Roots, iterations and logarithms of formal automorphisms

Roots, iterations and logarithms of formal automorphisms

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Praagman, C. (1986). Roots, iterations and logarithms of formal automorphisms. (Memorandum COSOR; Vol. 8601). Technische Universiteit Eindhoven.

Document status and date: Published: 01/01/1986

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EINDHOVEN UNIVERSITY OF TECHNOLOGY

Department of Mathematics and Computing Science

Memorandum COSOR 86 - 01 ROOTS, ITERATIONS AND LOGARITHMS

of formal automorphisms

by

C. Praagman

Eindhoven, The Netherlands February 1986

C. Praagman

Abstract.

In this paper it is proved that having a logarithm is e1uivalent to having roots of arbitrary order in the group

of automorphisms of a formal power series ring, and in algebraic

INTRODUCTION

The question whether an automorphism of a complex formal power series

ring has an iteration, or may be embedded in a one parameter group has

received much attention lately. The original setting concerned the

possibility of embedding an automorphism in a complex analytic one

dimensional Lie group (LEWIS [4, section 4] , STERNBERG [12, section 2]).

In the recent terminology this is expressed as having a complex analytic

iteration. At this moment the equivalence of the following statements

has been proven:

THEOREM 1

Let

F

be an automorphism of the complex formal power series

ring £[[x

1, ••• ,x

m

]].Then the following statements are equivalent:

i

F

has a complex analytic iteration.

ii

F

is conjugate to an automorphism in smooth normal form.

iii

F

has a real continuous iteration.

iv

F

has a rational continuous iteration.

v

F

is the exponential of a derivation of

c[ [xl"" ,x m]]

vi

F

has pairwise commuting roots of aU orders.

vii

F

is conjugate to an automorphism in normal form which has roots

of aU orders in normal form.

The notions mentioned in this theorem are explained in the sequel.

The proofs of the various equivalences may be found in: REICH

-SCHWAIGER [11, satz 4] for

i

~

ii,

BUCHER [1] for

i

~

iii,

PRAAG-MAN [7, Theorem 5] for

i

~

ii

~

iii

~

iv

~

v,

PRAAGMAN [8, Theorem 3] for

iv

~ V, MEHRING [5, satz 1.10] for

i

~

vi,

and PRAAGMAN [10, theorem 6] for V ~

vii.

In the first section I shall prove that the condition pairwise

having roots of all orders is equivalent to having a rational iteration,

which "lies between"

iv

and

vi.

In the remainder of t.."te paper these results are generalized to algebraic subgroups of the automorphism

group of a complete local ring.

§l ROOTS AND ITERATIONS

DEFINITIONS.

Denote the ring of formal power series over C in m

indeterminates, f:[[x1, ... ,x

m]] by

0,

let

m

be its maximal ideal, and equip 0 with the topology of coefficientwise convergence:

. a

I

v x for n -+ 00 if and only

a

is taken over all m-tuples a

i f v (n) -+

a m

E :NO' JN_O

Let L be the ring of all filtration preserving C-linear mappings: if h h

L E L then Lm c m for all h E IN. Equip L with the topology of

pointwise convergence, then L is complete and A = AutcO and V= Der£(O,m) 00

are closed subspaces. The map exp:

L

-+

L,

exp L

=

I

Ln/n! is well-n=O

defined and maps

V

into A (PRAAGMAN [7, section 2]).

LIE STRUCTURE.

In fact A has the structure of an infinite dimensional

complex analytic Lie group, and

V

of its Lie algebra. Let G be an arbitrary group containing ~, if necessary with some topological

or differentiable structure.

Definition.

Let F

E

A. Then

i

F

has a Logarithm

if F

E

exp(V).

ii

F

has a G-iteration

if there exists a

A E

Hom(G,A) such that

A(1}

=

F.

iii

F is called

divisible

if F has a

~ ~-iteration

for all h

E

IN (i.e. if F has roots of all orders).

3

-Note that a complex analytic iteration is an analytic C-iteration in the above terminology and so on. A weaker form of THEOREM 1 in this terminology would be ifF

has a logarithm if and only if

F

has

a Q-iteration."

DIVISIBLE

AUTOMORPHISMS.

In the proof of LEMMA 1 below I need some results from algebraic geometry. To be able to apply these I have to define group homomorphisms Pk: A -+ Gll(k)(:' where l(k)

=

dimcOlm

k •

Define Pk by (PkF) x a k Fxa mod k . Then P

k (A) is an algebraic

mod

m

=

m

subgroup of Gll(k)C:'

LEMMA 1.

Let

F E A

be divisible, and let

h E 1>1.

There exists a

divisible

h-th

root

of

F ,

i.e.

aGE

A,

divisible, satisfying

Gh F.

Proof.

Let k E ] \ I . Assume there exists a divisible G

k E A such that PkF (note that this holds trivially for k = 1, take G

1 = I). Define:

Z

n

Pk+l~n

=

~nPk+l obviously, and since P_k+₁Z_n is an algebraic subset of P

k+1

A

i t follows that Pk+l~nZn is a constructible subset of Pk+1

A

(HUMPREYS [3, theorem 4.4J). One easily verifies that ~ l Z 1 c ~ Z

n+ n+ n n so the Pk+l~nZn form a nested sequence of constructible sets in P

k+1

A.

Hence

n

P W Z is nonempty (OORT [6, lemma 2]), and there exists

nElN k+l n n

a G_k+₁ in n~N~nZn' Clearly G_k+₁ is divisible and satisfies P

k+1Gk+1 = P

h h Gh lim G

k, then PkG

=

PkGk

=

PkF, so

k~

Define G

=

F. To see that

G is divisible, fix n E IN and define

W {T E A

I

Tn

=

G

k} .

n,k

W is nonempty for all k and W k 1 C W k· Pl(W k) is a constructible

n,k n, + n, n,

subset of PiA , and hence as above

n

W k is nonempty. Let TEn W

k n, k n,k

n

then T = G and G is divisible.

COROLLARY.

Let

F E A .F

is divisible

if

and only if

F

has a f{J-iteration.

Proof.

1.(1) F, and by induction A(l/n!) is a divisible n-th root of A

01

(n-1) !) •

If p/q E Q define A(p/q) A(p(q-l)!/q!), then A E Hom(Q,A).

So together with THEOREM lone has:

5

-§2

ALGEBPAIC SUBGROUPS OF FORMAL AUTO'1ORPHISMS

ALGEBRAIC SUBGROUPS.

In this section I have gathered a number of

well-known results that will be needed in §3. First a defenition: a closed subgroup

H

of A is called

algebraic

if PkH is an algebraic group for all k E IN.

Examples.

1. Let

I

be an ideal in

0,

then

AI

{T E A

I

T

I

= I} is algebraic:

for all k PkAI is the intersection of PkA with the algebraic group leaving I/mk

n

I invariant.

2.

A(I)

=

{T

E

A lTV

=

v mod

I

for all v

EO}

is an algebraic subgroup of

A

(l) •

3. T = {T E A

I

TxCt

=

\/Ct

for all Ct E

JN~}

is algebraic. In fact

T""pT _{- k} (~*)m for all k ~ 2 by sending T, defined by Tx.

~

4. Let G be a linear algebraic group contained in Gl C. Sending m

to x

1, ••• ,xm identifies G with a subgroup of A.

t.x., to

~ ~

, .•. , e

NORMAL FORMS AND JORDAN DECOMPOSITION.

Let F

E

A. Then there exists

.a TEA such that T-1FT is in normal form: there exist Al, .•• ,A

m E C* such that:

-1 Ct T FTx

=

m

where f = ACt for all Ct, and f = 0 if either 8 < Ct lexicographically

CtCt

CtB

or ACt

i

AS • (PRAAGMAN [7, theorem 3]). Define sF, the topologically semisimple part of P, by spTxCt

=

ACtTxCt and up, the topologically nilpotent part of P, by upTxCt =

A-

Ct

L

f aTxB. Then [sF,up ]

=

B Ct",

by Tx.

=

t.x., then L E L defined

by Lxa

=

r

laSXS commutes with T if and only if ta

~

t

B

implies

B

laB 0). Since PksF • PkuF is the Jordan decomposition of pkF, it s

follows that if F E H , an algebraic subgroup of

A,

then P

k F and pkuF

E

PkH (HUMPREYS [3, theorem 15.3]) and hence sF and u F

E

H

(compare PRAAGMAN [7, theorem 2]).

ALGEBRAIC LIE ALGEBRA'S.

Let H be an algebraic subgroup of

A,

and identify its Lie algebra h with a subalgebra of

V.

(Then h is an algebraic sub Lie algebra of

V) .

Examples.

1. Let I be an ideal in

O.

Then

VI =

{D E

Vi

DI c I} is the Lie algebra associated to

AI .

2.

V(I)

=

{D E

Vi DO

c

I}

is the Lie algebra of

A(I).

3. t = {D

E

VI

Dxa = daXa for all a

E

~~}

is the Lie algebra of

T.

Since exp: Pkh ~ PkH for all k (VARADARAJAN [13, section 2.10]), clearly exp

h

c H.

7

-§3 ITERATIONS AND LOGARITHMS IN ALGEBRAIC SUBGROUPS

PROBLEM FORMULATION.

Let H be an algebraic subgroup of

A,

and hits

Lie algebra. p

E

H is called

divisible in

H if it has roots of all orders wi thin H: 'v'n EN: 3G E H such that Gn = P. The question

considered in this section is whether exp h consists of all elements

divisible in H, as was proved for

A

in

§1.

The problem is tackled in two steps. Pirst I prove that having a logarithm is equivalent to

having a rational iteration in H , and then a slight modification

of LEMMA 1 yields that being divisible in H is equivalent to having

a rational iteration.

u

Concentrating on the first problem, note that for P

E

H , log P

00

L

(I - p) j

/j

is welldefined, log uF E V and exp log up = up j=l

(PRAAGMAN [7, theorem 4]). And since Pk log up log Pk P, and u

log Pkup

E

Pkh(HUMPREYS

[3,

section

15.1])

it follows that log up

E

h. So if one finds a D

E

h such that exp D = , and [D, log uF] = 0, then D + log uF is a logarithm of P in h.

Purther note that all properties that are considered here are

invariant under internal automorphisms of

A:

exp (T-1DT) = T-1(expD)T;

-1 -1

if A E Hom (~,H) is an iteration of F, then T AT E Hom (~,T HT) is

an iteration of T-1pT E T-1HT, an algebraic subgroup conjugated to

the algebraic subgroup H. Since [A(t) ,ACS)] = 0 for all s,t E Q

-1

there exists a T such that T A(t)T is in normal form for all

t E~. (PRAAGMAN [7, lemma

2]).

Containing these two simplifications of the problem, it becomes

clear that special attention should be given to algebraic subgroups

DIAGONlZABLE GROUPS.

Let

H

be a connected algebraic subgroup of

T,

h h

then

H

~ (C*) for some h ~ m, and

h

~ £ (HUMPREYS [3, theorem 16.2]),

h m

and there exists aTE Hom«£*) I(~*) ) such that for all FE

H

there

h h m

exists apE (£*) such that Fx.

=

t. (p)x., and similarly cr E Hom(e ,£ )

l l l

such that for all 0 E

h

there exists a

v

E eh such that Ox.

=

o(v)x .•

l l

Moreover exp o(v) t(exp v),

I f F E

H,

then FX i

h h

ti (p)x

i for some p E (C*) . Take a vEe such that exp

v

=

p, and define 0 E

h

by Ox.

l 0. (v)x., then (exp O)x. l l l

=

exp 0i (v)x

i

=

ti(P)xi = FXi , So exp h = H and any choice of log p will do.

RATIONAL CHARACTERS.

To determine which choice of the logarithm will

be the right one, I will use the following lemma, or more precisely its corollary:

LEMMA 2. Hom(~i£*)

is

torsionfree,

Proof.

Let A E Hom

(~,C*)

satisfy Ak(t) 1 for all t E ~. Then

A(~) is a finite subgroup of C*. But ~ being divisible, so is A(~),

hence A(f2) = 1.

Corollary.

Every finitely generated subgroup of Hom(~,~*) is free.

CHOOSING THE LOGARITHM.

The following proposition yields the right

choice of the logarithm: a derivation which commutes with maps commuting with the iteration.

PROPOSITION 1.

Let

F E T,

and let

A E Hom(~,T)

be an iteration for

F.

9

-satisfying

[L, A(t)]

o

for aU

t E qJ.

Proof.

Let H be the algebraic closure of the subgroup of

T

generated by the set {A (t) It E ~}. H is connected since A (t) has roots of all orders in H for all t. Let A(t) be given by A(t)x.

1- A. (t)x., then 1.

1-A. E Hom (Q,~*). Let

A

be the free subgroup of Hom (~,~*) generated

1-by A

1, ... ,Am, and let Pl, •.• ,Ph be a free set of generators of A. Choose v 1" • .,v

h such that exp v i P i (1), and let y (1) , ••• ,y (m) E Zl d be defined by Ai

=

py(i). Define DE

h

by

Dx,

1. <v, y (i) > X. 1- with <a, S> =

I

a. S .• 1.

1-Then clearly exp D = F. Now if [L,A(t)]

o

for all t, and Lx a

I

1 SX 8

,

then Aa

#

A8 implies laS =

O.

Since <a ,lJ>

'f

<8, lJ>, where 8 a

lJ.

=

<v,)' (i» implies that Aa

#

AS I it follows immediately that

1-[L, A(t) ]

=

0 for all t implies [L,D] =

o.

Remark.

Note that D E

h.

LOGARITHMS

J

ITERATIONS .••

The theorem which follows is:

THEOREM 3.

Let

F E H,

an algebraic subgroup of

A •

Then

FE exp

h

if

and

only if there exists a rational iteration

A E

Hom

(qJ,H)of Fin

H.

Proof.

If F

=

exp D, then clearly A E Hom(~,C*) defined by A(t)

=

exp tD is an iteration of F in H, so assume A E Hom(Q,H) with A(l) F is given. Let E Hom(f2,H) be defined by (sA)(t) = s(A(t», then sA

is a rational iteration of sF. Since [sA (s), sA (t)]

=

a

for all s,t E Q there exists aTE

A

such that T AT s -1 E

Hom(qJ,T).

Let D E

h

such that exp T DT -1 T -1 sFT, and [T DT,L] -1 =

a

for all L with [T-1 sA(t)T,L] =

a

for all t E ~. So from [sA(t) ,UF]

a

follows that [D,uF ]

=

0 and hence [D, log uF]

=

O. Therefore

u u .

exp(D + log F) = exp D • exp log F

=

F.

,.,AND

ROOTS,

Note that in LEMMA 2 only the algebraicity of Pk

A

played a role. The argument works equally well for subgroups H

with PkH algebraic. So from this modification of LEMMA 2 and THEOREM

3 immediately follows:

THEOREM 4.

Let

F

E H ,

an algebraic subgroup of

A.

Then

F E exp

h

if

and only if

F

is divisible in

H.

Remark.

Note that THEOREM 4 in particular holds for linear algebraic groups.

- 11

-§4

MISCELLANEOUS REMARKS

BEHAVIOR UNDER r·t)RPH ISMS

I Let

G

be a (possibly infinite dimensional)

complex Lie group,

H

an algebraic subgroup of

A,

and q> a morphism of Lie groups from

H

onto

G.

Then dq> maps

h

onto

g,

and q> • exp = exp • dq>

(VARADARAJAN 13,[theorem 2.10.3]). Clearly q> maps divisible

automorphisms onto divisible elements of

G,

but under circumstances it is also the other way around:

LEMMA 3.

Let

kerq>

be an aLgebraic subgroup of H, and Zet

g

be divisibLe

in

G.

Then there exists an

F,

divisibLe in H, such that

q>(F) = g.

Proof·

(sketch). For all n E :IN let V = {T E q> -1 (g)

I

3T' E

H

such n

nl

that (T') . = T }. Then PkVn is constructible for all k and all n ,

v i e V , so using the same argument as in LEMMA 1, the intersection n+ n

of the V is nonempty, which yields the desired F.

n

As an immediate consequence it follows that divisibility in

G

is equivalent to having an automorphism. Consider the special case

where

G

=

Aut R, R

a/I.

Then the exact sequence

o -+

A

<I)

-+

AI

-+ Aut R -+ 0

yields with LEMMA 3 the results of PRAAGMAN [9, chapter Vld] :

divisibility in Aut R is equivalent with having a logarithm, and

even more: the same holds for algebraic subgroups of Aut R, the

images of algebraic subgroups of

AI'

THEOREM 5.

Let

I

be an ideaL

of

0, and

R

0/1. Then divisibiLity

in any aLgebraic subgroup of

Aut R

is equivaLent to having a

PROAFFINE ALGEBRAIC GROUPS.

Let K be the ~-algebra of polynomials

m

in the variables X. , i E {1, ... ,m}, et E NO' K

l.et = U ~ [ u X. ].

lEN letl~ l.et

Then (A,K) defines the structure of a proaffine algebraic group

over € (HOCHSCHILD-MOSTOW

[2,

section

2]),

and a similar construction is possible for Aut R, and the algebraic subgroups of

A,

coincide with those defined by this proaffine algebraic structure.

The question considered here is whether the results of §3 extend

to proaffine algebraic groups.

From HOCHSCHILD-MOSTOW [2, theorem 2.1] it follows that any proaffine

algebraic group is isomorphic to a projective limit of linear

algebraic groups and morphisms, and vice versa:

PROPOSITION 2. Let G be group_ Then G may be equipped with the

structure of a proaffine algebraic group if and only if there

exists a partially ordered set

r,

linear algebraic groups G ,

y

for all y E r, and for all y,o E r with 0 < y morphisms of

algebraic groups TIc :G

o -+ Gy satisfying for n < 0 < y: TIncTIoy

y

such that G = lim G

yEr y

TI , ny

One verifies without any difficulty that

g,

the Lie algebra of G is isomorphic to the inverse limit of the Lie algebras of the group

G :

9

=

lim

9 _

If x

E G

or

9

x is its image in G (g ) under the

y + . y y y y

canonical proj ection _ Further exp:

9

-+ _{G satisfies exp (lim x )}

+ y

lim (exp x )_

+ Y

As before g EGis said to have a logarithm if g

E

exp

g,

to

have a rational iteration if there exists a

A

E Hom(~,G) such that A(!)

=

g, and to be divisible if for all n E N there exists an f

E

G such that fn

n n

if for all n,m E N

g. As a new concept call g

normaZZy divisibZe

Sf ]

=

0 where Sx is defined as

- 13

-Finally, define Ux as lim Ux , and let cr(g)

=

U cr(g ) be the spectrum

+- y y Y

of g: cr(g ) is the set of eigenvalues of g considered as a linear

y y

map.

THEOREM 6.

Let

G

be a complex proaffine algebraic group,

G

~ lim

G .

.

yEr y

Suppose there exists

ayE r

such that for all

g

E G

the subgroup

I

(g)

of

f:*

generated bya(g) equals

L(g ),

the subgroup generated

y

by

a(g ).

Then

g E

G

has a logarithm i f and only i f it is normally

y

divisib

Proof.

Let fn = g and [Sf sf] = O. Then the algebraic subgroup

n n' m

H

of G generated by the sf is isomorphic to

H •

Since g is divisible

n y

Sg lies in the component of identity HO of H. HO:;: (c*)h for some hEN, and Sg -+ (Al, ••• ,A

h) E (c*)h. Now any choice of log Ai yields a d

E

h

with exp d = Sg, and [d, ug ]

=

O. Hence g

=

exp

u

(d + log g).

THEOREM 7.

Let

r

eN

as an ordered set. Then

g

EGis divisible

if and only i.f

g

has a rational iteration.

Proof.

Completely analogous to the proof of the COROLLARY of

LEMMA 1.

Remarks.

1. Combination of THEOREMS 6 and 7 does not yield an extension

of THEOREM 5. Any lim

G

I with a(g) = a(g ) may be embedded in

+- n n

an automorphism group of a complete local ring.

2. In fact one could take any totally ordered set

r

in THEOREM 7, since

r

always contains a countable subset

r'

such that lim

yEr

lim. To prove this, however, something about the ordinality

of the set of all linear algebraic groups over C should be said,

15

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