Varieties of elements of given order in simple algebraic groups
Claude Marion
Dipartimento di Matematica, Università degli Studi di Padova, Padova, Italy
[email protected]
Abstract.
Given a positive integer u and a simple algebraic group G defined over an algebraically closed field K of characteristic p, we derive properties about the subvariety G[u] of G consisting of elements of G of order dividing u. In particular, we determine the dimension of G[u], completing results of Lawther [7] in the special case where G is of adjoint type. We also apply our results to the study of finite simple quotients of triangle groups, giving further insight on a conjecture we proposed in [10] as well as proving that some finite quasisimple groups are not quotients of certain triangle groups.
Contents
-
1 Introduction
-
2 Preliminary results on centralizers
-
3 Proofs of Propositions 2 and 3
-
4 Proof of Theorem 4 for G of exceptional type
-
5 Some properties of Spinn(K)
-
6 Upper bounds for du(G) for G of classical type
-
6.1 G is of type Aℓ
-
6.2 G is of type Cℓ
-
6.3 G is of type Bℓ
-
6.4 G is of type Dℓ
-
7 Proofs of Theorems 4 and 5
-
8 Proof of Proposition 7
-
9 Proof of Theorem 9
-
10 Proof of Proposition 10
-
11 Some tables
1 Introduction
Let G be a reductive algebraic group defined over an algebraically closed field K of characteristic p (possibly equal to 0), C be a conjugacy class of G and u be a positive integer. In 2007 Guralnick [4] proved the following result:
Theorem 1**.**
[4, Theorem 1.1]**.
Given a reductive algebraic group G, a conjugacy class C of G and a positive integer u, the set {g∈G:gu∈C} is a finite union of conjugacy classes of G.
In this paper, we concentrate our attention to the case where G is connected and C={1} is the trivial conjugacy class of G. For a positive integer u, we let
[TABLE]
be the subvariety of G consisting of elements of G of order dividing u and set ju(G)=dimG[u]. We also let du(G) be the minimal dimension of a centralizer in G of an element of G of order dividing u. We are merely interested in determining ju(G) for every positive positive integer u (when G is a simple algebraic group).
For completeness, we begin by proving Guralnick’s result in the case where G is connected and C={1}. In the statement below, given g∈G, we let gG denote the conjugacy class of g in G.
Proposition 2**.**
Let G be a connected reductive algebraic group defined over an algebraically closed field K of characteristic p. Let u be a positive integer. Then the number of conjugacy classes of G of elements of order dividing u is finite. In particular G[u] is a finite union of conjugacy classes of G. Moreover dimG[u]=maxg∈G[u]dimgG and codim G[u]=du(G).
Given a simple algebraic group G defined over an algebraically closed field K of characteristic p, we denote by Gs.c. (respectively, Ga.) the simple algebraic group over K of simply connected (respectively, adjoint) type having the same Lie type and Lie rank as G. We prove the following result partially proved in [7, Theorem 3.11]:
Proposition 3**.**
Let G be simple algebraic group defined over an algebraically closed field K of characteristic p. Let u be a positive integer. Then
[TABLE]
In [7], given a positive integer u and a simple algebraic group G defined over an algebraically closed field K of characteristic p, Lawther gives a lower bound for du(G). Moreover, he proves that this lower bound is equal to du(G) in the case where G is of adjoint type. We establish the following result completing Lawther’s result. Recall that the Coxeter number of a simple algebraic group is defined to be the quotient of the number of roots in the root system associated to G by the number of such roots which are simple. Concretely, h=ℓ+1, 2ℓ, 2ℓ, 2ℓ−2, 6, 12, 12, 18 or 30 according respectively as G is of type Aℓ, Bℓ, Cℓ, Dℓ, G2, F4, E6. E7 or E8.
Theorem 4**.**
Let G be a simple algebraic group defined over an
algebraically closed field K of characteristic p. Let h be the
Coxeter number of G and u be a positive integer. Write h=zu+e and u=qv, where z and
e are nonnegative integers with 0≤e≤u−1 and q, v are
coprime positive integers such that q is a power of p and p does not divide v.
The following assertions hold.
- (i)
If G is of type Aℓ then
[TABLE]
except if p=2, u is even, h=zu and z is odd in
which case
[TABLE]
2. (ii)
If G is of type Bℓ then
[TABLE]
except if
- (a)
p=2, u≡2mod4, h=zu, and z≡u/2mod4 or z≡2mod4 in which case
[TABLE]
2. (b)
p=2, u≡4mod8, h=zu and z is odd in which
case
[TABLE]
3. (iii)
If G is of type Cℓ then
[TABLE]
except if p=2 and u is even in which case
[TABLE]
4. (iv)
If G is of type Dℓ then
[TABLE]
except if
- (a)
p=2, u=2, h=zu and z≡u/2mod4 in which
case
[TABLE]
2. (b)
p=2, u≡2mod4, u>2, h=zu and z≡u/2mod4 in which case
[TABLE]
3. (c)
p=2, u≡2mod4, e=u−2=0 and
z≡1mod4 in which case
[TABLE]
4. (d)
p=2, u≡4mod8, z is odd and h=zu
in which case
[TABLE]
5. (v)
If G is of exceptional type then du(Gs.c.)=du(Ga.) except if G is of type E7, p=2 and u∈{2,6,10,14,18} in which case
[TABLE]
We note that for G of exceptional type the only cases where the simply connected and the adjoint groups are not abstractly isomorphic is when G is of type E6 or E7 and p=3 or p=2 respectively.
Recall that if a simple algebraic group G over an algebraically closed field K of characteristic p is not of simply connected type nor of adjoint type then either G is abstractly isomorphic to SLn(K)/C where C⩽Z(SLn(K)), or G is of type Dℓ, p=2 and G is abstractly isomorphic to SO2ℓ(K) or a half-spin group HSpin2ℓ(K) where ℓ is even in the latter case. For simple algebraic groups that are neither of simply connected nor adjoint type we obtain the following result.
Theorem 5**.**
Let G be a simple algebraic group, neither of simply connected nor adjoint type, defined over an algebraically closed field K of characteristic p. Let u be a positive integer. Write u=qv where q and v are coprime positive integers such that q is a power of p and p does not divide v. Then the following assertions hold.
- (i)
If du(Gs.c.)=du(Ga.) then du(G)=du(Ga.)
2. (ii)
Suppose du(Gs.c.)=du(Ga.).
- (a)
If G=SLℓ+1(K)/C where C≤Z(SLℓ+1(K)), then du(G)=du(Ga.) if C has an element of order 2, otherwise du(G)=du(Gs.c.).
2. (b)
If G=SO2ℓ(K) where p=2, then du(G)=du(Ga.).
3. (c)
If G=HSpin2ℓ(K) where p=2 and ℓ is even, then du(G)=du(Gs.c.).
In [7] Lawther showed that if G is a simple algebraic group of rank ℓ with Coxeter number h over an algebraically closed field and u is a positive integer with u≥h then du(G)=ℓ. As a corollary of Theorems 4 and 5 we obtain the following result:
Corollary 6**.**
Let G be a simple algebraic group of rank ℓ over an algebraically closed field K of characteristic p. Let h be the Coxeter number of G and let u be a positive number. Then
- (i)
If u>h then du(G)=ℓ.
2. (ii)
Suppose u=h then du(G)=ℓ if G is (abstractly isomorphic to a group) of adjoint type, or p=2, or h is odd (and then G is of type A), or G=SLℓ+1(K)/C and C has an element of order 2 where C≤Z(SLℓ+1(K)), or G≅SO2ℓ(K) and p=2, or G is of type B or D and h≡0mod8, or G=E6(K), otherwise du(G)=ℓ+2.
Given a simple algebraic group G over an algebraically closed field, we also derive a further property of
[TABLE]
Proposition 7**.**
Let G be a simple algebraic group over an algebraically closed field K of characteristic p.
Then
[TABLE]
is a decreasing function of u.
We now apply our results to the study of finite quasisimple images of triangle groups. Recall that a finite group is quasisimple if it is perfect and the quotient by its centre is simple.
Every finite quasisimple group being 2-generated, given a finite quasisimple group G0 and a triple (a,b,c) of positive integers, a natural question to consider is whether G0 can be generated by two elements of orders dividing respectively a and b and having product of order dividing c. A finite group generated by two such elements is called an (a,b,c)-group. Equivalently, an (a,b,c)-group is a finite quotient of the triangle group T=Ta,b,c with presentation
[TABLE]
When investigating the finite (nonabelian) quasisimple quotients of T, we can assume that 1/a+1/b+1/c<1 as otherwise T is either soluble or T≅T2,3,5≅Alt5 (see [2]). The group T is then a hyperbolic triangle group. Without loss of generality, we will further assume that a≤b≤c and call (a,b,c) a hyperbolic triple of integers. (Indeed, Ta,b,c≅Ta′,b′,c′ for any permutation (a′,b′,c′) of (a,b,c).)
Recall that a finite quasisimple group G0 of Lie type occurs as the derived subgroup of the fixed point group of a simple algebraic group G, when char(K)=p is prime, under a Steinberg endomorphism F, i.e. G0=(GF)′. We use the standard notation G0=(GF)′=G(q) where q=pr for some positive integer r. (We include the possibility that G(q) is of twisted type.)
Given a simple algebraic group G and a triple (a,b,c) of positive integers, we say that (a,b,c) is rigid for G if the sum ja(G)+jb(G)+jc(G) is equal to 2dimG. When the latter sum is less (respectively, greater) than 2dimG, we say that (a,b,c) is reducible (respectively, nonrigid)
for G.
In [10] we conjectured the following finiteness result in the rigid case (first formulated for hyperbolic triples (a,b,c) of primes):
Conjecture 8**.**
Let G be a simple algebraic group defined over an algebraically closed field K of prime characteristic p and let (a,b,c) be a rigid triple of integers for G. Then there are only finitely many quasisimple groups G(pr) (of the form (GF)′ where F is a Steinberg endomorphism) that are (a,b,c)-generated.
In [6, Theorem 1.7], Larsen, Lubotzky and the author proved that Conjecture 8 holds except possibly if p divides abcd where d is the determinant of the Cartan matrix of G.
In a future paper [5] we will make further progress on Conjecture 8 by, in particular, completely settling it in the case where the finite group (GF)′ is simple. A crucial ingredient therein is the classification of the hyperbolic triples (a,b,c) of integers as reducible, rigid or nonrigid for a given simple algebraic group G. We determine this classification below.
Theorem 9**.**
The following assertions hold.
- (i)
The reducible hyperbolic triples (a,b,c) of integers for simple algebraic groups G of simply connected or adjoint type are exactly those given in Table 1. In particular there are no reducible hyperbolic triples of integers for simple algebraic groups of adjoint type.
2. (ii)
The rigid hyperbolic triples of integers (a,b,c) for simple algebraic groups G of simply connected type are exactly those given in Table 2.
3. (iii)
The rigid hyperbolic triples of integers (a,b,c) for simple algebraic groups G of adjoint type are exactly those given in Table 3.
4. (iv)
The classification of the reducible and the rigid hyperbolic triples of integers for SOn(K) is the same as for PSOn(K).
5. (v)
The classification of the reducible and the rigid hyperbolic triples of integers for HSpin2ℓ(K) is the same as for Spin2ℓ(K).
6. (vi)
If C⩽Z(SLn(K)) contains an involution then the classification of the reducible and the rigid hyperbolic triples of integers for SLn(K)/C is the same as for PSLn(K). Otherwise, the classification of the reducible and the rigid hyperbolic triples of integers for SLn(K)/C is the same as for SLn(K).
As Theorem 9 gives the list of the rigid triples of integers for a given simple algebraic group, it puts Conjecture 8 into a very concrete context. Concerning the reducible case, we prove the following nonexistence result.
Proposition 10**.**
Let G be a simple algebraic group defined over an algebraically closed field K of prime characteristic p and let (a,b,c) be a reducible triple of integers for G. Then a quasisimple group G(pr) (of the form (GF)′ where F is a Steinberg endomorphism) is never an (a,b,c)-group.
Theorem 9 which gives inclusively the list of the reducible triple of integers for a given simple algebraic group is one of the ingredients of the proof of Proposition 10. As an immediate corollary of Theorem 9 and Proposition 10 we obtain the following result.
Corollary 11**.**
If (G,p,(a,b,c)) is as in Table 1 then (GF)′=G(pr) is never an (a,b,c)-group.
Unless otherwise stated, we let G be a connected reductive algebraic group defined over an algebraically closed field K of characteristic p with maximal torus T. If T′ is a torus of G of dimension r we sometimes write T′=Tr.
Given a positive integer u, we let G[u] be the subvariety of G consisting of elements of order dividing u, we set
[TABLE]
and we let du(G) be the minimal dimension of a centralizer in G of an element of G of order dividing u.
For G simple, we fix some more notation as follows. We let Φ be the root system of G with respect to T and set Π={α1,…,αℓ}⊂Φ to be a set of simple roots of G, where ℓ is the rank of G.
We let h=∣Φ∣/rank(G)=∣Φ∣/ℓ be the Coxeter number of G. Recall that h=ℓ+1, 2ℓ, 2ℓ, 2ℓ−2, 6, 12, 12, 18 or 30 according respectively as G is of type Aℓ, Bℓ, Cℓ, Dℓ, G2, F4, E6. E7 or E8. Also we denote by Gs.c. (respectively, Ga.) the simple algebraic group over K of simply connected (respectively, adjoint) type having the same Lie type and Lie rank as G.
Unless otherwise stated given a positive integer u, we also write:
u=qv where q and v are coprime positive integers such that q is a power of p and p does not divide v,
h=zu+e where z,e are nonnegative integers such that 0≤e≤q−1,
h=αv+β where α,β are nonnegative integers such that 0≤α≤v−1,
α=γq+δ where γ,δ are nonnegative integers such that 0≤δ≤q−1.
An easy check yields γ=z. Note that if e=0 then β=δ=0. Furthermore, if β=0 then δ=0 if and only if e=0.
Finally for an nonnegative integer r, we let ϵr∈{0,1} be [math] if r is even, otherwise ϵr=1, and set σr∈{0,1} to be 1 if r=0, otherwise σr=0.
The outline of the paper is as follows. In §2 we give some preliminary results on centralizers. In §3 we give detailed proofs for Propositions 2 and 3. In §4, given a positive integer u, we determine du(G) for G a simple simply connected algebraic group of exceptional type over an algebraically closed field K. This establishes Theorem 4 for G of exceptional type. In §5 we recall some properties of the spin groups and determine when a preimage under the canonical map Spinn(K)→SOn(K) of a semisimple element of SOn(K) of a given order is also of that order. In §6 given a positive integer u, we determine precise upper bounds for du(G) where G is a simple algebraic group of classical type over an algebraically closed field. In §7 given a positive integer u, we determine du(G) for G of classical type, completing the proofs of Theorems 4 and 5. In §8 we prove that for a given simple algebraic group G over an algebraically closed field, du(G), seen as a function of u, is decreasing. This establishes Proposition 7. In §9 we classify the reducible and the rigid hyperbolic triples of integers for simple algebraic groups, establishing Theorem 9. In §10 we prove Proposition 10. Finally in §11 we collect Tables 4-8 which appear later in the paper.
Acknowledgements. The author thanks the MARIE CURIE and PISCOPIA research fellowship scheme and the University of Padova for their support. The research leading to these results has received funding from the European Comission, Seventh Framework Programme (FP7/2007-2013) under Grant Agreement 600376.
2 Preliminary results on centralizers
In this section, unless otherwise stated, G denotes a connected reductive group defined over an algebraically closed field K of characteristic p. Given a positive integer u, recall that du(G) is defined to be the minimal dimension of a centralizer in G of an element of G of order dividing u. We first recall some generalities on the centralizer in G of an element of G.
Lemma 2.1**.**
Let G be a connected reductive group defined over an algebraically closed field K of characteristic p. Let g be an element of G with Jordan decomposition g=xy where x is unipotent and y is semisimple. Let H=CG(y)0. The following assertions hold.
- (i)
We have x∈CG(y)0, CG(g)=CCG(y)(x) and CG(g)0=CCG(y)0(x)0.
2. (ii)
The group H=CG(y)0 is reductive and so H=H1⋯HrT′ for some nonnegative integer r and simple groups Hi and central torus T′. Morever each Hi is closed and normal in H, [Hi,Hj]=1 for i=j, and Hi∩H1…Hi−1Hi+1⋯Hr is finite for each i.
3. (iii)
Write x=x1⋯xrt where xi∈Hi and t∈T′. Then
[TABLE]
4. (iv)
dimCG(g)=dimCCG(y)0(x)=rank(T′)+∑i=1rdimCHi(xi).**
Proof.
We first consider part (i). Since G is connected reductive we have x∈CG(y)0 - see for example [9, Proposition 14.7]. It is well-known that CG(g)=CCG(y)(x). As CG(g)⩽CG(y), we have CG(g)0⩽CG(y)0. It now follows that CG(g)0⩽CCG(y)0(x) and so CG(g)0=CCG(y)0(x)0.
We consider part (ii). Since G is connected reductive and y is a semisimple element of G, CG(y)0 is connected reductive - see for example [9, Theorem 14.2]. Since H=CG(y)0 is connected reductive, we have H=[H,H]Z(H)0 where Z(H)0 is a central torus and [H,H] is semisimple - see for example [9, Corollary 8.22]. Part (ii) follows - see for example [9, Theorem 8.21]. Part (iii) is now an easy consequence of part (ii).
We now consider part (iv). We have
[TABLE]
Part (iv) now follows from parts (i)-(iii).
∎
We now give some properties of du(G). In the statement below, we adopt the notation of Lemma 2.1(ii) for the connected component of the centralizer of a semisimple element in a connected reductive algebraic group.
Proposition 2.2**.**
Let G be a connected reductive algebraic group over an algebraically closed field K of characteristic p. Let u be a positive integer. Write u=qv where u and v are coprime positive integers with q a power of p and p does not divide v. Then the following assertions hold.
- (i)
[TABLE]
2. (ii)
Let g=xy be the Jordan decomposition of an element g∈G where x and y are respectively the unipotent and semisimple parts of g. If g has order dividing u and du(G)=dimCG(g), then
[TABLE]
3. (iii)
If y is a semisimple element of G with CG(y)0=H1…HrT′ (as in Lemma 2.1(ii)) then
[TABLE]
4. (iv)
[TABLE]
Proof.
We first consider part (i). By definition du(G)=ming∈G[u] dimCG(g). Let g be any element of G[u]. Write g=xy where x and y are respectively the unipotent and semisimple parts of g. Clearly x∈G[q] and y∈G[v]. Following Lemma 2.1(i), we in fact have x∈(CG(y)0)[q] and CG(g)0=CCG(y)0(x)0. It follows that
[TABLE]
This establishes part (i).
We now consider part (ii). Note that x∈(CG(y)0)[q] and y∈G[v]. Since by assumption du(G)=dimCG(g) and by Lemma 2.1(iv) dimCG(g)=dimCCG(y)0(x), we obtain
[TABLE]
By part (i) du(G)=minl∈G[v]dq(CG(l)0) and so dq(CG(y)0)≥du(G).
However if dq(CG(y)0)>du(G) then dimCCG(y)0(x)>du(G), contradicting (2.1). Hence du(G)=dq(CG(y)0) as claimed.
We now consider part (iii). Recall that by Lemma 2.1(ii), CG(y)0 is connected reductive.
For x∈(CG(y)0)[q], write x=x1…xrt where xi∈Hi for 1≤i≤r and t∈T′.
By Lemma 2.1(iv) we have
[TABLE]
Hence
[TABLE]
where to obtain the final equality we used Lemma 2.1(ii)-(iii).
Part (iii) follows.
Finally part (iv) follows from parts (i) and (iii).
∎
The result below is the main ingredient in the proof of Proposition 3.
Lemma 2.3**.**
Let G be any group, C a normal subgroup of G, and ϕ:G→G/C be the canonical surjective map. Let g,k be any elements of G and set
[TABLE]
The following assertions hold.
- (i)
Either Gg,k=∅ or Gg,k is a coset of CG(g).
2. (ii)
We have CG/C(gC)=⋃k∈Cϕ(Gg,k).
3. (iii)
Suppose G is a simple algebraic group defined over an algebraically closed field K of characteristic p and ϕ is an isogeny. Then
[TABLE]
In particular,
dimCG/C(gC)=dimCG(g). Moreover if q is a power of p and g is a semisimple element of G then
[TABLE]
Proof.
We first consider part (i). Note that if g and gk are not conjugate then Gg,k=∅. Suppose that g and gk are conjugate, say l−1gl=gk for some l∈G.
Let x be any element of Gg,k. Then
[TABLE]
Hence lx−1g(lx−1)−1=g and xl−1∈CG(g). It follows that x∈CG(g)l and so Gg,k⊆CG(g)l.
Suppose now that x∈CG(g)l. Then x=yl for some y∈CG(g). Hence
[TABLE]
and so
x∈Gg,k. It follows that CG(g)l⊆Gg,k. We obtain Gg,k=CG(g)l, proving part (i).
We now consider part (ii). Suppose first that xC is any element of ϕ(Gg,k) where k∈C. Without loss of generality, x∈Gg,k. Hence
x−1gx=gk. Since k∈C, we obtain (xC)−1⋅gC⋅xC=gC and so xC∈CG/C(gC). Hence ⋃k∈Cϕ(Gg,k)⊆CG/C(gC).
Suppose now that xC∈CG/C(gC). Then
(xC)−1⋅gC⋅xC=gC and so (gC)−1⋅(xC)−1⋅gC⋅xC=C. It follows that there exists k∈C such that g−1x−1gx=k and so x∈Gg,k. Hence xC∈ϕ(Gg,k) with k∈C. This shows that CG/C(gC)⊆⋃k∈Cϕ(Gg,k). We obtain CG/C(gC)=⋃k∈Cϕ(Gg,k), proving part (ii).
Finally we consider part (iii). As G is simple and ϕ is an isogeny, ker ϕ=C is a finite central subgroup of G.
Also following part (ii), ϕ(CG(g))=CG(g)/C is a subgroup of CG/C(gC) of finite index and so CG/C(gC)0⩽(CG(g)/C)0. On the other hand, as CG(g)/C⩽CG/C(gC), we have (CG(g)/C)0⩽CG/C(gC)0. Hence
[TABLE]
Note that as CG(g)0 is connected so is the factor group CG(g)0/(C∩CG(g)0) - see for example [9, Proposition 1.10]. By the second isomorphism theorem, CG(g)0/(C∩CG(g)0) can be seen as a connected subgroup of CG(g)/C. As this subgroup is of finite index, we obtain
[TABLE]
As C is a finite group, it now follows from (2.2) that dimCG/C(gC)=dimCG(g). To conclude suppose that g is a semisimple element of G. By Lemma 2.1 CG(g)0 is a connected reductive group. By [7] given a simple algebraic group H defined over K, dq(H) is independent of the isogeny type of H (see Lemma 2.4 below).
It follows that dq(CG(g)0/(C∩CG(g)0))=dq(CG(g)0) and so by (2.2) dq(CG(g)0)=dq(CG/C(gC)0), as required.
∎
For matter of clarity we record [7, Lemma 2.5 and Corollary 2.6]:
Lemma 2.4**.**
Let G be a simple algebraic group defined over an algebraic closed field of characteristic p. Let q be a power of p. Then dq(G) does not depend on the isogeny type of G.
Furthermore, let α be a nonnegative integer and write α=γq+δ where 0≤δ≤q−1 and let ς∈{0,1}. Then the following assertions hold:
- (i)
dq(Aα−ς)=γ2q+(2γ+1)(α−ς−γq+1)−1.
2. (ii)
dq(B⌈2α⌉−ςϵα)=2γ2q+(2γ+1)(⌈2α⌉−ςϵα−2γq)+⌈2γ⌉ϵq.
3. (iii)
dq(C⌈2α⌉−ςϵα)=2γ2q+(2γ+1)(⌈2α⌉−ςϵα−2γq)+⌈2γ⌉ϵq.
4. (iv)
dq(D⌈2α+1⌉−ςϵα+1)=2γ2q+(2γ+1)(⌈2α+1⌉−ςϵα+1−2γq)+⌈2γ⌉ϵq−γ−ϵγ+2ςϵγ(α+1)σα−γq.
Corollary 2.5**.**
Let m be a natural number. The following assertions hold:
- (i)
If m+1=γq+δ with 0≤γ≤q−1, then dq(Am+1)−dq(Am)=2γ+1.
2. (ii)
If 2m+1=γq+δ with 0≤γ≤q−1, then dq(Bm+1)−dq(Bm)=2γ+1.
3. (iii)
If 2m+1=γq+δ with 0≤γ≤q−1, then dq(Cm+1)−dq(Cm)=2γ+1.
4. (iv)
If 2m=γq+δ with 0≤γ≤q−1, then dq(Dm+1)−dq(Dm)=2γ+1−2ϵγσδ.
Finally we record the following result:
Lemma 2.6**.**
[3, 6.17]**.
Let G be a simple algebraic group of rank ℓ over an algebraically closed field K and let x∈G. Then
[TABLE]
3 Proofs of Propositions 2 and 3
In this section we prove Propositions 2 and 3. We first give a detailed proof of Proposition 2.
Proof of Proposition 2.
Write u=qv where q and v are coprime positive integers such that q is a power of p and p does not divide v. Let g be an element of G of order dividing u. Let g=xy be the Jordan decomposition of g where x and y denote respectively the unipotent and semisimple parts of g. In particular the order of x divides q and that of y divides v. As G is connected, every semisimple element of G lies in a maximal torus of G, see for example [9, Corollary 6.11]. Since all maximal tori of G are conjugate, see for example [9, Corollary 6.5], it follows that there are only finitely many conjugacy classes of G of semisimple elements of order dividing v. Let y1,…,yr be representatives of the conjugacy classes of G of semisimple elements of order dividing v. For 1≤i≤r, CG(yi)0 is a connected reductive algebraic group
(see Lemma 2.1(ii)) and so by [8] CG(yi)0 has only finitely many unipotent classes. Let xi,1,…,xi,ti be representatives of the unipotent classes of CG(yi)0.
Now there exists 1≤i≤r such that y is conjugate in G to yi. Let k be an element of G such that kyk−1=yi. Then
[TABLE]
Note that kxk−1∈CG(yi). Indeed
[TABLE]
Since G is a connected reductive group and kxk−1⋅yi is the Jordan decomposition of kxk−1⋅yi, it follows that kxk−1 is a unipotent element of CG(yi)0 - see Lemma 2.1(i). Hence there exist l∈CG(yi)0 and 1≤j≤ti such that l⋅kxk−1⋅l−1=xi,j. Also as l∈CG(yi)0, we have lyil−1=yi. Hence
[TABLE]
and so g is conjugate to xi,jyi. It follows that there are at most ∑i=1rti conjugacy classes of elements of G of order dividing u. It follows that G[u] consists of the finite union of conjugacy classes of G of elements of order dividing u. Hence
[TABLE]
Since for any element g∈G, codim gG=dimCG(g) - see for example [3, Proposition 1.5], we obtain
[TABLE]
We can now prove Proposition 3.
Proof of Proposition 3.
First, by [7], we have du(Ga.)≤du(G). It remains to show that du(G)≤du(Gs.c.). Note that the result is trivial unless G is neither of adjoint nor simply connected type.
Let ϕ:Gs.c.→G be an isogeny. Let g be any element of Gs.c. of order dividing u such that du(Gs.c.)=dimCGs.c.(g). Then ϕ(g) is an element of G of order dividing u and by Lemma 2.3 dimCG(ϕ(g))=dimCGs.c.(g). Hence dimCG(ϕ(g))=du(Gs.c.). Since du(G)≤dimCG(ϕ(g)) we obtain du(G)≤du(Gs.c.). □
4 Proof of Theorem 4 for G of exceptional type
In this section, we determine du(G) for G a simple algebraic group of exceptional type defined over an algebraically closed field K of characteristic p. In particular we prove Theorem 4 for G of exceptional type (see Propositions 4.2 and 4.3 below). Recall that a simple algebraic group of exceptional type is either simply connected or adjoint. Following the result of Lawther [7] giving du(G) for G of adjoint type, we suppose that G is of simply connected type. Without loss of generality,
we assume that G is of type E6 or E7 and p=3 or 2 respectively, as
these are the only cases where the simply connected and the adjoint groups are
not abstractly isomorphic.
Let T be a maximal torus of G with corresponding root system Φ and set Π={α1,…,αℓ}⊂Φ to be a set of simple roots of G, where ℓ is the rank of G. Note that ℓ=6 or ℓ=7 according respectively as G=E6 or G=E7.
Given a positive integer u, we write u=qv where q and v are positive coprime integers with q a power of p and p does not divide v.
Let y be a semisimple element of G of order dividing v, i.e. y∈G[v]. Since every semisimple element of G belongs to a maximal torus of G and all maximal tori are conjugate, we can assume without loss of generality that y∈T.
Now
[TABLE]
where
Ψ={α∈Φ:α(y)=1} and Uα is the root subgroup of G corresponding to α.
Also we can write
[TABLE]
where for l∈K∗=K∖{0} and 1≤i≤ℓ, hαi(l) is the image of (l00l−1)∈SL2(K) under the canonical map SL2(K)→⟨Uαi,U−αi⟩, k is an element of K∗ of order v, and 0≤ci≤v−1 is an integer for all 1≤i≤ℓ.
Given α∈Φ we then have
[TABLE]
where
[TABLE]
and (,) denotes the inner product of the Euclidean space spanned by Φ.
Lemma 4.1**.**
Let G be a simple simply connected algebraic group of type X=E6 or E7 defined over an algebraically closed field K of characteristic p. Let u be a positive integer. Then du(G)=du(Ga.) except possibly if G=E6, p=3 and u≡0mod3, or G=E7, p=2 and u≡0mod2.
Proof.
Set C=Z(G) and note that
[TABLE]
Also note that if (X,p)∈{(E6,3),(E7,2)} then G is abstractly isomorphic to Ga. and so du(G)=du(Ga.) for every positive integer u. We therefore assume that (X,p)∈{(E6,3),(E7,2)}. Then C=Z(G)=⟨c⟩ is cyclic group of order 3 or 2 according respectively as X=E6 or E7. Let ϕ:G→Ga. be the surjective canonical map. Let gC be an element of Ga. of order dividing u such that dimCGa.(gC)=du(Ga.).
Suppose u=0mod∣C∣. Then g∈G has order dividing u, and by Lemma 2.3 dimCG(g)=dimCGa.(gC). Hence dimCG(g)=du(Ga.) and du(G)≤du(Ga.). It now follows from Proposition 3 that du(G)=du(Ga.).
∎
Proposition 4.2**.**
Suppose G is a simple simply connected algebraic group of type E6 defined over an algebraic closed field K of characteristic p. Let u be a positive integer. Then du(G)=du(Ga.).
Proof.
Note that by [7] du(G)≥du(Ga.) and du(Ga)=6 for u≥h=12.
By Lemma 4.1 we can assume without loss of generality that p=3 and u≡0mod3. Write u=qv where q and v are coprime positive integers such that q is a power of p and p does not divide v.
We first suppose that u≤h=12.
Assume u=3. Then q=1 and u=v=3. Let
[TABLE]
where k∈K∗ is an element of order 3. Then y3 is a semisimple element of G of order 3. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y3)0)=d1(A2A2A2)=24. Now by [7] du(Ga.)=24. We now deduce from Proposition 2.2(i) that du(G)=du(Ga.)=24.
Assume u=6. Then q=1 and u=v=6, or q=p=2 and v=3. Suppose first that q=1.
Let
[TABLE]
where k∈K∗ is an element of order 6. Then y6 is a semisimple element of G of order 6. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y6)0)=d1(A1A1A1T3)=12.
Suppose now that q=2 and v=3.
Consider the semisimple element y3 of G of order 3 defined above. We have CG(y3)0=A2A2A2 and dq(CG(y3)0)=d2(A2A2A2)=12.
Now by [7] du(Ga.)=12. We now deduce from Proposition 2.2(i) that du(G)=du(Ga.)=12.
Assume u=9. Then q=1 and u=v=9. Let
[TABLE]
where k∈K∗ is an element of order 9. Then y9 is a semisimple element of G of order 9. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y9)0)=d1(A1T5)=8. Now by [7] du(Ga.)=8. We now deduce from Proposition 2.2(i) that du(G)=du(Ga.)=8.
Assume u=12. Then q=1 and u=v=12, or q=4, p=2 and v=3. Suppose first that q=1.
Let
[TABLE]
where k∈K∗ is an element of order 12. Then y12 is a semisimple element of G of order 12. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y12)0)=d1(T6)=6.
Suppose now that q=4, p=2 and v=3.
Consider the semisimple element y3 of G of order 3 defined above. We have CG(y3)0=A2A2A2 and dq(CG(y3)0)=d4(A2A2A2)=6.
By Proposition 2.2(i) we obtain d12(G)=d12(Ga.)=6.
We now suppose that u>h=12. If v≥h then let
[TABLE]
where k∈K∗ is an element of order v. Then y is a semisimple element of G of order v. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y)0)=dq(T6)=6 and so by Proposition 2.2(i) du(G)=6.
We finally suppose that u>h=12 and v<h. As we are assuming that u≡0mod3 and p=3 we have v≡0mod3. Hence v∈{3,6,9}.
Assume v=3. As u>12 we have q≥5. Now y3 is a semisimple element of G of order 3 with CG(y3)0=A2A2A2. As q≥3, we have
dq(CG(y3)0)=dq(A2A2A2)=6.
Assume v=6. As u>12 we have q≥3. Now y6 is a semisimple element of G of order 6 with CG(y3)0=A1A1A1T3. As q≥2, we have
dq(CG(y6)0)=dq(A1A1A1T3)=6.
Assume v=9. As u>12 we have q≥2. Now y9 is a semisimple element of G of order 9 with CG(y9)0=A1T5. As q≥2, we have
dq(CG(y9)0)=dq(A1T5)=6.
Hence in the case where u>h and v<h, by Proposition 2.2(i), we get du(G)=6.
∎
Proposition 4.3**.**
Suppose G is a simple simply connected algebraic group of type E7 defined over an algebraic closed field K of characteristic p. Let u be a positive integer. The following assertions hold.
- (i)
We have du(G)=du(Ga.) unless p=2 and u∈{2,6,10,14,18}.
2. (ii)
Suppose p=2. Then d2(G)=du(Ga.)+6 and du(G)=du(Ga.)+2 for u∈{6,10,14,18}. More precisely,
[TABLE]
Proof.
Recall that by [7] du(G)≥du(Ga.), du(Ga)=7 for u≥h=18, and
[TABLE]
By Lemma 4.1 we can assume without loss of generality that p=2 and u≡0mod2, as otherwise du(G)=du(Ga.). Write u=qv where q and v are coprime positive integers such that q is a power of p and p does not divide v. Note that q is odd and v≡0mod2.
We first suppose that u≤h=18.
Assume u=2. Then q=1 and u=v=2. By [1, Table 6], d2(G)=69. Let
[TABLE]
where k∈K∗ is an element of order 2. Then y2 is a semisimple element of G of order 2. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y2)0)=d1(A1D6)=69=d2(G).
Assume u=4. Then q=1 and u=v=4. By [7] and [1, Table 6], d4(G)=d4(Ga)=33. Let
[TABLE]
where k∈K∗ is an element of order 4. Then y4 is a semisimple element of G of order 4. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y4)0)=d1(A3A3A1)=33=d4(G).
Assume u=6. Then q=1 and u=v=6, or q=p=3 and v=2. Suppose first that q=p=3 and v=2. By [1, Table 6] a semisimple element y of G of order 2 is
such that CG(y)0=A1D6 or E7. It follows from Proposition 2.2(i) that d6(G)=d3(A1D6)=23.
Suppose now that q=1. By [1, Table 6], d6(G)=23. Let
[TABLE]
where k∈K∗ is an element of order 6. Then y6 is a semisimple element of G of order 6. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y6)0)=d1(A3A1A1T2)=23=d6(G).
Assume u=8. Then q=1 and u=v=8. Let
[TABLE]
where k∈K∗ is an element of order 8. Then y8 is a semisimple element of G of order 8. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y8)0)=17. Since, by [7], d8(Ga)=17, we deduce from Proposition 2.2(i) that d8(G)=17.
Assume u=10. Then q=1 and u=v=10, or q=p=5 and v=2. Suppose first that q=p=5 and v=2. By [1, Table 6] a semisimple element y of G of order 2 is
such that CG(y)0=A1D6 or E7. It follows from Proposition 2.2(i) that d10(G)=d5(A1D6)=15.
Suppose now that q=1. By (4.1) and (4.2), a semisimple y element of G of order 10 is such that CG(y)0 is generated by at least four root subgroups Uα where α is a positive root of G.
It follows from [7] that d10(G)>d10(Ga.)=13. Hence by Lemma 2.6 d10(G)≥15. Let
[TABLE]
where k∈K∗ is an element of order 10. Then y10 is a semisimple element of G of order 10. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence
[TABLE]
It now follows from Proposition 2.2(i) that d10(G)=15.
Assume u=12. Then q=1 and u=v=12, or q=p=3 and v=4. Suppose first that q=p=3 and v=4. Consider the semisimple element y4 of G defined above. Then CG(y4)0=A3A3A1 and dq(CG(y4)0)=d3(A3A3A1)=11. Hence by Proposition 2.2(i), we obtain d12(G)≤11.
Suppose now that q=1 and u=v=12. Let
[TABLE]
where k∈K∗ is an element of order 12. Then y12 is a semisimple element of G of order 12. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y12)0)=11.
Since, by [7], d12(Ga)=11, we deduce that d12(G)=11 for all p.
Assume u=14. Then q=1 and u=v=14, or q=p=7 and v=2. Suppose first that q=p=7 and v=2. By [1, Table 6] a semisimple element y of G of order 2 is
such that CG(y)0=A1D6 or E7. It follows from Proposition 2.2(i) that d14(G)=d7(A1D6)=11.
Suppose now that q=1. By (4.1) and (4.2), a semisimple y element of G of order 14 is such that CG(y)0 is generated by at least two root subgroups Uα where α is a positive root of G. We deduce that CG(s)0=A1T6. It follows from [7] that d14(G)>d14(Ga.)=9. Hence by Lemma 2.6 d14(G)≥11. Let
[TABLE]
where k∈K∗ is an element of order 14. Then y14 is a semisimple elements of G of order 14. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence
[TABLE]
It now follows from Proposition 2.2(i) that d14(G)=11.
Assume u=16. Then q=1 and u=v=16. Let
[TABLE]
where k∈K∗ is an element of order 16. Then y16 is a semisimple element of G of order 16. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y16)0)=9. Since, by [7], d16(Ga)=9, we deduce from Proposition 2.2(i) that d16(G)=9.
Assume u=18. Then q=1 and u=v=18, or q=9, p=3 and v=2. Suppose first that q=9, p=3 and v=2. By [1, Table 6] a semisimple element y of G of order 2 is
such that CG(y)0=A1D6 or E7. It follows from Proposition 2.2(i) that d18(G)=d9(A1D6)=9.
Suppose now that q=1. By (4.1) and (4.2), a semisimple y element of G of order 18 is such that CG(y)0 is generated by at least one root subgroup Uα where α is a positive root of G. We deduce that CG(y)0=T7. It follows from [7] that d18(G)>d18(Ga.)=7. Hence by Lemma 2.6 d18(G)≥9. Let
[TABLE]
where k∈K∗ is an element of order 18. Then y18 is a semisimple element of G of order 18. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence
[TABLE]
It now follows from Proposition 2.2(i) that d18(G)=9.
We now suppose that u>h=18. Recall that p=2 and v≡0mod2. Assume v>h. Note that v≥20. Let
[TABLE]
where k∈K∗ is an element of order v. Then y is a semisimple element of G of order v≥20. Moreover by (4.1) and (4.2) we get
[TABLE]
Hence dq(CG(y)0)=dq(T7)=7 and so by Proposition 2.2(i) du(G)=7.
We finally suppose that u>h=18 and v≤h. Since v≡0mod2 we have v∈{2,4,6,8,10,12,14,16,18}.
Assume v=2. As u≥20 we have q≥10. Now y2 is a semisimple element of G of order 2 with CG(y2)0=A1D6. As q≥10, we have
dq(CG(y2)0)=dq(A1D6)=7.
Assume v=4. As u≥20 we have q≥5. Now y4 is a semisimple element of G of order 4 with CG(y4)0=A3A3A1. As q≥4, we have
dq(CG(y4)0)=dq(A3A3A1)=7.
Assume v=6. As u≥20 and p=2 we have q≥5. Now y6 is a semisimple element of G of order 6 with CG(y6)0=A3A1A1T2. As q≥4, we have
dq(CG(y6)0)=dq(A3A1A1T2)=7.
Assume v=8. As u≥20 we have q≥3. Now y8 is a semisimple element of G of order 8 with CG(y8)0=A2A1A1T3. As q≥3, we have
dq(CG(y8)0)=dq(A2A1A1T3)=7.
Assume v=10. As u≥20 and p=2 we have q≥3. Now y10 is a semisimple element of G of order 10 with CG(y10)0=A2A1T4. As q≥3, we have
dq(CG(y10)0)=dq(A2A1T4)=7.
Assume v∈{12,14}. As u≥20 and p=2 we have q≥3. Now y12 and y14 are semisimple elements of G of respective orders 12 and 14 with CG(y12)0=A1A1T5 and CG(y14)0=A1A1T5 . As q≥2, we have
dq(CG(y12)0)=dq(CG(y14)0)=dq(A1A1T5)=7.
Assume v∈{16,18}. As u≥20 and p=2 we have q≥3. Now y16 and y18 are semisimple elements of G of respective orders 16 and 18 with CG(y16)0=A1T6 and CG(y18)0=A1T6. As q≥2, we have
dq(CG(y16)0)=dq(CG(y18)0)=dq(A1T6)=7.
Hence in the case where u>h and v≤h, by Proposition 2.2(i), we get du(G)=7.
∎
5 Some properties of Spinn(K)
Let K now denote an algebraically closed field K of
characteristic p=2. We give a characterization of semisimple elements of Spinn(K) of a given order (see Lemma
5.1 below). Before doing so we recall some properties of
the group Spinn(K) and the canonical surjective map Spinn(K)→SOn(K), where n≥7 is a positive
integer. Write n=2ℓ or n=2ℓ+1 for some integer ℓ≥1, according respectively as n is even or odd.
Let V be the natural module for SOn(K) and let
[TABLE]
(according
respectively as n is even or odd) be a standard basis of V. That
is, (ei,fi)=1, (ei,ei)=(fi,fi)=0 for 1≤i≤ℓ,
(ei,fj)=(ei,ej)=(fi,fj)=0 for i=j such that 1≤i,j≤ℓ and if n is odd then (d,d)=1 and (d,ei)=(d,fi) for 1≤i≤ℓ. Here (,):V×V→K denotes the non-degenerate symmetric bilinear form
associated to V.
Let T0(V,K) be the K-algebra of polynomials in
e1,f1,…,eℓ,fℓ if n is even (respectively, in
e1,f1,…,eℓ,fℓ,d if n is odd). Also let I(V,K) be
the ideal of T0(V,K) generated by {vv−(v,v):v∈V} and let
C0(V,K)=T0(V,K)/I(V,K).
Let ∗:C0(V,K)→C0(V,K) be the K-linear map such
that for any even positive integer r
and elements x1,…,xr∈B, we have the equality
(x1…xr)∗=xr…x1 (in C0(V,K)). Note that for x in
V, we have x∗x=xx∗=xx=(x,x).
We can now define the spin group Spinn(K) as an abstract
group, namely Spinn(K) consists of the elements t of
C0(V,K) such that:
[TABLE]
and the map
[TABLE]
has determinant 1.
Given an element t of Spinn(K), consider the map ϕt:V→V
defined by ϕt(x)=txt−1 for x in V. Then for any x in V,
[TABLE]
and so ϕt is an element of SOn(K). In fact the map
ϕ:
[TABLE]
is a surjective homomorphism with kernel equal to {1,−1}.
We can now characterize a preimage of a semisimple element of SOn(K) of a given order under the canonical surjective map Spinn(K)→SOn(K), where n≥7 is a positive integer and K is an
algebraically closed field of characteristic p=2.
Lemma 5.1**.**
Let g be an element of SOn(K) defined over an
algebraically closed field K of characteristic p=2.
Suppose g is a semisimple element of SOn(K) of order u, and let ω be a u-th root of unity of K. Let ±w be
any of the two preimages of g under the canonical surjective map
ϕ:Spinn(K)→SOn(K). The following
assertions hold.
- (i)
The order of w in Spinn(K) is divisible by u.
2. (ii)
There exits t∈{±w} of order u if and only if u
is odd, or u is even and the number of eigenvalues of ϕ(t) of
the form ωi with i odd is divisible by 4.
3. (iii)
If neither w nor −w has order u, then u is even and
±w has order 2u.
Proof.
This is a classical result which follows from the properties of the
canonical surjective map ϕ:Spinn(K)→SOn(K) described above. We give a sketch of the argument.
We first consider part (i). Let g be any semisimple element of SOn(K) of order u.
Let w be a preimage of g under ϕ (so that −w is the other
preimage of g under ϕ) and let m be the order of w. We have
gm=ϕ(w)m=ϕ(wm)=ϕ(1)=1. Hence u, which is the order of
g, divides the order of a preimage of g under ϕ. This
establishes part (i).
We now consider parts (ii) and (iii). Let V be the natural module
for SOn(K). There is a standard basis B of V,
where B=(e1,f1,…,eℓ,fℓ) or (e1,f1,…,eℓ,fℓ,d) according respectively as n is even or odd, such
that the matrix of g with respect to B is diagonal
with entries di satisfying diu=1 for 1≤i≤n. Note
that if 1≤j≤ℓ then d2j−1=d2j−1 is a power of
ω, and if n is odd then dn=1. Moreover without loss of
generality d1=ω and d2=ω−1. For 1≤j≤ℓ, let gj be the element of SOn(K) such that, with
respect to B, gj is diagonal and
(gj)2j−1,2j−1=dj, (gj)2j,2j=dj+1,
(gj)k,k=1 for k∈{2j−1,2j}. Note that
g=g1…gℓ. In the decomposition
g=g1…gℓ delete the elements gj with 1≤i≤ℓ such that gj=1, and renumbering the elements gj
where 1≤j≤ℓ, if necessary, write
g=g1…gι where 1≤ι≤ℓ and
g1,…,gι are not the identity element.
Let 1≤j≤ι and let 1≤k≤ι be such that
(gj)2k−1,2k−1 and (gj)2k,2k are both not 1. Write
θj=(gj)2k−1,2k−1. Note that θj=1 is a power
of ω and θj−1=(gj)2k,2k. Moreover
gj(ek)=θjek, gj(fk)=θj−1fk and gj
fixes every other element of B. Write
θj=ωlj for some positive integer lj.
For a nonsingular element x in V, let Rx be the reflection:
[TABLE]
An easy check yields
[TABLE]
Let
[TABLE]
Then
xj∈C0(V,K) and xjxj∗=16(1−θj)2θj−1. Let
τj∈K be such that τj2=θj−1=ω−lj,
furthermore if u and lj are both odd take
τj=ω(u−lj)/2, and if lj is even take
τj=ω−lj/2. Let
[TABLE]
Then tj is an element of Spinn(K) and the image of tj
under the canonical surjective map ϕ:Spinn(K)→SOn(K) is gj. Furthermore, for any positive integer m,
[TABLE]
and so
[TABLE]
Finally, let t=t1…tι. Then t∈Spinn(K) and
ϕ(t)=g=g1…gι. Now for 1≤i,j≤ι,
ti and tj commute and so
[TABLE]
It follows from (5.1) that tu=1 if and only if u
is odd or u is even and the number of eigenvalues of ϕ(t)=g of
the form ωi with i odd is divisible by 4. Otherwise
tu=−1 and t2u=1.
Note that the other preimage of g under ϕ is −t and
(−t)u=tu if u is even, otherwise (−t)u=−tu. Parts (ii) and
(iii) now follow from part (i).
∎
6 Upper bounds for du(G) for G of classical type
Let G be a simple algebraic group of classical type defined over an algebraically closed field K of characteristic p. Let u be a positive integer. In this section, we give an upper bound for du(G).
Since by Proposition 3 du(G)≤du(Gs.c.) we assume without loss of generality that G is of simply connected type.
Unless otherwise stated, we use the notation introduced at the end of §1. Recall, we let h be the Coxeter number of G and write:
u=qv where q and v are coprime positive integers such that q is a power of p and p does not divide v,
h=zu+e where z,e are nonnegative integers such that 0≤e≤q−1,
h=αv+β where α,β are nonnegative integers such that 0≤α≤v−1,
α=γq+δ where γ,δ are nonnegative integers such that 0≤δ≤q−1.
An easy check yields γ=z. Also if e=0 then β=δ=0. Furthermore, if β=0 then δ=0 if and only if e=0.
Finally for an nonnegative integer r, we let ϵr∈{0,1} be [math] if r is even, otherwise ϵr=1, and set σr∈{0,1} to be 1 if r=0, otherwise σr=0.
Moreover, we also let ω be a v-th root of 1 in K and consider the following diagonal blocks M1, M2, M3, M4, M5, M6, M7, M8 and M9 where:
[TABLE]
[TABLE]
[TABLE]
[TABLE]
and if β≥2 is even then
[TABLE]
[TABLE]
and if v≥4 is even then
[TABLE]
[TABLE]
[TABLE]
We also write (1) for I1=diag(1) and (−1) for −I1=diag(−1). More generally given a positive integer i, we write (ωi) for diag(ωi) Finally, given some nonnegative integers r1,…,r11 and some nonnegative integers s1,…,sv−1we denote by
[TABLE]
the diagonal matrix consisting of ri blocks Mi for 1≤i≤9, Ir10, −Ir11 and ωjIsj for 1≤j≤v−1.
6.1 G is of type Aℓ
Let G=(Aℓ)s.c. with ℓ≥1 defined over an algebraically closed field K of characteristic p. Here ∣Φ∣=ℓ(ℓ+1) and h=ℓ+1.
Lemma 6.1**.**
Let G=(Aℓ)s.c. be defined over an algebraically closed field of characteristic p. Let u=qv be a positive integer where q and v are coprime positive integers such that q is a power of p and p does not divide v. Write h=zu+e=αv+β and α=zq+δ where z, e, α, β, δ are nonnegative integers such that e<u, β<v, and δ<q.
Let y be the element of Gs.c. of order v defined in Table 4 (see §11). Then
CG(y)0 and dq(CG(y)0) are given in Table 4 and dq(CG(y)0) is an upper bound for du(G).
Proof.
We first determine CG(y)0.
Suppose that ϵv=1 or (ϵv,ϵα)=(0,0). As any nontrivial eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α+ϵβ eigenvalues equal to 1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤⌊β/2⌋, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=AαβAα−1v−βTv−1.
Suppose that (ϵv,ϵα,ϵβ)=(0,1,1). As any nontrivial eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤(β−1)/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=AαβAα−1v−βTv−1.
Suppose that (ϵv,ϵα,ϵβ)=(0,1,0) and β≥2. As any eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤(β−2)/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=AαβAα−1v−βTv−1.
Suppose finally that (ϵv,ϵα)=(0,1) and β=0. As any eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α−1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=AαAα−1v−2Aα−2Tv−1.
Note that by Proposition 2.2(i), du(G)≤dq(CG(y)0) and so dq(CG(y)0) is an upper bound for du(G). It remains to calculate dq(CG(y)0).
By Lemma 2.4,
[TABLE]
[TABLE]
and
[TABLE]
where for determining dq(Aα−2) we do the Euclidean division of α−1 by q:
[TABLE]
Assume first that (ϵv,ϵα)=(0,1) or β>0. We then have
[TABLE]
By Proposition 2.2(iii), (6.1) and (6.2)
[TABLE]
Assume now that (ϵv,ϵα)=(0,1) and β=0. Since β=0, recall that e=0 if and only if δ=0. We then have
[TABLE]
By Proposition 2.2(iii), (6.1), (6.2) and (6.3)
[TABLE]
∎
6.2 G is of type Cℓ
Let G=(Cℓ)s.c be a simply connected group of type Cℓ with ℓ≥2 defined over an algebraically closed field K of characteristic p. Here ∣Φ∣=2ℓ2 and h=2ℓ.
Lemma 6.2**.**
Let G=(Cℓ)s.c be defined over an algebraically closed field of characteristic p.
Let u=qv be a positive integer where q and v are positive integers such that q is a power of p. Write h=zu+e=αv+β and α=zq+δ where z, e, α, β, δ are nonnegative integers such that e<u, β<v, and δ<q.
Let y be the element of G of order v defined in Table 5 (see §11). Then
CG(y)0 and dq(CG(y)0) are given in Table 5 and dq(CG(y)0) is an upper bound for du(G).
Proof.
Since h=2ℓ, note that if ϵv=1 then ϵα=ϵβ and ϵu=ϵq, whereas if ϵv=0 then ϵu=0, ϵq=1 and ϵβ=0. We first determine CG(y)0.
Suppose that ϵv=1. As any eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α+ϵβ eigenvalues equal to 1, [math] eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤⌊β/2⌋, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=Aα⌊2β⌋Aα−12v−1−⌊2β⌋C⌈2α⌉T2v−1.
Suppose that (ϵv,ϵα)=(0,0). As any eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=Aα2βAα−12v−1−2βC2α2T2v−1.
Suppose that (ϵv,ϵα)=(0,1) and β≥2. As any eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤(β−2)/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=Aα2β−1Aα−12v−2βC2α+12T2v−1.
Suppose finally that (ϵv,ϵα)=(0,1) and β=0. As any eigenvalue of y can be paired with its inverse, y is an element of G of order v. Also y has α−1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, and every other eigenvalue of y occurs with multiplicity α. Therefore CG(y)0=Aα−12v−1C2α+1C2α−1T2v−1.
By Proposition 2.2(i), du(G)≤dq(CG(y)0) and so dq(CG(y)0) is an upper bound for du(G). It remains to calculate dq(CG(y)0).
Note that by Lemma 2.4,
[TABLE]
[TABLE]
[TABLE]
and if α is odd then
[TABLE]
Assume first that ϵv=1. Recall that ϵu=ϵq and ϵα=ϵβ. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that (ϵv,ϵα)=(0,0). Recall that ϵq=1 and ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that (ϵv,ϵα)=(0,1) and β≥2. Recall that ϵq=1 and ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume finally that (ϵv,ϵα)=(0,1) and β=0. Recall that ϵq=1. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
∎
6.3 G is of type Bℓ
Let G=(Bℓ)s.c be a simply connected group of type Bℓ with ℓ≥3 defined over an algebraically closed field K of characteristic p. Here ∣Φ∣=2ℓ2, h=2ℓ and G=Spinh+1(K) if p=2, otherwise G=SOh+1(K). Given an element g in G, we let g be the image of g under the canonical surjective map G→SOh+1(K).
Lemma 6.3**.**
*Let G=(Bℓ)s.c where ℓ≥3 be defined over an algebraically closed field K of characteristic p. Let u=qv be a positive integer where q and v are positive integers such that q is a power of p and p does not divide v. Write h=zu+e=αv+β and α=zq+δ where z, e, α, β, δ are nonnegative integers such that e<u, β<v, and δ<q. Consider the following conditions:
- (a)
ϵv=1.
2. (b)
v=2.
3. (c)
(ϵv,ϵα)=(0,1), v>2 and (vmod4,βmod4,ℓmod4)∈{(0,0,0),(0,2,3),(2,0,3),(2,2,2)}.
4. (d)
(ϵv,ϵα)=(0,1)* and (vmod4,βmod4,ℓmod4)∈{(0,2,1),(2,2,0)}.*
5. (e)
(ϵv,ϵα)=(0,1), β>0 and (vmod4,βmod4,ℓmod4)∈{(0,0,2),(2,0,1)}.
6. (f)
(ϵv,ϵα)=(0,1), v>2, β=0 and (vmod4,ℓmod4)∈{(0,2),(2,1)}.
7. (g)
(ϵv,ϵα)=(0,0), v>2 and (vmod4,βmod4,ℓmod4)∈{(0,0,0),(0,2,3),(2,0,0),(2,2,3)}.
8. (h)
(ϵv,ϵα)=(0,0)* and (vmod4,βmod4,ℓmod4)=(2,2,1).*
9. (i)
(ϵv,ϵα)=(0,0), β>0 and (vmod4,βmod4,ℓmod4)∈{(0,2,1),(2,0,2)}.
10. (j)
(ϵv,ϵα)=(0,0), β>0 and (vmod4,βmod4,ℓmod4)∈(0,0,2).
11. (k)
(ϵv,ϵα)=(0,0), v>2, β=0 and (vmod4,ℓmod4)=(2,2).
Let y be the element of SOh+1(K) of order v defined in Table 6 (see §11). Then y has a preimage y in G of order v, and
CSOh+1(y)0 and dq(CG(y)0)=dq(CSOh+1(K)(y)0) are given in Table 6. Furthermore dq(CG(y)0) is an upper bound for du(G).
Proof.
Note that by Lemma 2.3(iii), dq(CG(y)0)=dq(CSOh+1(K)(y)0).
Since h=2ℓ, note that if ϵv=1 then ϵα=ϵβ and ϵu=ϵq, whereas if ϵv=0 then ϵu=0, ϵq=1 and ϵβ=0. Note also that if ℓ≡2mod4, v≡0mod4 and β=0 then α is odd, as otherwise on one hand we would have 2ℓ≡4mod8 and on the other 2ℓ=αv≡0mod8, a contradiction. We first determine CSOh+1(K)(y)0.
Suppose that case (a) holds so that v is odd. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Furthermore, since v is odd, by Lemma 5.1 it follows that y has a preimage y in G of order v. Also y has α+1+ϵβ eigenvalues equal to 1, [math] eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤⌊β/2⌋, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα⌊2β⌋Aα−12v−1−⌊2β⌋B⌈2α⌉T2v−1.
Suppose that case (b) holds so that v=2. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Now the number N of eigenvalues ωi=(−1)i of y with i odd is equal to ℓ, ℓ−1, ℓ+2 or ℓ+1 according respectively as ℓmod4 is 0, 1, 2 or 3. Since N is divisible by 4, it follows from Lemma 5.1 that y has a preimage y in G of order v. Also
[TABLE]
Suppose that case (c) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (c), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα2βAα−12v−1−2βB2α−1D2α+1T2v−1.
Suppose that case (d) or (e) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (d) or (e), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+2 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where if case (d) holds then i is any integer with 1≤∣i∣≤(β/2−1) and if case (e) holds then i any integer with 2≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα2β−1Aα−12v−2βB2α+1D2α+1T2v−1.
Suppose that case (f) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (f), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+2 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤v/2−1, and every other eigenvalue of y occurs with multiplicity α−1. Therefore CSOh+1(K)(y)0=Aα−12v−2Aα−2B2α+1D2α+1T2v−1.
Suppose that case (g) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (g), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα2βAα−12v−1−2βB2αD2αT2v−1.
Suppose that case (h) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (h), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤β/2, (α+1) eigenvalues equal to ω2v−1, (α+1) eigenvalues equal to ω−(2v−1) and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα2βAα−12v−2β−1B2αD2αT2v−1.
Suppose that case (i) or (j) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (i) or (j), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+2 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where if case (i) holds then i is any integer with 1≤∣i∣≤(β/2−1) and if case (j) holds then i any integer with 2≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα2β−1Aα−12v−2βB2αD2α+2T2v−1.
Suppose that case (k) holds. As any eigenvalue of y different from 1 can be paired with its inverse, y is an element of SOh+1(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumption of case (k), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+2 eigenvalues equal to −1, α eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤v/2−2, and every other eigenvalue of y occurs with multiplicity α−1. Therefore CSOh+1(K)(y)0=Aα−12v−2Aα−2B2αD2α+2T2v−1.
By Proposition 2.2(i), du(G)≤dq(CG(y)0) and so dq(CG(y)0) is an upper bound for du(G). It remains to calculate dq(CSOh+1(y)0)=dq(CG(y)0).
It follows from Lemma 2.4 that
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
if α is odd then
[TABLE]
[TABLE]
and if α is even then
[TABLE]
and
[TABLE]
Assume first that case (a) holds so that ϵv=1. Recall that ϵu=ϵq and ϵα=ϵβ. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (b) holds so that v=2, ϵu=ϵv=0 and ϵq=1. Recall that α=ℓ and β=0. Note also that e=0 if and only if δ=0. We have
[TABLE]
If α≡0mod4 then z is even if δ=0, and by Proposition 2.2(i)
[TABLE]
If α≡1mod4 then by Proposition 2.2(iii)
[TABLE]
If α≡2mod4 then z is even if δ=0, and by Proposition 2.2(iii)
[TABLE]
If α≡3mod4 then by Proposition 2.2(iii)
[TABLE]
Assume that case (c) holds. Recall that ϵq=ϵα=1 and ϵu=ϵv=ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (d) or case (e) holds. Recall that ϵq=ϵα=1 and ϵu=ϵv=ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (f) holds. Recall that ϵq=ϵα=1, ϵu=ϵv=ϵβ=0 and β=0. Also δ=0 if and only if e=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (g) or case (h) holds. Recall that ϵq=1 and ϵu=ϵv=ϵα=ϵβ=0. Also if δ=0 then z must be even.
We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (i) or case (j) holds. Recall that ϵq=1 and ϵu=ϵv=ϵα=ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (k) holds. Recall that ϵq=1, ϵu=ϵv=ϵα=ϵβ=0 and β=0. Also δ=0 if and only if e=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
∎
6.4 G is of type Dℓ
Let G=(Dℓ)s.c be a simply connected group of type Dℓ with ℓ≥4 defined over an algebraically closed field K of characteristic p. Here ∣Φ∣=2ℓ(ℓ−1), h=2ℓ−2 and G=Spinh+2(K) if p=2, otherwise G=SOh+2(K). Given an element g in G, we let g be the image of g under the canonical surjective map G→SOh+2(K).
Lemma 6.4**.**
*Let G=(Dℓ)s.c where ℓ≥4 be defined over an algebraically closed field K of characteristic p. Let u=qv be a positive integer where q and v are positive integers such that q is a power of p and p does not divide v. Write h=zu+e=αv+β and α=zq+δ where z, e, α, β, δ are nonnegative integers such that e<u, β<v, and δ<q. Consider the following conditions:
- (a)
ϵv=1.
2. (b)
v=2.
3. (c)
(ϵv,ϵα)=(0,1), v>2 and (vmod4,βmod4,ℓmod4)∈{(0,0,1),(0,2,0),(2,0,0),(2,2,3)}.
4. (d)
(ϵv,ϵα)=(0,1)* and (vmod4,βmod4,ℓmod4)=(0,2,2).*
5. (e)
(ϵv,ϵα)=(0,1), v>2 and (vmod4,βmod4,ℓmod4)=(2,2,1).
6. (f)
(ϵv,ϵα)=(0,1), β>0 and (vmod4,βmod4,ℓmod4)=(0,0,3).
7. (g)
(ϵv,ϵα)=(0,1), v>2, β>0, β=v−2 and (vmod4,βmod4,ℓmod4)=(2,0,2).
8. (h)
(ϵv,ϵα)=(0,1), v>2, β=v−2 and (vmod4,βmod4,ℓmod4)=(2,0,2).
9. (i)
(ϵv,ϵα)=(0,1), β=0 and (vmod4,βmod4,ℓmod4)=(0,0,3).
10. (j)
(ϵv,ϵα)=(0,1), v>2, β=0 and (vmod4,βmod4,ℓmod4)=(2,0,2).
11. (k)
(ϵv,ϵα)=(0,0), v>2 and (vmod4,βmod4,ℓmod4)∈{(0,0,1),(0,2,0),(2,0,1),(2,2,0)}.
12. (l)
(ϵv,ϵα)=(0,0)* and (vmod4,βmod4,ℓmod4)=(2,2,2).*
13. (m)
(ϵv,ϵα)=(0,0)* and (vmod4,βmod4,ℓmod4)=(0,0,3).*
14. (n)
(ϵv,ϵα)=(0,0)* and (vmod4,βmod4,ℓmod4)=(0,2,2).*
15. (o)
(ϵv,ϵα)=(0,0), v>2, β>0 and (vmod4,βmod4,ℓmod4)=(2,0,3).
16. (p)
(ϵv,ϵα)=(0,0), v>2, β=0 and (vmod4,ℓmod4)=(2,3).
Let y be the element of SOh+2(K) of order v defined in Table 7 (see §11). Then y has a preimage y in G of order v, and
CSOh+2(y)0 and dq(CG(y)0)=dq(CSOh+2(K)(y)0) are given in Table 7. Furthermore dq(CG(y)0) is an upper bound for du(G).
Proof.
Note that by Lemma 2.3(iii), dq(CG(y)0)=dq(CSOh+2(K)(y)0).
Since h=2ℓ−2, note that if ϵv=1 then ϵα=ϵβ and ϵu=ϵq, whereas if ϵv=0 then ϵu=0, ϵq=1 and ϵβ=0. We first determine CSOh+2(K)(y)0.
Suppose that case (a) holds so that v is odd. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Furthermore, since v is odd, by Lemma 5.1 it follows that y has a preimage y in G of order v. Also y has α+2−ϵβ eigenvalues equal to 1, [math] eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤⌈β/2⌉, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+1(K)(y)0=Aα⌈2β⌉Aα−12v−1−⌈2β⌉D⌈2α+1⌉T2v−1.
Suppose that case (b) holds so that v=2. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Now the number N of eigenvalues ωi=(−1)i of y with i odd is equal to ℓ, ℓ−1, ℓ+2 or ℓ+1 according respectively as ℓmod4 is 0, 1, 2 or 3. Since N is divisible by 4, it follows from Lemma 5.1 that y has a preimage y in G of order v. Also
[TABLE]
Suppose that case (c) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (c), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2α+1D2α+1T2v−1.
Suppose that case (d) or (e) holds. Note that if case (d) holds then h≡2mod8 and so β<v−2 as otherwise h=(α+1)v−2 which, under the assumptions of case (d), is equal to 6 modulo 8, a contradiction. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (d) or (e), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where if case (d) holds then i is any integer with 1≤∣i∣≤β/2−1 or ∣i∣=v/2−2 and if case (e) holds then i any integer with 1≤∣i∣≤β/2−1 or ∣i∣=v/2−1, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2α+1D2α+1T2v−1.
Suppose that case (f) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (f), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where if β=v−4 then i is any integer with 2≤∣i∣≤β/2 or ∣i∣=v/2−2 and if β=v−4 then i is any integer with 1≤∣i∣≤v/2−3 or ∣i∣=v/2−1, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2α+1D2α+1T2v−1.
Suppose that case (g) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (g), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤β/2 or ∣i∣=v/2−1, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2α+1D2α+1T2v−1.
Suppose that case (h) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (h), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+3 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤v/2−2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2v−2Aα−1D2α+1D2α+3T2v−1.
Suppose that case (i) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (i), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α−1 eigenvalues equal to −1, α eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤v/2−1, and every other eigenvalue of y occurs with multiplicity α+1. Therefore CSOh+2(K)(y)0=AαAα−12v−2D2α+1D2α−1T2v−1.
Suppose that case (j) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (j), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+1 eigenvalues equal to 1, α+1 eigenvalues equal to −1, α+1 eigenvalues equal to ω, α+1 eigenvalues equal to ω−1, α eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤v/2−2, α−1 eigenvalues equal to ω2v−1 and α−1 eigenvalues equal to ω−(2v−1). Therefore CSOh+2(K)(y)0=AαAα−12v−3Aα−2D2α+1D2α+1T2v−1.
Suppose that case (k) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (k), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+2 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2αD2α+2T2v−1.
Suppose that case (l) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (l), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α+2 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2−1 or ∣i∣=v/2−1, and every other eigenvalue of y occurs with multiplicity α. Therefore CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2αD2α+2T2v−1.
Suppose that case (m) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (m), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also if β=v−4 then y has α+2 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤β/2 or ∣i∣=v/2−2, and every other eigenvalue of y occurs with multiplicity α. On the other hand if β=v−4 then y has α+2 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2−1 or ∣i∣=β/2+1, and every other eigenvalue of y occurs with multiplicity α. Therefore in both cases CSOh+2(K)(y)0=Aα2βAα−12v−1−2βD2αD2α+2T2v−1.
Suppose that case (n) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (n), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also if β=v−2 then y has α+2 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤β/2−1 or ∣i∣=v/2−2, and every other eigenvalue of y occurs with multiplicity α. On the other hand if β=v−2 then y has α+2 eigenvalues equal to 1, α+2 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤v/2−2, and every other eigenvalue of y occurs with multiplicity α. Therefore
[TABLE]
Suppose that case (o) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (o), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also if β=v−2 then y has α+2 eigenvalues equal to 1, α eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 2≤∣i∣≤β/2 or ∣i∣=v/2−1, and every other eigenvalue of y occurs with multiplicity α. On the other hand if β=v−2 then y has α+2 eigenvalues equal to 1, α+2 eigenvalues equal to −1, α+1 eigenvalues equal to ωi where i is any integer with 1≤∣i∣≤v/2−2, and every other eigenvalue of y occurs with multiplicity α. Therefore
[TABLE]
Suppose that case (p) holds. As any eigenvalue of y can be paired with its inverse, y is an element of SOh+2(K) of order v. Let N be the number of eigenvalues ωi of y with i odd. Then
[TABLE]
From the assumptions of case (p), it follows that N is divisible by 4. Hence
by Lemma 5.1, y has a preimage y in G of order v.
Also y has α eigenvalues equal to 1, α+2 eigenvalues equal to −1, and every other eigenvalue of y occurs with multiplicity α. Therefore
[TABLE]
By Proposition 2.2(i), du(G)≤dq(CG(y)0) and so dq(CG(y)0) is an upper bound for du(G). It remains to calculate dq(CSOh+2(y)0)=dq(CG(y)0).
It follows from Lemma 2.4 that
[TABLE]
[TABLE]
[TABLE]
[TABLE]
if α is odd then
[TABLE]
[TABLE]
and if α is even then
[TABLE]
Assume first that case (a) holds so that ϵv=1. Recall that ϵu=ϵq and ϵα=ϵβ. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (b) holds so that v=2, ϵu=ϵv=0 and ϵq=1. Recall that α=ℓ−1 and β=0. Note also that e=0 if and only if δ=0. We have
[TABLE]
If α≡0mod2 then z is even if δ=0, and by Proposition 2.2(iii)
[TABLE]
If α≡1mod4 then by Proposition 2.2(iii)
[TABLE]
If α≡3mod4 then by Proposition 2.2(iii)
[TABLE]
Assume that case (c), (d), (e), (f) or (g) holds. Recall that ϵq=ϵα=1 and ϵu=ϵv=ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (h) holds. Recall that ϵq=ϵα=1, ϵu=ϵv=ϵβ=0 and β=v−2. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (i) holds. Recall that ϵq=ϵα=1, ϵu=ϵv=ϵβ=0 and β=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (j) holds. Recall that ϵq=ϵα=1, ϵu=ϵv=ϵβ=0 and β=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (k), (l), (m), (n) or (o) holds, and that β=v−2 if in cases (n) or (o). Recall that ϵq=1 and ϵu=ϵv=ϵα=ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume that case (n) or (o) holds, and β=v−2. Recall that ϵq=1 and ϵu=ϵv=ϵα=ϵβ=0. We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
Assume finally that case (p) holds. Recall that ϵq=1, ϵu=ϵv=ϵα=ϵβ=0 and β=0.
We have
[TABLE]
By Proposition 2.2(iii),
[TABLE]
∎
7 Proofs of Theorems 4 and 5
Recall that in §4 we proved Theorem 4 for G of exceptional type, see Propositions 4.2 and 4.3.
In this section given a positive integer u and a simple algebraic group G of classical type over an algebraically closed field K of characteristic p, we determine du(G).
We begin with the case where G is of simply connected type and complete the proof of Theorem 4. We will then consider G of neither simply connected type nor adjoint type and prove Theorem 5.
Proof of Theorem 4.
As noticed at the beginning of this section, following Propositions 4.2 and 4.3 we can assume that G is simply connected of classical type. By [7, Theorem 1], where Lawther determines du(Ga.) and shows that du(G)≥du(Ga.), and Lemmas 6.1, 6.2, 6.3 and 6.4, where the upper bounds for du(G) are determined accordingly respectively as G is of type Aℓ, Cℓ, Bℓ, or Dℓ, we get that du(G)=du(Ga.) except possibly if one of the cases in Table 8 (see §11) holds.
It remains to show that in the cases appearing in Table 8 the upper bounds given for du(G)−du(Ga.) are in fact the precise values for du(G)−du(Ga.). As usual, we let h be the Coxeter number of G and write: h=zu+e=αv+β and α=zq+δ where z, e, α, β, δ are nonnegative integers such that e<u, β<v, and δ<q. Noting that in the cases appearing in Table 8 u is even, we set s=v/2.
Suppose first that G is of type Aℓ. By Table 8, we have to consider the case where p=2, u is even, h=zu and z is odd. Since e=0, we have β=δ=0. Also since p=2 and z is odd, q and α are also odd.
Since 0≤du(G)−du(Ga.)≤2, by Lemma 2.6 it suffices to show that du(G)−du(Ga.)>0.
Under the assumptions on h, u and z, adapting the proof of of [7, Proposition 3.3], we obtain that du(Ga.) is attained for an element of Ga. whose semisimple part y of order v has centralizer satisfying
CGa.(y)0=Aα−1vTv−1. Moreover du(Ga.)=dq(Aα−1vTv−1).
By the assumptions on h, u and z it follows that y is not the image of an element of SLh(K) of order dividing v under the canonical map SLh(K)→PSLh(K). Hence du(G)−du(Ga.)>0 and du(G)=du(Ga.)+2 as required.
Suppose now that G is of type Bℓ. By Table 8, we have to consider the case where p=2, u is even, h=zu and z is odd unless u≡2mod4 and z≡2mod4; moreover if z is odd then either u≡2mod4 and z≡u/2mod4, or u≡4mod8. Since e=0, we have β=δ=0 and α>0. Also since p=2, q is also odd.
Write s=v/2 and let a and b be nonnegative integers such that ℓ−1=as+b with 0≤b≤s−1. Under the assumptions on h, u and z we have a=α−1 and b=s−1.
Since 0≤du(G)−du(Ga.)≤2, by Lemma 2.6 it suffices to show that du(G)−du(Ga.)>0. Under the assumptions on h, u and z, adapting the proof of [7, Proposition 3.6], we obtain that du(Ga.) is attained for an element of Ga. whose semisimple part y of order v has centralizer satisfying
[TABLE]
Moreover
[TABLE]
By the assumptions on h, u and z it follows from Lemma 5.1 that y is not the image of an element of Spinh+1(K) of order dividing v under the canonical map Spinh+1(K)→SOh+1(K). Hence du(G)−du(Ga.)>0 and du(G)=du(Ga.)+2 as required.
Suppose now that G is of type Cℓ. By Table 8, we have to consider the case where p=2 and u is even.
Let y∈G[v] be of order v and such that Z=CG(y)0 is of minimal dimension. We first show that dq(Z)−du(Ga.)=2⌈z/2⌉.
There are essentially three possibilities for the structure of Z=CG(y)0. Indeed either y has no eigenvalues equal to 1 or −1 in which case Z has no C factors, or y has some eigenvalues equal to 1 or −1 but not both in which case Z has one C factor, or y has some eigenvalues equal to 1 and −1 in which case Z has two C factors.
Using Corollary 2.5 and arguing in a similar way as in the proof of [7, Proposition 3.4] we can assume that either Z=AabAa−1s−1−bTs−1 where a,b are nonnegative integers such that ℓ=a(s−1)+b and 0≤b<s−1, or Z=Aa⌊2b⌋Aa−1s−1−⌊2b⌋C⌈2a⌉Ts−1 where a,b are nonnegative integers such that 2ℓ=a(2s−1)+b and 0≤b<2s−1, or Z=Aab−ϵa(1−σb)Aa−1s−1−b+ϵa(1−σb)C⌈2a⌉C⌈2a⌉−ϵaσbTs−1 where a,b are nonnegative integers such that ℓ=as+b and 0≤b<s.
Suppose first that Z=AabAa−1s−1−bTs−1 where a,b are nonnegative integers such that ℓ=a(s−1)+b and 0≤b<s−1. Note that v>2 and so s>1. Write a=cq+d where c,d are nonnegative integers such that 0≤d<q. It follows from Lemma 2.4 that
[TABLE]
An easy check yields h=zu+e=c(u−2q)+f where f=d(v−2)+2b satisfies 0≤f≤u−2q−2. In particular c≥z. Write c=z+j where j≥0.
[7, Theorem 1] which gives the value of du(Ga.) now yields
[TABLE]
We claim that z≥j−1. Suppose not. Then c≥2z+2 and so we get
[TABLE]
Since v>2, we have u≥4q and so we get 0≥1, a contradiction. Therefore z≥j−1 as claimed. It follows that
[TABLE]
Suppose now that Z=Aa⌊2b⌋Aa−1s−1−⌊2b⌋C⌈2a⌉Ts−1 where a,b are nonnegative integers such that 2ℓ=a(2s−1)+b and 0≤b<2s−1. Write a=cq+d where c,d are nonnegative integers such that 0≤d<q. It follows from Lemma 2.4 that
[TABLE]
An easy check yields h=zu+e=c(u−q)+f where f=d(v−1)+b satisfies 0≤f≤u−q−1. In particular c≥z. Write c=z+j where j≥0.
[7, Theorem 1] which gives the value of du(Ga.) now yields
[TABLE]
We claim that z≥j−1. Suppose not. Then c≥2z+2 and so we get
[TABLE]
We have u≥2q and so we get 0≥1, a contradiction. Therefore z≥j−1 as claimed. It follows that
[TABLE]
Suppose finally that Z=Aab−ϵa(1−σb)Aa−1s−1−b+ϵa(1−σb)C⌈2a⌉C⌈2a⌉−ϵaσbTs−1 where a,b are nonnegative integers such that ℓ=as+b and 0≤b<s. Writing a=cq+d where c,d are nonnegative integers such that 0≤d<q one can show that c=z and Lemma 2.4 yields dq(Z)−du(Ga.)=2⌈z/2⌉ (see the proof of [7, Proposition 3.4]) for more details.
We have in fact showed that if G is simply connected of type Cℓ, p=2, v is even and y is a semisimple element of G of order v having centralizer Z of minimal dimension then dq(Z)=dqv(Ga.)+2⌈z/2⌉ where z=⌊h/(qv)⌋. Since
[TABLE]
and for any divisor r of v we have ⌊rh/(qv)⌋≥⌊h/qv⌋ and, by [7, Lemma 1.3], dqv/r(Ga.)≥dqv(Ga.), we deduce that du(G)=dq(Z)=du(Ga.)+2⌈z/2⌉.
Suppose finally that G is of type Dℓ. By Table 8, we have to consider the case where p=2, z is odd, and u≡2mod4 or u≡4mod8. Moreover if u≡4mod8 then h=zu. Finally if u≡2mod4 then h=zu and z≡u/2mod4, or e=u−2=0 and z≡1mod4. Since v is even, β is also even. Also since p=2, q is odd. Finally under the assumptions on h, u and z, α is odd.
Write s=v/2 and let a and b be nonnegative integers such that ℓ−1=as+b with 0≤b≤s−1. We have a=α is odd and b=β/2.
Suppose first that u=2 so that q=1 and v=2. Corollary 2.5 and Lemma 5.1 yield that an element y of G of order u with centralizer of minimal dimension has centralizer with connected component D(z+3)/2D(z−1)/2 and du(G)=du(Ga.)+4.
Suppose now that u>2. Then 0≤du(G)−du(Ga.)≤2, and so by Lemma 2.6 it suffices to show that du(Gs.c.)−du(Ga.)>0.
Under the assumptions on h, u and z, adapting the proof of [7, Proposition 3.5], we obtain that du(Ga.) is attained for an element of Ga. whose semisimple part y of order v has centralizer satisfying
[TABLE]
Moreover
[TABLE]
By the assumptions on h, u and z, it follows from Lemma 5.1 that y is not the image of an element of Spinh+2(K) of order dividing v under the canonical map Spinh+2(K)→SOh+2(K). Hence du(G)−du(Ga.)>0 and du(Gs.c.)=du(Ga.)+2 as required.
□
Given a positive integer u, we can now determine du(G) for G a simple algebraic group of neither simply connected nor adjoint type and prove Theorem 5.
Proof of Theorem 5. Note that since G is neither simply connected nor adjoint, G is of classical type. We first consider part (i). If du(Gs.c.)=du(Ga.) then Proposition 3 yields du(G)=du(Ga.).
We now consider part (ii) and assume that du(Gs.c.)=du(Ga.). In particular, by Theorem 4, p=2 and u is even (and so v is also even). By Proposition 3 we have du(Ga.)≤du(G)≤du(Gs.c.).
(a) Suppose first that G=SLℓ+1(K)/C where C≤Z(SLℓ+1(K)) is a central subgroup of Gs.c.. Note that C is finite and cyclic and write C=⟨c⟩. Since du(Gs.c.)=du(Ga.) Theorem 4 yields h=zu=zqv where z is odd.
Assume (u,∣C∣)=1. Let gC∈G be an element of order dividing u. Then gu∈C and say the eigenvalue of gu is cl for some 0≤l≤∣C∣−1.
Since (u,∣C∣)=1, there is a positive integer j such that ju≡1mod∣C∣. Let k be a positive integer such that k≡−jlmod∣C∣ and set g′=g⋅diag(ck,…,ck). Then g′ is an element of Gs.c. of order dividing u, g′C=gC and by Lemma 2.3 dimCGs.c.(g′)=dimCG(g′C). It follows that du(G)=du(Gs.c.).
Assume now that (u,∣C∣)>1. We claim that du(G)=du(Gs.c.) if ∣C∣ is odd, otherwise du(G)=du(Ga.). By Lemma 2.6, Proposition 3 and Theorem 4, du(G)≡ℓmod2, du(Ga.)≤du(G)≤du(Gs.c.) and du(Gs.c)−du(Ga.)=2. Hence either du(G)=du(Ga.) or du(G)=du(Gs.c.).
Suppose du(G)=du(Ga.). Let gC be an element of G of order dividing u such that dimCG(gC)=du(Ga.). Write g=xy where x and y are respectively the unipotent and semisimple parts of g. As gu∈C and (p,∣C∣)=1, the order of x divides q and the order of y divides v∣C∣. By Lemma 2.3 du(Ga.)=dimCGs.c.(g) and so yv=1, as otherwise g has order dividing u and du(Ga)=du(Gs.c.), a contradiction.
Furthermore, y must have v distinct eigenvalues, each repeated zq times. These v distinct eigenvalues must be ω1+jk,ω−(1+jk) where 0≤j≤(v−2)/2, k>1 is a divisor of ∣C∣ and ω∈K is a primitive kv-th root of unity. Considering 1=yv∈C we deduce that ωv=ω−v and so ωv has order 2. Hence 2 divides ∣C∣. It follows that if ∣C∣ is odd then du(G)=du(Gs.c.).
Suppose that ∣C∣ is even. Consider g a semisimple element of Gs.c. having v distinct eigenvalues, each repeated zq times, equal to ω1+2j,ω−(1+2j) where 0≤j≤(v−2)/2, ω∈K is a primitive 2v-th root of unity. Then gC is an element of G of order v and CG(gC)0=Aα−1vTv−1. It follows that dq(CG(gC)0)=du(Ga.) and so du(G)=du(Ga.).
(b) Suppose G=SO2ℓ(K). Using the notation of Lemma 6.4 and Theorem 4, we let ω∈K be a primitive v-th root of 1 and y∈G be a semisimple element of order v such that
[TABLE]
Then
[TABLE]
An easy check yields that dq(CG(y))=du(Ga.) and so du(G)=du(Ga.).
(c) Finally suppose that G=HSpin2ℓ(K) where p=2 and ℓ is even.
Write H=Gs.c. and let C=Z(H)=⟨c1,c2⟩, C1=⟨c1⟩, C2=⟨c2⟩ where c1,c2 are of order 2 and H/C=Ga., H/C1=G and H/C2=SO2ℓ(K).
Suppose first that u=2. Let hC1 be an element of H/C1 of order 2 satisfying
[TABLE]
By Lemma 2.3 we have
[TABLE]
Since hC1 has order 2, either h2=1 or h2=c1. Also note that hC has order dividing 2. If hC is trivial then h∈C and so by (7.1) dimCH/C1(hC1)=dimH/C1, a contradiction. Hence hC has order 2.
We claim that h2=c1. Suppose not. Then hC is an element of H/C of order 2 and hC2 is an element of H/C2 of order 4. It follows that hC2 has ℓ eigenvalues equal to ω and ℓ eigenvalues equal to ω−1 where ω∈K is a primitive fourth root of unity. Hence CH/C2(hC2)=Aℓ. Using (7.1), it follows that d2(G)>d2(Gs.c.), contradicting d2(G)≤d2(Gs.c.).
Hence h2=1 and so by (7.1) d2(G)≥d2(Gs.c.). Therefore d2(G)=d2(Gs.c.).
Suppose now that u>2. Since by Theorem 4 du(Gs.c.)−du(Ga.)=2, in order to show that du(G)=du(Gs.c.) it is enough by Lemma 2.6 to prove that du(G)=du(Ga.). Suppose for a contradiction that du(G)=du(Ga.). Let hC1 be an element of G=H/C1 of order u such that
[TABLE]
By Lemma 2.3 we have
[TABLE]
Since hC1 has order u, hu∈C1 and so hu=1 or hu=c1. Also hC has order dividing u whereas hC2 has order dividing u or 2u.
Suppose hC2 has order dividing u. Then hu∈C1∩C2=1 and so h is an element of Gs.c. of oder dividing u. Hence by (7.2), du(Gs.c.)=du(Ga.), a contradiction.
Suppose hC2 has order dividing 2u but (hC2)u=C2. Then hu=c1 and (hC2)u=c1C2. Let h=xy be the Jordan decomposition of h where x is unipotent and y is semisimple. Note that (yC2)v=c1C2. Without loss of generality, yC2 is diagonal and (yC2)i,i(yC2)i+1,i+1=1 for every odd i with 1≤i≤2ℓ−1. Note that no eigenvalue of yC2 is equal to 1 or −1. Let d be a semisimple element of H such that dC2 is diagonal, (dC2)i,i(dC2)i+1,i+1=1 for every odd i with 1≤i≤2ℓ−1 and dC2 has ℓ eigenvalues equal to ω and ℓ eigenvalues equal to ω−1 where ω∈K is a primitive 2v-th root of unity. Then (dC2)v=c1C2 and (dyC2)v=C2.
Hence dyC2 is an element of H/C2 of order dividing v and CH/C2(dyC2)≅CH/C2(yC2).
Since ℓ is even, it follows from Lemma 5.1 that d is a semisimple element of H of order dividing 2v and so dy is a semisimple element of H of order dividing v.
Now by Lemma 2.3
[TABLE]
and so we obtain
[TABLE]
Hence du(Gs.c.)=du(Ga.), a contradiction.
□
8 Proof of Proposition 7
Given a simple algebraic group G defined over an algebraically closed field K of characteristic p, we now show that du(G):N→N is a decreasing function of u.
We need to show that for every positive integer u, we have du(G)−du+1(G)≥0. Now by Proposition 3, we have du(Ga.)≤du(G)≤du(Gs.c.) for every positive integer u. Hence the conclusion of the proposition will follow at once after we show that for every positive integer u, we have
[TABLE]
If G is of exceptional type then the result follows at once from [7, Theorem 1] and Theorem 4. We therefore assume that G is of classical type.
Note that if u=1 then du(Ga.)=d1(Ga.)=dimG≥du+1(Gs.c.) and so (8.1) holds.
Suppose now that u≥h. Since u≥h we have du(Ga.)=ℓ. Also as u+1>h, Theorem 4 yields du+1(Gs.c.)=du+1(Ga)=ℓ. Therefore du(Ga.)=du+1(Gs.c.)=ℓ and so (8.1) holds.
We can therefore assume that 2≤u≤h−1. In particular z≥1. Write h=z′(u+1)+e′ where z′, e′ are nonnegative integers such that 0≤e′≤u. An easy check gives 1≤z′≤z. Write z′=z−j where 0≤j≤z−1. Then e′=e−z+j(u+1). Note that if j=1 then z≥2. Also z−e≤j(u+1)≤u+z−e. We claim that either z>2j−1 or j=1. Suppose otherwise. Then as z≥1 we get j≥2 and z≤2j−1. Using the latter inequality and the fact that j(u+1)≤u+z−e, we get j≤1−e/(u−1) and so j≤1, a contradiction establishing the claim.
Suppose that G is of type A. By Theorem 4 we have du+1(Gs.c.)≤du+1(Ga.)+2 and so
[TABLE]
Let D=du(Ga.)−du+1(Ga). Note that Lemma 2.6 yields du(Ga.)≡du+1(Ga.)mod2 and so D≡0mod2. Using [7, Theorem 1], we have
[TABLE]
Recalling that z≥2 if j=1, we note that if j=1 then D>0. Also if z>2j−1 then D>0.
Hence D>0 in all cases.
As D≡0mod2 we deduce that D≥2 and so (8.2) yields du(Ga.)−du+1(Gs.c.)≥0. Hence (8.1) holds.
Suppose that G is of type C. By Theorem 4 we have du+1(Gs.c.)≤du+1(Ga.)+2ϵu⌈(z−j)/2⌉ and so
[TABLE]
Let D=du(Ga.)−du+1(Ga). Note that Lemma 2.6 yields du(Ga.)≡du+1(Ga.)mod2 and so D≡0mod2. Using [7, Theorem 1], we have
[TABLE]
Recalling that z≥2 if j=1, we note that if j=1 then D≥0 and moreover D≥(z2+ϵz)/2≥z+ϵz=2⌈z/2⌉ for u odd. Also if z>2j−1 then D≥0 and moreover D≥2⌈z/2⌉ for u odd.
In all cases we therefore have D≥0 and moreover D≥2⌈z/2⌉ for u odd.
Now (8.3) yields du(Ga.)−du+1(Gs.c.)≥0. Hence (8.1) holds.
Suppose that G is of type B. By Theorem 4 we have du+1(Gs.c.)≤du+1(Ga.)+2ϵu and so
[TABLE]
Let D=du(Ga.)−du+1(Ga). Note that Lemma 2.6 yields du(Ga.)≡du+1(Ga.)mod2 and so D≡0mod2. Using [7, Theorem 1], we have
[TABLE]
Arguing as in the case where G is of type C, we deduce that D≥0 and moreover D≥2 for u odd. Now (8.4) yields du(Ga.)−du+1(Gs.c.)≥0. Hence (8.1) holds.
Suppose that G is of type D. Since u≥2, by Theorem 4 we have du+1(Gs.c.)≤du+1(Ga.)+2ϵu and so
[TABLE]
Let D=du(Ga.)−du+1(Ga). Note that Lemma 2.6 yields du(Ga.)≡du+1(Ga.)mod2 and so D≡0mod2. Using [7, Theorem 1], we have
[TABLE]
Arguing as in the case where G is of type B, we deduce that D≥0 and moreover
D≥2 for u odd. We deduce that du(Ga.)−du+1(Gs.c.)≥0. Hence (8.1) holds.
□
9 Proof of Theorem 9
In this section we prove Theorem 9. We proceed in two steps, first in the special case where G is of simply connected type (see Proposition 9.1), and then we consider other types (see Proposition 9.2).
Proposition 9.1**.**
Let G be a simple simply connected algebraic group over an algebraic closed field K of prime characteristic p. The classification of the hyperbolic triples (a,b,c) of integers for G is as given in Theorem 9.
Proof.
As usual we let h be the Coxeter number of G and for a positive integer u, we let z and e be the nonnegative integers such that h=zu+e and 0≤e≤u−1.
Also du(G) denotes the minimal dimension of the centralizer of an element of G of order dividing u. By Proposition 2 du(G) is the codimension of the subvariety G[u] of G consisting of elements of order dividing u. Writing ju(G)=dimG[u], we have du(G)=dimG−ju(G).
Let (a,b,c) be a hyperbolic triple of integers. Let S(a,b,c)=da(G)+db(G)+dc(G) and D(a,b,c)=S(a,b,c)−dimG. Saying that (a,b,c) is reducible (respectively, rigid, nonrigid) for G amounts to saying that D(a,b,c) is greater than (respectively equal to, less than) 0.
Recall the partial order we put on the set of hyperbolic triples of integers. Given two hyperbolic triples (a,b,c) and (a′,b′,c′) of integers, we say that (a,b,c)≤(a′,b′,c′) if and only if a≤a′, b≤b′ and c≤c′. Among all hyperbolic triples, exactly three are minimal: (2,3,7), (3,3,4) and (2,4,5).
By Proposition 7, if (a,b,c) is nonrigid for G then (a′,b′,c′) is nonrigid for G for every (a′,b′,c′)≥(a,b,c).
For a nonnegative integer p, we let θp∈{0,1} be such that θp=1 if p is odd or p=0, otherwise θp=0.
Suppose first that G is of exceptional type. By Theorem 4 and [7, Theorem 1], every hyperbolic triple (a,b,c) of integers is nonrigid for G unless G is of type G2 and (a,b,c)∈{(2,4,5),(2,5,5)} in which case (a,b,c) is rigid for G.
Suppose now that G is of classical type. In a first step we show that if ℓ≥11, ℓ≥10, ℓ≥15, or ℓ≥9 accordingly respectively as G is of type A, B, C or D, then every hyperbolic triples of integers is nonrigid for G.
Suppose first that G is of type A. Note that dimG=h2−1.
By Theorem 4 and [7, Theorem 1], we have
[TABLE]
Let g(e)=u(h−e)(h+e)+e−1+2θp(1−ϵu). Then g′(e)=uu−2e, and g′(e)>0 if and only if e<u/2. Hence du(G)≤F(u) where
[TABLE]
In particular, for any hyperbolic triple (a,b,c) of integers, we have S(a,b,c)≤F(a)+F(b)+F(c).
Suppose (a,b,c)=(2,3,7). We have S(2,3,7)≤F(2)+F(3)+F(7). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −42h2+2θp+1 is negative for h≥12, it follows that (2,3,7) is nonrigid for G provided h≥12, that is ℓ≥11. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,3,7) is nonrigid for G with ℓ≥11.
Suppose (a,b,c)=(2,4,5). We have S(2,4,5)≤F(2)+F(4)+F(5). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −20h2+4θp+43 is negative for h≥10, it follows that (2,4,5) is nonrigid for G provided h≥10, that is ℓ≥9. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,4,5) is nonrigid for G with ℓ≥9.
Suppose (a,b,c)=(3,3,4). We have S(3,3,4)≤F(3)+F(3)+F(4). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −12h2+2θp+21 is negative for h≥6, it follows that (3,3,4) is nonrigid for G provided h≥6, that is ℓ≥5. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(3,3,4) is nonrigid for G with ℓ≥5.
Suppose now that G is of type B. Note that dimG=h(h+1)/2.
By Theorem 4 and [7, Theorem 1], we have
[TABLE]
Let g(e)=2u(h−e)(h+e+1)+2e+1+2θp(1−ϵu). Then g′(e)=2uu−2e−1, and g′(e)>0 if and only if e<(u−1)/2. Hence du(G)≤F(u) where
[TABLE]
In particular, for any hyperbolic triple (a,b,c) of integers, we have S(a,b,c)≤F(a)+F(b)+F(c).
Suppose (a,b,c)=(2,3,7). We have S(2,3,7)≤F(2)+F(3)+F(7). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −3364h2+4h−797+2θp is negative for h≥20, it follows that (2,3,7) is nonrigid for G provided h≥20, that is ℓ≥10. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,3,7) is nonrigid for G with ℓ≥10.
Suppose (a,b,c)=(2,4,5). We have S(2,4,5)≤F(2)+F(4)+F(5). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −1604h2+4h−359+4θp is negative for h≥16, it follows that (2,4,5) is nonrigid for G provided h≥16, that is ℓ≥8. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,4,5) is nonrigid for G with ℓ≥8.
Suppose (a,b,c)=(3,3,4). We have S(3,3,4)≤F(3)+F(3)+F(4). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −964h2+4h−203+2θp is negative for h≥10, it follows that (3,3,4) is nonrigid for G provided h≥10, that is ℓ≥5. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(3,3,4) is nonrigid for G with ℓ≥5.
Suppose now that G is of type C. Note that dimG=h(h+1)/2.
By Theorem 4 and [7, Theorem 1], we have
[TABLE]
Let
[TABLE]
Then
[TABLE]
and if u is even then g′(e)>0 if and only if e<(u−2)/2, else g′(e)>0 if and only if e<(u−1)/2. Hence du(G)≤F(u) where
[TABLE]
In particular, for any hyperbolic triple (a,b,c) of integers, we have S(a,b,c)≤F(a)+F(b)+F(c).
Suppose (a,b,c)=(2,3,7). We have S(2,3,7)≤F(2)+F(3)+F(7). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −84h2−20h−236 is negative for h≥30, it follows that (2,3,7) is nonrigid for G provided h≥30, that is ℓ≥15. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,3,7) is nonrigid for G with ℓ≥15.
Suppose (a,b,c)=(2,4,5). We have S(2,4,5)≤F(2)+F(4)+F(5). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −40h2−14h−121 is negative for h≥22, it follows that (2,4,5) is nonrigid for G provided h≥22, that is ℓ≥11. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,4,5) is nonrigid for G with ℓ≥11.
Suppose (a,b,c)=(3,3,4). We have S(3,3,4)≤F(3)+F(3)+F(4). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −24h2−2h−59 is negative for h≥10, it follows that (3,3,4) is nonrigid for G provided h≥10, that is ℓ≥5. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(3,3,4) is nonrigid for G with ℓ≥5.
Suppose finally that G is of type D. Note that dimG=(h+1)(h+2)/2.
By Theorem 4 and [7, Theorem 1], we have
[TABLE]
Let
[TABLE]
Then
[TABLE]
and if u is even then g′(e)>0 if and only if e<(u−2)/2, else g′(e)>0 if and only if e<(u−3)/2. Hence du(G)≤F(u) where
[TABLE]
In particular, for any hyperbolic triple (a,b,c) of integers, we have S(a,b,c)≤F(a)+F(b)+F(c).
Suppose (a,b,c)=(2,3,7). We have S(2,3,7)≤F(2)+F(3)+F(7). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −84h2+24h−276+4θp is negative for h≥16, it follows that (2,3,7) is nonrigid for G provided h≥16, that is ℓ≥9. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,3,7) is nonrigid for G with ℓ≥9.
Suppose (a,b,c)=(2,4,5). We have S(2,4,5)≤F(2)+F(4)+F(5). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −40h2+18h−109+6θp is negative for h≥12, it follows that (2,4,5) is nonrigid for G provided h≥12, that is ℓ≥7. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(2,4,5) is nonrigid for G with ℓ≥7.
Suppose (a,b,c)=(3,3,4). We have S(3,3,4)≤F(3)+F(3)+F(4). Now
[TABLE]
Hence
[TABLE]
and
[TABLE]
Since −24h2+6h−75+2θp is negative for h≥10, it follows that (3,3,4) is nonrigid for G provided h≥10, that is ℓ≥6. Also every hyperbolic triple (a′,b′,c′) of integers with (a′,b′,c′)≥(3,3,4) is nonrigid for G with ℓ≥6.
Finally to classify hyperbolic triples (a,b,c) of integers for G with ℓ≤10, ℓ≤9, ℓ≤14, ℓ≤8 accordingly respectively as G is of type A, B, C, D one can use Theorem 4 and [7, Theorem 1] to find da(G), db(G) and dc(G) and compute D(a,b,c). Note that if u>h then du(G)=ℓ.
∎
Proposition 9.2**.**
Let G be a simple algebraic group over an algebraic closed field K of prime characteristic p. The classification of hyperbolic triples (a,b,c) of integers for G is as given in Theorem 9.
Proof.
The case where G is of simply connected type is treated in Proposition 9.1. We therefore suppose that G is not of simply connected type.
We denote by Gs.c. (respectively, Ga.) the simple algebraic group of simply connected (respectively, adjoint) type having the same Lie type and Lie rank as G.
For an integer u, let du(G) denote the minimal dimension of the centralizer of an element of G of order dividing u. By Proposition 2 du(G) is the codimension of the subvariety G[u] of G consisting of elements of order dividing u. Writing ju(G)=dimG[u], we have du(G)=dimG−ju(G).
Let (a,b,c) be a hyperbolic triple of integers. Let S(a,b,c)=da(G)+db(G)+dc(G) and D(a,b,c)=S(a,b,c)−dimG. Recall that saying that (a,b,c) is reducible (respectively, rigid, nonrigid) for G amounts to saying that D(a,b,c) is greater than (respectively equal to, less than) 0.
By Proposition 3, given a positive integer u, we have du(G)≤du(Gs.c.).
It follows that every hyperbolic triple of integers which is nonrigid for Gs.c. is nonrigid for G. In particular, by Proposition 9.1, we can now assume that G is of classical type. By the proof of Proposition 9.1 if G is such that ℓ≤11, ℓ≤10, ℓ≤15, ℓ≤9 accordingly respectively as G is of type A, B, C, D then every hyperbolic triple of integers is nonrigid for G. For G of small rank, one can use [7, Theorem 1] and Theorems 4 and 5 to find da(G), db(G) and dc(G), compute D(a,b,c) and classify a given hyperbolic triple (a,b,c) of integers for G. Note that if u>h then du(G)=ℓ.
∎
10 Proof of Proposition 10
We consider Proposition 10. Let G be a simple algebraic group over an algebraically closed field K. The main ingredient in the proof of Proposition 10 is the following result proved in [10] combined with the classification given in Theorem 9 of the reducible and the rigid hyperbolic triples of integers for simple algebraic groups
if p is a bad prime for G or G is of exceptional type. Recall that p is said to be bad for G if G is of type Bℓ, Cℓ, Dℓ and p=2, or of type G2, F4, E6, E7 and p∈{2,3}, or of type E8 and p∈{2,3,5}. A prime p is said to be good for G if it is not bad for G. Also in the statement below and from now on, by an irreducible subgroup of a classical group G, we mean a subgroup acting irreducibly on the natural module for G.
Proposition 10.1**.**
[10, Proposition 2.1]**.
Suppose that G is of classical type and p is a good prime for G. If g1, g2, g3 are elements of G such that g1g2g3=1 and ⟨g1,g2⟩ is an irreducible subgroup of G then
[TABLE]
We can now prove Proposition 10.
Proof of Proposition 10.
If G is of classical type and p is a good prime for G then the result follows from Proposition 10.1. The remaining cases follow from Theorem 9 which shows that there are no reducible hyperbolic triples of integers for G if G is of symplectic or orthogonal type and p=2, or if G is of exceptional type.
□
11 Some tables
In this section we collect Tables 4-8 of the paper.
Tables 4-7 appear in §6 and Table 8 appears in §7.
[FIGURE:]
[FIGURE:]