Quasi-translations and singular Hessians

0.25 MB·English

by  Michiel de Bondt

Quasi-translations and singular Hessians
5
1
Michiel de Bondt
0
2
January 22, 2015
n
a
J

1 Abstract
2
In 1876 in [8], the authors Paul Gordan and Max N¨other classify all
] homogeneouspolynomialshinatmostfivevariablesforwhichtheHessian
G
determinant vanishes. For that purpose, they study quasi-translations
A which are associated with singular Hessians.
We will explain what quasi-translations are and formulate some ele-
.
h mentarypropertiesofthem. Additionally,weclassifyallquasi-translations
t
a with Jacobian rank one and all so-called irreducible homogeneous quasi-
m translations with Jacobian rank two. Thelatter isan important result of
[8]. Using these results, we classify all quasi-translations in dimension at
[
most three and all homogeneous quasi-translations in dimension at most

1 four.
v Furthermore,wedescribetheconnectionofquasi-translationwithsin-
8 gular Hessians, and as an application, we will classify all polynomials in
6
dimension twoandallhomogeneouspolynomialsindimensionsthreeand
1
four whose Hessian determinant vanishes. More precisely, we will show
5 thatuptolinearterms,thesepolynomialscanbeexpressedinn−1linear
0
forms,wherenisthedimension,accordingtoaninvalidtheoremofHesse.
.
1 In the last section, we formulate some known results and conjectures
0 in connection with quasi-translations and singular Hessians.
5
1
: Key words: Quasi-translation, Hessian, determinant zero, homogeneous, lo-
v
cally nilpotent derivation, algebraic dependence, linear dependence.
i
X
r MSC 2010: 14R05, 14R10,14R20, 13N15.
a
1 Quasi-translations
LetAbeacommutativeringwithQandnbeapositivenaturalnumber. Write
x
1
 x2 
x= .. =(x1,x2,...,xn)
 . 
 
 x 
n
 
1
for the identity map from An to An, thus x ,x ,...,x are variables that cor-
1 2 n
respond to the coordinates of An. A translation of An is a map
x +c
1 1
 x2+c2 
x+c= .
.
 . 
 
 xn+cn 
 
where c An is a fixed vector. For a translation x+c, we have that x c is
∈ −
the inverse polynomial map, because
(x +c ) c x
1 1 1 1
−
 (x2+c2) c2   x2 
(x c) (x+c)= . − = . =x
− ◦  ..   .. 
   
 (xn+cn) cn   xn 
 −   
is the identity map.
Inspiredbythis property,wedefine a quasi-translation asapolynomialmap
x +H x +H (x)
1 1 1 1
 x2+H2   x2+H2(x) 
x+H = . = .
. .
 .   . 
   
 xn+Hn   xn+Hn(x) 
   
such that x H is the inverse polynomial map of x+H, i.e.
−
x +H (x) H (x+H)
1 1 1
−
 (cid:0)x2+H2(x)(cid:1) H2(x+H) 
(x−H)◦(x+H)= (cid:0) (cid:1)...− =x
 
 x +H (x) H (x+H) 
n n n
 − 
(cid:0) (cid:1)
The differencebetweenatranslationx+canda quasi-translationx+H is that
c A for all i, while H A[x]=A[x ,x ,...,x ] for all i.
i i 1 2 n
∈ ∈
Foraregulartranslationx+c,wehavethatapplyingitmtimescomesdown
tothetranslationx+mc. Iff C[x]isaninvariantofaregulartranslationx+c,
∈
i.e. f(x+c)=f(x), then f(x+(m+1)c)=f(x+mc) follows by substituting
x=x+mc,whencebyinductiononm, f(x+mc)=f(x)forallm N. Below,
∈
we show similar results for quasi-translations.
Since (x H) (x+H)=x, we see that
− ◦
(x+mH) (x+H)= (m+1)x m(x H) (x+H)
◦ − − ◦
=((cid:0)m+1)(x+H) mx=(cid:1) x+(m+1)H
−
Henceitfollowsbyinductiononmthatapplyingx+H mtimescomesdownto
the map x+mH indeed. Consequently, if f A[x] is an invariant of a x+H,
∈
i.e. f(x+H)=f(x), then
f(x+mH)=f((x+H) (x+H))=f(x)
◦···◦
2
follows for each m N by applying f(x+H)=f(x) m times. Thus
∈
f(x+H)=f(x)= f(x+mH)=f(x) for all m N (1.1)
⇒ ∈
With some techniques ‘on the shelf’, we can improve (1.1) to the following.
Lemma 1.1. Assume that x+H is a quasi-translation over A. Then
f(x+H)=f(x)= f(x+tH)=f(x) (1.2)
⇒
where t is a new indeterminate.
Proof. Letd:=degf andwritef(x+tH) f =cdtd+cd−1td−1+ +c1t+c0.
− ···
Since f(x+mH) f =0 for all m 0,1,2,...,d on account of (1.1),
− ∈{ }
1 0 02 0d c 0
0
···
 1 1 12 1d   c1   0 
···
1 2 22 2d c 0
2 =
 ···     
 ... ... ... ... ... · ...   ... 
     
 1 d d2 dd   cd   0 
 ···     
The matrix on the left hand side is of Hadamard type and hence invertible,
becauseA Q. Consequently,c =c =c = =c =0. Hence f(x+tH)=
0 1 2 d
⊇ ···
f, as desired.
But GordanandNo¨ther did notuse the termquasi-translation,andcharac-
terizedquasi-translationsinanotherway. Todescribeit,wedefinetheJacobian
matrix of a polynomial map H =(H ,H ,...,H ):
1 2 m
∂ H ∂ H ∂ H
∂x1 1 ∂x2 1 ··· ∂xn 1
JH := ∂∂x1...H2 ∂∂x2...H2 ·.·.·. ∂∂xn...H2 
 
 ∂ H ∂ H ∂ H 
 ∂x1 m ∂x2 m ··· ∂xn m 
Notice that H is a row vector if H is a single polynomial, i.e. m=1.
J
IfweseeH asacolumnvector,wecanevaluatethematrixproduct H H,
J ·
and the characterization of quasi-translations x + H by Gordan and No¨ther
comes down to
H H =(01,02,...,0n) (1.3)
J ·
Here,the powersofzeroare onlytakento indicate the lengthofthe zerovector
on the right.
Gordan and No¨ther derived the following under the assumption of (1.3).
Take a polynomial f A[x] such that f H =0. By the chain rule
∈ J ·
f(x+tH) H =( f) (I +t H) H
x=x+tH n
J · J | · J ·
∂
=( f) H = f(x+tH)
x=x+tH
J | · ∂t
3
where In is the unit matrix of size n. Here, x=··· means substituting for x.
| ···
Since f H =0, it follows from the above that
J ·
∂
f(x+tH) f(x) H = f(x+tH) (1.4)
J − · ∂t
(cid:0) (cid:1)
Suppose that t divides the right hand side of (1.4) exactly r < times. Then
∞
t divides f(x+tH) f(x) more than r times. Hence t divides the left hand
−
side of (1.4) more than r times as well, which is a contradiction. So both sides
of (1.4) are zero. Since the right hand side of (1.4) is zero and A Q, we get
⊇
f(x+tH)=f, and that is what Gordan and No¨ther derived from (1.3).
Lemma 1.2. Assume H is a polynomial map over A, such that H H =
J ·
(01,02,...,0n). Then
f H =0= f(x+tH)=f(x) (1.5)
J · ⇒
for every f A[x].
∈
With lemmas 1.1 and 1.2, we can prove the following.
Proposition 1.3. Let A be a commutative ring with Q and H :An An be a
⇒
polynomial map. Then the following properties are equivalent:
(1) x+H is a quasi-translation,
(2) H(x+tH)=H, where t is a new indeterminate,
(3) H H =(01,02,...,0n).
J ·
Furthermore,
f(x+H)=f(x) f(x+tH)=f(x) f H =0 (1.6)
⇐⇒ ⇐⇒J ·
holds for all f A[x], and additionally
∈
( H)x=x−tJH =( H)+t( H)2+t2( H)3+ (1.7)
J | J J J ···
if any of (1), (2) and (3) is satisfied.
Proof. The middle hand side of (1.6) gives the left hand side by substituting
t=1 andthe righthandside bytakingthe coefficientoft1. Hence (1.6)follows
from (1) and lemma 1.1 and (3) and lemma 1.2.
By taking the Jacobian of (2), we get ( H) (I +t H) = H,
x=x+tH n
J | · J J
which gives (1.7) after substituting t = t. Therefore, it remains to show that
−
(1), (2) an (3) are equivalent.
(1) ⇒ (2) Assume (1). Since x=(x H) (x+H)=x+H H(x+H), we
− ◦ −
see that H(x+H)=H, and (2) follows by taking f =H in lemma 1.1.
i
4
(2) ⇒ (1) Assume (2). Then
(x tH) (x+tH)=(x+tH) tH(x+tH)
− ◦ −
=x+tH tH =x (1.8)
−
which gives (1) after substituting t=1.
(2) ⇒ (3) Assume (2). By taking the coefficient of t1 of (2), we get (3).
(3) ⇒ (2) Assume (3). By taking f =H in lemma 1.2, we get (2).
i
Notice that(1.8)tells usthatin somesense, x+tH is aquasi-translationas
well.
Remark 1.4. Let D be the derivation n H ∂ . Then one can easily verify
i=1 i∂xi
that P
f H =0 Df =0
J · ⇐⇒
and
H H =0 D2x =D2x = =D2x =0
1 2 n
J · ⇐⇒ ···
Hencequasi-translationscorrespondtoaspecialkindoflocallynilpotentderiva-
tions. Furthermore, invariants of the quasi-translation x+H are just kernel
elements of D.
In addition, we can write exp(D) and exp(tD) for the automorphisms cor-
responding to the maps x+H and x+tH 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.
2 Singular Hessians and Jacobians
Now that we havehad some introduction aboutquasi-translations,it is time to
show how they are connected to singular Hessians. For that purpose, we define
the Hessian matrix of a polynomial h C[x] as follows:
∈
∂ ∂ h ∂ ∂ h ∂ ∂ h
∂x1∂x1 ∂x2∂x1 ··· ∂xn∂x1
 ∂ ∂ h ∂ ∂ h ∂ ∂ h 
Hh:= ∂x1∂...x2 ∂x2∂...x2 ·.·.·. ∂xn...∂x2 
 
 ∂ ∂ h ∂ ∂ h ∂ ∂ h 
 ∂x1∂xn ∂x2∂xn ··· ∂xn∂xn 
A Hessian is a Jacobian which is symmetric with respect to the main diagonal.
Hence each dependence between the columns of a Hessian is also a dependence
between the rows of it.
For generality,we considerJacobianswhich are not necessaryHessians. As-
sume G : Cn Cn is a polynomial map and det G = 0. Let rkM denote
→ J
the rank of a matrix M. It is known that rk H = trdeg C(H), see e.g.
C
J
5
Proposition 1.2.9 of either [5] of [1]. Hence there exists a nonzero polynomial
R C[y]=C[y ,y ,...,y ] such that
1 2 n
∈
R(G ,G ,...,G )=0 (2.1)
1 2 n
Here, y ,y ,...,y are also variables that correspondto the coordinates of Cn.
1 2 n
Define
∂R
H := (2.2)
i
(cid:16)∂yi(cid:17)(cid:12)y=G
(cid:12)
for all i. By taking the Jacobian of (2.1), w(cid:12)e get
0= (R(G))=( R) G=Ht G
y y=G
J J J | ·J ·J
where Ht = (H H H ) is the transpose of H. Hence H is a dependence
1 2 n
···
between the rows of G. If G is a Hessian, H is also a dependence between
J J
thecolumnsof G. Henceitseemsagoodideatoassumethat G H =0. But
forthe sakeofgJenerality,weonly assumethat G H˜ =0 forsJome· polynomial
vector H˜ that does not need to be equal to H.J ·
If we write R for the transpose of R, then H =( R)(G), and by the
y y y
∇ J ∇
chain rule
H H˜ = ( R)(G) H˜ =( R) G H˜ =0 (2.3)
y y y y=G
J · J ∇ · J ∇ | ·J ·
whence x+H is a quasi-translation on account of (3) (1) of 1.3 if H = H˜.
⇒
Thus we have proved the following.
Proposition 2.1. Assume H is defined as in (2.2) for each i, with R is as
i
in (2.1), where G : Cn Cn is a polynomial map. Then H is a dependence
→
between the rows of G.
If H˜ is a dependJence between the columns of G, then H H˜ = 0. In
J J ·
particular, if H is a dependence between the columns of G, then x+H is a
J
quasi-translation.
ButH definedasabovemaybethezeromap. Thisishowevernotthecaseif
we choose the degree of R as small as possible, without affecting (2.1), because
∂ R is nonzero for some i and of lower degree than R itself.
∂yi
Example 2.2. Let p=x2x +x x x +x2x , h=pr and R=y y y2. Then
1 3 1 2 4 2 5 3 5− 4
R( h)=(rpr−1x2)(rpr−1x2) (rpr−1x x )2 =0
∇ 1 2 − 1 2
and R=(0,0,y , 2y ,y ). So
y 5 4 3
∇ −
x+H :=x+( R)( h)=x+rpr−1(0,0,x2, 2x x ,x2)
∇y ∇ 2 − 1 2 1
is a quasi-translation. Indeed, x , x and p are invariants of x+H.
1 2
6
Example 2.3. Let n 6 be even and
≥
h=(Ax1 Bx2)2+(Ax3 Bx4)2+ +(Axn−1 Bxn)2
− − ··· −
Then (By2i−1+Ay2i)y=∇h =0 for all i, because
|
∂h ∂h
B =2AB(Ax2i−1 Bx2i)= A
∂x2i−1 − − ∂x2i
for all i. Hence (B2y22i−1−A2y22i)|y=∇h =0 for all i as well, and also
1
R:= B2(y2+y2+ +y2 ) A2(y2+y2 +y2)2
4AB 1 3 ··· n−1 − 2 4··· n
(cid:0) (cid:1)
satisfies R( h)=0. Now one can compute that
∇
∂R( h)
∂y1 ∇ B(Ax1 Bx2)
 ∂∂∂yRR2((∇hh))   BA((AAxx1−−BBxx2)) 
x+H :=x+ ∂∂y∂∂∂∂∂nyyyRRRn−34(1(...∇∇∇(∇hhh))) = BAA((AA(Axxnnx−−3311...−−−−BBBxxx44)nn)) 
is a quasi-translation, and h H =(01,02,...,0n).
H ·
Now let a:=x x x x , b:=x x x x , and G:=( h) . Then
1 4 2 3 3 6 4 5 A=a,B=b
− − ∇ |
R(G)=0aswell. Thus H˜ :=( R)(G)=H isa dependence between
y A=a,B=b
∇ |
the rows of
G= ( h)+( h) a+( h) b
J H JA∇ ·J JB∇ ·J A=a,B=b
(cid:0) (cid:1)(cid:12)
(cid:12)
Since
(x2i−1x2j x2ix2j−1) H˜
J − ·
=(x2i−1H˜2j x2iH˜2j−1)+(x2jH˜2i−1 x2j−1H˜2i)
− −
=(ax2i−1 bx2i)(ax2j−1 bx2j)+(bx2j ax2j−1)(ax2i−1 bx2i)
− − − −
=0
for all i,j, we have a H˜ = b H˜ =0. Consequently,
J · J ·
G H˜ =( h) H˜ +( h) a H˜ +( h) b H˜
A=a,B=b a b
J · H | · J ∇ ·J · J ∇ ·J ·
= ( h) H =(01,02,...,0n)
H · A=a,B=b
(cid:0) (cid:1)(cid:12)
(cid:12)
Thus x+H˜ is a quasi-translation as well. In fact, it appears in [4, Th. 2.1] as
a homogeneous quasi-translationwithout linear invariants.
7
3 Hesse’s theorem
A polynomial is homogeneous of degree d if all its terms have degree d. A
homogeneous polynomial of degree one is called a linear form. We call a quasi-
translationx+H homogeneousifthereexistsad Nsuchthateachcomponent
∈
H of H is either zero or homogeneous of degree d.
i
When Gordan and No¨ther published their article in 1876, it was two years
ago that Otto Hesse had died. The role of Hesse is, that the starting point
of Gordan and No¨ther was a wrong theorem of Hesse, which they proved in
dimensions two, three and four, and disproved for all dimensions greater than
four.
Hesse’s (invalid) theorem. Assume n 2 and h is a homogeneous polyno-
≥
mial in x ,x ,...,x over C whose Hessian matrix is singular. Then there is a
1 2 n
constant vector c=(c ,c ,...,c ) such that
1 2 n
∂h ∂h ∂h
c +c + +c =0 (3.1)
1 2 n
∂x ∂x ··· ∂x
1 2 n
or equivalently h c=0.
J ·
Gordan and No¨ther first observe that (3.1) is equivalent to the existence of
a linear transformationwhich makes h a polynomial in less than n variables,in
other words, the existence of an invertible matrix T such that
h(Tx) C[x1,x2,...,xn−1]
∈
In order to see that, we first compute the Jacobian matrix of h(Tx) by way of
the chain rule, where means substituting the vector Tx for x.
x=Tx
|
h(Tx) = h(x) T (3.2)
J J x=Tx·
(cid:0) (cid:1) (cid:0) (cid:1)(cid:12)
If the i-th column of T is equal to c, we have(cid:12)
n
∂ ∂h
h(Tx)= h(x) c= c
∂x J x=Tx· (cid:18) j∂x (cid:19)(cid:12)
i (cid:0) (cid:1)(cid:12) Xj=1 j (cid:12)x=Tx
(cid:12) (cid:12)
(cid:12)
whence for all c C,
0
∈
n
∂ ∂h
h(Tx)=c c =c (3.3)
0 j 0
∂x ⇐⇒ ∂x
i Xj=1 j
By taking i = n and c = 0 in (3.3), we get the above remark of Gordan and
0
No¨ther.
WeshallproveHesse’stheoremindimensionstwo,three,andfour. Ifwedrop
theconditionthathishomogeneous,wecanproveHesse’stheoremindimension
two, but only for polynomials h without linear terms. For polynomials h with
linear terms, we can only get
∂h ∂h ∂h
c +c + +c =c C
1 2 n 0
∂x ∂x ··· ∂x ∈
1 2 n
8
whichholdsforhomogeneoushindimensiononeaswell(h=x ishomogeneous,
1
but does not satisfy Hesse’s theorem in dimension one). So we will prove the
following.
Theorem 3.1. Hesse’s theorem is true in dimension n 4, and for non-
≤
homogeneous polynomials in dimension n 2.
≤
Before we prove our affirmative results about Hesse’s theorem, we take a
lookatthe effects ofaffinetransformationsofh andofremovinglineartermsof
h.
Viewthevectorsx=(x ,x ,...,x ),c=(c ,c ,...,c )andc˜=(c˜ ,c˜ ,...,
1 2 n 1 2 n 1 2
c˜ ) as column matrices of variables and constants respectively. Let T be an
n
invertible matrix of size n over C. Then
h(Tx+c)
is an affine transformation of h. Write Mt for the transpose of a matrix M.
Then c˜tx, which is the product of a row matrix and a column matrix, is a
matrix with only one entry, and we associateit with the value of its only entry.
If we define
h˜ :=h(Tx+c) c˜tx
−
then it appears that det h˜ = 0, if and only if det h = 0. To see this, notice
H H
that on account of (3.2),
h˜ = h(Tx+c) c˜tx =( h) T c˜t
x=Tx+c
J J − J | · −
(cid:0) (cid:1)
The transpose of this equals
h˜ =Tt ( h) c˜
x=Tx+c
∇ · ∇ | −
where f stands for the transpose of f. Taking the Jacobian of the above,
∇ J
we get
h˜ = h(Tx+c) c˜tx = Tt ( h) c˜ =Tt ( h) T
x=Tx+c x=Tx+c
H H − J · ∇ | − · H | ·
(cid:0) (cid:1) (cid:0) (cid:1)
which is a singular matrix, if and only if h is. If det h=0, then R( h)=0
for some nonzero R C[y] on account ofH(2.1), and simHilarly R˜( h˜)=∇0.
We shall describe∈a natural connection between R and R˜. If∇R( h)=0 for
∇
some R C[y], then
∈
R (Tt)−1 (Tt ( h) c˜)+c˜ = R( h) =0
x=Tx+c
(cid:16) (cid:0) · ∇ | − (cid:1)(cid:17) (cid:16) ∇ (cid:17)(cid:12)(cid:12)x=Tx+c
thus R˜ :=R((Tt)−1(y c˜)) satisfies R˜( h˜)=0. (cid:12)
− ∇
If we define H as in (2.2) with G = h for each i, then we get H =
i
∇
( R)( h), and x+H is a quasi-translation on account of corollary 2.1. In a
y
∇ ∇
similar manner, x+H˜ is a quasi-translation if H˜ = ( R˜)( h˜), and again we
y
∇ ∇
describe a natural connection. By the chain rule,
JyR˜ =JyR (Tt)−1(y−c˜) =(JyR)|y=(Tt)−1(y−c˜)·(Tt)−1
(cid:0) (cid:1)
9
whence
yR˜ =T−1( yR)y=(Tt)−1(y−c˜)
∇ ∇ |
Combining this with h˜ =Tt ( h) +c˜, we get that
x=Tx+c
∇ · ∇ |
H˜ =( R˜)( h˜)=T−1H(Tx+c)
y
∇ ∇
thus H˜ is a linear conjugation of H if c=0. In general,
x+H˜ =x+T−1H(Tx+c)=T−1 Tx+c+H(Tx+c) c
(cid:16)(cid:0) (cid:1)− (cid:17)
isanaffinelylinearconjugationofx+H. Inthenextsection,weshallshowthat
affinely linear conjugations of quasi-translations are again quasi-translations.
We are now ready to prove the following.
Theorem 3.2. Assume h C[x] and R=0 satisfies (2.1) and is of minimum
∈ 6
degree. Define H =( R)( h).
y
∇ ∇
If h is homogeneous, then R and H are homogeneous as well.
If the dimension of the linear span of the image of H is at most one, then
degR=1 and x+H is a regular translation.
Proof. If h is homogeneous, then (2.1) is still satisfied if R is replaced by any
homogeneous component of it. Since R was chosen of minimum degree, we see
that R and hence also H is homogenous if h is homogeneous.
Assume that the dimension of the linear span of the image of H is at most
one. Then there are n 1 independent vectors c(i) Cn such that
− ∈
(c(1))tH =(c(2))tH = =(c(n−1))tH =0
···
Suppose that degR = 1. Since (c(i))tH = ((c(i))t R)( h) for each i and R is
6 ∇ ∇
ofminimumdegree,wehave(c(i))t R=0foreachi. Takeaninvertiblematrix
T over C such that the i-th colum∇n of (Tt)−1 equals c(i) for each i n 1.
≤ −
Then ( R) c(i) =0 for each i n 1. By applying (3.3) with R(y) instead of
h(x) anJd c ·=0, we obtain ∂ R≤ (T−t)−1y =0 for all i n 1, so
0 ∂yi ≤ −
(cid:0) (cid:1)
R˜(y):=R (Tt)−1y C[y ]
n
∈
(cid:0) (cid:1)
Notice that for h˜ := h(Tx), we have R˜( h˜) = R (Tt)−1Tt h = 0,
∇ ∇ x=Tx
whence (cid:0) (cid:1)(cid:12)
∂ (cid:12)
˜h
∂x
n
isalgebraicoverC. Thisisonlypossibleifthisderivativeisaconstantc ,which
0
means that the left hand side of (3.3) holds for i = n. Thus R can be chosen
of degree one, and since that is the minimum possible degree for R, we have
degR=1 and H is constant.
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