50.2. Result. [Sylvester, 1853, pp.511], [Procesi, 1978] The roots of P are all real

˜ ˜ ˜

if and only if the matrix B(P ) ≥ 0. Then we have ∆k (P ) := ∆k (a1 , . . . , an ) ≥ 0

˜

for 1 ¤ k ¤ n. The rank of B(P ) equals the number of distinct roots of P , and its

signature equals the number of distinct real roots.

50.3. Proposition. Let now P be a smooth curve of polynomials

P (t)(x) = xn ’ a1 (t)xn’1 + · · · + (’1)n an (t)

with all roots real and distinct for t = 0. Then P is smoothly solvable near 0.

This is also true in the real analytic case and for higher dimensional parameters,

and in the holomorphic case for complex roots.

d

Proof. The derivative dx P (0)(x) does not vanish at any root, since they are dis-

tinct. Thus, by the implicit function theorem we have local smooth solutions x(t)

of P (t, x) = P (t)(x) = 0.

50.4. Splitting Lemma. Let P0 be a polynomial

P0 (x) = xn ’ a1 xn’1 + · · · + (’1)n an .

If P0 = P1 · P2 , where P1 and P2 are polynomials with no common root. Then for P

near P0 we have P = P1 (P )·P2 (P ) for real analytic mappings of monic polynomials

P ’ P1 (P ) and P ’ P2 (P ), de¬ned for P near P0 , with the given initial values.

Proof. Let the polynomial P0 be represented as the product

P0 = P1 .P2 = (xp ’ b1 xp’1 + · · · + (’1)p bp )(xq ’ c1 xq’1 + · · · + (’1)q cq ).

Let xi for i = 1, . . . , n be the roots of P0 , ordered in such a way that for i = 1, . . . , p

we get the roots of P1 , and for i = p + 1, . . . , p + q = n we get those of P2 . Then

(a1 , . . . , an ) = φp,q (b1 , . . . , bp , c1 , . . . , cq ) for a polynomial mapping φp,q , and we get

σ n = φp,q —¦ (σ p — σ q ),

det(dσ n ) = det(dφp,q (b, c)) det(dσ p ) det(dσ q ).

From (50.1) we conclude

(xi ’ xj ) = det(dφp,q (b, c)) (xi ’ xj ) (xi ’ xj )

1¤i<j¤n 1¤i<j¤p p+1¤i<j¤n

which in turn implies

det(dφp,q (b, c)) = (xi ’ xj ) = 0,

1¤i¤p<j¤n

so that φp,q is a real analytic di¬eomorphism near (b, c).

50.4

50.7 50. Applications to perturbation theory of operators 539

50.5. For a continuous function f de¬ned near 0 in R let the multiplicity or order

of ¬‚atness m(f ) at 0 be the supremum of all integers p such that f (t) = tp g(t)

near 0 for a continuous function g. If f is C n and m(f ) < n then f (t) = tm(f ) g(t)

where now g is C n’m(f ) and g(0) = 0. If f is a continuous function on the space

of polynomials, then for a ¬xed continuous curve P of polynomials we will denote

by m(f ) the multiplicity at 0 of t ’ f (P (t)).

The splitting lemma (50.4) 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 + a2 (t)xn’2 ’ · · · + (’1)n an (t)

is smoothly solvable with smooth roots t ’ xi (t), and that all roots of P (0) are

equal. Then for (k = 2, . . . , n)

˜

m(∆k ) ≥ k(k ’ 1) min m(xi ).

1¤i¤n

m(ak ) ≥ k min m(xi ).

1¤i¤n

This result also holds in the real analytic case and in the smooth case.

Proof. This follows by (50.1.2) for ∆k and by ak (t) = σk (x1 (t), . . . , xn (t)).

50.6. Lemma. Let P be a polynomial of degree n with all roots real. If a1 = a2 = 0

then all roots of P are equal to zero.

x2 = s2 (x) = σ1 (x) ’ 2σ2 (x) = a2 ’ 2a2 = 0.

2

Proof. From (50.1.1) we have 1

i

50.7. Multiplicity lemma. For an integer r ≥ 1 consider a C nr curve of poly-

nomials

P (t)(x) = xn + a2 (t)xn’2 ’ · · · + (’1)n an (t)

with all roots real. Then the following conditions are equivalent:

(1) m(ak ) ≥ kr for all 2 ¤ k ¤ n.

˜

(2) m(∆k ) ≥ k(k ’ 1)r for all 2 ¤ k ¤ n.

(3) m(a2 ) ≥ 2r.

Proof. We only have to treat r > 0.

(1) implies (2): From (50.1.1) we have m(˜k ) ≥ rk, and from the de¬nition of

s

˜ ˜

∆k = det(Bk ) we get (2).

˜

(2) implies (3) since ∆2 = ’2na2 .

(3) implies (1): From a2 (0) = 0 and lemma (50.6) it follows that all roots of the

polynomial P (0) are equal to zero and, then, a3 (0) = · · · = an (0) = 0. There-

fore, m(a3 ), . . . , m(an ) ≥ 1. Under these conditions, we have a2 (t) = t2r a2,2r (t)

and ak (t) = tmk ak,mk (t) for k = 3, . . . , n, where the mk are positive integers and

50.7

540 Chapter X. Further Applications 50.8

a2,2r , a3,m3 , . . . , an,mn are continuous functions, and where we may assume that

either mk = m(ak ) < ∞ or mk ≥ kr.

Suppose now indirectly that for some k > 2 we have mk = m(ak ) < kr. Then we

put

m := min r, m3 , . . . , mn < r.

3 n

We consider the following continuous curve of polynomials for t ≥ 0:

¯

Pm (t)(x) := xn + a2,2r (t)t2r’2m xn’2

’ a3,m3 (t)tm3 ’3m xn’3 + · · · + (’1)n an,mn (t)tmn ’nm .

¯

If x1 , . . . , xn are the real roots of P (t) then t’m x1 , . . . , t’m xn are the roots of Pm (t),

¯

for t > 0. So for t > 0, Pm (t) is a family of polynomials with all roots real. Since

¯

by theorem (50.2) the set of polynomials with all roots real is closed, Pm (0) is also

a polynomial with all roots real.

¯

By lemma (50.6), all roots of the polynomial Pm (0) are equal to zero, and for those

k with mk = km we have nr ’ mk ≥ kr ’ mk ≥ 1, thus ak,mk is C nr’mk ⊆ C 1 and

ak,mk (0) = 0, therefore m(ak ) > mk , a contradiction.

50.8. Algorithm. Consider a smooth curve of polynomials

P (t)(x) = xn ’ a1 (t)xn’1 + a2 (t)xn’2 ’ · · · + (’1)n an (t)

with all roots real. The algorithm has the following steps:

(1) If all roots of P (0) are pairwise di¬erent, P is smoothly solvable for t near

0 by (50.3).

(2) If there are distinct roots at t = 0 we put them into two subsets which splits

P (t) = P1 (t).P2 (t) by the splitting lemma (50.4). We then feed Pi (t) (which

have lower degree) into the algorithm.

(3) All roots of P (0) are equal. We ¬rst reduce P (t) to the case a1 (t) = 0 by

replacing the variable x by y = x ’ a1 (t)/n. Then all roots are equal to 0,

so m(a2 ) > 0.

˜

(3a) If m(a2 ) is ¬nite then it is even since ∆2 = ’2na2 ≥ 0, m(a2 ) = 2r, and by

the multiplicity lemma (50.7) ai (t) = ai,ir (t)tir (i = 2, . . . , n) for smooth

ai,ir . Consider the following smooth curve of polynomials

Pr (t)(x) = xn + a2,2r (t)xn’2 ’ a3,3r (t)xn’3 + · · · + (’1)n an,nr (t).

If Pr (t) is smoothly solvable and xk (t) are its smooth roots, then xk (t)tr

are the roots of P (t), and the original curve P is smoothly solvable, too.

Since a2,2m (0) = 0, not all roots of Pr (0) are equal, and we may feed Pr

into step 2 of the algorithm.

(3b) If m(a2 ) is in¬nite and a2 = 0, then all roots are 0 by (50.6), and thus the

polynomial is solvable.

(3c) But if m(a2 ) is in¬nite and a2 = 0, then by the multiplicity lemma (50.7)

all m(ai ) for 2 ¤ i ¤ n are in¬nite. In this case we keep P (t) as factor of

50.8

50.9 50. Applications to perturbation theory of operators 541

the original curve of polynomials with all coe¬cients in¬nitely ¬‚at at t = 0

after forcing a1 = 0. This means that all roots of P (t) meet of in¬nite order

of ¬‚atness (see (50.5)) 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 Pr (t),

hence by (50.9) bounded, so x(t) = tr’1 .ty(t) and t ’ ty(t) is continuous

at t = 0.

This algorithm produces a splitting of the original polynomial

P (t) = P (∞) (t)P (s) (t),

where P (∞) has the property that each root meets another one of in¬nite order at

t = 0, and where P (s) (t) is smoothly solvable, and no two roots meet of in¬nite

order at t = 0, if they are not equal. Any two choices of smooth roots of P (s) di¬er

by a permutation.

The factor P (∞) may or may not be smoothly solvable. For a ¬‚at function f ≥ 0

consider:

x4 ’ (f (t) + t2 )x2 + t2 f (t) = (x2 ’ f (t)).(x ’ t)(x + t).

Here the algorithm produces this factorization. For f (t) = g(t)2 the polynomial

is smoothly solvable. There exist smooth functions f (see (25.3) or [Alekseevsky,

Kriegl, Losik, Michor, 1997, 2.4]) such that x2 = f (t) is not smoothly solvable, in

fact not C 2 -solvable. Moreover, in loc. cit. one ¬nds a polynomial x2 +a2 (t)x’a3 (t)

with smooth coe¬cient functions a2 and a3 which is not C 1 -solvable.

50.9. Lemma. For a polynomial

P (x) = xn ’ a1 (P )xn’1 + · · · + (’1)n an (P )

˜ ˜

with all roots real, i.e. ∆k (P ) = ∆k (a1 , . . . , an ) ≥ 0 for 1 ¤ k ¤ n, let

x1 (P ) ¤ x2 (P ) ¤ · · · ¤ xn (P )

be the roots, increasingly ordered.

Then all xi : σ n (Rn ) ’ R are continuous.

Proof. We show ¬rst that x1 is continuous. Let P0 ∈ σ n (Rn ) be arbitrary. We have

to show that for every µ > 0 there exists some δ > 0 such that for all |P ’ P0 | < δ

there is a root x(P ) of P with x(P ) < x1 (P0 )+µ and for all roots x(P ) of P we have

x(P ) > x1 (P0 ) ’ µ. Without loss of generality we may assume that x1 (P0 ) = 0.

We use induction on the degree n of P . By the splitting lemma (50.4) for the

C 0 -case, we may factorize P as P1 (P ) · P2 (P ), where P1 (P0 ) has all roots equal

to x1 = 0 and P2 (P0 ) has all roots greater than 0, and both polynomials have

coe¬cients which depend real analytically on P . The degree of P2 (P ) is now

smaller than n, so by induction the roots of P2 (P ) are continuous and thus larger

than x1 (P0 ) ’ µ for P near P0 .

50.9

542 Chapter X. Further Applications 50.10

Since 0 was the smallest root of P0 we have to show that for all µ > 0 there exists

a δ > 0 such that for |P ’ P0 | < δ any root x of P1 (P ) satis¬es |x| < µ. Suppose

there is a root x with |x| ≥ µ. Then we get a contradiction as follows, where n1 is

the degree of P1 . From

n1

(’1)k ak (P1 )xn1 ’k

’xn1 =

k=1

we have

n1 n1 n1

µk 1’k

k 1’k 1’k

µ ¤ |x| = ¤ |ak (P1 )| |x|

(’1) ak (P1 )x < µ = µ,

n1

k=1 k=1 k=1

provided that n1 |ak (P1 )| < µk , which is true for P1 near P0 , since ak (P0 ) = 0.

Thus, x1 is continuous.

Now we factorize P = (x ’ x1 (P )) · P2 (P ), where P2 (P ) has roots x2 (P ) ¤ · · · ¤

xn (P ). By Horner™s algorithm (an = bn’1 x1 , an’1 = bn’1 + bn’2 x1 , . . . , a2 =

b2 + b1 x1 , a1 = b1 + x1 ), the coe¬cients bk of P2 (P ) are continuous, and so we may

proceed by induction on the degree of P . Thus, the claim is proved.

50.10. Theorem. Consider a smooth curve of polynomials

P (t)(x) = xn + a2 (t)xn’2 ’ · · · + (’1)n an (t)

with all roots real, for t ∈ R. Let one of the two following equivalent conditions be

satis¬ed:

(1) If two of the increasingly ordered continuous roots meet of in¬nite order

somewhere then they are equal everywhere.

˜

(2) Let k be maximal with the property that ∆k (P ) does not vanish identically

˜

for all t. Then ∆k (P ) vanishes nowhere of in¬nite order.

Then the roots of P can be chosen smoothly, and any two choices di¬er by a per-

mutation of the roots.

Proof. The local situation. We claim that for any t0 , without loss of generality

t0 = 0, the following conditions are equivalent:

(1) If two of the increasingly ordered continuous roots meet of in¬nite order at

t = 0 then their germs at t = 0 are equal.

˜

(2) Let k be maximal with the property that the germ at t = 0 of ∆k (P ) is not

˜

0. Then ∆k (P ) is not in¬nitely ¬‚at at t = 0.

(3) The algorithm (50.8) never leads to step (3c).

(3) ’ (1) Suppose indirectly that two of the increasingly ordered continuous non-

equal roots meet of in¬nite order at t = 0. Then in each application of step (2) these

two roots stay with the same factor. After any application of step (3a) these two

roots lead to nonequal roots of the modi¬ed polynomial which still meet of in¬nite

50.10

50.11 50. Applications to perturbation theory of operators 543

order at t = 0. They never end up in a factor leading to step (3b) or step (1). So

they end up in a factor leading to step (3c).

(1) ’ (2) Let x1 (t) ¤ · · · ¤ xn (t) be the continuous roots of P (t). From (50.1.2)

we have

˜ (xi1 ’ xi2 )2 . . . (xi1 ’ xin )2 . . . (xik’1 ’ xik )2 .

(4) ∆k (P (t)) =

i1 <i2 <···<ik

˜ ˜

The germ of ∆k (P ) is not 0, so the germ of one summand is not 0. If ∆k (P ) were

in¬nitely ¬‚at at t = 0, then each summand was in¬nitely ¬‚at, there were two roots

among the xi which met of in¬nite order, thus by assumption their germs were

equal, so this summand vanished.

˜

(2) ’ (3) Since the leading ∆k (P ) vanishes only of ¬nite order at zero, P has

exactly k di¬erent roots o¬ 0. Suppose indirectly that the algorithm (50.8) leads to

step (3c). Then P = P (∞) P (s) for a nontrivial polynomial P (∞) . Let x1 (t) ¤ · · · ¤

xp (t) be the roots of P (∞) (t) and xp+1 (t) ¤ · · · ¤ xn (t) those of P (s) . We know

that each xi meets some xj of in¬nite order and does not meet any xl of in¬nite

order, for i, j ¤ p < l. Let k (∞) > 2 and k (s) be the number of generically di¬erent

roots of P (∞) and P (s) , respectively. Then k = k (∞) + k (s) , and an inspection of

˜

the formula for ∆k (P ) above leads to the fact that it must vanish of in¬nite order

at 0, since the only non-vanishing summands involve exactly k (∞) many generically

di¬erent roots from P (∞) .

The global situation. From the ¬rst part of the proof we see that the algorithm

(50.8) allows to choose the roots smoothly in a neighborhood of each point t ∈ R,

and that any two choices di¬er by a (constant) permutation of the roots. Thus, we

may glue the local solutions to a global solution.

50.11. Theorem. Consider a curve of polynomials

P (t)(x) = xn ’ a1 (t)xn’1 + · · · + (’1)n an (t), t ∈ R,

with all roots real, where all ai are of class C n . Then there is a di¬erentiable curve

x = (x1 , . . . , xn ) : R ’ Rn whose coe¬cients parameterize the roots.

That this result cannot be improved to C 1 -roots is shown in [Alekseevsky, Kriegl,

Losik, Michor, 1996, 2.4].

Proof. First we note that the multiplicity lemma (50.7) remains true in the C n -

case for n > 2 and r = 1 in the following sense, with the same proof:

If a1 = 0 then the following two conditions are equivalent

(1) ak (t) = tk ak,k (t) for a continuous function ak,k , for 2 ¤ k ¤ n.

(2) a2 (t) = t2 a2,2 (t) for a continuous function a2,2 .

In order to prove the theorem itself, we follow one step of the algorithm. First we

1

replace x by x + n a1 (t), or assume without loss of generality that a1 = 0. Then we

choose a ¬xed t, say t = 0.

50.11

544 Chapter X. Further Applications 50.12

If a2 (0) = 0 then it vanishes of second order at 0, for if it vanishes only of ¬rst order

˜

then ∆2 (P (t)) = ’2na2 (t) would change sign at t = 0, contrary to the assumption

that all roots of P (t) are real, by (50.2). Thus, a2 (t) = t2 a2,2 (t), so by the variant

of the multiplicity lemma (50.7) described above we have ak (t) = tk ak,k (t) for

continuous functions ak,k , for 2 ¤ k ¤ n. We consider the following continuous

curve of polynomials

P1 (t)(x) = xn + a2,2 (t)xn’2 ’ a3,3 (t)xn’3 · · · + (’1)n an,n (t).

with continuous roots z1 (t) ¤ · · · ¤ zn (t), by (50.9). Then xk (t) = zk (t)t are

di¬erentiable at 0 and are all roots of P , but note that xk (t) = yk (t) for t ≥ 0, but

xk (t) = yn’k (t) for t ¤ 0, where y1 (t) ¤ · · · ¤ yn (t) are the ordered roots of P (t).

This gives us one choice of di¬erentiable roots near t = 0. Any choice is then given

by this choice and applying afterwards any permutation of the set {1, . . . , n} which

keeps the function k ’ zk (0) invariant .

If a2 (0) = 0 then by the splitting lemma (50.4) for the C n -case we may factor

P (t) = P1 (t) . . . Pk (t), where the Pi (t) have again C n -coe¬cients, and where each

Pi (0) has all roots equal to ci , and where the ci are distinct. By the arguments

above, the roots of each Pi can be arranged di¬erentiably. Thus P has di¬erentiable

roots yk (t).

But note that we have to apply a permutation on one side of 0 to the original roots,

in the following case: Two roots xk and xl meet at zero with xk (t) ’ xl (t) = tckl (t)

with ckl (0) = 0 (we say that they meet slowly). We may apply to this choice an

arbitrary permutation of any two roots which meet with ckl (0) = 0 (i.e. at least of

second order), and we get thus every di¬erentiable choice near t = 0.

Now we show that we can choose the roots di¬erentiable on the whole domain R.

We start with the ordered continuous roots y1 (t) ¤ · · · ¤ yn (t). Then we put

xk (t) = yσ(t)(k) (t),

where the permutation σ(t) is given by

σ(t) = (1, 2)µ1,2 (t) . . . (1, n)µ1,n (t) (2, 3)µ2,3 (t) . . . (n ’ 1, n)µn’1,n (t) ,

and where µi,j (t) ∈ {0, 1} will be speci¬ed as follows: On the closed set Si,j of all t

where yi (t) and yj (t) meet of order at least 2 any choice is good. The complement

of Si,j is an at most countable union of open intervals, and in each interval we

choose a point, where we put µi,j = 0. Going right (and left) from this point we

change µi,j in each point where yi and yj meet slowly. These points accumulate

only in Si,j .

50.12. Theorem. The real analytic case. Let P be a real analytic curve of

polynomials

P (t)(x) = xn ’ a1 (t)xn’1 + · · · + (’1)n an (t), t ∈ R,

50.12

50.13 50. Applications to perturbation theory of operators 545

with all roots real.

Then P is real analytically solvable, globally on R. All solutions di¬er by permuta-

tions.

By a real analytic curve of polynomials we mean that all ai (t) are real analytic in t,

and real analytically solvable means that we may ¬nd xi (t) for i = 1, . . . , n which

are real analytic in t and are roots of P (t) for all t. The local existence part of this

theorem is due to [Rellich, 1937, Hilfssatz 2], his proof uses Puiseux-expansions.

Our proof is di¬erent and more elementary.

Proof. We ¬rst show that P is locally real analytically solvable near each point

t0 ∈ R. It su¬ces to consider t0 = 0. Using the transformation in the introduction

we ¬rst assume that a1 (t) = 0 for all t. We use induction on the degree n. If n = 1

the theorem holds. For n > 1 we consider several cases:

The case a2 (0) = 0. Here not all roots of P (0) are equal and zero, so by the

splitting lemma (50.4) we may factor P (t) = P1 (t).P2 (t) for real analytic curves of

polynomials of positive degree, which have both all roots real, and we have reduced

the problem to lower degree.

The case a2 (0) = 0. If a2 (t) = 0 for all t, then by (50.6) all roots of P (t) are 0,

and we are done. Otherwise 1 ¤ m(a2 ) < ∞ for the multiplicity of a2 at 0, and

˜

by (50.6) all roots of P (0) are 0. If m(a2 ) > 0 is odd, then ∆2 (P )(t) = ’2na2 (t)