PACIFIC JOURNAL OF MATHEMATICS

Vol. 104, No 1, 1983

CENTRALIZERS OF IRREGULAR ELEMENTS

IN REDUCTIVE ALGEBRAIC GROUPS

JOHN F. KURTZKE, JR.

Let G be a reductive linear algebraic group defined over an algebrai-

cally closed field K. An element x G G i s called regular if dim ZG(x) is
the rank of G — which is the smallest possible dimension for a central-
izer — otherwise x is called irregular. T. A. Springer has shown that if
x 6 G ί s regular, then ZG(x)° is abelian. In this paper, we show that
when char K is good for (7, the converse is also true — if x G G is
irregular, then ZG(χ)° is nonabelian. In the course of the proof of this,
we show that if G is a classical group or G2 (char K good), then
dim Z(ZG(x)°) is at most rank G. Further, if G — Sp2n(K) (charK =£
2), then Z(ZG(x)°) consists of polynomials in x.

Introduction and statement of results. In this paper, all groups will be

linear algebraic groups defined over an algebraically closed field K. If G is
a group, then G° denotes its identity component and g or £(G) denotes
the Lie algebra of G. If G is reductive (we require that reductive groups be
connected), then by " the simple components of G", we mean the simple
components of the semisimple group G/Rad G where Rad G is the radical
of G.

Let G be a reductive group. An element x E G i s called regular if

dim ZG{x) is minimal in (dim ZG(y) \y E G}; otherwise, x is called
irregular. The regular elements are dense in G; the identity element is
always irregular. In GLn(K), a diagonalizable element is regular if and
only if all its eigenvalues are distinct. It is well known that if x E G is
regular, then
dim ZG(x) = rank G

and if y E G is irregular, then

dim ZG(y) > 2 + rank G

(cf. [17]). If

dim ZG(y) - 2 + rank G,

then y is called subregular.

T. A. Springer has shown that if x E G is regular, then ZG(x)° is

abelian. [13] Under some very mild restrictions on char K (char if = 0 or
char if > 7), we shall prove the converse; namely, that if ZG(x)° is
133
134 JOHN F. KURTZKE, JR.

abelian, then x is regular — or, equivalently, if x E G is irregular, then

ZG(x)° is nonabelian. This last statement is the one that we will actually
prove.

Our proof uses the classification of nilpotent conjugacy classes in a

simple Lie algebra, and since this classification is only valid in good
characteristics, we must require that char K be good for G. The usual way
of defining good characteristic for a reductive group G (or Lie algebra g)
is to define a bad prime as one that divides a coefficient in the highest
root. Then all other primes and zero are called good. If G is of type An9
then all characteristics are good. So char K is good for G
(a) if G has a component of type Bn, C n , or Dn, then char K ¥=2
(b) if G has a component of type G 2 , F4, E6, or E79 then char K φ 2
or 3
(c) if G has a component of type Es, then charΛΓ φ 2, 3, or 5.
Hence, if char K — 0 or char K > 7, then char K is always good.

THEOREM A. Let G be a reductive group with char K good for G. Then

x G G is regular if and only if ZG(x)° is abelian.

Proving Theorem A is the main object of this paper. We will first

make some reductions (§2), then handle the classical groups (§3) and
finally the exceptional groups (§§4 and 5). In the course of this, we will
derive some stronger results for the classical groups. We say that a
reductive group G is of classical type if none of its components is an
exceptional group.

THEOREM B. 1. Let G be a group of classical type with char KΦ2. Let

JCGG. Then
dim z(zG(xf) <
Equality holds if and only if x is regular.
2. Let G- G2 with char Kφ2or3and let x G G. Then

dim Z(zc(x)°) <rankG.

Equality holds only in two cases.
(i) x is regular, or
(ii) x is subregular, but not semisimple.

The first part is proved in §3; the second part in §5. Note that
dim Z(ZG(x)°) can equal rank G even if x is irregular (we suspect that G2
 REDUCTIVE ALGEBRAIC GROUPS 135

is the only simple group in which this can happen). For the symplectic
groups, we can improve Theorem B:

THEOREM C. Let G = Sp2n(K) with chaτK φ 2 and let x E G. Then

Z(ZG(x)°) contains only polynomials in x.

This is proved in §3. If G = PGL2(C) and

x =ι
0]
o ij
we see that x is regular, but ZG(x) is nonabelian — of course ZG(x)° is
abelian. This happens because ZG(x) is not connected, and that happens
because G is not simply connected (as x is semisimple). Now if G is
semisimple and simply connected, and if x E G is regular semisimple,
then ZG(x) is connected and thus abelian. In her dissertation [9], B. Lou
proved that if u is a regular unipotent element in a semisimple group G,
then ZG(u) is abelian (charK arbitrary). Therefore, if G is a semisimple,
simply connected group, and if x E G is regular, then ZG(x) is abelian. So
as a corollary to Theorem A we have:

THEOREM D. Let G be a semisimple, simply connected group with

char K good for G. Then x E G is regular if and only if ZG(x) is abelian.

As general references to algebraic groups and Lie algebras, we cite

Humphreys' books [7, 7a] which we follow for basic definitions and
conventions. For conjugacy classes and centralizers, the reader is invited
to consult the Springer-Steinberg article [16] or Steinberg's Tata notes [18].
The classification of unipotent and nilpotent conjugacy classes is de-
scribed in papers of Dynkin [4], Elkington [6], and Bala-Carter [1, 2]. The
extension of the Dynkin classification to all good characteristics is de-
scribed by Springer-Steinberg [16] for the classical groups, by Jeurissen
and Stuhler [19] for G2, and by Pommerening [10, 11] for the other
exceptional groups. Dynkin, Elkington, and Bala-Carter also have infor-
mation about the centralizers of nilpotent elements, which can also be
found in an article of Elasvili [5].

2. Reductions. In this section, we show that in order to prove Theorem

A, it suffices to prove:

THEOREM α. Let Q be a simple Lie algebra with char K good for Q . If

X E Q is nilpotent and irregular, then %Q(X) is nonabelian.
136 JOHN F. KURTZKE, JR.

The rest of this paper will then be devoted to proving Theorem α. The
first thing we notice is that if G is reductive and t E G is semisimple, then
ZG(t)° is well in hand:

PROPOSITION 2.1. Let G be a reductive group and let t E G be semisim-

ple. Let T be a maximal torus for G containing t and let Ψ be the resulting
set of roots. Then
1. ZG(t)° is generated by T and the one parameter unipotent subgroups
Ua for those a E Ψ for which a(t) = 1. Moreover, ZG(t)° is reductive and
has the same rank as G.
2. ZG{t) is generated by ZG(t)° and the n E NG{T)/T such that
ntn~x = t.
3. / is regular if and only ifa(t) Φ 1 for all a E Ψ
4. dim Z(ZG(t)°) < rank G
Equality occurs if and only if t is regular.

This is rather well known and the proof can be found in [18] and [3].
If we were dealing with real compact semisimple Lie groups, then we
would be finished here. It is the unipotent elements of G that are the
obstacle.

If G is reductive, and x E G i s arbitrary, then we can write x — su

with s and u commuting semisimple and unipotent elements of G respec-
tively (Jordan decomposition). If we let H — ZG(s)°, we see that u E H
and ZG(x)° = ZH(u)°. Because the ranks of G and H are the same, we see
that x is regular in G if and only if u is regular in H. So we have:

REMARK 2.2. To prove Theorem A, it suffices to assume that the

element x is unipotent.

The next step is to reduce the problem to the case where the group G
is simple. Further we will only have to look at isogeny classes of simple
groups instead of isomorphism classes.

PROPOSITION 2.3. Let G be a reductive group, R a closed normal

subgroup contained in the center, and x E G. Let π: G -» G/R — H be the
canonical quotient map with TΓ(Λ ) = x. Then
1. R Π (G, G) is finite and R° is a torus.
2. dim ZG(x) = dim R + dim ZH(x).
Thus x is regular in G if and only ifx is regular in H.
3. dim Z(ZG(x)°) = dim R + dim Z(ZH(x)°).
Thus the conclusions of Theorems A and B hold for G if and only if they
hold for H.
 REDUCTIVE ALGEBRAIC GROUPS 137

Proof. l^This is standard; see [7aJ, for example.

2. Let C = ZH(x) and C = π~\C). We have to show that dimC =
1
dim Zc(x). Now y E C if and only if xyx'V" BΛ. As R Π (G, G) is
λ x
finite, so is R Π {xyx~ y~ \ y E G). Let {r,,... ,rp) be that set and choose
r λ λ λ x
yt E G so that ^ ^ ' V Γ ~ / Now xyx~ y~ = xzx~ z~ precisely when
1

]
z~ 7 E Z G (x), so the yi are determined mod Z (JC). Therefore, C —
G
a n d c z a s
=i-V, ^ ( ^ ) ° = 0 c ( * ) ° desired. 0
Let Zλ = ZCZ^ίJc) ) and Zj = w ' ί Z , ) . Clearly

We want to show that the first inclusion is an equality. Choose y E Zv

Now (y9 ZG(x)°) is a finite set, contains 1 (as j> commutes with x) and is
connected. Hence y E Z(ZG(x)°). As Zλ is connected, we have y E
Z ( Z ( J C ) ° ) ° as desired.
G
D

REMARK 2.4. 1. Let G be reductive. Then the conclusions of Theorems

A and B hold for G if and only if they hold for G/Rad G. So we can
assume in Theorems A and B that G is semisimple.
2. Let G and H be isogenous semisimple groups. Then the conclusions
of Theorems A and B hold for G if and only if they hold for H.
3. To prove Theorems A and B, we may assume that G is simple and
x is unipotent. Further, it is enough to prove these theorems for one group
in each simple isogeny class.

Proof. 1. Let R = Rad G and apply Proposition 2.3.

2. Let πx: G -> G and π2: G -» i/ be the universal covers of G and //.
Now apply Proposition 2.3.
3. We can assume that G is adjoint and semisimple. Then G = Π[=1G7
is a direct product of simple groups. If x E G, then x = JC, x r with
xf E G, . So Z G (x)° = Πί=1ZG|(xf.)°. D

The next task is to transfer all this to the Lie algebra. If x E G and
X E g , we want to use results about iQ(X) to understand ZG(x)°. If
char^ - 0, then t(ZG(x)°) = %Q(x) and e(Z c (Jr)°) = 8 f l W — if * and
X are semisimple, then these hold no matter what char K is. When j£ = C,
then there are maps (log and exp) from G to g and vice versa which are
power series and local isomorphisms of manifolds. So when K — C and
logx = X, then we have £(Z c (x)°) = iq(x) = 8 f l ( * ) = β(Z<y(*) 0 )
which is as nice a relation among centralizers as one could want. We
would like this to happen in our case — G reductive and x unipotent.
Because we have room in the isogeny classes to skip around and because
char K is good (by assumption), we can actually do this.
138 JOHN F. KURTZKE, JR.

PROPOSITION 2.5 (Richardson). Let G be a reductive group with char #

good for G. Let x G G and X G g = t(G). Then

and

t(ZG(X)°) =
The proof is in [12]. Next, we let U = {x G G | x is unipotent}, the
unipotent variety, and n ^ l J Γ G g l X i s nilpotent}, the nilpotent variety.

PROPOSITION 2.6. (Springer). Let G be a semisimple, simply connected

group with ch&ΐ K good /or G. ΓΛe« /Aere w # G-equiυariant bijective
morphismf: V -* n which topologically is a homeomorphism.

The proof is in [15]. By G-equivariant, we mean that for g G G and

v G V, Ad g(f(v)) — f(gvg~λ). Thus/preserves centralizers and iΐf(x) =
X, then δ Q (x) = £(Z G (x)°) = t(ZG(X)°) = aβ(AΓ). It now makes sense
to define regular, subregular, and irregular elements in reductive Lie
algebras. In good characteristics, the results are the same as in reductive
groups.
The final thing we have to notice is that in going from a group to its
Lie algebra, the center can only get larger. That is, β(Z(G)°) C j(g).
Thus, in our case, if iQ(X) is nonabelian, then so is ZG(X)° and ZG(x)°.

REMARK 2.7. To prove Theorem A, it is sufficient to prove Theorem

α.

PROPOSITION 2.8. (Steinberg, Springer). 1. Let G be a reductive group

with char K arbitrary. Let x G G be unipotent and let Ψ be a set of roots for
G. Choose a base for Ψ so that x is supported by positive roots, i.e.
ε c
x = Πa>o a( a) where ca ^ %- Then x is regular if and only if ca Φ 0 for a
simple. Further, all regular unipotent elements are conjugate.
2. Let Q be a reductive Lie algebra with char K good for g. Let X G g be
nilpotent and let Ψ be a set of roots for g. Choose a base for ψ so that X is
supported by positive roots, i.e. X — Σa>obaea with ba G K. Then X is
regular if and only ifbaφ 0 for a simple. All regular nilpotent elements are
conjugate.

Part 1 is proved in [17]; part 2 in [14]. In particular Π α G Δ ε α (l) and

Σ α G Δ ε α a r e regular (Δ is a base of Ψ). These are called the standard
regular and nilpotent elements respectively.
 REDUCTIVE ALGEBRAIC GROUPS 139

3. The classical groups. In this section, we will prove Theorems A, B,

and C for the classical groups. The next result is well-known.

PROPOSITION 3.1. Let G = GLn(K) or SLn{K) and let x G G. Then,

1. Z(ZG(x)) = {polynomials in x} Π G. Therefore,
dim Z(ZG(x)) < rank G
and equality holds if and only if x is regular.
2. Z(ZG(x)°) consists of polynomials in x. Therefore,
dim z(ZG(x)°) <rankG

and equality holds if and only if x is regular.

3. The following statements are equivalent:
(a) JC is regular.
(b) ZG(x) consists of polynomials in x.
(c) ZG(x)° consists of polynomials in x.
(d) ZG(x) is abelian.
(e) ZG(x)° is abelian.
(f) The minimal polynomial of x is the characteristic polynomial of x.
(g) Different blocks have different eigenvalues in the Jordan form of x.

This more than takes care of groups of type An. For the rest of this
section, we will be concerned with the orthogonal and symplectic groups,
so we assume that char K ^ 2. At this point, we want to describe explicitly
the map of Proposition 2.6. Let G = SOn(K) or Spn(K) and g = Zon(K)
or §>pn{K), respectively, Consider
f(x) = (1 - x)(l + x)'1
(1 is the n X n identity matrix). This map takes unipotent elements of G to
nilpotent elements of g ([20]). Its inverse is formally the same:

If x is unipotent and .Y nilpotent, t h e n / a n d / " 1 are polynomials in x and

X, respectively. Thus there is no problem in going from G to g and vice
versa.

PROPOSITION 3.2. Let g = &on(K) or $pn(K) with chaτK^2. Let

i G g {not necessarily nilpotent) and let k be the degree of the minimal
polynomial of x. Then
\k
dim {{polynomials in x) Π g) = —

where [n] is the greatest integer in n.

140 JOHN F. KURTZKE, JR.

Proof. If s is the form, then l E g means that

X=-s-λX<s
so

X< = (_1)<(X')' 5

and X1 E g if and only if / is odd. D

PROPOSITION 3.3. 1. Let g = S\)2n(K) or $o2n+ι(K) with charKΦ 2.

// X E g is a regular nilpotent element, then X is regular in §>l2n(K) or
%\ln+x(K), respectively. Therefore,
iQ(X) = {polynomials in X} Π g.

2. Let g = %oln(K) with char K φ 2. If X E g is a regular nilpotent

element, then X is subregular in §>l2n(K). Hence, dim {{polynomials in
X) Π g) = n — 1 < rank g and {polynomials in I } Π g < hQ(X)>

Proof. We use etj to denote the matrix with 1 in the (i, j) position and
zeroes elsewhere. The argument for the first part consists of writing down
the standard regular nilpotent element and seeing that it is regular in
§>l2n(K) in the symplectic case and regular in §I 2 w + 1 (A") in the orthogo-
nal case. Then, Proposition 3.2 shows that {polynomials in X} Π g is the
whole of hQ(X). For part 2, let g = §>o2n(K) and let X be the standard
regular nilpotent element, i.e.
n-\
X — 2J (£/,,+1 ~~e/7 + /+l,« + /) + ^Λ-l,2/i ~ e
n2n-\'
ι=l

Now the rank of X is 2 n — 2 and if we look at the n + 1st basis vector, we

ln x
see that t ~ is the minimal polynomial of X, thus X is subregular in
D

Ifg = §>o2n(K) and X i s the standard regular nilpotent element, then

an example of a Y E 3 β ( X) which is not a polynomial in X is:

~ e2n,n+\

Then X 7 = 7X = 0. Thus is Y where a polynomial in X, it would have to

be a multiple of X2n~2, which cannot happen for two reasons: First,
2 2n 2
Y Φ 0; and second, X ~ £ g.
If XG$ln(K) is nilpotent, then the conjugacy class of X (via
SLn(K)) is determined by the Jordan form of X, and thus by the sizes of
the Jordan blocks. When we say that

X~kl9k2,...,kr
 REDUCTIVE ALGEBRAIC GROUPS 141

we mean that the Jordan form of X has blocks of sizes kλ X kλ, k2X k2,
etc. The usual conventions is that ki >: ki+x for all /. Then a nilpotent
conjugacy class in %ln(K) is uniquely represented by a partition of n. The
regular class is n, and the subregular class is n — 1, 1.
Let X E §>ln(K) be nilpotent with Jordan form kx,k2, ,&5, and let
ry be the number of fcz that are equal to j . Springer and Steinberg have
shown that X is conjugate (via SLn(K)) to an element of §>on(K) if and
only if rj is even when j is even, and X is conjugate to an element of
%pn(K) if and only if η is even wheny is odd. In §$n(K) and §>o2n+ι(K)
the Jordan form determines the conjugacy class; in §>o2n(K), two different
conjugacy classes can have the same Jordan form, but then they are
conjugate via O2n(K) [16]. So to investigate i^(X) for X nilpotent and g
orthogonal or symplectic it is sufficient to exhaust the possible Jordan
forms for X.

LEMMA 3.4. Let V be a finite dimensional vector space over K (char K φ

2), and let Q = 3p(V) or §o(K). / / l E g is nilpotent, then we can
orthogonally decompose V and X:

where Xi E g • = §£(?)) or §o(J^ ) respectively and either

(a) Xt is regular in §> I( Vt) and Q^or
(b) Xj has Jordan form λ, λ where λ is odd in the symplectic case and
even in the orthogonal case.
Further, 3(3fl[X.)) = [polynomials in Xt} Π g ; ,

PROPOSITION 3.5. Let g = So^ίΓ) or %pn(K) with charK Φ 2 and let

X E g be nilpotent. Then

%(iQ(X)) — {polynomials in X) Π g

except in the case where g = §>o2m(K) and X has Jordan form λ, μ where λ
and μ are unequal and odd. In that case, {polynomials in X) Π g has
codimension one in h(hq(X)). In any event,

( g
) <rankg

and equality holds if and only if X is regular.

The proofs of 3.4 and 3.5 are straightforward. Note that 3.5 is
sufficient to prove Theorem A for the classical groups. Let us say
something about the exceptional case — which includes the case where
I E §>o2m(K) is regular. For convenience, we change the form to sum of
142 JOHN F. KURTZKE, JR.

squares. Let

0
X =
0 X2_

where Xx €Ξ §>oλ(K) and X2 €= So^K) are regular nilpotent. Let

0 A
W =
Y-A' 0.
where XλA = 0 and Λ * 2 = 0. Then W<Ξι(%Q(X)) but W is not a
polynomial in X.

THEOREM 3.7. Assume char AT ^ 2

1. If G — Spn(K) and x E G, ί/ιefl Z(ZG(x)°) consists of polynomials
in x.
2.IfG = SOn(K) orSpn(K) andx E G,

G (x)°) < r a n k G

equality holds if and only if x is regular. Therefore x E G is regular if

and only if ZG(x)° is abelian.

Proof. Let x = su be the Jordan decomposition oί x, H — ZG(s)°9

ί) = 3 g O),/the Cayley map, and N = f(u).

Then V = Kn = Vλ + F 2 + + F r + V[ + V^ + - + V'r where

= λ,I ands L, = λ"1/.
(a) If no λz is a λy, then

Now Z c ( x ) ° = Zff(N)° and the Lie algebra of this group is i^N). So

3(Si,(^)) — 8(Λ) θ ({polynomials in TV) Π ή) = ({polynomials in s} U
{polynomials in N}) Π ί). Therefore Z(ZG(x)°) consists of polynomials
in x, because .s and N are polynomials in x9 and because for any affine
group A,

(b) If some λf. = λ~ι, then one gets a pair of symplectic algebras in
hQ(s) — if λ f = ± 1 , then one gets one symplectic algebra. Proposition 2.6
allows us to use the same argument as above. This completes the sym-
plectic case.
 REDUCTIVE ALGEBRAIC GROUPS 143

2. G — SOn{K). The situation here is similar to the symplectic case —

everything is easy unless λi = λ"1 and then orthogonal algebras occur. For
then if N restricted to one of these algebras has just two unequal odd
blocks, then — a n d only t h e n — an element can appear in Z(ZG(x)°)
which is not a polynomial in x. Clearly the only time that dim Z(ZG(x)°)
could equal rank G is when x is regular. D

So now we have proved Theorem C and Theorems A and B for the

classical groups. To finish the proofs of Theorems A and B, we need only
work with the exceptional groups and algebras — this will be done in the
next two sections.

4. Semiregular elements. Let G be a reductive group and g its Lie

algebra. A unipotent element x G G (nilpotent element i G g ) is called
semiregular if whenever a semisimple element t E G centralizes x (or X),
then / E Z(G). Regular elements are semiregular; however, we shall use
semiregular mean semiregular and irregular. If chaxK is good, then
semiregular elements occur only in groups of type Dn, E69 EΊ, and £ 8 . In
this section we will prove that if g is E6, E7, or 2s8, char AT is good for g,
and X E g is semiregular, then

<rankg.

Thus, iq(X) is nonabelian. Unfortunately, our only proof of this is

computational, so we omit the details which can be found in [8].
E6 has only one semiregular class, while EΊ and Es each have two.
These classes are usually denoted by E6(aλ), E7(aλ), E7(a2)9 E%(ax)9 and
Es(a2). We use Elkington's paper [6] for representatives of these classes.
For our numbering of the roots, the reader should take a glance at Table
1.
First, we handle E6(aλ). Here we take

PROPOSITION 4.1. Let g = E6 with char K φ 2 or 3 and let X be as

above. Then,

dim i(%Q(X)) - 3 < rank g

Next, we move to EΊ. A representative for EΊ{ax) is

144 JOHN F. KURTZKE, JR.

This is the subregular class. A representative for EΊ(a2) is

γ
= eaχ

PROPOSITION 4.2. Lei g = EΊ with char K =£2 or 3 and let X and Y be

as above. Then,

dim 3(3<,(*)) = 5 < rank g

and

( ( ) ) = 4<rankg.

Finally, g = Es. A representative for Es(ax) is

X=eΛι + e«2 + ea3 + ^«4+«5 + e«6 + ^ 7 + ^ + e« 5 + Λ 8 ;

A'is subregular. A representative for Es(a2) is

dimje(y)=12[4].

PROPOSITION 4.3. Le/ g = £ 8 with char i ί ^ 2, 3, or 5 α«J feί X and Y

be as above. Then

a ( a g ( ) ) = 5 <rankg

and

dim5(5 f l (7)) = 4 < r a n k g .

This completes the case of the semiregular elements. The next section
deals with the rest of the irregular elements.

5. Non-semiregular elements. In this section we shall prove that if g is

an exceptional Lie algebra with charK good for g and if l E g is
nilpotent and neither regular non semiregular, then δ g (X) is nonabelian.
This will complete the proof of Theorem A. We will also prove Theorem B
for G 2 , which will complete the proof of Theorem B.
Let t be a root system. A subset Σ c t is called an integrally
closed subsystem of Ψ if Σ itself is a root system and if when-
ever nλaλ + +nrar E Ψ where nt G Z and at E Σ, then nλaλ + •
E Σ.
 REDUCTIVE ALGEBRAIC GROUPS 145

Let G be a reductive group and Gx a reductive subgroup of G such

that G, contains a maximal torus T of G. Then we call G, an integrally
closed subgroup of G if the root system of Gλ relative to T is an integrally
closed subsystem of the root system of G relative to T. We define
integrally closed subalgebra in the analogous manner, although we must
be careful in extending the roots to a larger Cartan subalgebra. Further,
we need char K to be good for the Lie algebra g.
If Σ is an integrally closed subsystem of Ψ, then Σ is called a maximal
subsystem if there are no integrally closed subsystems properly between Σ
and Ψ. From this, we define maximal integrally closed reductive subgroup
and subalgebra.

PROPOSITION 5.1. Let G be α reductive group and let x E G be

unipotent. Then there is a reductive integrally closed subgroup Gλ of G with
the same rank as G such that x is regular or semiregular in Gx.

The proof is in [16] — note that it holds in all characteristics. The Lie
algebra analogue is:

PROPOSITION 5.2. Let Q be a semisimple Lie algebra with char K good

for Q and let X E g be nilpotent. Then there is a semisimple integrally closed
subalgebra g j of Q such that X is regular or semiregular in g 1 . Further,
rank g, < rank g.

The problem we will run into is the case where the ranks are the same.
Anyway, the problem of determining nilpotent conjugacy classes in g
takes on a tractable form. First determine all the integrally closed subsys-
tems of a given root system. Then determine which root systems can have
semiregular elements. Finally, solve the conjugacy problem: Given Xi
regular or semiregular in g, (i — 1,2) integrally closed subalgebras of g,
then when are Xλ and X2 conjugate in g?
This was first done by Dynkin [4] in characteristic zero. The task of
showing that the Dynkin results hold in all good characteristics has been
completed by Pommerening [10, 11] (for a more complete list of citations,
see §1).

Dynkin's method for determining the integrally closed subsystems of

a root system Ψ is as follows. Take the lowest root, -μ, and attach it to
the Dynkin diagram for Ψ in the usual manner. That is, if a is a simple
root, then a and -μ are joined by

(-μ,-μ)(a, a)
146 JOHN F. KURTZKE, JR.

lines. If -μ and a have different lengths, then one draws an arrow from
the longer root to the shorter root. This gives the extended Dynkin
diagram of Ψ. Table 1 shows the extended diagrams for all the irreducible
root systems. From the extended diagram, one deletes one or more nodes.
This gives the diagram of an integrally closed subsystem of Ψ. Repeating
this process ultimately yields all the integrally closed subsystems of Φ.

PROPOSITION 5.3. (Dynkin). All the maximal integrally closed subsys-

tems are obtained by omitting a root whose coefficient in the highest root is
either one or a prime.

We proceed on the inductive assumption that we have proved Theo-

rem α for simpler systems — or if one prefers, induction is by dimension.
Thus is X is regular nilpotent in QX where Qj is an integrally closed
subsystem of g which is not maximal, then we know that there is a
semisimple subalgebra g' such that QX C g' C g is a chain of integrally
closed subalgebras. By induction, i^(X) is nonabelian. Of course if X
semiregular in g, then we already know (by §§3 and 4) that
dims(δ g (Jθ) < r a n k g .
So we are looking at the situation of a simple Lie algebra g (with
char K good) and J ί G g , C g where X is regular nilpotent in g, an
integrally closed subalgebra of g.

PROPOSITION 5.4. Let g be a simple Lie algebra with char K good for g.
Suppose X E g is nilpotent such that X is regular or semiregular in g 1? an
integrally closed subalgebra of g. Suppose further that rank gj < rank g.
Then

Therefore dim h(i%(X)) < rank g.

Proof. We have to show that if Y E g — g1 centralizes X, then there is

a W E 3g( JQ such that [Y,W]Φ 0. Let Ψ be the root system for g,. Since
3β (X) is nilpotent, we can choose orderings for Ψ and Φj such that
iQi(X) QQΪ = sρan{eγ | γ E % + }
and Ψj+ C t
+
. We use Δ to denote the base of Ψ in this ordering, and we
write
1
Δά uβ^β "^ Zi υaΓLa ' Zi " j δ ^
)β<0 αGΔ )β>0

with aβ, ba E K and we label the three pieces of Y as Y~ , 7°, and 7 + .

 REDUCTIVE ALGEBRAIC GROUPS 147

If there is a root δ E Ψ — Ψ, such that aδ Φ 0, then there is an H E ί)

(the Cartan subalgebra for g spanned by the // α ) such that [H, eδ] = eδ
and [//, g j] = 0. Then /f will commute with Xbut not with Y.
That leaves us with the case that both Γ~ and Γ + E g,. As the
Killing form is nondegenerate, we can write

Y° = HX+ H2
where Ht E ί), if} E gj and [i/ 2 , g j = 0. But then Y~ +Hλ + Y+ G
%Q[(X) c g^ , so Y~ = Hλ = 0. So what is really the case is that
Y= Y+ +H2
where Y+ E g^ and H2 is orthogonal to g^ Because H2 φ 0 (otherwise F
belongs to g^, there is a positive root δ E Ψ — Ψ{ such that δ is not
orthogonal to H2. We take δ to have maximal height; that is

(ii) If ht v > ht δ, then [H2, ev] = 0.

We now claim that eδ is the W that we want. First we must show that
eδ centralizes X:
[H2,[eδ,X]]=0
by maximality of δ and the fact that X is supported by positive roots. But
[H2,[es,X]]=-[es,[X,H2]] -[X,[H2,es]\ =[X,λes]
because H2 centralizes X. Therefore [X, λeδ] = 0 and so eδ E %Q(X).
Next we show [eδ, Y] φ 0. [ea, Y] = [e8, H2+ Y+] = -λeδ + higher
root vectors. Hence [eδ, Y] φ 0 and s o Γ ί g(3g(ΛΓ)).
Finally, if Λ'is regular in g 1? then

dimg(8 f l l (A r )) = rankg,

and if X is semiregular in g 1? then

In any event, then

dim a ( δ g ( X ) ) < r a n k g . D

The case where g, and g have the same rank is much harder. In fact, it
does not seem to be true that 3(3 g (X)) C g, in that case. However we are
still hopeful that it will turn out that dim g(g g ( JίQ) < rank g. What we did
was to compute just enough of %Q(X) to show that it was nonabelian.
Before we get to that, there is still one last thing that can be done.
148 JOHN F. KURTZKE, JR.

PROPOSITION 5.5. Let $ be a simple Lie algebra with char K good for g.
Suppose X E g is irregular and nilpotent such that X is regular in a maximal
integrally closed semisimple subalgebra g' which is not simple. Then 3 fl (X) is
nonabelian.

Proof. Let Ψ b e a root system for g with base Δ. Then a base Δ' for
Ψ\ the root system for g' is {-μ} U (Δ — {/?}) for some simple root β (μ
is the highest root). As Ψ' is reducible, there are at least two simple roots
α, and α 2 that are connected to β in the extended diagram (the case where
one of these is -μ is permitted). So ignoring root lengths and possibly
other roots connected to /?, we have

o—o—o
aλ β a2

It does no harm to assume that Xis standard regular in g', that is,

As Δ' is reducible, Δ' = Δ', ύ Δ'2 with α;. G Δ;.. Write X = Xx + X2 where
X
i ~ 2 β e Δ ;^«
Now let Y = e_ai_β + ae_ai_β where a = -(Na^aι_β/N _β) φ 0.
Then [X, Y] = 0 and clearly [X, Λ)] = 0 (i = 1,2).
a ι < ί β β
So 5gg ( Jf) is nonabelian. D

What remains is the following case: g is a simple Lie algebra (char K

good), g, C g is a maximal integrally closed subalgebra of g with g{
simple; X is regular nilpotent in g, such that X is not conjugate to an
element in a "smaller" subalgebra (i.e. one that is integrally closed but not
maximal). These are the possibilities: (cf. Table 1)
(a) g = G 2
X= e_μ + eβ(typeA2)
(b) g = F 4
X= e_μ + ea + eβ + eγ(type B4)
(c) g = E6
Any irreducible integrally closed subsystem which is maximal is E6, so
nothing happens here.
(d) Z = EΊ
y i 3 4 5 e n ^
We will show that X is conjugate to an element semiregular in a system of
type E6, so nothing happens here either.
 REDUCTIVE ALGEBRAIC GROUPS 149

(e) Q = Es
(i) X = e_μ + eaχ + ea2 + ea^ + ea^ + eas + eae + eaη (type As)
(ii) X=e_μ + eΛχ + eai + ea^ + ea4 + eas + eaβ + ea% (type D%).
First we tackle g = G2. X — e_μ + eβ. Dynkin has shown that X is
subregular, so dim %Q(X) = 4. Now X is conjugate to Y = ^_M + e α to
which we could apply Proposition 5.5, but that would not help us with
Theorem B. The following result can be checked by computation.

PROPOSITION 5.6. Let g = G2 with char^Γ φ 2 or 3, and let X—e_μ +

eβ. Then a basis for ig(X) is {X, e_a, e_2a_β, e_3a_β) and a basis for

In G 2 , there are no irregular semiregular elements. The only other

integrally closed subsystem of rank 2 is Aλ + Ax whose regular element is
the Y above. By Dynkin's tables X and Y are conjugate, so dim δ(δ g (Γ))
= 2. Note that for X and 7, the double centralizer has dimension 2 which
equals the rank.

PROPOSITION 5.7. Let G-G2 with char K^2or3.Ifx£G, then

dim Z( ZG(x)°) < rank G = 2

Equality can occur only when

1. x is regular, or
2. x is subregular, but not semisimple.

Proof. If x is subregular semisimple, then ZG(x)° is reductive of rank

2 and dimension 4, hence dim Z(ZG(x)°) — 1. If x — su is the Jordan
decomposition of a subregular element x which is neither unipotent nor
semisimple, then s and u belong to Z(ZG(x)°) and so dim Z(ZG(x)°) = 2.
Proposition 5.6 handles subregular unipotent elements and Proposition
5.4 handles everything else. D

This finishes the proof of Theorem B and proves Theorem A for G2.
We now move to F4 = g. The element X regular in a system of type B4 is
subregular in F4. Thus dim lQ{X) = 6.

PROPOSITION 5.8. Let g = F4 with char K ψ 2 or 3 and let

X= e_μ + ea + eβ + ey.

Then the following six elements form a basis for 8β(-Y):

X, e_8, e_β_2y_28, ea+β+y + aeβ+2y + be_μ,

^~ Ce-a-2β-2y-28-> a n e
" -a-β-y-8 " ^ ^e-β-2y-δ^
150 JOHN F. KURTZKE, JR.

for appropriate a, b, c, d E K. Moreover,

<3<rankg.

The proof is computational. To find the exact values of a, b, c, and d

(and then the exact dimension of s(5Q(X))), one must compute a number
of structure constants. In any event, a — ±2 and b, c, and d are ± 1. More
detail can be found in [8]. This proves Theorem A for F4.
As we have said, E6 has no irreducible maximal integrally closed
subsystems, so the proof of Theorem A is complete for E6 as well. To
handle E7, we have to describe the Dynkin diagram of a nilpotent element
in a simple Lie algebra g.
If g is simple and X E g is nilpotent, then there is a {X, Y, H) C g
with X and Ύ nilpotent and H semisimple so that [X9Y] = H, [H, X] =
2X and [H9Y] = -2Y. By taking the Dynkin diagram for g and putting
the number a^H) at the ai node, we get the Dynkin diagram for X. By
applying the Weyl group W to the base, we can always arrange matters so
that the numbers at{H) are 0, 1, or 2. A nilpotent element is regular if and
only if it has a 2 at every node in its diagram.
In our case, g = EΊ (char K φ 2 or 3), and X is regular in a subsystem
of type AΊ, with base α,, α 2 , «3, α 4 , α 5 , α 6 , -μ. Thus aλ{H) = a2(H) =
• = -μ(H) = 2. Since μ has an α 7 term, we can see that aΊ(H) = -16.
If we write the diagram of X with just the numbers at(H)9 we have
222222
-16
If w, is reflection in α, , then by applying w79 we get
2 2 2 - 14 2 2
16
Continuing, we eventually get
020202
0
Consulting the Dynkin tables (in [4] or [6]), we see that this is the diagram
for E6(aλ). That is, Jf is conjugate to an element which is semiregular in a
subsystem of type E6. So by Propositions 4.1 and 5.4, we see

Thus we have proven:

PROPOSITION 5.9. Let g = E6 or E7 with char K Φ 2 or 3. // X E g is

nilpotent and irregular, then iQ(X) is nonabelian.
 REDUCTIVE ALGEBRAIC GROUPS 151

That finishes the proof of Theorem A for E6 and E7, leaving us only
with g = Es. Unfortunately, the regular elements in the A% and Z>8
subsystems are not conjugate to anything in a smaller subsystem. Hence,
what we do is compute a large enough part of the centralizers to show that
they are nonabelian. This is presented in [8] with all the details.
To simplify notation, we will use the 8-tuple (nιn2n3n4n5n6nΊns) to
denote the root β = Σf=1/!,•«,• when ni > 0. For - β , we write
-(nιn2n3n4n5n6n7ns). Since ni is at most 6, this is unambiguous. The
proofs of the next two results are simply computations.
First we handle the As subsystem. So

Dynkin's tables tell us that dim iq(X) = 24.

PROPOSITION 5.10. Let g = E% with char K φ 2, 3, or 5 and let X be as

above. Let
e
~ ^-(12345423) ~*~ ^(11111100) {0\ 111110)

and
— ^(12345422) ^-(11111101) "*~ ^-(01112101) ^-(01111 111)

" ^-(00112111) ^-(00012211)*

Then Y and Z both centralize X but not each other and so iQ(X) is
nonabelian.

Lastly, we have the Ds case. So

From Dynkin's tables, we see that dim iQ{X) = 16 ([4] or [6]).

PROPOSITION 5.11. Let g = £ 8 with char K φ 2, 3, or 5 and let X be as

above. Let

Y ~ ^-(12345422) ~~ e(\ 1111101) ~*~ e(0\ 112101) ~~ e(00122101)

and

Z — ^-(12344322) ~ ^(11122101) ' ^(01222101)*

Then Y and Z both centralize X, but not each other. Therefore, iQ(X) is
nonabelian.

This finishes the proof of Theorem A.

152 JOHN F . KURTZKE, JR.

TABLE 1

Extended Dynkin diagrams for the irreducible root systems

1 2 3 n-l n

-μ

Λ. μ = α + 2α 2 + 2α
1 2 5 n-1 n

-μ 1 2 3 n-1 n

n-1

μ= + 2α
O- n-2
1 2 n-2
Vl+αn

μ = 3 α + 2]3
N
α jB -μ

O O
μ = 2α+ 3β+4Y+ 2δ
-μ α β Y

+ 3α + 2α^ + α^

-μ

-μ
E
7
2a^

-μ

1 2
3 4 5 6 7 + 6a5+4a6+2a7+3a8
 REDUCTIVE ALGEBRAIC GROUPS 153

ACKNOWLEDGEMENTS. This paper is a revision of the author's Ph.D.

dissertation at U.C.L.A. under the direction of Professor Robert Stein-
berg. The author is very much indebted to Professor Steinberg and wishes
to thank him here. The author also wishes to thank Professor Mario
Borelli for reading an earlier version of this paper, and the referee for his
suggestions.

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Received January 25, 1980 and in revised form November 30, 1981.

UNIVERSITY OF NOTRE DAME

NOTRE DAME, IN 46556