula substitute y j for f j. Then we obtain In other words, h j (x) is f 1 h 1 (x) + + f nh n (x) = 0. h (x) = g y j (f 1,, f n ). Moreover it is not difficult to see f k1 h 1 (x) + + f knh n (x) = 0 k. In other words, ( (h 2 ) 1,, h n ) f x k x l = 0. 49

50 Recall that ( 2 f ) (h 1,, h n ) x k x l = 0. If we assume that ( 2 f/ x k x l ) has corank 1, we are done!! Namely, h j (x) = c(x)h j(x) is a function of f 1,, f n. ( Corank 1 is not essential.) 50

51 To summarize the observation of Gordan-Noether: If f(x) is a form with zero Hessian, then there exists a self-vanishing system h = (h 1,, h n ) such that f(x) Sol(h; R) and moreover f(x) Sol( x j h; R), j = 1, 2,, n. (Shortly we prove the second part.) We state it as a theorem: Theorem Let f(x) be a form with zero Hessian. Then there exists an SVS h = (h 1,, h n ) such that Then f Sol(h; R) and f Sol( x h; R), j = 1, 2,, n. j 51

52 Proof. We know the first assertion: f(x) Sol(h; R). (Sorry I have to use an odd notation f = (f 1,, f n ).) This says that the vector product h f is zero. I.e., h 1 f h n f n = 0. Apply the differential operator to x k Then we get ( ) xk h We know that h h f = 0. f + h x k f = 0. So x f = 0. k ( ) xk h f = 0. 52

53 page47 Now we can prove Gordan-Noether s result. Theorem of Gordan-Noether Suppose that f(x) is a form with zero Hessian. Then a variable can be eliminated in f(x) by a birational transformation. Proof. Let h = (h 1,, h n ) be a self-vanishing system such that f(x) Sol(h; R). Then f(x) = f(s 1, s 2,, s n 1, 0), where s i = x i h i h n x n, i = 1, 2,, n 1. This shows that f is a polynomial in n 1 rational functions. We have to show that K(s 1,, s n 1, x n ) = K(x 1,, x n ). 53

54 s 1 s 2. s n 1 x n x 1 x 2. x n 1 x n h 1 h n h 2 h n = h n 1 h n h 1 h n h 2 h n = h n 1 h n x 1 x 2. x n 1 x n s 1 s 2. s n 1 x n.. 54

55 We already knew that h i (x 1,, x n 1, x n ) = h i (s 1,, s n 1, 0), i = 1, 2,. QED. Remark Suppose that f(x) is a form with zero Hessian. Then f(x) does not degenerate if we set x n = 0 provided that h n (x) 0. This is the end of Part I. 55

56 This is the start of Part II. If we denote f = (f 1,, f n ). It is a polynomial vector. If we denote f = f(x), it is a polynomial. When we write f = (f 1,, f n ), most of the time f i are partials of a polynomial f(x). At times f i are just homogeneous forms of the same degree. These should be distinguished from context. The difference is that a linear transformation of the variables induces a linear transformation of the partials, but does not in the other case. 56

57 Proposition FCAE. Let f j = f(x)/ x j. 1. f 1,, f n are linearly dependent. 2. A variable can be eliminated from f(x) by means of a linear change of the variables. FCAE. 1. f 1,, f n are algebraically dependent. 2. The Jacobian (f 1,...,f n ) vanishes identically. (x 1,...,x n ) (In this statement, f j do not have to be the partials of a polynomial.) 57

58 Proposition Let R = K[x 1,..., x n ]. Let (f 1, f 2,..., f n ) be a vector of forms in R. Then ( ) fi rank K(x) = tr.deg K K(f 1, f 2,..., f n ). x j In particular the following conditions are equivalent. 1. f 1,..., f r are algebraically dependent. 2. The rank of Jacobian matrix ( f i x j ) is < r. 3. tr.deg K K(f 1, f 2,..., f r ) < r. 58

59 I think Gordan and Noether took this theorem for granted. I am not certain what definition they had when they said about the dimension of a variety. Even without referring to Hessian, there were many new things to me in their paper. 59

60 6 The observation of Gordan and Noether Revisited Assume f 1,..., f n are polynomials of the same degree. (Do not have to be the partials of an f(x).) Assume that these are algebraically dependent. Let ϕ : K[y 1,, y n ] K[x 1,..., x n ] be the homomorphism defined by y j f j. Let g(y) be a homogeneous polynomial in the kernel of ϕ of the smallest degree. Let h j (x 1, x 2,..., x n ) = g y i (f 1,..., f n ), h j (x 1, x 2,..., x n ) = 1 GCD(h 1, h 2,..., h n )h j (x 1,..., x n ). 60

61 We call the vector (h 1, h 2,..., h n ) the system of polynomials associated to g(y), and (h 1,..., h n ) the reduced system of polynomials associated g(y). Theorem 1. (h 1, h 2,, h n ) is a syzygy of (f 1, f 2,, f n ). 2. (h 1, h 2,, h n ) is a syzygy of ( f 1 for all j = 1, 2. x j, f 2 x j,, f n x j ), 3. ( h 1 x, h 2 j x,, h n j x ) is a syzygy of (f j 1, f 2,, f n ), for all j = 1, 2. Did you know this? This is easy to prove, but I did not 61

62 know about this fact, not until I saw their paper. 62

63 Example K[u, v, w] f = (f 1, f 2, f 3 ) = (u 4, u 2 vw, v 2 w 2 ). Let ϕ : K[y 1, y 2, y 3 ] K[u, v, w]. Then kerϕ = (g(y 1, y 2, y 3 ) = y 2 2 y 1y 3 ). 1. (h 1, h 2, h 3 ) = ( v 2 w 2, 2u 2 vw, u 4 ) is syzygy of (u 4, u 2 vw, v 2 w 2 ). 2. (h 1, h 2, h 3 ) is a syzygy of (4u 3, 2uvw, 0), and (0, u 2 w, 2vw 2 ) and (0, u 2 v, 2v 2 w). 63

64 Gordan-Noether applied this to the partials of a form with zero Hessian. In the next page I show such an example. 64

65 page61 Example f = u 3 x + u 2 vy + v 3 z. (We assume x 1 = u, x 2 = v, x 3 = x, x 4 = y, x 5 = z.) (f 1, f 2, f 3, f 4, f 5 ) = (3u 2 x+2uvy, u 2 z+3v 2 z, u 3, u 2 v, v 3 ) g(y 1, y 2, y 3, y 4, y 5 ) = y 3 4 y2 3 y 5 (h 1, h 2, h 3, h 4, h 5 ) = (0, 0, 2u3 v 3, 3u 4 v 2, u 6 ) (h 1, h 2, h 3, h 4, h 5 ) = (0, 0, 2v 3, 3uv 2, u 3 ) The differential equation is 0 F x F x 2 2v 3 F x 3 + 3uv 2 F x 4 u 3 F x 5 = 0. (6) 65

66 f satisfies not only this diff equation, but it satisfies: and 0 F x F x 2 6v 2 F x 3 + 6uv F x 4 0 F x 5 = 0 0 F x F x 2 0 F x 3 + 3u F x 4 3u 2 F x 5 = 0 Gordan-Noether called the class of functions (6) die Functionen Φ. Yamada called (h 1,..., h n ) a self-vanishing system. The vector (h 3, h 4, h 5 ) is a syzygy of f 3, f 4, f 5. I.e., h 3 f 3 + h 4 f 4 + h 5 f 5 = 0. 66

67 Now we consider forms with zero Hessian where n 5. 7 Rational map defined by a self-vanishing system h. Let f(x) = f(x 1,..., x n ) be a form with zero Hessian. Let f j = f x j and let h = (h 1,..., h n ) be a self-vanishing system of forms associated to f. Let Z : P n 1 (x) P n 1 (y) be the rational map defined by the correspondence x = (x 1 : : x n ) (h 1 : : h n ). Let W be the image of Z and T the fundamental locus of Z in P n 1 (x). So T is defined by the equations h 1 (x) = h 2 (x) = = h n (x) = 0. The algebraic set W P n 1 (y) is defined 67

68 by the kernel of ϕ : K[y] K[x], y j h j, j = 1, 2, Lemma Let h = (h 1,..., h n ) be a self-vanishing system of forms in K[x 1,..., x n ]. Then, rank ( ) hi x j In particular Krull.dim K[h 1,..., h n ] n/2. n/2. (This lemma is independent of forms with zero Hessian.) Proof. Consider the morphism of the affine space defined by x j h j (x). ϕ : A n A n, Since h is self-vanishing, the composition is zero: ϕ 2 = 0. Let J ϕ = ( h i x j ) be the 68

69 Jacobian matrix. Since we have Jϕ 2 = 0, the rank cannot exceed n/2. QED. If n 5, then dimw 1. Recall the definition of T, W, etc. Z : P n 1 (x) P n 1 (y) is the rational map defined by h = (h 1,, h n ). W = Proj(K[(h 1,, h n )), the image of Z. T = Proj(K[x 1,, x n ]/(h 1,, h n )), the fundamental locus of Z. 69

70 Proposition FCAE. (a) deg (h j ) = 0, i.e., h is a constant vector. (b) dim W = 0, i.e., W is a one-point set. (c) T is empty, i.e., Z is a morphism. Suppose (a) is not true. Then we have h j (h) = 0. This shows T is not empty. Thus (c) (a). (a) (b) (c) are trivial. In this case a variable can be eliminated from f(x) by 70

71 a linear transformation of the variables. Indeed we have h 1 f h n f n = 0. In other words the partials of f(x) has a linear relation. 71

72 Proposition If dim W 1, then n/2 Krull.dim K[x]/(h 1,..., h n ) n 2, or equivalently, n/2 1 dim T n 3. Proof. Since height (h 1,, h n ) 2, we have Krull dimension n 2. Consider the ring extension S := K[h 1,, h n ] R := K[x 1,, x n ]. We have shown that dims n/2. (This is a consequence of h j (h 1,, h n ) = 0, j, a very telling argument!) 72

73 So the dim of the fiber is n/2. QED. If n = 4, then dimt = 1. If n = 5, then 1 dimt 2. Before we go to the next theorems, recall that W is the image of the rational map: P n 1 (x) P n 1 (y) (x) (h) When dimw = 1, we have very satisfying theorems. 73

74 8 Proof for n = 4 and n = 5 Theorem Assume that dim W = 1. Let i : P n 1 (y) P n 1 (x) be the natural map y j x j. Then L(i(W )) T, where L(i(W )) is the linear closure of i(w ) in P n 1 (y). Theorem With the same notation and assumption, we have h j (x) Sol(L(i(W )); R) for all j = 1, 2,..., n. 74

75 As long as dimw > 0, we have It is simply that i(w ) T. h j (h 1,, h n ) = 0 j. The first theorem says that you can prove that we can prove that the linear closure of i(w ) is contained in T (under the assumption dimw = 1). Problem: Is L(i(W )) T not true if we drop dim W = 1? We have to assume dim W 0. The second theorem says that h is a Gordan-Noether type. Recall the example I showed you before: 75

76 Example 2 Let h j K[x] be homogeneous polynomials (of the same degree). Suppose that h = (h 1,..., h n ) satisfy the following conditions. 1. h 1 = = h r = 0, for some integer r; 1 r < n. 2. The polynomials h r+1,..., h n are functions only in x 1,..., x r. Then h is a self-vanishing system of forms. This should be called Gordan-Noether type. Now we deal with forms with zero Hessian for n = 4, 5. 76

77 9 Meaning of Sol(i(L(W )); R) Assume n = 4. If W is a point, the partials of f has a linear relation. (So Hesse s claim holds.) Assume dimw > 0. Then from n/2 1 dim T n 3 we have dimt = 1. Hence the linear space i(l(w )) T P 3 is one dimensional. This means dim K Kh 1 + Kh 2 + Kh 3 + Kh 4 = 2 we may assume that h 1 = h 2 = 0. The linear space L(W ) can be assumed {(0, 0,, )}. Recall that we used g(y 1, y 2, y 3, y 4 ) such that g(f 1, f 2, f 3, f 4 ) = 0 to define h = (h 1, h 2, h 3, h 4 ). Now h 1 = h 2 = 0 means that g is a form in two variables. g Reason: y (f 1 1,, f 4 ) = 0 means that g y (y 1 1,, y 4 ) = 77

78 0, because we chose g to have the smallest degree. So it does not contain the variable y 1. By the same reason, it does not contain y 2. Thus g involves only two variables. So it is a linear form, because it has to be an irreducible polynomial. Hence the partials of f has a linear relation. Now assume n = 5. In this case we also have dimw = 1, but this time dimt = 1, 2. This means that diml(w ) = 1, 2, since L(i(W )) T. If diml(w ) = 1, we may assume h 1 = h 2 = h 3 = 0, and L(W ) = {(0, 0, 0,, )}. Since h j Sol(L(i(W )); R), h 4, h 5 are polynomials only in x 1, x 2, x 3. Since h 1 = h 2 = h 3 = 0, as in the case g n = 4, y = g 1 y = g 2 y = 0. So g is a polynomial only 3 in the two variables. It has to be a linear form and it is a 78

79 linear relation among the partials of f because g involves only two variables. We are left with the case, diml(w ) = 2, we may assume h 1 = h 2 = 0, L(W ) = {(0, 0,,, )}. So we may assume that h 3, h 4, h 5 are polynomials only in x 1, x 2. Recall the fact: (1) f Sol(h; R) (2) f Sol( x j h; R), j = 1,, n. Note (1) follows from (2). Since h 1 = h 2 = 0, and since h 3, h 4, h 5 are functions only in x 1, x 2, (2) maybe 79

80 rewitten as ( h3 h 4 h 5 x 1 x 1 x 1 h 3 h 4 h 5 x 2 x 2 x 2 ) f 3 = f 4 f 5 Let A = (a ij (x)) be the matrix: ( ) ( h3 1/g A = /g 2 ( ) 0 0 h 4 h 5 x 1 x 1 x 1 h 3 h 4 h 5 x 2 x 2 x 2 where g 1 is the GCD of the 1st row and g 2 is the GCD of the 2nd row. We still have A f 3 f 4 f 5 = ( 0 0 ). ), 80

81 We claim that if f K[x 1,, x 5 ] is homogeneous of degree d with respect to x 3, x 4, x 5, then f is: f K[x 1, x 2 ] d. where a 13 a 14 a 15 = a 23 a 24 a 25 x 3 x 4 x 5 To prove it, note that (f 3, f 4, f 5 ) is determined by the matrix A up to a multiple of an element in K[x 1, x 2, x 3, x 4, x 5 ]. In other words, (f 3, f 4, f 5 ) = M(δ 1, δ 2, δ 3 ), where M K[x 1,, x 5 ]. Take a vector product with (x 3, x 4, x 5 ). Then, since f 81

82 is homogeneous with respect to x 3, x 4, x 5, we have (degf)f = x 3 f 3 + x 4 f 4 + x 5 f 5 = M. Notice that Sol( x j h; R), j = 1, 2. Hence M Sol( x j h; R), j = 1, 2. By induction we have M = M d 1, M K[x 1, x 2 ]. Thus f = M d. Even if we do not assume that f is homogeneous w.r.t. x 3, x 4, x 5, we have shown that f Sol( x j h; R), j = 1, 2. f K[x 1, x 2 ][ ]. QED. In the above proof we used the fact ranka = trans.deg K(h 3, h 4, h 5 ) = 2. 82

83 Even if we do not assume dimw = 1, it is possible that h j (x) Sol(i(L(W )); R) for all j = 1, 2,..., n. is satisfied. Gordan-Noether (in 6 f-problem of their paper) describes all possible form F with zero Hessian under the assumption h j (x) Sol(i(L(W )); R) for all j = 1, 2,..., n. (I do not think they are quite right in the description of these forms.) The condition h j (x) Sol(i(L(W )); R) for all j = 1, 2,..., n. is exactly what I said Gordan-Neother type 83

84 Example 2 Let h j K[x] be homogeneous polynomials (of the same degree). Suppose that h = (h 1,..., h n ) satisfy the following conditions. 1. h 1 = = h r = 0, for some integer r; 1 r < n. 2. The polynomials h r+1,..., h n are functions only in x 1,..., x r. Then h is a self-vanishing system of forms. 84

85 Questions 1. What are other types of SVS s arising from forms with zero Hessian. 2. Provide an integrity basis for the subring µ Sol( h; R), where µ = dimw. x j j=1 for h of Gordan-Noether type. 3. Prove that if f is a form with zero Hessian properly containing n variables, prove that the Gorenstein algebra that corresponds to it is not a complete intersection. 4. If f is a symmetric function properly containing n 85

86 variables, then prove that the Hessian of f does not vanish. 86

87 Thank you for listening. 87

88 [1] P. Gordan and M. Nöther, Ueber die algebraischen Formen, deren Hesse sche Determinante identisch verschwindet, Math. Ann. 10 (1876), [2] T. Maeno and J. Watanabe, Illinois J. Math., vol. 53, no. 2 (2009), [3] J. Watanabe, A remark on the Hessian of homogeneous polynomials, in The Curves Seminar at Queen s Volume XIII, Queen s Papers in Pure and Appl. Math., Vol. 119, 2000, [4] H. Yamada, On a theorem of Hesse P. Gordan and M. Noether s theory, Unpublished. 88