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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
1 Why are we interested in homogeneous polynomials
with zero Hessian?(f such that
∣∣∣ ∂2f∂xk∂xl
∣∣∣ = 0)
R = K[x1, x2, · · · , xn]
A polynomial ring.
A = R/I = ⊕Adi=0
A graded Artinian Gorenstein algebra.
∃F ∈ R; degree F = d such that
I = AnnR(F ) := {p(x1, · · · , xn) ∈ R|p(∂1, · · · , ∂n)F = 0},2
Remark
I contains no linear forms. ⇔ F is a form in n vari-
ables properly. (No variable can be eliminated by a
linear change of variables.)
Definition
A has the strong Lefschetz property if
∃L = ξ1x1 + · · · + ξnxn, (ξi ∈ K) such that
×Ld−2i : Ai → Ad−i,
is bijective for i = 0, 1, 2, · · · .
3
Proposition
Assume that F is a form in n variables properly.
FCAE:
1.×Ld−2 : A1 → Ad−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
2 A history of Hessian
In 1851, 1856
Otto Hesse (1811 - 1874) 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
1875
Moritz Pasch proved that Hesse’s claim is true for
ternary and quaternary cubics.
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6
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[x1, · · · , x5], they determined all homogeneous
forms with zero Hessian.
7
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.
− − − − − − − − − − − − − − − − − − −−
1985
Professor Hiroshi Yamada (1932-2009) 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
the unfinished paper “Yamada’s Notes.”
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1995
J. Watanabe gave a 15-minuet talk at Northeastern
University: AMS Regional meeting
Let A = ⊕di=0Ai be a Gorenstein ring.
(∀) Ld−2 : A1 → Ad−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
− − − − − − − − − − − − − − − − − − −−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. 171-178. What I wrote is :
Ld−2 : A1 → An−1 is not bijective ∀L⇔
F has zero Hessian
− − − − − − − − − − − − − − − − − − −−10
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.”
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I knew the existence of the following paper only two
weeks ago.2004
Cristoph Lossen:
“When does the Hessian determinant vanish
identically?”
Bull Braz Math Soc, New Series 35(1), 71-82.
11
− − − − − − − − − − − − − − − − − − −−
I knew the existence of the following paper only a week
ago.2008
Alice Garbagnati and Flavia Repetto:
“A geometric approach to Gordan-Noether’s and
Franchetta’s contribution of a question posed by Hesse”
ArXiv Math 0802-0905v1[math.AG].
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2009
12
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}.
− − − − − − − − − − − − − − − − − − −−
2012
We are writing a preprint “On the theory of13
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
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 vari-
able can be eliminated by means of a birational trans-
formation.
To understand the paper of Gordan-Noether, it is help-
ful to know some examples:
15
Example 1
f =∏
1≤i<j≤n
(xi − xj)
has zero Hessian, but R/I has the strong Lefscehtz
property. This is not a contradiction, because a vari-
able can be eliminated by means of a linear transfor-
mation.
y1 = x1 − x2, y2 = x2 − x3, · · · , yn−1 = xn−1 − xn.
yn = x1 + · · · + xn. The partials has the relation:
f1 + · · · + fn = 0.
AnnRf is generated by the elementary symmetric
functions.
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Example 2
f = u2x + uvy + v2z
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 7→ u, v 7→ v, x 7→ x − v2z/u2, y 7→ y +
vz/y, z 7→ 0, then
f = u2
(x − (
v2
u2)z
)+ uv
(y + (
v
u)z
)17
In other words, z has disappeared and f has been re-
duced to
u2x′ + uvy′.
It seems that Hesse did not known this example. It is
clear that Gordan and Noether knew this was the sim-
plest counter example to Hesse’s claim.
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Geometric meaning of u2x + uvy + v2z
H. Nasu showed me: V (u2x + uvy + v2z) ⊂ P4 is the
generic projection of the image of the Segre embedding
P1 × P2 → P5.
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Example 3
Let m1, · · · ,m10 be the 10 monomials of degree 3 in
K[u, v, w]. Let xi be 10 new variables and let
f = m1x1 + m2x2 + · · · + m10x10.
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
1 13 12 13 1.
This is also an example of non-unimovdal Gorenstein
Hilbert series.20
(As I said, )
u2x + uvy + v2z
is the simplest counter example to Hesse’s claim.
u2x + uvy + v2z + w3
is another such example in SIX variables.
(u2x + uvy + v2z)w
is another such example in SIX variables.
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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 deter-
mined.
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Gordan-Noether discovered that a form with zero Hes-
sian 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
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Hessian?
2.What are the solutions of the system of partial diff.
equations. In other words, how and what are the poly-
nomials with zero Hessian?
Let me start with a partial differential equation without
saying anything about Hessians.
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3 Partial differential equations
In the polynomial ring R = K[x1, x2, . . . , xn], we
consider the partial differential equation:
h1(x)∂
∂x1f + h2(x)
∂
∂x2f + · · · + hn(x)
∂
∂xnf = 0,
where hi = hi(x) ∈ R are polynomials.
Put h = (h1, · · · , hn). The set of solutions in R is de-
noted by Sol(h;R). It is a subring of R.
We will always assume that hi 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.
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Consider the case hi(x) are constants. So hi(x) = ai ∈K, a = (a1, · · · , an) = 0. Then the differential equation
a1∂
∂x1f + a2
∂
∂x2f + · · · + an
∂
∂xnf = 0 (1)
is essentially
∂
∂xnf = 0.
Then in this case Sol(a;R) = K[x1, · · · , xn−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
In fact the algebra Sol(a;R) can be described as follows:
Sol(a;R) = K[∆ij|1 ≤ i < j ≤ n],
where
∆ij =
∣∣∣∣∣ai ajxi xj
∣∣∣∣∣ .If an = 0, then
∆1n = a1xn − anx1
∆2n = a2xn − anx2
...
∆n−1,n = an−1xn − anxn−1
is a basis of Sol(a;R).Any ∆ij is a linear combination
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of ∆in and ∆jn.
∆ij =1
an(aj∆in − ai∆jn).
1 ≤ i < j ≤ n − 1.
In short, we may regard An as a union of lines and
Sol(a;R) is the set of functions which take the same
values on the lines.
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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 an = 0, then Sol(a;R) is gener-
ated by ∆1n,∆2n,∆n−1,n as an algebra over K.
3. f(x) = f(x + ta). ∀t ∈ K′, where K′ is any exten-
sion field.
4. If f is not a constant, f(a) = 0 (Moreover, if
deg f = 1, then ∂∂xj
f(a) = 0 ∀j.)
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We consider a system of linear differential equations: a(1)1
∂∂x1
f + a(1)2
∂∂x2
f + · · · + a(1)n
∂∂xn
f = 0
a(2)1
∂∂x1
f + a(2)2
∂∂x2
f + · · · + a(2)n
∂∂xn
f = 0(2)
We may assume that the matrix (a(i)j )i=1,2;j=1,··· ,n has
rank two. Thus the system is essentially
∂
∂xn−1f =
∂
∂xnf = 0.
Then the set of solutions in R is:
Sol(a(1), a(2);R) = K[x1, x2, · · · , xn−2].
If we want to describe the subring without making a
30
linear change of variables, then we can say
Sol(a(1), a(2);R) = K[∆ijk|1 ≤ i < j < k ≤ n],
where
∆ijk =
∣∣∣∣∣∣∣∣a(1)i a
(1)j a
(1)k
a(2)i a
(2)j a
(2)k
xi xj xk
∣∣∣∣∣∣∣∣ .If
∣∣∣∣∣a(1)1 a(1)2
a(2)1 a
(2)2
∣∣∣∣∣ = 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
Consider the system of linear equations:a(1)1
∂∂x1
f + a(1)2
∂∂x2
f + · · · + a(1)n
∂∂xn
f = 0
a(2)1
∂∂x1
f + a(2)2
∂∂x2
f + · · · + a(2)n
∂∂xn
f = 0...
a(r)1
∂∂x1
f + a(r)2
∂∂x2
f + · · · + a(r)n
∂∂xn
f = 0
(3)
(The rank of the coefficient matrix is r.)
The set of solutions is:
Sol(a(1), · · · , a(r);R) = K[∆j1j2···jr|1 ≤ j1 ≤ j2 · · · ≤ jr+1 ≤ n],
32
where
∆j1j2···jrjr+1=
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
a(1)j1
a(1)j2
· · · a(1)jr+1
a(2)j1
a(2)j2
· · · a(2)jr+1
... ... ...
a(r)j1
a(r)j2
· · · a(r)jr+1
xj1 xj2 · · · xjr+1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣.
In any case we can find n− r linearly independent ele-
ments, such that they generate the subring Sol(a(1), · · · , a(r);R)
33
In the above argument all coefficients are constants.
We would have the same result if we treat the coeffi-
cients as indeterminates INDEPNDENT of x1, · · · , xn.u(1)1
∂∂x1
f + u(1)2
∂∂x2
f + · · · + u(1)n
∂∂xn
f = 0
u(2)1
∂∂x1
f + u(2)2
∂∂x2
f + · · · + u(2)n
∂∂xn
f = 0...
u(r)1
∂∂x1
f + u(r)2
∂∂x2
f + · · · + u(r)n
∂∂xn
f = 0
(4)
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 determi-
nantal ideal.
34
We want to introduce notation.
If L ⊂ Kn such that L = ⟨a(1), · · · , a(r)⟩, we write
Sol(L;R) for Sol(a(1), · · · , a(r);R).
If a = (a1 : · · · : an) ∈ Pn−1, then the value
a1∂f
∂x1+ · · · + an
∂f
∂xn
is determined up to a scalar multiple. So it makes sense
to speak of solutions of the differential equation. If L ⊂Pn−1 is a linear subspace, we use the same notation
Sol(L;R) = Sol(a(1), · · · , a(r);R).
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This is the end of this section. We will define a self-vanishing system.
Now we consider
h1∂
∂x1F + · · · + hn
∂
∂xnF = 0,
where hj = hj(x).
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4 Self-vanishing system and some properties
Caution: In this context “system” is a “vector.”
Definition
Suppose that h = (h1, · · · , hn) is polynomial vector.
(hj = hj(x) ∈ K[x1, . . . , xn]). Then h is
a “self-vanishing system (SVS)” if hj ∈ Sol(h;R) ∀j.
In other words h is an SVS if hj (∀j) is a solution of
the differential equation:
h1∂
∂x1F + · · · + hn
∂
∂xnF = 0
37
Example 1
A constant vector h = (a1, a2, . . . , an) ∈ Kn is obvi-
ously a self-vanishing system.
Example 2
Let hj ∈ K[x] be homogeneous polynomials (of the
same degree). Suppose that h = (h1, . . . , hn) satisfy
the following conditions.
1. h1 = · · · = hr = 0, for some integer r; 1 ≤ r < n.
2. The polynomials hr+1, . . . , hn are functions only in
x1, . . . , xr.
Then h is a self-vanishing system of forms.
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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 Hes-
sian.
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Self-vanishing system behaves like a constant vector.
Theorem
Suppose that h = (h1(x), · · · , hn(x)) is an SVS.
FCAE.
1. f(x) = f(x + th). ∀t ∈ K′, where K′ is any exten-
sion field.
2. f(x) ∈ Sol(h;R).
In particular ∆ij =
∣∣∣∣∣hi hj
xi xj
∣∣∣∣∣ ∈ Sol(h;R)
Proof. Let y = (y1, · · · , yn) be a set of variables. Define
40
the operator Dyx by
Dyx = y1∂
∂x1+ · · · + yn
∂
∂xn.
Then we have
f(x + yt) = f(x) +1
1!Dyxf(x)t + · · · +
1
d!Dd
yxf(x)td.
Define the operator Dhx by
Dhx = h1∂
∂x1+ · · · + hn
∂
∂xn.
If h is self-vanishing, then we have
f(x + ht) = f(x) +1
1!Dhxf(x)t + · · · +
1
d!Dd
hxf(x)td.
QED.
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Corollary
Suppose that h = (h1(x), · · · , hn(x)) is a self-
vanishing system. Let f(x) ∈ Sol(h;R)(= the set of
solutions). Then
1. f(h1, · · · , hn) = 0.
2. In particular hj(h1, · · · , hn) = 0 ∀j.
3. f(x) = f(s1, s2, · · · , sn−1, 0), where
si(x) = xi −hi(x)
hn(x)xn, (1 ≤ i ≤ n).
(Assumed that hn = 0.)
4. f(x)g(x) ∈ Sol(h;R) ⇒ f(x), g(x) ∈ Sol(h;R).
42
Proof. 1. Look at the d times polarization of f .
3. In the expression f(x) = f(x + th) substitute
t = −xn
hn.
Then, since ith component of (x + th) is
xi −hi
hnxn,
(x + th) = (s1, s2, · · · , sn−1, 0).
4. This follows from f(x) = f(x+ th) ⇔ f ∈ Sol(h;R).
QED.
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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 co-
efficients are an SVS.
44
Let f ∈ K[x1, . . . , xn] be a homogeneous polynomial.
Let fj = ∂f∂xj
. Assume that the Hessian∣∣∣ ∂2f∂xk∂xl
∣∣∣ = 0.
Then there exits a vector (h1, · · · , hn) such that
(h1, · · · , hn)
(∂2f
∂xk∂xl
)= 0. (5)
(We will assume that GCD(h1, · · · , hn) = 1. )
(h1, · · · , hn)
(∂2f
∂xk∂xl
)x1...
xn
= 0.
This shows
(h1, · · · , hn)
f1...
fn
= 0.
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Hence
f(x) ∈ Sol(h;R).
We already knew that, for each k,
(h1, · · · , hn)
f1k...
fnk
= 0.
(Just look at the kth column of (5).)
Thus
fk(x) ∈ Sol(h;R).
We have to prove that hj(x) ∈ Sol(h;R) ∀j.This can be done by showing that there exists a poly-
46
nomial c(x) such that c(x)hj(x) is a polynomial of
f1(x), · · · , fn(x).
In fact suppose that c(x)hj(x) is a polynomial in f1, · · · , fn.Then
c(x)hj(x) ∈ K[f1, · · · , fn] ⊂ Sol(h;R).
Hence hj(x) ∈ Sol(h;R).
It remains to prove that c(x)hj(x) is a polynomial in
f1, · · · , fn for some c(x).
47
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Another way to obtain the vector (h1, · · · , hn).
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Note that the Hessian∣∣∣ ∂2f∂xk∂xl
∣∣∣ is the Jacobian of the
partials f1, · · · , fn of f .
So there is an algebraic relation among the partials;
g(y1, · · · , yn) ∈ K[y1, · · · , yn]
such that g(f1, · · · , fn) = 0. g is homogeneous. So we
have
y1∂g
∂y1+ · · · + yn
∂g
∂yn= (deg)g
(Choose g so that the degree is the smallest.)
48
In this formula substitute yj for fj. Then we obtain
f1h′1(x) + · · · + fnh
′n(x) = 0.
In other words, h′j(x) is
h′(x) =∂g
∂yj(f1, · · · , fn).
Moreover it is not difficult to see
fk1h′1(x) + · · · + fknh
′n(x) = 0 ∀k.
In other words,
(h′1, · · · , h
′n)
(∂2f
∂xk∂xl
)= 0.
49
Recall that
(h1, · · · , hn)
(∂2f
∂xk∂xl
)= 0.
If we assume that (∂2f/∂xk∂xl) has corank 1, we are
done!! Namely, h′j(x) = c(x)hj(x) is a function of f1, · · · , fn.
(“Corank 1” is not essential.)
50
To summarize the observation of Gordan-Noether:
If f(x) is a form with zero Hessian, then there exists a
self-vanishing system
h = (h1, · · · , hn)
such that f(x) ∈ Sol(h;R) and moreover f(x) ∈ Sol( ∂∂xj
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 = (h1, · · · , hn) such that Then f ∈Sol(h;R) and f ∈ Sol( ∂
∂xjh;R), j = 1, 2, · · · , n.
51
Proof. We know the first assertion: f(x) ∈ Sol(h;R).
(Sorry I have to use an odd notation f = (f1, · · · , fn).)This says that the vector product h · f is zero. I.e.,
h1f1 + · · · + hnfn = 0.
Apply the differential operator∂
∂xkto
h · f = 0.
Then we get(
∂∂xk
h)· f + h · ∂
∂xkf = 0.
We know that h · ∂∂xk
f = 0. So(
∂∂xk
h)· f = 0.
52
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 = (h1, · · · , hn) be a self-vanishing system
such that f(x) ∈ Sol(h;R). Then
f(x) = f(s1, s2, · · · , sn−1, 0),
where si = xi −hihn
xn, i = 1, 2, · · · , n − 1. This shows
that f is a polynomial in n − 1 rational functions. We
have to show that
K(s1, · · · , sn−1, xn) = K(x1, · · · , xn).
53
s1s2...
sn−1
xn
=
1 0 · · · 0 −h1hn
0 1 · · · 0 −h2hn
. . .
0 0 · · · 1 −hn−1hn
0 0 · · · 0 1
x1
x2...
xn−1
xn
.
x1
x2...
xn−1
xn
=
1 0 · · · 0 h1hn
0 1 · · · 0 h2hn
. . .
0 0 · · · 1hn−1hn
0 0 · · · 0 1
s1s2...
sn−1
xn
.
54
We already knew that
hi(x1, · · · , xn−1, xn) = hi(s1, · · · , sn−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 xn = 0 provided
that hn(x) = 0.
This is the end of Part I.
55
This is the start of Part II.
If we denote f = (f1, · · · , fn). It is a polynomial vector.
If we denote f = f(x), it is a polynomial.
When we write f = (f1, · · · , fn), most of the time fiare partials of a polynomial f(x).
At times fi 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
Proposition
• FCAE. Let fj = ∂f(x)/∂xj.
1. f1, · · · , fn are linearly dependent.
2. A variable can be eliminated from f(x) by means
of a linear change of the variables.
• FCAE.
1. f1, · · · , fn are algebraically dependent.
2. The Jacobian∂(f1,...,fn)∂(x1,...,xn)
vanishes identically.
(In this statement, fj do not have to be the partials
of a polynomial.)
57
Proposition
Let R = K[x1, . . . , xn]. Let (f1, f2, . . . , fn) be a vector
of forms in R. Then
rankK(x)
(∂fi
∂xj
)= tr.degK K(f1, f2, . . . , fn).
In particular the following conditions are equivalent.
1. f1, . . . , fr are algebraically dependent.
2. The rank of Jacobian matrix (∂fi∂xj) is < r.
3. tr.degK K(f1, f2, . . . , fr) < r.
58
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
6 The observation of Gordan and Noether Revisited
Assume f1, . . . , fn 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[y1, · · · , yn] → K[x1, . . . , xn]
be the homomorphism defined by yj 7→ fj. Let g(y)
be a homogeneous polynomial in the kernel of ϕ of the
smallest degree.
Let
h′j(x1, x2, . . . , xn) =
∂g
∂yi(f1, . . . , fn),
hj(x1, x2, . . . , xn) =1
GCD(h′1, h
′2, . . . , h
′n)
h′j(x1, . . . , xn).
60
We call the vector (h′1, h
′2, . . . , h
′n) the system of poly-
nomials associated to g(y), and (h1, . . . , hn) the reduced
system of polynomials associated g(y).
Theorem
1. (h1, h2, · · · , hn) is a syzygy of (f1, f2, · · · , fn).
2. (h1, h2, · · · , hn) is a syzygy of (∂f1∂xj, ∂f2∂xj
, · · · , ∂fn∂xj),
for all j = 1, 2 · · · .
3. (∂h1∂xj
, ∂h2∂xj
, · · · , ∂hn∂xj
) is a syzygy of (f1, f2, · · · , fn),for all j = 1, 2 · · · .
Did you know this? This is easy to prove, but I did not
61
know about this fact, not until I saw their paper.
62
Example
K[u, v, w] ⊃ f = (f1, f2, f3) = (u4, u2vw, v2w2).
Let ϕ : K[y1, y2, y3] → K[u, v, w]. Then kerϕ =
(g(y1, y2, y3) = y22 − y1y3).
1. (h1, h2, h3) = (−v2w2, 2u2vw,−u4) is syzygy of
(u4, u2vw, v2w2).
2. (h1, h2, h3) is a syzygy of
(4u3, 2uvw, 0), and
(0, u2w, 2vw2) and
(0, u2v, 2v2w).
63
− − − − − − − − − − − − − − − − − − − − − − −
Gordan-Noether applied this to the partials of a form
with zero Hessian. In the next page I show such an
example.
− − − − − − − − − − − − − − − − − − − − − − −
64
page61
Example
f = u3x + u2vy + v3z. (We assume x1 = u, x2 =
v, x3 = x, x4 = y, x5 = z.)
(f1, f2, f3, f4, f5) = (3u2x+2uvy, u2z+3v2z, u3, u2v, v3)
g(y1, y2, y3, y4, y5) = y34 − y23y5
(h′1, h
′2, h
′3, h
′4, h
′5) = (0, 0,−2u3v3, 3u4v2,−u6)
(h1, h2, h3, h4, h5) = (0, 0,−2v3, 3uv2,−u3)
The differential equation is
0∂F
∂x1+ 0
∂F
∂x2− 2v3
∂F
∂x3+ 3uv2
∂F
∂x4− u3 ∂F
∂x5= 0. (6)
65
f satisfies not only this diff equation, but it satisfies:
0∂F
∂x1+ 0
∂F
∂x2− 6v2
∂F
∂x3+ 6uv
∂F
∂x4− 0
∂F
∂x5= 0
and
0∂F
∂x1+ 0
∂F
∂x2− 0
∂F
∂x3+ 3u
∂F
∂x4− 3u2 ∂F
∂x5= 0
Gordan-Noether called the class of functions (6) “die
Functionen Φ.”
Yamada called (h1, . . . , hn) a self-vanishing system.
The vector (h3, h4, h5) is a syzygy of f3, f4, f5. I.e.,
h3f3 + h4f4 + h5f5 = 0.
66
Now we consider forms with zero Hessian where n ≤ 5.
7 Rational map defined by a self-vanishing system h.
Let f(x) = f(x1, . . . , xn) be a form with zero Hessian.
Let fj = ∂f∂xj
and let h = (h1, . . . , hn) be a self-vanishing
system of forms associated to f . Let
Z : Pn−1(x) → Pn−1(y)
be the rational map defined by the correspondence x =
(x1 : · · · : xn) 7→ (h1 : · · · : hn). Let W be the image
of Z and T the fundamental locus of Z in Pn−1(x). So
T is defined by the equations h1(x) = h2(x) = · · · =
hn(x) = 0. The algebraic set W ⊂ Pn−1(y) is defined
67
by the kernel of
ϕ : K[y] → K[x], yj → hj, j = 1, 2, · · ·
Lemma
Let h = (h1, . . . , hn) be a self-vanishing system of
forms in K[x1, . . . , xn]. Then, rank(∂hi∂xj
)≤ n/2.
In particular Krull.dim K[h1, . . . , hn] ≤ n/2.
(This lemma is independent of forms with zero Hessian.)
Proof. Consider the morphism of the affine space
ϕ : An → An,
defined by xj → hj(x). Since h is self-vanishing, the
composition is zero: ϕ2 = 0. Let Jϕ = (∂hi∂xj
) be the
68
Jacobian matrix. Since we have J2ϕ = 0, the rank cannot
exceed n/2. QED.
If n ≤ 5, then dimW ≤ 1.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Recall the definition of T,W , etc.
Z : Pn−1(x) → Pn−1(y)
is the rational map defined by h = (h1, · · · , hn).
W = Proj(K[(h1, · · · , hn)), the image of Z.
T = Proj(K[x1, · · · , xn]/(h1, · · · , hn)), the fundamen-
tal locus of Z.
69
Proposition
FCAE.
(a) deg (hj) = 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 hj(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
a linear transformation of the variables. Indeed we have
h1f1 + · · · + hnfn = 0.
In other words the partials of f(x) has a linear relation.
71
Proposition
If dim W ≥ 1, then
n/2 ≤ Krull.dim K[x]/(h1, . . . , hn) ≤ n − 2,
or equivalently,
n/2 − 1 ≤ dim T ≤ n − 3.
Proof. Since height (h1, · · · , hn) ≥ 2, we have “Krull
dimension ≤ n − 2.” Consider the ring extension
S := K[h1, · · · , hn] → R := K[x1, · · · , xn].
We have shown that dimS ≤ n/2. (This is a consequence
of “hj(h1, · · · , hn) = 0, ∀j,” a very telling argument!)
72
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:
Pn−1(x) → Pn−1(y)
(x) 7→ (h)
When dimW = 1, we have very satisfying theorems.
73
8 Proof for n = 4 and n = 5
Theorem
Assume that dim W = 1. Let
i : Pn−1(y) → Pn−1(x)
be the natural map yj → xj. Then L(i(W )) ⊂T , where L(i(W )) is the linear closure of i(W ) in
Pn−1(y).
Theorem
With the same notation and assumption, we have
hj(x) ∈ Sol(L(i(W ));R) for all j = 1, 2, . . . , n.
74
As long as dimW > 0, we have
i(W ) ⊂ T.
It is simply that
hj(h1, · · · , hn) = 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
Example 2
Let hj ∈ K[x] be homogeneous polynomials (of the
same degree). Suppose that h = (h1, . . . , hn) satisfy
the following conditions.
1. h1 = · · · = hr = 0, for some integer r; 1 ≤ r < n.
2. The polynomials hr+1, . . . , hn are functions only in
x1, . . . , xr.
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
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 ⊂ P3 is
one dimensional. This means dimKKh1+Kh2+Kh3+
Kh4 = 2 we may assume that h1 = h2 = 0.
The linear space L(W ) can be assumed {(0, 0, ∗, ∗)}.Recall that we used g(y1, y2, y3, y4) such that g(f1, f2, f3, f4) =
0 to define
h = (h1, h2, h3, h4).
Now h1 = h2 = 0 means that g is a form in two variables.
Reason: ∂g∂y1
(f1, · · · , f4) = 0 means that ∂g∂y1
(y1, · · · , y4) =
77
0, because we chose g to have the smallest degree. So it
does not contain the variable y1. By the same reason, it
does not contain y2.
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 as-
sume h1 = h2 = h3 = 0, and L(W ) = {(0, 0, 0, ∗, ∗)}.Since hj ∈ Sol(L(i(W ));R), h4, h5 are polynomials only
in x1, x2, x3. Since h1 = h2 = h3 = 0, as in the case
n = 4, ∂g∂y1
= ∂g∂y2
= ∂g∂y3
= 0. So g is a polynomial only
in the two variables. It has to be a linear form and it is a
78
linear relation among the partials of f because g involves
only two variables.
We are left with the case, dimL(W ) = 2, we may as-
sume h1 = h2 = 0, L(W ) = {(0, 0, ∗, ∗, ∗)}. So we may
assume that h3, h4, h5 are polynomials only in x1, x2.
Recall the fact:
(1) f ∈ Sol(h;R)
(2) f ∈ Sol( ∂∂xj
h;R), j = 1, · · · , n.
Note (1) follows from (2). Since h1 = h2 = 0, and
since h3, h4, h5 are functions only in x1, x2, (2) maybe
79
rewitten as (∂h3∂x1
∂h4∂x1
∂h5∂x1
∂h3∂x2
∂h4∂x2
∂h5∂x2
)f3f4f5
=
(0
0
)
Let A = (aij(x)) be the matrix:
A =
(1/g1 0
0 1/g2
)(∂h3∂x1
∂h4∂x1
∂h5∂x1
∂h3∂x2
∂h4∂x2
∂h5∂x2
),
where g1 is the GCD of the 1st row and g2 is the GCD
of the 2nd row.
We still have
A
f3f4f5
=
(0
0
).
80
We claim that if f ∈ K[x1, · · · , x5] is homogeneous of
degree d with respect to x3, x4, x5, then f is:
f ∈ K[x1, x2]∆d.
where
∆ =
∣∣∣∣∣∣∣a13 a14 a15a23 a24 a25x3 x4 x5
∣∣∣∣∣∣∣To prove it, note that (f3, f4, f5) is determined by the
matrix A up to a multiple of an element in K[x1, x2, x3, x4, x5].
In other words, (f3, f4, f5) = M(δ1, δ2, δ3), where M ∈K[x1, · · · , x5].
Take a vector product with (x3, x4, x5). Then, since f
81
is homogeneous with respect to x3, x4, x5, we have
(degf)f = x3f3 + x4f4 + x5f5 = M∆.
Notice that ∆ ∈ Sol( ∂∂xj
h;R), j = 1, 2.
Hence M ∈ Sol( ∂∂xj
h;R), j = 1, 2.
By induction we have M = M ′∆d−1, M ′ ∈ K[x1, x2].
Thus f = M ′∆d.
Even if we do not assume that f is homogeneous w.r.t.
x3, x4, x5, we have shown that
f ∈ Sol(∂
∂xjh;R), j = 1, 2. ⇒ f ∈ K[x1, x2][∆].
QED. In the above proof we used the fact
rankA = trans.deg K(h3, h4, h5) = 2.
82
Even if we do not assume dimW = 1, it is possible that
hj(x) ∈ Sol(i(L(W ));R) for all j = 1, 2, . . . , n.
is satisfied.
Gordan-Noether (in §6 “f-Problem” of their paper) de-
scribes all possible form F with zero Hessian under the
assumption
hj(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
hj(x) ∈ Sol(i(L(W ));R) for all j = 1, 2, . . . , n.
is exactly what I said “Gordan-Neother type”
83
Example 2
Let hj ∈ K[x] be homogeneous polynomials (of the
same degree). Suppose that h = (h1, . . . , hn) satisfy
the following conditions.
1. h1 = · · · = hr = 0, for some integer r; 1 ≤ r < n.
2. The polynomials hr+1, . . . , hn are functions only in
x1, . . . , xr.
Then h is a self-vanishing system of forms.
84
Questions
1.What are other types of SVS’s arising from forms with
zero Hessian.
2. Provide an integrity basis for the subringµ∩
j=1
Sol(∂
∂xjh;R), where µ = dimW.
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 al-
gebra that corresponds to it is not a complete inter-
section.
4. If f is a symmetric function properly containing n
85
variables, then prove that the Hessian of f does not
vanish.
86
Thank you for listening.
87
参考文献
[1] P. Gordan and M. Nother, Ueber die algebraischen
Formen, deren Hesse’sche Determinante identisch ver-
schwindet, Math. Ann. 10 (1876), 547-568.
[2] T. Maeno and J. Watanabe, Illinois J. Math., vol.
53, no. 2 (2009), 591-603.
[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, 171-178.
[4] H. Yamada, On a theorem of Hesse —P. Gordan
and M. Noether’s theory —, Unpublished.
88