The Curtis-Steinberg-Tits presentation was discovered by Curtis [13] (and independently by Tits) adapting work done by Steinberg, and offers a presentation of central extensions of groups of Lie type. For our purposes, we will use a reduced presentation described in [3, Section 4.2].
Theorem 5.1.1 (Curtis-Steinberg-Tits presentation). [3, Theorem 4.2] Fix Φ as the set of roots of a simple Lie algebra L of rank n≥2, and Φ+ a positive system of roots. For 1 ≤ i < j ≤ n, let Φi,j denote the rank 2 subsystem spanned by the
i-th and j-th fundamental roots, and set Ψ = S
i<jΦi,j. Also set Υi,j = {(α, β) ∈
Φi,j×Φi,j|α6=±β}andΥ =Si<jΥi,j. LetFpe be a finite field andB ={b1, . . . , be}
a basis of Fpe as a Fp-vector space. Then a presentation for a central extension of
the simple group of Lie type Gcorresponding to the Lie algebra L over Fpe is given
by generators{yα(bt) :α∈Ψ, t= 1, . . . , e} satisfying the following relations:
yα(bt)p= 1 α∈Ψ,1≤t≤e [yα(bt), yα(bu)] = 1 α∈Ψ,1≤t < u≤e [yα(bt), yβ(bu)] = Y i,j>0 yiα+jβ(Ci,j,α,βbitbuj) (α, β)∈Υ,1≤t, u≤n
where the Ci,j,α,β are the structure constants described below, and where for λ =
Pe
t=1λtbt∈Fpe with 0≤λt< p, we define yα(λ) := Qe
t=1yα(bt)λt.
Remark 5.1.2. All terms in the product in the third set of relations in Theorem 5.1.1 commute, except when the group is G2(q). We will not be considering this group in this thesis, but details on how to define the product in this case can be found in [51].
We sketch the construction of theCi,j,α,β - the full description of these con-
diagrams of the formBnorDn, we can assume that whenαandβ are fundamental roots at least one ofiorj is equal to 1.
Leteα and eβ denote two root vectors in the Lie algebraLcorresponding to
roots α and β respectively, and supposeα+β is also a root. Then it follows that [eα, eβ] is a scalar multiple of eα+β where [ , ] denotes the Lie bracket ofL. Thus
we can write [eα, eβ] = Nα,βeα+β. (The integers Nα,β are known as the structure
constants for the Lie algebra). It is the usual convention to set Nα,β = 0 ifα+β is
not a root. With these structure constants, we then make the following definitions: • Ci,1,α,β = 1i! i−1 Q t=0 Nα,tα+β. • C1,j,α,β= (−1)jj1! j−1 Q t=0 Nβ,tβ+α.
Example 5.1.3. LetL have Dynkin diagramA2 as depicted below, and letα1, α2 denote the fundamental roots.
α1 α2
Since the two roots have a single edge connecting them, we have in the notation of Theorem 5.1.1
Ψ = Φ1,2 ={±α1,±α2,±(α1+α2)}.
We can consider the underlying Lie algebra directly, or use the Magma function
LieConstant C, to determine that the structure constants can be chosen as below, withi= 1 and j= 2: C1,1,αi,αj = 1 C1,1,−αj,−αi = 1 C1,1,αj,−αi−αj = 1 C1,1,αi+αj,−αj = 1 C1,1,−αi,αi+αj = 1 C1,1,−αi−αj,αi = 1 C1,1,αj,αi =−1 C1,1,−αi,−αj =−1 C1,1,−αi−αj,αj =−1 C1,1,−αj,αi+αj =−1 C1,1,αi+αj,−αi =−1 C1,1,αi,−αi−αj =−1
By convention we take the remaining structure constants to be 0.
Example 5.1.4. Let L have Dynkin diagram Dn, as depicted below, and let
α1 α2 . . . αn−2
αn−1
αn
Then for each pair of fundamental rootsαi,and αj, there is either a single
edge connecting them, or no edge. If there is no edge connecting αi to αj, then
Φi,j = {±αi,±αj} and we do not need any structure constants to determine the
relations in Theorem 5.1.1, since the subgroup generated by{yα(b) :α∈Φi,j, b∈Fq}
will be isomorphic to SL2(q)×SL2(q) with corresponding Dynkin diagramA1×A1. If there is a single edge (and without loss of generality i < j) then we can use the structure constants as given in Example 5.1.3, since the subgroup generated by {yα(b) : α ∈Φi,j, b ∈Fq} will be isomorphic to SL3(q) with corresponding Dynkin diagramA2.
Example 5.1.5. Let L have Dynkin diagram B2 as depicted below. Let α1, α2 denote the fundamental roots.
α1 α2
Due to the double edge connectingα1 and α2, we get that Ψ ={±α1,±α2,±(α1+α2),±(α1+ 2α2)}.
Taking i = 1 and j = 2 in the below table, we can derive the structure constants viaMagmaas in Example 5.1.3:
C1,1,αi,αj = 1 C1,1,−αj,αi = 1 C1,1,αj,αi+αj = 2 C1,1,−αi−αj,−αj = 2 C1,1,αj,−αi−αj = 2 C1,1,αi+αj,−αj = 2 C1,1,αi+αj,−αi−2αj = 1 C1,1,αi+2αj,−αi−αj = 1 C1,1,−αi,αi+αj = 1 C1,1,−αi−αj,αi = 1 C1,1,−αj,αi+2αj = 1 C1,1,−αi−2αj,αj = 1 C2,1,αj,αi = 1 C2,1,−αj,−αi = 1 C2,1,αj,−αi−2αj = 1 C2,1,−αj,αi+2αj = 1 C1,2,αi,−αi−αj = 1 C1,2,αi,αi+αj = 1 C1,2,αi+2αj,−αi−αj = 1 C1,2,−α1−2α2,α1+α2 = 1 C1,1,αj,αi =−1 C1,1,αi,−αj =−1 C1,1,αi+αj,αj =−2 C1,1,−αj,−αi−αj =−2 C1,1,−αi−αj,αj =−2 C1,1,−αj,αi+αj =−2 C1,1,−αi−2αj,αi+αj =−1 C1,1,−αi−αj,αi+2αj =−1 C1,1,αi+αj,−αi =−1 C1,1,αi,−αi−αj =−1 C1,1,αi+2αj,−αj =−1 C1,1,αj,−αi−2αj =−1 C1,2,αi,αj =−1 C1,2,−αi,−αj =−1 C1,2,−αi−2αj,αj =−1 C1,2,αi+2αj,−αj =−1 C2,1,−αi−αj,−αi =−1 C2,1,αi+αj,−αi =−1 C2,1,αi−αj,αi+2αj =−1 C2,1,αi+αj,−αi−2αj =−1
Example 5.1.6.LetLhave Dynkin diagramBnas depicted below. Letα1, α2, . . . , αn
denote the fundamental roots.
α1 α2 . . . αn−1 αn
Then, similarly to Example 5.1.4, we can use the structure constants from Examples 5.1.3 and 5.1.5 to find the structure constants, depending on the number of edges connectingαi and αj.
For the twisted groups, we require a separate presentation; the one we will use is also due to Steinberg (see for instance [20]), although we will again use a shortening of this presentation described in [3].
Theorem 5.1.7. [3, Section 6.1] LetG be a twisted group of type2An(q) for n >1
odd, 2Dn(q) for n ≥ 4, or 2E6(q), with associated twisted root systems of type
Cn−1 2 +1,
Dn−1 or F4 respectively. Define Ψand Υ as in Theorem 5.1.1. Let q=pe
and B denote a basis of Fq2 as a Fp-vector space, with B = {b1, . . . , b2e} chosen
such that {b1, . . . , be} is a basis for Fq as an Fp-vector space. Then we have a
presentation of a central extension of G with the following form: it has generators
{yα(bt) : α∈Ψ, t= 1, . . . , kαe} (wherekα= 1 if α is a long root and kα= 2 ifα is
a short root), satisfying the following relations:
yα(bt)p = 1 α∈Ψ,1≤t≤kαe
[yα(bt), yα(bu)] = 1 α∈Ψ,1≤t < u≤kαe
and, for (α, β)∈Υ , 1≤t≤kαe, 1≤u≤kβe:
[yα(bt), yβ(bu)] = 1 α+β /∈Ψ
[yα(bt), yβ(bu)] =yα+β(α,βbtbu) α, β, α+β all long or all short
[yα(bt), yβ(bu)] =yα+β(α,β(bt¯bu+ ¯btbu)) α, β short, α+β long
[yα(bt), yβ(bu)] =yα+β(α,βbtbu)yα+2β(ηα,βbtbu¯bu) α, α+ 2β long, β, α+β short
where the α,β and ηα,β take the value ±1 and for λ =P2t=1e λtbt ∈ Fp2e with 0 ≤
λt< p, we defineyα(λ) :=Q2t=1e yα(bt)λt.
Example 5.1.8. Let L have Dynkin diagram A3 = D3 as depicted below. Let
α1, α2, α3 denote the fundamental roots. Note the unusual numbering of the roots below - this is to keep the notation consistent when we generalise toDn later.
α2 α1 α3
Lhas a graph automorphismγ interchanging the roots α2 andα3. Suppose
Lis a Lie algebra overFq2 with field automorphismσ of order 2. The twisted group consists of those matrices which are fixed by the automorphism σγ (see [9] or [49] for more information on the construction of the twisted groups). The twisted group 2D
3 has underlying Lie algebraB2, with fundamental rootsβ1, β2.
β1 β2
b∈F∗
q; hence we can take yβ1(b) =yα1(b) for b∈Fq. Similarly, σγ fixes yβ2(a) :=
yα2(a)yα3(a
q) for any a∈
F∗q2.
It turns out that we can derive the structure constants from the constants used in Example 5.1.5; see Remark 5.1.10.
Example 5.1.9.LetLhave Dynkin diagramDn, with fundamental rootsα1, . . . , αn.
Again, this has a graph automorphism γ of order 2 interchanging αn−1 and αn,
giving rise to the twisted group 2D
n with underlying Dynkin diagram Bn−1 with fundamental roots β1, . . . , βn−1. We can derive the corresponding generators as before: yβi(b) =yαi(b) ∀i≤n−2,∀b∈F ∗ q yβn−1(r) =yαn−1(r)yαn(r q) ∀r∈ F∗q2
To determine the constants α,β and ηα,β from Theorem 5.1.7, we will again use
results from smaller dimensions. For relations not involvingyβn−1(r), the generators involved are identical to the ones used in Theorem 5.1.1 applied to Dn and so we can directly find α,β in this case. For relators involving yβn−1(r), the structure constants will be the same as those discussed in Example 5.1.8.
Remark 5.1.10. It follows from [52, Table 0A8] that there is an embedding of
Bn−1 inside 2Dn. This is obtained by the restriction of the group of Lie type
corresponding to2Dn(which is overFq2) to elements over the fieldFq. In particular, it follows from [52, p. 165] that we can obtain the commutator relations forBn−1 from the commutator relations from2D
n. However, where the structure constants
are concerned, the converse is also true; since the structure constants only depend on the roots and not the field, the structure constants for 2Dn and Bn−1 must coincide. Hence, we can use the structure constants as discussed in Example 5.1.5 and Example 5.1.6 for the twisted groups.
5.1.4 Diagonal automorphisms of Ω