Last revised 8:00 p.m. May 16, 2016
A simple way to compute structure constants
Bill Casselman
University of British Columbia
cass@math.ubc.ca
The standard technique for computing the structure constants of semi-simple Lie algebras is used in the
computer program MAGMA and described by [Cohen-Murray-Taylor:2005]. It relies on the additive structure
of roots. In [Casselman:2015] I explained a reasonably efficient way to compute the structure constants of
semi-simple Lie algebras by implementing an idea originally found in [Tits:1966a]. Tits’ idea was to replace
the additive structure by features of the normalizer of a maximal torus. This provided some mathematical
structure to the problem of computing structure constants that was missing in the standard approach. In
practice, computation based on this method went fairly rapidly and seemed at least roughly comparable
in efficiency to reported runs of the standard computation. There were, however, a number of rather ugly
and presumably inefficient formulas involved in this new algorithm. Recently, it was suggested by Robert
Kottwitz (in conversation with me in May of 2014 and expanded by him later that year), that an observation
of his about choosing bases of semi-simple Lie algebras makes it possible to bypass the nastiest parts in a
more efficient manner. In this paper I’ll explain how this goes.
Kottwitz’ principal observations can be briefly summarized. Suppose
G = a simple, connected, simply connected, complex group
B = Borel subgroup
T = maximal torus in B
Σ = associated root system
∆ = associated simple roots
W = Weyl group.
Let g, etc. be the corresponding Lie algebras. The root spaces gγ all have dimension one. Fix for each α in
∆ a spanning element eα . The triple (B, T, {eα }) make up a frame for G. The set of all frames is a principal
homogeneous space for the adjoint quotient of G. (The notion originated in work by French mathematicians.
In French the term is ‘épinglage’, which some translate literally into the noun ‘pinning’. But ‘frame’ is the
term adopted in the English translation of Bourbaki’s treatise on Lie algebras.)
Because G is simply connected, the coroot lattice X∗ (T ) may be identified with the lattice spanned by the
simple coroots α∨ .
Chevalley has defined integral structures on g and G. If N (Z) is the group of integral points in the normalizer
N = NG (T ), it fits into a well understood extension
1 −→ T (Z) = {±1}∆ −→ N (Z) −→ W −→ 1 .
[Tits:1966b] described this extension precisely enough to enable computations in it, and [Langlands-Shelstad:1987] gave an explicit formula for a defining 2-cocycle. This extension certainly does not split. But now
let VZ be the direct sum of non-trivial root spaces in gZ . Let S(Z) be the subgroup of transformations in
GL(VZ ) that act as ±1 on each root space. It may be identified with Hom(Σ, ±1). The group T (Z) embeds
∨
into S(Z): α∨ (x) goes to (xhγ,α i )γ . This embedding gives rise to an extension
1 −→ S(Z) = {±1}Σ −→ Next (Z) −→ W −→ 1 .
The extension also acts on VZ . Kottwitz has shown that this new extension does split, and he gives a simple
splitting w 7→ ẅ . It has the property that if wλ = λ then ẅ acts as the identity on gλ . This allows one to
specify a natural choice of Chevalley basis, once one fixes one eγ for each W -orbit in Σ.
Computing structure constants
2
Some variant of the method will work for a large class of Kac-Moody root systems, but I do not see how it
can be extended to all of them. A necessary requisite for extending the method to Kac-Moody algebras can
be found in [Carbone et al.:2015], which classifies conjugacy classes of simple roots.
Other formulas found in [Cohen-Murray-Taylor:2005], particularly those describing the action of N (Z) on a
semi-simple Lie algebra, also become simpler. I’ll not discuss that here.
Since this was written, a new construction of bases of a semi-simple Lie algebra has appeared in [Geck:2016].
It is not at all apparent how it relates to the construction in this paper.
Contents
1.
2.
3.
4.
5.
Chevalley bases
Tits’ idea
Kottwitz’ splittings
Computation
References
For g in G, x in g, I’ll write
g ♦ x = Ad(g)x .
I’ll usually refer to [Tits:1966a] as [T].
1. Chevalley bases
I fix once and for all a frame (B, T, {eα }). This is not significant, since all are equivalent under inner
automorphisms.
The frame determines embeddings of SL2 into G, one for each simple root. Let ια be the one parametrized
by α. Each of these takes the diagonal matrices into T and the differential of ια takes
◦
1
◦
◦
to eα . The associated embedding of C× is the coroot α∨ , and is independent of the choice of eα . Let hα be
the image under dια of
1
◦
◦
−1
.
The image e−α of
◦
◦
−1
◦
is the unique element of g−α such that
[e−α , eα ] = hα .
There are other conventions regarding the choice of e−α , but Tits’ choice is exactly what is needed to make
his analysis of structure constants work.
TITS’ SECTION.
The group N (Z) fits into a short exact sequence
1 −→ T (Z) = (±1)∆ −→ N (Z) −→ W −→ 1 ,
and [Tits:1966b] shows how to define a particularly convenient section. Define ṡα to be the image under ια
of
◦
1
−1
◦
.
Computing structure constants
3
It lies in the normalizer of T . Suppose w in W to have the reduced expression w = s1 . . . sn . Then the
product
ẇ = ṡ1 . . . ṡn
depends only on w, not the particular expression. The defining relations for this group, given those for W ,
are
(xy)˙ = ẋẏ (ℓ(xy) = ℓ(x) + ℓ(y))
ṡα2 = α∨ (−1) (α ∈ ∆) .
CHEVALLEY’S FORMULA. Given eγ 6= 0 in gγ , there exists a unique e−γ in g−γ such that hγ = [e−γ , eγ ] lies
in T and satisfies hγ, hγ i = 2. For sl2 this might be the triple
◦
◦
−1
◦
,
1
◦
◦
−1
,
◦
1
◦
◦
.
There exists a unique automorphism θ of g acting as −I on t and taking each eα (α ∈ ∆) to e−α . For each
other root γ there will exist eγ in gγ such that eθγ = e−γ . It is unique up to sign, and only up to sign. Any set
{eγ } invariant under θ up to sign is called a Chevalley basis (with respect to the given frame). If it is actually
invariant, as it is here, I’ll call it a strict Chevalley basis.
For such a basis, Chevalley proved that if λ, µ, ν are roots with λ + µ + ν = 0 then
[eλ , eµ ] = ±(pλ,µ + 1)e−ν .
Here pλ,µ is the least p such that µ − pλ is a root. I refer to [Chevalley:1955] for the original result and to
[Casselman:2015] for a proof extracted from [T], which works uniformly for all Kac-Moody groups. Tits’
choice of the e−α and the symmetry of his triples simplify things quite a bit.
Determining the sign in this equation has always seemed rather mysterious. Of course there is no one
formula, since the choice of a Chevalley basis is not canonical. But it has never been very clear what is going
on. Changing one eγ to −eγ forces a lot of other sign changes without apparent pattern. The situation has
now been cleared up somewhat by Kottwitz, who has explained to me how to choose an almost canonical
Chevalley basis. I’ll discuss that in §3.
In §4 I’ll show how Kottwitz’ basis simplifies the computation of the signs in Chevalley’s formula. The
starting point, at least, is the same as it was in [Casselman:2015], in which I have already outlined the
principal ingredients of a recipe for the constants. One of the troublesome points in the earlier approach was
a somewhat arbitrary choice of Chevalley basis.
2. Tits’ idea
I have fixed the frame {eα }, but I have not extended it to a Chevalley basis. In fact, I am eventually going to
specify a strict Chevalley basis that is related to the original frame but does not agree with it for the simple
roots.
Suppose given some fγ 6= 0 in gγ ∩ gZ . I have indicated above how it determines an embedding of SL2 into
G and elements hγ , f−γ making up a copy of sl2 . We also get then an element σγ in N (Z), the image of
◦
1
−1
◦
.
Thus σγ depends only on the choice of fγ (and T , which we have fixed). If γ = α and fα = eα , then σα = ṡα .
In any case its image in W will be sγ , the reflection corresponding to γ . The basic observation of Tits ([T],
Computing structure constants
4
Proposition 1) is that each of the items f±λ , σλ determine the other two. The choice of sign for any one of
these determines a change of sign in the others. This relationship is summarized in the equation
exp(fγ ) exp(f−γ ) exp(fγ ) = σγ .
In other words, the choice of a Chevalley basis is equivalent to a certain choice of elements σγ in the normalizer
N (Z) = NG (T ) ∩ G(Z). Note that σγ−1 = γ ∨ (−1)σγ . The equation above is invariant with respect to a sign
swap. Hence σγ = σ−γ .
There are two integral elements ±fγ in gγ , and likewise two γ ∨ (±1) σγ in N (Z). Let Wγ (Z) be the inverse
image of sγ in N (Z), so that |Wγ (Z)| = 2.
One practical consequence of Tits’ observation is this:
2.1. Lemma. Suppose w to be in N (Z), wλ = µ. Then
w ♦ fλ = ε fµ
if and only if
wσλ w−1 = µ∨ (ε) σµ .
Proof. Since
exp(εfµ ) exp(εf−µ ) exp(εfµ ) = µ∨ (ε)σµ .
I define a Tits triple to be a set of roots λ, µ, ν whose sum is 0.
2.2. Lemma. ([T], Lemme 1 of §2.5)) Suppose (λ, µ, ν) to be a Tits triple. The following are equivalent:
(a) kλk ≥ kµk = kνk;
(b) sλ µ = −ν ;
(c) hµ, λ∨ i = −1.
Any triple may be rotated to one of these.
This is more or less equivalent to the basic fact that two roots of a Tits triple cannot have greater length than
a third. I’ll call such a triple of roots (λ, µ, ν) ordered.
For an ordered triple
σλ σµ σλ−1 = ν ∨ (±1)σν .
The following is the basis of computation of structure constants by Tits’ method.
2.3. Proposition. There exists a function ε(σλ , σµ , σν ), defined on all products Wλ (Z) × Wµ (Z) × Wν (Z)
whenever (λ, µ, ν) is a Tits triple, such that
[fλ , fµ ] = ε(σλ , σµ , σν ) (pλ,µ + 1)f−ν .
This function satisfies some basic properties ([T], §2.9).
(εa)
(εb)
(εc)
(εd)
replacing σλ by σλ−1 changes the sign;
it is skew-symmetric in any pair;
it is invariant under cyclic rotation of the arguments.
If λ, µ, ν are an ordered triple with σλ σµ σλ−1 = σν then
ε(σλ , σµ , σν ) = (−1)pλ,µ .
Computing structure constants
5
The first two are immediate, but the third is not quite so. Together, these mean that we can apply a permutation
to any triple to reduce to a special case, but what is now needed is one explicit formula in that special case—i.e.
to pin down signs. That is what the last does. It follows from an analysis (in [T], §1.3) of the action of copies
of SL2 (Z) on the root spaces determined by root strings in g.
2.4. Lemma. Suppose (λ, µ, ν) to be an ordered Tits triple. The following are equivalent:
(a) σλ σµ σλ−1 = ν ∨ (ε) σν ;
(b) σλ ♦ fµ = ε f−ν ;
(c) [fλ , fµ ] = ε (−1)pλ,µ (pλ,µ + 1) f−ν .
One more thing I need is due to Tits, but formulated more explicitly in Chapter 4 (Theorem 4.1.2 (ii)) of
[Carter:1972] and as Lemma 2.5 in [Casselman:2015]:
2.5. Proposition. If (λ, µ, ν) is a Tits triple then
pµ,ν + 1
pν,λ + 1
pλ,µ + 1
=
=
kνk2
kλk2
kµk2
In other words, pλ,µ satisfies a twisted cyclic symmetry.
3. Kottwitz’ splittings
Kottwitz’ contribution is to remove nearly all annoying ambiguity in choosing a Chevalley basis.
Our choice of frame gives us a map w 7→ ẇ from W back to N (Z) ⊂ Next (Z). How can it be modified to
become a homomorphism?
We are looking for a splitting of the sequence
1 −→ S(Z) −→ Next (Z) −→ W −→ 1 .
This will be of the form
w 7−→ ẅ = ẇ · τw ,
with each τw in S(Z). Thus for each root β we shall be given a factor τw (β) = ±1. The map w 7→ ẅ will be a
homomorphism if and only if
(a) 1̈ = 1
(b) s̈α ẍ = (sα x)¨ if sα x > x
(c)
s̈α s̈α = 1 .
These translate to properties of τw :
(a′ ) τ1 = 1
(b′ ) τsα (yβ)τy (β) = τsα y (β) (sα y > y, for all β)
∨
(c′ ) (−1)hβ,α
i
= τsα (sα β) · τsα (β) .
In these, α is an arbitrary simple root.
I now give Kottwitz’ solution of this problem. Let
Rw = {λ > 0 | wλ < 0} .
Thus ℓ(xy) = ℓ(x) + ℓ(y) if and only if
(3.1)
Rxy = Ry ⊔ y −1 Rx ,
Computing structure constants
6
and in particular
R1 = ∅
Rsα = {α}
Rsα w = Rw ⊔ {w−1 α} (w−1 α > 0) .
Following Kottwitz, I require the factor τw (β) to be of the form (−1)F (w,β) with F of the form
F (w, β) =
(3.2)
X
hhβ, γii .
γ∈Rw
The summands are yet to be specified, and everything in this formula is to be taken modulo 2. Since R1 = ∅
and an empty sum is 0, condition (a) above is immediate.
What about condition (b)? Suppose x = sα y > y . It must be shown that the cocycle condition
F (sα y, β) = F (sα , yβ) + F (y, β)
holds. First of all, note that
F (sα , β) = hhβ, αii
since Rsα = {α}. Also
F (x, β) =
X
hhβ, γii = hhβ, y −1 αii +
γ∈Rx
X
hhβ, γii
γ∈Ry
whereas
F (sα , yβ) + F (x, β) = hhyβ, αii +
X
hhβ, γii .
γ∈Ry
Therefore (b) will be satisfied if W -invariance holds:
hhwβ, wγii = hhβ, γii
for all w in W .
Condition (c)? We have
s̈α ♦ eβ = (−1)hhβ,αii ṡα ♦ eβ .
Since ṡ2α = α∨ (−1) we thus require that
hhsα β, αii + hhβ, αii = hβ, α∨ i .
If hβ, α∨ i = 0 and hence sα β = β this imposes no condition (since everything is modulo 2). Otherwise
hβ, α∨ i and hsα β, α∨ i will be of different signs, so it is natural to set
(3.3)
hhβ, γii =
hβ, γ ∨ i if hβ, γ ∨ i > 0
0
if hβ, γ ∨ i < 0.
This satisfies the condition of W -invariance.
The condition that w 7→ ẅ be a homomorphism imposes no extra condition in the case that hβ, γ ∨ i = 0,
but one more requirement on this homomorphism will do so. I require now, for reasons that will become
apparent soon, that
ẅ ♦ eβ = eβ
if wβ = β . To guarantee that this occurs, it suffices to assume that β lies in the closed positive Weyl chamber.
Then the w fixing β are generated by simple root reflections, so we need to require only that s̈α vβ = vβ
Computing structure constants
7
(vβ ∈ gβ ) for simple roots α with hβ, α∨ i = 0. Consideration of the representation of SL2 corresponding to
the root string tells us that
ṡα ♦ eβ = (−1)pα,β eβ .
Therefore
s̈α ♦ eβ = (−1)hhβ,αii (−1)pα,β eβ
and so we require
(3.4)
hhβ, γii = pγ,β
if hβ, γ ∨ i = 0 .
Equations (3.3) and (3.4) define the terms hhβ, γii completely. In summary:
3.5. Theorem. (Kottwitz) Let

hβ, γ ∨ i if this is positive


pγ,β
if hβ, γ ∨ i = 0
hhβ, γii =

 0
otherwise.
X
F (w, β) =
hhβ, γii
γ∈Rw
τw = ((−1)F (w,β ) .
Then
ẅ = ẇ · τw
is a splitting homomorphism of Next (Z).
In addition, if wγ = γ then Ad(ẅ) is the identity on gγ .
If the root system is simply laced or equal to G2 then sλ β = β implies that pλ,β = 0. Therefore the non-trivial
case occurs only for systems Bn , Cn , or F4 .
Remark. Lemma 2.1A of [Langlands-Shelstad:1987] exhibits the 2-cocycle defining the extension N (Z)
determined by Tits’s splitting w 7→ ẇ. Explicitly,
ẋẏ = κ(x, y)(xy)˙ with κ(x, y) =
Y
γ ∨ (−1) .
γ>0
x−1 γ<0
y −1 x−1 γ<0
Does Kottwitz’ splitting make arguments of Langlands and Shelstad any simpler?
3.6. Corollary. There exists a Chevalley basis {eγ } of VZ such that ẅ ♦ eγ = ewγ for all roots γ and w in W
such that γ , wγ > 0.
I’ll call such a basis semi-canonical. There are several possibilities, two for each W -orbit in Σ.
Proof. The W -orbits are the sets of all roots of the same length. Pick one simple root α in each orbit, and let
eα = eα be the corresponding element in the frame chosen at the beginning. If λ = wα is a positive root with
α equal to one of these, define
eλ = ẅ ♦ eα .
The definition of F (sα , β) in the case when hβ, α∨ i = 0 insures that this is a valid definition.
If γ > 0, set e−γ = eθγ .
The last condition is certainly necessary in order to be sure we have a Chevalley basis, but is it redundant?
In fact, we can specify exactly how ẅ and θ interact. The following is straightforward:
Computing structure constants
8
3.7. Lemma. For all β , γ
hhβ, γii + hh−β, γii = hβ, γ ∨ i
This is to be interpreted modulo 2, of course. Now let
h(w, β) =
X
hβ, γ ∨ i .
γ∈Rw
We deduce easily:
3.8. Lemma. For v in gβ
(ẅ ♦ v)θ = (−1)h(w,β) ẅ ♦ v θ .
Proof. Because ṡα commutes with θ.
Kottwitz has a more intuitive formulation. The height of a root is defined by the formula
X
X
ht
λα α =
λα .
∆
Induction on the length of w, combined with (3.1), will prove:
3.9. Lemma. For w in W and root λ
ht(wλ) − ht(λ) = h(w, λ) .
Applied to wα this implies Kottwitz’ formulation:
3.10. Proposition. For any root γ
eθγ = (−1)ht(γ)−1 e−γ .
In particular, if α is simple then
eθα = e−α .
The specification for negative roots is therefore not redundant. A basis defined for all roots by the condition
ewα = ẅ ♦ eα
would not be invariant under θ.
Example. For any simply laced root system we have
F (sα , β) =
n
hβ, α∨ i if hβ, α∨ i > 0
0
otherwise.
This applies in particular to G = SL3 . Take as the standard simple roots α, β , and let γ = α + β . Then
hβ, α∨ i = −1
hγ, α∨ i = 1
hα, β ∨ i = −1
hγ, β ∨ i = 1 .
Computing structure constants
9
Recall that ei,j is the matrix with a single non-zero entry 1 at (i, j). Thus e1,2 and e2,3 span the root spaces
for α, β . Fix them to be eα and eβ . The corresponding elements of N (Z) are

◦
ṡα = −1
1
◦
◦
◦


◦
1
1
◦
◦
ṡβ =  ◦
◦
−1
1 .
◦

◦

◦
A straightforward calculation shows that
ṡβ ♦ e1,2 = −e1,3
ṡα ♦ e1,3 = −e2,3 .
Recall that hα, β ∨ i < 0 while hγ, α∨ i > 0. If we start with eα = e1,2 we get
eα = e1,2 = eα
eγ = s̈β eα
= (−1)0 ṡβ ♦ e1,2
= −e1,3
eβ = s̈α eγ
= (−1)1 ṡα ♦ (−e1,3 )
= −e2,3 = −eβ .
In summary:
3.11. Proposition. Suppose G = SL3 . If eα = eα then eβ = −eβ .
We see that a semi-canonical basis does depend on the choice of starting eα .
In order to specify these, we fix one simple root α in each W -orbit, and set eα = eα . Then we choose the other
eβ for β in ∆. The W -orbits are in bijection with possible root lengths. For finite-dimensional Lie algebras
simple roots of the same length lie in simply laced segments in the Dynkin diagram. For every simple root
α, let
d(α) = the distance from α to the selected initial element in its segment.
This is well defined since the Dynkin diagram is a tree. Any two neighbours in the Dynkin diagram of
the same length lie in the simple root system of a copy of SL3 . The following is hence a consequence of
calculations in the previous section:
3.12. Corollary. For α in ∆ let cα = (−1)d(α) . Then
eα = cα eα
σα = α∨ (cα ) ṡα .
Computing structure constants
10
4. Computation
I’ll now sketch how the computation of structure constants goes.
The computation starts with the Cartan matrix (hα, β ∨ i). The first thing to do is to construct the positive
roots by building what I call the root graph. Its nodes are the positive roots, and there is an oriented edge
λ → µ if µ = sα λ for some α in ∆, with hλ, αi < 0. Thus the height of µ is greater than that of λ. The base of
the graph is made up of the simple roots, and as one builds the graph one can also build a spanning forest in
it (in one of several possible ways). The roots of the forest are in ∆, and at the end we have a way to traverse
all roots. Associated to every root, among other things, are an array (hλ, α∨ i) and a norm kλk2 .
Next we choose the eα as explained in the last section, starting with initial elements for each W -orbit of roots.
Kottwitz’ basis makes computation simple because the following is simple:
4.1. Lemma. Suppose α in ∆, sα λ = µ. If
∨
Cα,λ = (−1)hhλ,αii chλ,α
α
i
then
σα ♦ eλ = Cα,λ eµ .
Proof. Suppose λ > 0. Then
s̈α eλ = eµ
= (−1)hhλ,αii ṡα ♦ eλ
(definition)
∨
i
σα ♦ eλ (Corollary 3.12)
ṡα ♦ eλ = chλ,α
α
∨
i
(−1)hhλ,αii eµ .
σα ♦ eλ = chλ,α
α
For negative roots, we start with
∨
∨
s̈α e−λ = s̈α eθλ = (−1)hλ,α i eθµ = (−1)hλ,α i e−µ
and apply Lemma 3.7.
What we want eventually is a good way to find the ε such that
[eλ , eµ ] = ε · (pλ,µ + 1) · e−ν
whenever λ, µ, ν are a Tits triple. Because of Lemma 2.2, we can rotate the trio to make it an ordered triple.
Since σ−λ = σλ , it therefore suffices to calculate ε(σλ , σµ , σν ) for ordered Tits triples with λ > 0.
I start by looking at ordered triples (α, µ, ν) with α in ∆. We make a list of all these by scanning through Σ,
and for each µ we find all α in ∆ such that
hµ, α∨ i = −1 .
In this case, µ + α = sα µ is a root, say −ν . We have hhµ, αii = 0, so by Lemma 4.1
σα ♦ eµ = cα e−ν .
and
[eα , eµ ] = (−1)pα,µ · cα · (pα,µ + 1) e−ν .
Keep in mind that computing pα,µ for ordered triples is easy. If hµ, α∨ i = −1 then pα,µ is 0 if µ − α is not a
root, and is 1 if it is (in which case we are dealing with G2 ).
Computing structure constants
11
At the end of this phase we have a list of all ordered Tits triples (α, µ, ν) with α in ∆.
We now progress to all ordered Tits triples with λ positive by pushing λ up the root tree. If we have
[eλ , eµ ] = Nλ,µ · e−ν
and λ → sα λ is an edge in the tree, we apply σα to both sides to get
[σα ♦ eλ , σα ♦ eµ ] = Nλ,µ · σα ♦ e−ν .
Use Lemma 4.1 to calculate structure constants for the new triple (sα λ, sα µ, sα ν).
At the end we have the structure constants for all ordered Tits triples with λ positive. Given an arbitrary Tits
triple, we can rotate it to make it an ordered triple, so that we can calculate ε. But we can also calculate pλ,µ
since this also possesses by Proposition 2.5 a twisted cyclic invariance.
The amount of storage required in this method is roughly proportional to the number of Tits triples. As
reported in [Cohen-Murray-Taylor:2005], this is of order n3 , where n is the rank of the system. I don’t think
the order can be reduced, but a bit more careful technical considerations will probably reduce the constant
of proportionality.
5. References
1. Lisa Carbone, Alexander Conway, Walter Freyn, and Diego Penta, ‘Weyl group orbits on Kac-Moody root
systems’, arXiv.1407:3375, 2015.
2. Bill Casselman, ‘On Chevalley’s formula for structure constants’, Journal of Lie Theory 25 (2015), 431–441.
3. ——, ‘Structure constants of Kac-Moody Lie algebras’, to appear in [Howe et al.:2015].
4. Roger W. Carter, Simple groups of Lie type, John Wiley & Sons, 1972.
5. Claude Chevalley, ‘Sur certains groupes simples’, Tôhoku Mathematics Journal 48 (1955), 14–66.
6. Arjeh Cohen, Scott Murray, and Don Taylor, ‘Computing in groups of Lie type’, Mathematics of Computation 73 (2004), 1477–1498.
7. Meinholf Geck, ‘On the construction of semi-simple Lie algebras and Chevalley grousp’, arxiv:1481674,
February 15, 2016.
8. Roger Howe et al. (editors), Symmetry: representation theory and its applications, in the series Progress
in Mathematics 257, Elsevier, 2015.
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