Choosing roots of polynomials smoothly

26 Pages·0.28 MB·English

by  Dmitri Alekseevsky

Toappearin:
IsraelJ.Math.
CHOOSING ROOTS OF POLYNOMIALS SMOOTHLY
8 Dmitri Alekseevsky
9 Andreas Kriegl
9 Mark Losik
1 Peter W. Michor

n
a Erwin Schr¨odinger International Institute
J of Mathematical Physics, Wien, Austria

7

February 1, 2008
]
A
C Abstract. Weclarifythequestionwhether forasmoothcurveofpolynomialsone
. can choose the roots smoothly and related questions. Applications to perturbation
h theoryofoperatorsaregiven.
t
a
m
[
Table of contents

1
v 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
6 2. Choosing differentiable square and cubic roots . . . . . . . . . . . . . 2
2 3. Choosing local roots of real polynomials smoothly . . . . . . . . . . . 7
0 4. Choosing global roots of polynomials differentiably . . . . . . . . . . . 11
1
5. The real analytic case . . . . . . . . . . . . . . . . . . . . . . . 15
0
8 6. Choosing roots of complex polynomials . . . . . . . . . . . . . . . . 16
9 7. Choosing eigenvalues and eigenvectors of matrices smoothly . . . . . . . 18
/
h
1. Introduction
t
a
m We consider the following problem. Let
:
v (1) P(t)=xn a (t)xn−1+ +( 1)na (t)
1 n
i − ··· −
X
be a polynomial with all roots real, smoothly parameterizedby t near 0 in R. Can
r
a wefindnsmoothfunctionsx (t),...,x (t)oftheparametertdefinednear0,which
1 n
are the roots of P(t) for each t? We can reduce the problem to a = 0, replacing
1
the variable x by with the variabley =x a (t)/n. We will say that the curve (1)
1
−
is smoothly solvable near t=0 if such smooth roots x (t) exist.
i
1991Mathematics Subject Classification. 26C10,47A55.
Keywords and phrases. smoothrootsofpolynomials.
Supportedby‘FondszurFo¨rderungderwissenschaftlichenForschung,ProjektP10037PHY’.
TypesetbyAMS-TEX
1
2 D.V. ALEKSEEVSKY, A. KRIEGL, M. LOSIK, P.W. MICHOR
We describe an algorithm which in the smooth and in the holomorphic case
sometimes allows to solve this problem. The main results are: If all roots are real
then they can always be chosen differentiable, but in general not C1 for degree
n 3; and in degree 2 they can be chosen twice differentiable but in general not
≥
C2. If they are arranged in increasing order, they depend continuously on the
coefficientsofthepolynomial,andifmoreovernotwoofthemmeetofinfiniteorder
intheparameter,thentheycanbechosensmoothly. Wealsoapplytheseresultsto
obtain a smooth 1-parameter perturbation theorem for selfadjoint operators with
compactresolventunderthe condition,thatno pairofeigenvaluesmeets ofinfinite
order.
We thank C. Fefferman who found a mistake in a first version of 2.4, to M.
and Th. Hoffmann-Ostenhof for their interest and hints, and to Jerry Kazdan for
arranging [15].
2. Choosing differentiable square and cubic roots
2.1. Proposition. The case n = 2. Let P(t)(x) = x2 f(t) for a function f
−
defined on an open interval, such that f(t) 0 for all t.
≥
Iff is smoothandisnowhereflat ofinfiniteorder, then smoothsolutions xexist.
If f is C2 then C1-solutions exist.
If f is C4 then twice differentiable solutions exist.
Proof. Suppose that f is smooth. If f(t ) > 0 then we have obvious local smooth
0
solutions f(t). If f(t ) = 0 we have to find a smooth function x such that
0
±
f = x2, a smooth square root of f. If f is not flat at t then the first nonzero
p 0
derivative at t has even order 2m and is positive, and f(t) = (t t )2mf (t),
0 0 2m
where f (t) := 1 (1−r)2m−1f(2m)(t +r(t t ))dr gives a smooth−function and
2m 0 (2m−1)! 0 − 0
f (t ) = 1 f(2m)(t ) > 0. Then x(t) := (t t )m f (t) is a local smooth
2m 0 (2m)! R 0 − 0 2m
solution. One can piece together these local solutions, changing sign at all points
p
where the first non-vanishing derivative of f is of order 2m with m odd. These
points are discrete.
Let us consider now a function f 0 of class C2. We claim that then x2 =f(t)
≥
admits a C1-rootx(t), globallyint. We considera fixedt . If f(t )>0then there
0 0
islocallyevenaC2-solutionx (t)= f(t). Iff(t )=0thenf(t)=(t t )2h(t)
± 0 0
± −
where h 0 is continuous and C2 off t with h(t ) = 1f′′(t ). If h(t ) > 0 then
≥ p0 0 2 0 0
x (t)= (t t ) h(t) is C2 off t , and
± 0 0
± −
p
x (t) x (t )
x′ (t )= lim ± − ± 0 = lim h(t)= h(t )= 1f′′(t ).
± 0 t→t0 t t0 t→t0± ± 0 ± 2 0
− p p q
If h(t ) = 0 then we choose x (t ) = 0, and any choice of the roots is then
0 ± 0
differentiable at t with derivative 0, by the same calculation.
0
One can piece together these local solutions: At zeros t of f where f′′(t) > 0
we have to pass through 0, but where f′′(t) = 0 the choice of the root does not
matter. The set t : f(t) = f′′(t) = 0 is closed, so its complement is a union of
{ }
open intervals. Choose a point in each of these intervals where f(t) > 0 and start
therewiththepositiveroot,changingsignsatpointswheref(t)=0=f′′(t): these
6
CHOOSING ROOTS OF POLYNOMIALS SMOOTHLY 3
points do not accumulate in the intervals. Then we get a differentiable choice of
a root x(t) on each of this open intervals which extends to a global differentiable
root which is 0 on t:f(t)=f′′(t)=0 .
{ }
Note that we have
f′(t) if f(t)>0
x′(t)= 2x(t)
( f′′(t)/2 if f(t)=0
±
In points t with f(t ) > 0 the soplution x is C2; locally around points t with
0 0 0
f(t )=0 and f′′(t )>0 the root x is C1 since for t=t near t we have f(t)>0
0 0 0 0
6
and f′(t)=0, so by l’Hospital we get
6
f′(t)2 2f′(t)f′′(t) f′′(t )
lim x′(t)2 = lim = lim = 0 =x′(t )2,
t→t0 t→t0 4f(t) t→t0 4f′(t) 2 0
andsincethe choiceofsigns wascoherent,x′ iscontinuousatt ; iff′′(t )=0then
0 0
x′(t )=0andx′(t) 0fort t forbothexpressions,bylemma2.2below. Thus
0 0
→ →
x is C1.
If moreover f 0 is C4, then the solution x from above may be modified
≥
to be twice differentiable. Near points t with f(t ) > 0 any continuous root
0 0
t x (t) = f(t) is even C4. Near points t with f(t ) = f′(t ) = 0 we have
± 0 0 0
7→ ±
f(t)=(t t )2h(t) where h 0 is C2. We may choose a C1-rootz with z2 =h by
− 0 p ≥
the argumentsabove,andthenx(t):=(t t )z(t) istwice differentiable att since
0 0
−
we have
x′(t) x′(t ) z(t)+(t t )z′(t) z(t )
0 0 0
− = − −
t t t t
0 0
− −
z(t) z(t ) f(4)(t )
=z′(t)+ − 0 2z′(t )= 0 .
0
t t → ± 4!
0 r
−
If f(t ) = f′′(t ) = f(4)(t ) = 0 then any C1 choice of the roots is twice differen-
0 0 0
tiable at t , in particular x(t)= t t z(t).
0 0
| − |
Now we can piece together this solutions similarly as above. Let y be a global
C1-rootoff,chosenasabovechangingsignonlyatpointstwithf(t)=0<f′′(0).
We put x(t)=ε(t)y(t), where ε(t) 1 will be chosen later. The set t:f(t)=
∈{± } {
f′′(t) = f(4)(t) = 0 has a countable union of open intervals as complements. In
}
each of these intervals choose a point t with f(t ) > 0, near which y is C4. Now
0 0
let ε(t ) = 1, and let ε change sign exactly at points with f(t) = f′′(t) = 0 but
0
f(4)(t) > 0. These points do not accumulate inside the interval. Then x is twice
differentiable. (cid:3)
2.2. Lemma. Let f 0 be a C2-function with f(t ) = 0, then for all t R we
0
≥ ∈
have
(1) f′(t)2 2f(t) max f′′(t +r(t t )):0 r 2 .
0 0
≤ { − ≤ ≤ }
Proof. If f(t)=0 then f′(t)=0 so (1) holds. We use the Taylor formula
1
(2) f(t+s)=f(t)+f′(t)s+ (1 r)f′′(t+rs)dr s2
−
Z0
4 D.V. ALEKSEEVSKY, A. KRIEGL, M. LOSIK, P.W. MICHOR
In particular we get (replacing t by t and then t +s by t)
0 0
1
(3) f(t)=0+0+ (1 r)f′′(t +r(t t ))dr (t t )2
0 0 0
− − −
Z0
(t t )2
− 0 max f′′(t +r(t t )):0 r 2
0 0
≤ 2 { − ≤ ≤ }
Now in (2) we replace s by εs (where ε=sign(f′(t))) to obtain
−
1
(4) 0 f(t εs)=f(t) f′(t)s+ (1 r)f′′(t εrs)dr s2
≤ − −| | − −
Z0
Let us assume t t and then choose (using (3))
0
≥
2f(t)
s(t):= t t .
smax f′′(t0+r(t t0)):0 r 2 ≤ − 0
{ − ≤ ≤ }
Note that we may assume f(t) > 0, then s(t) is well defined and s(t) > 0. This
choice of s in (4) gives
1 s(t)2
f′(t) f(t)+ max f′′(t εrs(t)):0 r 1
| |≤ s(t) 2 { − ≤ ≤ }
(cid:18) (cid:19)
1 s(t)2
f(t)+ max f′′(t r(t t )): 1 r 1
0
≤ s(t) 2 { − − − ≤ ≤ }
(cid:18) (cid:19)
2f(t)
= = 2f(t) max f′′(t +r(t t )):0 r 2
0 0
s(t) { − ≤ ≤ }
p
which proves (1) for t t . Since the assertion is symmetric it then holds for all
0
t. (cid:3) ≥
2.3. Examples. Iff 0isonlyC1,thentheremaynotexistadifferentiableroot
of x2 =f(t), as the fol≥lowing example shows: x2 =f(t):=t2sin2(logt) is C1, but
tsin(logt) is not differentiable at 0.
±
If f 0 is twice differentiable there may not exist a C1-root: x2 = f(t) =
≥
t4sin2(1) is twice differentiable, but t2sin(1) is differentiable, but not C1.
t ± t
If f 0 is only C3, then there may not exist a twice differentiable root of
x2 = f(≥t), as the following example shows: x2 = f(t) := t4sin2(logt) is C3, but
t2sin(logt) is only C1 and not twice differentiable.
±
2.4. Example. If f(t) 0 is smooth but flat at 0, in general our problem has
≥
no C2-root as the following example shows, which is an application of the general
curve lemma 4.2.15in [5]: Let h:R [0,1]be smooth with h(t)=1 for t 0 and
→ ≥
h(t)=0 for t 1. Then the function
≤−
∞
2n 1
f(t):= h (t t ). (t t )2+ , where
n − n 2n − n 4n
n=1 (cid:18) (cid:19)
X
1 1
h (t):=h n2 +t .h n2 t and
n n.2n+1 n.2n+1 −
(cid:18) (cid:18) (cid:19)(cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
n−1
2 2 1 1
t := + + + ,
n k2 k.2k+1 n2 n.2n+1
k=1(cid:18) (cid:19)
X
CHOOSING ROOTS OF POLYNOMIALS SMOOTHLY 5
is 0 and is smooth: the sum consists of at most one summand for each t, and
≥
the derivatives of the summands converge uniformly to 0: Note that h (t)=1 for
n
t 1 and h (t) = 0 for t 1 + 1 hence h (t t ) = 0 only for
| | ≤ n.2n+1 n | | ≥ n.2n+1 n2 n − n 6
r < t < r , where r := n−1 2 + 2 . Let c (s) := 2ns2+ 1 0 and
n n+1 n k=1 k2 k.2k+1 n 2n 4n ≥
H :=sup h(i)(t) :t R . Then
i {| | ∈ } P (cid:0) (cid:1)
1 1
n2sup (h c )(k)(t) :t R =n2sup (h c )(k)(t) : t +
{| n· n | ∈ } {| n· n | | |≤ n.2n+1 n2}
k
k 1 1
n2 n2iH sup c(k−i)(t) : t +
≤ i i {| n | | |≤ n.2n+1 n2}
i=0(cid:18) (cid:19)
X
k
k
n2i+2H sup c(j)(t) : t 2, j k
≤ i i {| n | | |≤ ≤ }
(cid:18)i=0(cid:18) (cid:19) (cid:19)
X
andsince c israpidlydecreasinginC∞(R,R)(i.e. p(n)c :n N is boundedin
n n
C∞(R,R) for each polynomial p) the right side of t{he inequality∈ab}ove is bounded
with respect to n N and hence the series h ( t )c ( t ) converges
uniformly in each d∈erivative, and thus representns ann ele−mnentnof f− n C∞(R,R).
P ∈
Moreover we have
1 2n
f(t )= , f′(t )=0, f′′(t )= .
n 4n n n 2n−1
Let us assume that f(t) = g(t)2 for t near sup t < , where g is twice differen-
n n ∞
tiable. Then
f′ =2gg′
f′′ =2gg′′+2(g′)2
2ff′′ =4g3g′′+(f′)2
2f(t )f′′(t )=4g(t )3g′′(t )+f′(t )2
n n n n n
thus g′′(t ) = 2n, so g cannot be C2, and g′ cannot satisfy a local Lipschitz
n
±
condition near limt . Another similar example can be found in 7.4 below.
n
According to [15], some results of this section are contained in Frank Warners
dissertation (around 1963, unpublished): Non-negative smooth functions have C1
square roots whose second derivatives exist everywhere. If all zeros are of finite
order there are smooth square roots. However, there are examples not possessing
a C2 square root. Here is one:
f(t)=sin2(1/t)e−1/t+e−2/t for t>0, f(t)=0 for t 0.
≤
This is a sum of two non-negative C∞ functions each of which has a C∞ square
root. But the second derivative of the square root of f is not continuous at the
origin.
In [6] Glaeser proved that a non-negative C2-function on an open subset of Rn
which vanishes of second order has a C1 positive square root. A smooth function
f 0 is constructedwhich is flat at 0 such that the positive square rootis not C2.
≥
In [4] Dieudonn´e gave shorter proofs of Glaeser’s results.
6 D.V. ALEKSEEVSKY, A. KRIEGL, M. LOSIK, P.W. MICHOR
2.5. Example. The case n 3. We will construct a polynomial
≥
P(t)=x3+a (t)x a (t)
2 3
−
with smooth coefficients a , a with all roots real which does not admit C1-roots.
2 3
Multiplying with other polynomials one then gets polynomials of all orders n 3
≥
which do not admit C1-roots.
Suppose that P admits C1-roots x , x , x . Then we have
1 2 3
0=x +x +x
1 2 3
a =x x +x x +x x
2 1 2 2 3 3 1
a =x x x
3 1 2 3
0=x˙ +x˙ +x˙
1 2 3
a˙ =x˙ x +x x˙ +x˙ x +x x˙ +x˙ x +x x˙
2 1 2 1 2 2 3 2 3 3 1 3 1
a˙ =x˙ x x +x x˙ x +x x x˙
3 1 2 3 1 2 3 1 2 3
We solve the linear system formed by the last three equations and get
a˙ a˙ x
3 2 1
x˙ = −
1
(x x )(x x )
2 1 3 1
− −
a˙ a˙ x
3 2 2
x˙ = −
2
(x x )(x x )
3 2 1 2
− −
a˙ a˙ x
3 2 3
x˙ = − .
3
(x x )(x x )
1 3 2 3
− −
We consider the continuous function
a˙3+a˙2a˙ a a˙3a
b :=x˙ x˙ x˙ = 3 2 3 2− 2 3.
3 1 2 3 4a3+27a2
2 3
For smooth functions f and ε with ε2 1 we let
≤
u:= 12a :=f2
2
−
v :=108a :=εf3.
3
thenallrootsofP arerealsincea 0and432(4a3+27a2)=v2 u3 =f6(ε2 1)
2 ≤ 2 3 − − ≤
0. We get then
4v˙3 27uu˙2v˙ +27u˙3v
11664b = −
3 v2 u3
−
f3ε˙3+9f2f˙ε˙2ε+27ff˙2ε˙(ε2 1)+27f˙3ε(ε2 3)
=4 − − .
ε2 1
−
Now we choose
∞
n
f(t):= h (t t ). (t t ) ,
n − n 2n − n
nX=1 (cid:16) (cid:17)
∞
1
ε(t):=1 h (t t ). ,
− n − n 8n
n=1
X
CHOOSING ROOTS OF POLYNOMIALS SMOOTHLY 7
where h and t are as in the beginning of 2.4. Then f(t ) = 0, f˙(t ) = n , and
n n n n 2n
ε(t )= 1 , hence
n 8n
f˙(t )3ε(t )(ε(t )2 3)
108b (t )= n n n − n3
3 n ε(t )2 1 ∼ →∞
n
−
is unboundedonthe convergentsequencet . So the rootscannotbe chosenlocally
n
Lipschitz, thus not C1.
3. Choosing local roots of real polynomials smoothly
3.1. Preliminaries. We recall some known facts on polynomials with real coeffi-
cients. Let
P(x)=xn a xn−1+ +( 1)na
1 n
− ··· −
be a polynomial with real coefficients a ,...,a and roots x ,...,x C. It
1 n 1 n
∈
is known that a = σ (x ,...,x ), where σ (i = 1,...,n) are the elementary
i i 1 n i
symmetric functions in n variables:
σ (x ,...,x )= x ...x
i 1 n j1 ji
1≤j1<X···<ji≤n
Denotebys theNewtonpolynomials n xi whicharerelatedtothe elementary
i j=1 j
symmetric function by
P
(1) s s σ +s σ + +( 1)k−1s σ +( 1)kkσ =0 (k n)
k k−1 1 k−2 2 1 k−1 k
− ··· − − ≤
Thecorrespondingmappingsarerelatedbyapolynomialdiffeomorphismψn,given
by (1):
σn :=(σ ,...,σ ):Rn Rn
1 n
→
sn :=(s ,...,s ):Rn Rn
1 n
→
sn :=ψn σn
◦
Note that the Jacobian (the determinant of the derivative) of sn is n! times the
Vandermondedeterminant: det(dsn(x))=n! (x x )=:n! Van(x),andeven
i>j i− j
the derivative itself d(sn)(x) equals the Vandermonde matrix up to factors i in the
Q
i-th row. We also have det(d(ψn)(x)) = ( 1)n(n+3)/2n! = ( 1)n(n−1)/2n!, and
− −
consequently det(dσn(x))= (x x ). We consider the so-called Bezoutiant
i>j j − i
Q s s ... s
0 1 n−1
s s ... s
1 2 n
B := . . . .
. . .
. . .
s s ... s 
 n−1 n 2n−2
 
Let B be the minor formed by the first k rows and columns of B. From
k
1 1 ... 1 1 x1 ... x1k−1
x x ... x 1 x ... xk−1
Bk(x)= ...1 ...2 ...n ·... ...2 2... 
x1k−1 x2k−1 ... xnk−1 1 xn ... xnk−1
   
8 D.V. ALEKSEEVSKY, A. KRIEGL, M. LOSIK, P.W. MICHOR
it follows that
(2) ∆ (x):=det(B (x))= (x x )2...(x x )2...(x x )2,
k k i1− i2 i1− in ik−1− ik
i1<i2X<···<ik
since for n k-matrices A one has det(AA⊤) = det(A )2, where
× i1<···<ik i1,...,ik
A is the minor of A with indicated rows. Since the ∆ are symmetric we
i1,...,ik P k
have ∆ =∆˜ σn for unique polynomials ∆˜ and similarly we shall use B˜.
k k k
◦
3.2. Theorem. (Sylvester’s version of Sturm’s theorem, see [14], [10]) The roots
of P are all real if and only if the symmetric (n n) matrix B˜(P) is positive
×
semidefinite; then ∆˜ (P) := ∆˜ (a ,...,a ) 0 for 1 k n. The rank of B
k k 1 n
≥ ≤ ≤
equals the number of distinct roots of P and its signature equals the number of
distinct real roots.
3.3. Proposition. Let now P be a smooth curve of polynomials
P(t)(x)=xn a (t)xn−1+ +( 1)na (t)
1 n
− ··· −
with all roots real, and distinct for t=0. Then P is smoothly solvable near 0.
This is also truein thereal analytic case and for higher dimensional parameters,
and in the holomorphic case for complex roots.
Proof. The derivative d P(0)(x) does not vanish at any root, since they are dis-
dx
tinct. Thus by the implicit function theorem we have local smooth solutions x(t)
of P(t,x)=P(t)(x)=0. (cid:3)
3.4. Splitting Lemma. Let P be a polynomial
0
P (x)=xn a xn−1+ +( 1)na .
0 1 n
− ··· −
If P =P P , where P and P are polynomials with no common root. Then for P
0 1 2 1 2
near P we·have P =P (P)P˙ (P) for real analytic mappings of monic polynomials
0 1 2
P P (P) and P P (P), defined for P near P , with the given initial values.
1 2 0
7→ 7→
Proof. Let the polynomial P be represented as the product
0
P =P .P =(xp b xp−1+ +( 1)pb )(xq c xq−1+ +( 1)qc ),
0 1 2 1 p 1 q
− ··· − − ··· −
letx for i=1,...,n be the rootsofP ,orderedinsuchawaythatfor i=1,...,p
i 0
we get the roots of P , and for i = p+1,...,p+q = n we get those of P . Then
1 2
(a ,...,a )=φp,q(b ,...,b ,c ,...,c ) for a polynomial mapping φp,q and we get
1 n 1 p 1 q
σn =φp,q (σp σq),
◦ ×
det(dσn)=det(dφp,q(b,c))det(dσp)det(dσq).
From 3.1 we conclude
(x x )=det(dφp,q(b,c)) (x x ) (x x )
i j i j i j
− − −
1≤i<j≤n 1≤i<j≤p p+1≤i<j≤n
Y Y Y
which in turn implies
det(dφp,q(b,c))= (x x )=0
i j
− 6
1≤i≤p<j≤n
Y
so that φp,q is a real analytic diffeomorphism near (b,c). (cid:3)
CHOOSING ROOTS OF POLYNOMIALS SMOOTHLY 9
3.5. For a continuous function f defined near 0 in R let the multiplicity or order
of flatness m(f) at 0 be the supremum of all integer p such that f(t) = tpg(t)
near 0 for a continuous function g. If f is Cn and m(f) < n then f(t) = tm(f)g(t)
where now g is Cn−m(f) and g(0) = 0. If f is a continuous function on the space
6
of polynomials, then for a fixed continuous curve P of polynomials we will denote
by m(f) the multiplicity at 0 of t f(P(t)).
7→
The splitting lemma 3.4 shows that for the problem of smooth solvability it is
enough to assume that all roots of P(0) are equal.
Proposition. Suppose that the smooth curve of polynomials
P(t)(x)=xn+a (t)xn−2 +( 1)na (t)
2 n
−··· −
is smoothly solvable with smooth roots t x (t), and that all roots of P(0) are
i
7→
equal. Then for (k =2,...,n)
m(∆˜ ) k(k 1) min m(x )
k i
≥ − 1≤i≤n
m(a ) k min m(x )
k i
≥ 1≤i≤n
This result also holds in the real analytic case and in the smooth case.
Proof. This follows from 3.1.(2) for ∆ , and from a (t)=σ (x (t),...,x (t)). (cid:3)
k k k 1 n
3.6. Lemma. Let P be a polynomial of degree n with all roots real. If a =a =0
1 2
then all roots of P are equal to zero.
Proof. From 3.1.(1) we have x2 =s (x)=σ2(x) 2σ (x)=a2 2a =0. (cid:3)
i 2 1 − 2 1− 2
3.7. Multiplicity lemma. PConsider a smooth curve of polynomials
P(t)(x)=xn+a (t)xn−2 +( 1)na (t)
2 n
−··· −
with all roots real. Then, for integers r, the following conditions are equivalent:
(1) m(a ) kr for all 2 k n.
k
≥ ≤ ≤
(2) m(∆˜ ) k(k 1)r for all 2 k n.
k
≥ − ≤ ≤
(3) m(a ) 2r.
2
≥
Proof. We only have to treat r >0.
(1) implies (2): From 3.1.(1) we have m(s˜ ) rk, and from the definition of
k
≥
∆˜ =det(B˜ ) we get (2).
k k
(2) implies (3) since ∆˜ = 2na .
2 2
−
(3) implies (1): From a (0) = 0 and lemma 3.6 it follows that all roots of the
2
polynomial P(0) are equal to zero and, then, a (0) = = a (0) = 0. There-
3 n
···
fore, m(a ),...,m(a ) 1. Under these conditions, we have a (t) = t2ra (t)
3 n 2 2,2r
≥
and ak(t) = tmkak,mk(t) for k = 3,...,n, where the mk are positive integers and
a ,a ,...,a are smoothfunctions, andwhere we may assume that either
2,2r 3,m3 n,mn
m =m(a )< or m kr.
k k k
∞ ≥
Suppose now indirectly that for some k > 2 we have m = m(a ) < kr. Then
k k
we put
m m
3 n
m:=min(r, ,..., )<r.
3 n
10 D.V. ALEKSEEVSKY, A. KRIEGL, M. LOSIK, P.W. MICHOR
We consider the following continuous curve of polynomials for t 0:
≥
P¯ (t)(x):=xn+a (t)t2r−2mxn−2
m 2,2r
a (t)tm3−3mxn−3+ +( 1)na (t)tmn−nm.
− 3,m3 ··· − n,mn
Ifx ,...,x aretherealrootsofP(t)thent−mx ,...,t−mx aretherootsofP¯ (t),
1 n 1 n m
for t > 0. So for t > 0, P¯ (t) is a family of polynomials with all roots real. Since
m
by theorem 3.2 the set of polynomials with all roots real is closed, P¯ (0) is also a
m
polynomial with all roots real.
By lemma 3.6 all roots of the polynomial P¯ (0) are equal to zero, and for
m
those k with m = km we have a (0) = 0 and, therefore, m(a ) > m , a
k k,mk k k
contradiction. (cid:3)
3.8. Algorithm. Consider a smooth curve of polynomials
P(t)(x)=xn a (t)xn−1+a (t)xn−2 +( 1)na (t)
1 2 n
− −··· −
with all roots real. The algorithm has the following steps:
(1) If all roots of P(0) are pairwise different, P is smoothly solvable for t near
0 by 3.3.
(2) Iftherearedistinctrootsatt=0weputthemintotwosubsetswhichsplits
P(t) = P (t).P (t) by the splitting lemma 3.4. We then feed P (t) (which
1 2 i
have lower degree) into the algorithm.
(3) All roots of P(0) are equal. We first reduce P(t) to the case a (t) = 0 by
1
replacing the variable x by y = x a (t)/n. Then all roots are equal to 0
1
−
so m(a )>0.
2
(3a) If m(a ) is finite then it is even since ∆˜ = 2na 0, m(a )=2r and by
2 2 2 2
− ≥
the multiplicity lemma 3.7 a (t)=a (t)tir (i=2,...,n) for smooth a .
i i,ir i,ir
Consider the following smooth curve of polynomials
P (t)(x)=xn+a (t)xn−2 a (t)xn−3 +( 1)na (t).
r 2,2r 3,3r n,nr
− ··· −
IfP (t)issmoothlysolvableandx (t)areitssmoothroots,thenx (t)tr are
r k k
the roots of P(t) and the original curve P is smoothly solvable too. Since
a (0)=0, not all roots of P (0) are equal and we may feed P into step
2,2m r r
6
2 of the algorithm.
(3b) If m(a ) is infinite and a = 0, then all roots are 0 by 3.6 and thus the
2 2
polynomial is solvable.
(3c) But if m(a ) is infinite and a = 0, then by the multiplicity lemma 3.7 all
2 2
6
m(a ) for 2 i n are infinite. In this case we keep P(t) as factor of the
i
≤ ≤
originalcurveofpolynomialswithallcoefficientsinfinitelyflatatt=0after
forcing a = 0. This means that all roots of P(t) meet of infinite order of
1
flatness (see 3.5) at t = 0 for any choice of the roots. This can be seen as
follows: If x(t) is any root of P(t) then y(t) := x(t)/tr is a root of P (t),
r
hence by 4.1 bounded, so x(t) = tr−1.ty(t) and t ty(t) is continuous at
7→
t=0.