CC BY
15
УДК 517.55
Singular Points of Complex Algebraic Hypersurfaces
Irina A. Antipova*
Institute of Space and Information Technologies Siberian Federal University Kirensky, 26, Krasnoyarsk, 660074
Russia
Evgeny N. Mikhalkin1" Avgust K. Tsikh*
Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041
Russia
Received 03.09.2018, received in revised form 22.10.2018, accepted 28.10.2018 We consider a complex hypersurface V given by an algebraic equation in к unknowns, where the set A С Zk of monomial exponents is fixed, and all the coefficients are variable. In other words, we consider a family of hypersurfaces in (C \ 0)k parametrized by its coefficients a = (aa)aeA € CA. We prove that when A generates the lattice Zk as a group, then over the set of regular points a in the A-discriminantal set, the singular points of V admit a rational expression in a.
Keywords: singular point, A-discriminant, logarithmic Gauss map. DOI: 10.17516/1997-1397-2018-11-6-670-679.
Introduction
By the general algebraic hypersurface (or the A-hypersurface) we mean the algebraic set V defined by the equation in k unknowns y = (yi,... ,yk) G (C \ 0)k:
f(yi,...,yk):= £ aayr ...yT =0. (1)
a=(a1,...,ak )eA
Here A c Zk is a fixed finite set while all coefficients aa are treated as independent variables. We assume that the set A generates the lattice Zk as a group. The set of polynomials (1) is identified with the space CA of sequences a = (aa)oeA of dimension N := #A. We can think about V as a family of hypersurfaces Va in (C \ 0)k parametrized by coefficients a G CA.
The aim of the present paper is to obtain explicit formulas for almost all singular points of the hypersurface V. Recall that a point y G V is said to be singular if the polynomial f in (1) and all its partial derivatives f ' 1 ,...,f'k vanish at y. In the classical case, when k =1, the following formulas were given in [1, Ch.1, Th.1.5]: if the equation
f(y) := adyd + ... + aiy + ao = 0 (2)
1 [email protected] ^ [email protected] © Siberian Federal University. All rights reserved
has a unique multiple root y = y(a) = y(a0,..., ad), and its multiplicity equals two, then y is given by rational expressions
Ai(a) A2(a) Ad(a)
y = - - -
Ao(a) Ai(a) '' Ad-i(a)' d A
where Aj = —— are derivatives of the discriminant A(a) of the polynomial f (y) in (2). Analogous daj
formulas for a unique root of multiplicity v > 2 are given in [2], where instead of using the discriminant A, the resultant of f and its derivative f(with respect to y) of order v — 1 is used. Actually, in this case, we keep in mind rational formulas for multiple roots over the discriminant cuspidal strata. These strata were studied in [3].
We prove that almost all singular points y(a) (actually, those that correspond to a belonging to the regular part of the discriminantal set) admit a rational representation (Theorem 3). In the last section we consider an example with comments how the type of a singular point y(a) e V depends of the singular type of a e VA.
As the main tool in our study, we use the Horn-Kapranov parametrization and that it is the inverse of the logarithmic Gauss mapping for the A-discriminantal hypersurface. Remark that the same concepts were used in [2], where we operated with the discriminant of the system, studied before in the paper [4].
1. A-discriminant and the reduced equation
Definition 1 ([1]). Let V° denote the set of all (aa) e CA such that the equation (1) has critical roots y e (C \ 0)k, i.e. roots at which the gradient of f vanishes:
f <»>=f <"=.■■=f <y»=a
The closure V° =: Va in CA is said to be the A-discriminantal set.
In the set VA is a hypersurface in CA, then by the A-discriminant one means an irreducible integral polynomial A a in coefficients a of f e CA which vanishes on Va.
The solution y = y(a) to the equation (1) is (k+1)-homogeneous (it satisfies k+1 homogeneity conditions), and the A-discriminant inherits this property. To see this, we consider the following action on the space CA of polynomials (1). For A = (A0, Ai,..., Ak) e (C \ 0)fc+1 we define it as follows
A : f (yi, ...,yk) ^ Aof (Aiyi,..., Ak yk).
Observe that the set VA is invariant under the A-action. In terms of coefficients (aa) of the polynomial f this action can be written in the following form:
aa ^ AoAa1 ... A'0k aa, a e A.
Here ai,... ,ak are the coordinates of a. In the toric part (C \ 0)A C CA the orbits of this action are the equivalence classes with respect to the (k + 1)-parametric subgroup defined by the immersion
(Ao,Ai,...,Afc) ^ AoAa1 ...Aak, a e A. - 671 -
Its injectivity follows from the fact that A generates Zk. Renumerating the elements of A as a1,..., aN we represent this immersion in the form
where A is the matrix
A =
(aa) = XA
( 1 1 .
«11 «21 .
\ aik a2k .
1
aN i aNk )
(3)
and
AA = (Xa ,...,Xa ) = (XqX?11 ... Xak
,XoX1N1 ...XakNk )
with a? being the columns of this matrix. Remark that we keep the notation A (which was used for the set of exponents a in (1)) for this extended matrix. Thus, an equivalence class can be written in the form Aa • g with the coordinate-wise multiplication. In order to parameterize all equivalence classes we represent them in the form of an m-parametric subgroup
g
where C is an m x N-matrix with m the N x N-matrix
zC, z G (C \ 0)m,
- N — k — 1. Choosing the matrix C in such a way that
(4)
A = (A
is unimodular (with determinant ±1), we conclude that the transform
1 : (X, z) ^ XA
is an automorphism of the complex torus (C \ 0)A. Thus, for such C the m-parametric subgroup g = zC parametrizes all equivalence classes modulo the subgroup AA. Denoting by ca the column of the matrix C indexed by an element a £ A, we arrive at the following reduced equation for (1):
f (y)
J2z°a ya
(5)
aeA
where the coefficients z — z~c ... Zfji , a G A, run over the m-parametric subgroup zC in (C \ 0)A. The discriminantal set of the equation (5) we denote by V'A and call it the reduced discriminantal set. The defining polynomial of V'A is obtained from the A-discriminantal polynomial A a. It is called the reduced discriminant.
By Kapranov's theorem [5] the reduced discriminantal set is birationally equivalent to the
projective space CPm 1. Moreover, there is an explicit formula
= (Bs)B, s G CPm~\
parametrizing VA. Clearly, then we get a parametrization of VA as
a = (aa)aEA = XA • (Bs)BC.
0
2. Parametrization of singular points
The matrix C, extending A in (4) defines a special matrix B, the so called Gale transform of A (see [1, p. 225]). Namely, the inverse of the matrix A can be represented in the following block form:
(!)-1 — (D\B),
where D and B are blocks with k +1 and m columns, respectively. They satisfy the relations
A ■ B =0, A ■ D — Ek+1.
Remark that we can use columns a of A to index the rows for B writing them as ba. With the help of B and D we can formulate the theorem on singular points of the reduced hypersurface (5).
The most convenient reductions of the equation (1) are associated with matrices C which contain k +1 zero columns at that the other m columns form the unit matrix. Such matrices can be used for extension of A to be unimodular if A has k +1 columns, say a0, a1..., ak, for which the columns
10 k 0 a — a , ... ,a — a
form a unimodular k x k-matrix 5. In this case the reduction of (1) is just a fixation of the coefficients: &q0 — — ... — aak = 1. We can use such a reduction when 5 is nondegenerate as well as in the case when 5 is unimodular.
After dividing by ya and denoting aj — a0 by a.j, j — 1,... ,N — 1 we can assume that the reduction has the following form
f (yu ...,yk ) = i+£ yr... y?k+e z*yak+iA ■■■yk
ak+i,k
(7)
i=l
i=l
where the matrix 5 — (a j), i,j — 1,... ,k is nondegenerate. Let b0, b1,..., bk be the first k + 1 rows of the matrix B. In this case we have the following statement.
Theorem 1. The vector-function y(s) — (y1(s),..., yk(s)) with the coordinates
k
*^ n (MF -
(b0, s
where XjV are the entries of the matrix 5~1, parameterizes the set of singular points of the reduced hypersurface (7).
Proof. Firstly, we consider the case when 5 is the unit matrix, i.e. when f is of the type
f (yi, ...yk ) := 1 + yi + ... + yk + £ Ziya^-1 ... yakk+i,k = 0
i=i
(8)
with an associated matrix
A
iii 0i0 0 0 1
000
1 1 1
0 ak+i,i ak+2,i
0 ak+i,2 ak+2,2
1 ak+i,k ak+2,k
1
aN-i,i aN — i ,2
aN-i,k J
(9)
0
Choose the dual matrix
B
( bo \
bi
bk
\ Em J
where
/ к к k \
bo = ( - 1 + I] ak+i,j; -1+1] ak+2,j; • • •; -1 + I] aN-i,j) v j=i j=i j=i 7
bi = ^ — ak+1,1; -ak+2, i; • • •; —«n-M)
bk = ( - «fc+i,fc; -ak+2,k; • • •; -«N-i,k) •
Due to the Horn-Kapranov formula (6) the discriminantal set of the equation (8) is given by the following parametrization
к
-i+ E ak + i,j
г, = s,(b0,s) j=1 (bi, s)-ak+i'1 ... (bk, s)-ak+ik ,i = 1, 2 ...,m, where s = (si,..., sm).
Lemma 1. The vector-function y(s) = (yi(s), • • .yk(s)) with coordinates
f N (bi ,s) , N (bk,s)
yi(s) = 7!-7i-.-iVk(s) -
satisfies the system of equations
(bo ,s)'-
df(yi, ■■■,Vk) dyi
(bo, s)
df (yi,---,yk) dyk
0.
(10)
Proof. Let us substitute y = y(s) into the equation (8) with the coefficients z = = (Bs)B. We get the following expression
(bi, s) (bk, s)
1 + lA-v + ... + TT^sL +
(bo, s)
(bo ,s)
m /
£ s
i=i \
+ > , I si(bo, s) (bi, s)-ak+i-1 ... (bk, s)-ak+ik
( (bk1s)_ \
\(bo,s)) . y (bo, s) J
a-k+i,k
(bo, s) + (bi, s) + ... + (bk, s) + si + ... + sn
(bo, s)
(bo, s)
The last sum vanishes, since
(b0, s) + (bi, s) + ... + (bk, s) = (si + ... + Sm).
Recall that the sum of all rows of the matrix B is equal to zero, and B consists of rows b0,... ,bk supplemented by the unit m x m-matrix. So, y(s) annihilates f (y) when z = (Bs)B. Similarly for the derivatives, one has as follows
%(yi(s),...yk(s)) 1 , 1 m 1 Л, v . m ) n
— =1 + (bos ak+ij si = (js) [(b>,s) + 2. ak+ijsiJ =0. J x ' i=i x J ' i=i
dyj
The last expression vanishes due to the property of vectors b?. So, the proof of Lemma 1 is completed. □
In order to continue the proof of Theorem 1 let us turn to the equation (7). We introduce the monomial change
xi = vT1 ... vtik, i = 1, 2,..., k,
which can be rewritten in the matrix form as x = yS. Since S is nondegenerate, one has
y = x
Let us write the matrix A = (aij) in the block form A = (S, S'). Then after the substitution
y = xs 1 in (7) we get
1 + Xi + J2 zi (xS~ S )i = 0,
(11)
where (xs 1s )i is the i-th coordinate of the vector xs 1s . The exponents in equation (11) form the k x W-matrix (Ek, S-1S'). This matrix supplemented by the row of units looks like (9) where the block S' is changed by S-1S':
.4:
/11 ... 1
0
1
1
Ek
S-1S'
0
The computation shows that the dual matrix to A is the matrix
( bo \ bi
B
bk
\ Em )
Further applying Lemma 1 we complete the proof of Theorem 1.
□
3. Rational expression for singular points
As it follows from the definition of the A-discriminantal set, the singular points of the hyper-surface which we consider coincide with the restrictions of solutions to the equation (1) on the A-discriminantal set, i.e. with . For the reduced equation (7) the singular points y(z)
are given by
v(z(s)) = y((Bs)B).
However, according to Kapranov's theorem [5] the parametrization z = (Bs)B is the inverse of the logarithmic Gauss map
Y : ^ CPm-1
of a reduced A-discriminantal set V'A. At the regular points z e regV'A this mapping can be written explicitly (see [6])
Y : z ^ (zi(A%i : ... : zm(A')Zm) = (si : ... : sm),
where for simplicity we write A' instead of AA. Therefore, by Theorem 1 we get the following statement.
Theorem 2. The singular points of the reduced hypersurface (7) over the set reg V'A admit in global coordinates z the following radical representation:
Vi (z) = II
f(bv(z)) ' Uo,Y (z)).
Xjv
j = 1, 2,
(12)
where XjV are the entries of the matrix 5 1. Now we can formulate the main result.
Theorem 3. Let the set A in (1) generate Zk as a group. The singular points y(a) of the hypersurface (1) over the set reg V a admit a rational representation.
, b
Proof. We consider an arbitrary reduction of the type (7) with fixed coefficients aaj0 =aa3k = 1. Let Bj' be the submatrix of the dual matrix B consisting of rows bajl, Then by Theorem 2 the singular points of the reduced hypersurface can be found in the following way:
ajk •
V(z)
({Bj ' ,Y(z))\ [i ,Y(z)) )
Consider all subsets J = {jo,j1,---,jk} C {1,..., N} for which the corresponding matrices 5J are nondegenerate. Then there exist integer numbers qJ such that
y^gj 5 j = Ek.
Consequently, we have
F Sj
V(a) = y k = V J
n / {Bj' ,7(z)A JJlqjSJ ^ ( {Bj',7(z)A V i ,Y(Z))J i ,Y(z)) J '
The last term is a rational expression in variables z. Since by Kapranov's theorem 7 is a birational map we get rationality of y(a) in variables a. □
4. Example
Let us consider the following polynomial equation
aoo + aioyi + aoiy2 + a3iyfy2 + a^^yl y3 = 0.
It is associated with the matrix
11111 A = I 0 10 3 6 0 0 113
which has the right annulator
B
( 3 8 \
-3 -6 -1 -3 10 01
k
V
The reduced equation looks as follows:
f = 1 + y 1 + y2 + z iy 3 y2 + Z2y 6 y| = 0. (13)
According to (6) the parametrization of the reduced A-discriminantal set V' = V'A for f is
zi = si(3si + 8s2)3(-3si - 6s2)-3(-si - 3s2)-1 = (3 +(3s+3(is)+ 3s), (14)
s(3 + 8s)8
z2 = s2(3si + 8s2)8(-3si - 6s2)-6(-si - 3s2)-3 = - (3 + 6s)6(i + 3s)3 , (15)
where s := — is an affine coordinate in CP1. After elimination of the parameter s in the si
system (14)-(15) we get the reduced A-discriminant A' = A'a:
A' = -262144z3 + 331776z1z3 + 331776z3zf - 61236z6z2 - 61236z2z3 - 19683z| - 398034z4z2+
+59049z7z2 + 19683z3z3 + 59049z^z2 - 19683z8 + 19683z?.
The matrix S for the equation (12) is the unit matrix, therefore by (12) we get the following formulas for singular points:
-3zi(A')Zl - 6z2(A')Z2 -zi(A')Zl - 3z2(A')Z2
Vi = —TTTR-—77Ä-> V2 =
3zi(A')zi +8z2(A')z2 ' * 3zi (A')zi + 8z2(A')z2
The derivatives
dz1 = (3 + 8s)2(4s + 1)2 dz2 = (3 + 8s)7(4s + 1)2 ~ds = - 9(1 + 2s)4(1 + 3s)2 , ~d! = - 243(1 + 2s)7(1 +3s)4
13
vanish when s = -— and s = - —. It means that VA has two singular points
48
( 1 \ (32 1024 \ ( 3 \ , s
< - ^ = (27, and < - s) = (0,0).
Elimination of the parameter s in the system (14)-(15) leads us to the A-discriminant
A' = AA:
A' = -262144z| + 331776z1z| + 331776z3z| - 61236z6z2 - 61236z2z| - 19683z4 - 398034z4z^ + +59049z7z2 + 19683z3z3 + 59049zfz^ - 19683z8 + 19683z?.
Consider the Taylor decomposition of A' at the point z(-4), i.e. by powers of p = z1 - and 1024 27
q = z2 -
729
A, 68719476736 3 536870912 2 4194304 2 3 486539264 4 , N
A' = -p3--p2 q +--pq - 32768q3 +--p4 + r(p, q),
1968^ 243 9 81 yl,Hh
where r(p, q) is a sum of monomials of degree ^ 4 except the monomial p4. Here the initial homogeneous cubic form is a cube power of an affine polynomial
32768 .
(115221 - 243z2 - 1024)3.
14348907
32
Consequently, in coordinates m = 1152z4 — 243z2 — 1024 and l = z4 — — the discriminant has the form
A' = am3 + bl4 + ..., a = 0, b = 0,
It means that z = (32/27,1024/729) is a cuspidal point of the type (4, 3) for the discriminant A'.
Now we have to study singular types of singular points of the complex curve (13) which are given by Theorem 2:
—3 — 6s —1 — 3s
yi(s) = TW, y2(s) = T+87.
At the singular points y(s) we have the following expression for the Hessian of f:
dff ( 92f )2 = (3 + 8s)2(1 + 4s)2 dy2 dy2 — Idy^J = — (1 + 2s)2(1 + 3s)2 .
Therefore, only y( — 4) = ( — |, — 4) is not a Morse point.
Consider the expression of the polynomial (13) at the point y( — 4):
f = —12(y2 + 1/4)2 — 4(yi + 3/2)(y2 + 1/4) — 1(yi + 3/2)2 + 16(y2 + 1/4)3+
+48(yi + 3/2)(y2 + 1/4)2 + f(yi + 3/2)2(y2 + 1/4) + 32/27(yi +3/2)3 — 448 (yi +3/2)3(y2 + 1/4) —
20 320
—64(yi + 3/2)(y2 + 1/4)3 — 80(yi +3/2)2(y2 + 1/4)2 — —(yi + 3/2)4 + -^(yi + 3/2)2(y2 + 1/4)3+
640 16 80
+—^(yi + 3/2)3(y2 + 1/4)2 + gj (yi + 3/2)5 + -(yi + 3/2)4(y2 + 1/4) + ....
After the change of variables
we get
3 u v 1 8u v
yi +— =---1--, y2 +— =---
y 2 15 30' y 4 45 180
f = — u2 +--v4 + r(u, v),
J 3 8201250 v '
where r(u,v) consists of monomials of weighted degree > 4 with respect to the weight (2,1). This means that y( — i) is a self-intersection point for the curve (13) with a common tangent line.
The first two authors were supported by the grant of Ministry of Education and Science of the Russian Federation (no. 1.2604-2017/PCh). The third author was supported by the grant of the Russian Federation Government for scientific research under the supervision of leading scientists at Siberian Federal University (no. 14.Y26.31.0006) and grant RFBR, no. 18-51-41011 Uzb
References
[1] I.Gelfand, M.Kapranov, A.Zelevinsky, Discriminants, resultants and multidimensional determinants, Birkhauser: Boston, 1994.
[2] I.A.Antipova, E.N.Mikhalkin, A.K.Tsikh, Rational expressions for multiple roots of algebraic equations, Matem sb, 209(2018), no 10, 3-30 (in Russian). Engl. transl. in Sb. Math., 209(2018), in press.
[3] E.N.Mikhalkin, A.K.Tsikh, Singular strata of cuspidal type for the classical discriminant, Sb. Math., 206(2015), no 2, 282-310.
[4] I.A.Antipova, A.K.Tsikh, The discriminant locus of a sistem of n Laurent polynomials in n variables, Izv. Math., 76(2012), no 5, 881-906.
[5] M.M.Kapranov, A characterization of A-discriminantal hypersurfaces in terms of the logarithmic Gauss map, Math. Ann, 290(1991), 277-285.
[6] G.Mikhalkin, Real algebraic curves, the moment map and amoebas, Ann. of Math., 151(2000), no. 2, 309-326.
Сингулярные точки комплексных алгебраических гиперповерхностей
Ирина А. Антипова
Институт космических и информационных технологий Сибирский федеральный университет Киренского, 26, Красноярск, 660074
Россия
Евгений Н. Михалкин Август К. Цих
Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, 660041
Россия
Рассматривается комплексная гиперповерхность V, заданная алгебраическим уравнением с к неизвестными и с переменными коэффициентами, причем множество A С Zk показателей мономов уравнения произвольное, но фиксированное. Таким образом, мы рассматриваем семейство гиперповерхностей, параметризованных наборами коэффициентов a = (aa)aeA € CA. Доказывается, что если A порождает решетку Zk как группу, то над множеством регулярных точек A-дискриминантного множества сингулярные точки гиперповерхности V рационально выражаются через коэффициенты a.
Ключевые слова: особая точка, A-дискриминант, логарифмическое отображение Гаусса.
