Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 4 July 2013

§7. Central extensions

Our object is to prove that if π,G, and G are as in §6, then (π,G) is a universal central extension of G in a sense to be defined. The reader is referred to the lecturer's paper in Colloque sur la théorie des groupes algébriques, Bruxelles, 1962 and for generalities to Schur's papers in J. Reine Angew. Math. 1904, 1907, 1911.

Definition: A central extension of a group G is a couple (π,G) where G is a group, π is a homomorphism of G onto G, and kerπ center of G.


(a) π,G,G as in §6.
(b) π:GG the natural homomorphism of one Chevalley group onto another constructed from a smaller weight lattice. E.g., π:SLnPSpn, and π:SpinnSOn.
(c) π:GG a topological covering of a connected topological group; i.e., π is a local isomorphism, carrying a neighborhood of 1 isomorphically onto one of G. We note that π is central since a discrete normal subgroup of a connected group is necessarily central. To see this, let N be a discrete normal subgroup of a connected group G and let nN. Since the map GN given by ggng-1, gG, has a discrete and connected image, gng-1=n for all gG.

Definition: A central extension (π,E) of a group G is universal if for any central extension (π,E) of G there exists a unique homomorphism φ:EE such that πφ=π, i.e., the following diagram is commutative:

E φ E π π G

We abbreviate universal central extension by u.c.e. We develop this property in a sequence of statements.

(i) If a u.c.e. exists, it is unique up to isomorphism.


If (π,E) and (π,E) are u.c.e. of G, let φ:EE and φ:EE be such that πφ=π and πφ=π. Now φφ:EE and π(φφ)=π. Hence φφ is the identity on E by the uniqueness of φ in the definition of a u.c.e. Similarly φφ is the identity on E.

(ii) If (π,E) is a u.c.e. of G then E=𝒟E and hence G=𝒟G, where 𝒟H is the derived group of H.


Consider the central extension (π,E) where E=E×E/𝒟E and π(a,b)=π(a), aE, bE/𝒟E. Now if φ1(a)=(a,1) and φ2(a)=(a,a+𝒟E), then πφi=π, i=1,2, and hence φ1=φ2. Thus, E/𝒟E=1 and E=𝒟E.

(iii) If G=𝒟G and (π,E) is a central extension of G, then E=C·𝒟E where C is a central subgroup of E on which π is trivial. Moreover, 𝒟E=𝒟2E.


We have π𝒟E=𝒟(πE)=𝒟G=G. Hence, E=C𝒟E where C=kerπ. Also, 𝒟E=𝒟(C𝒟E)=𝒟2E.

(iv) If G=𝒟G, then G possesses a u.c.e.


For each xG we introduce a symbol e(x). Let F be the group generated by {e(x),xG} subject to the condition that e(x)e(y)e(xy)-1 commutes with e(z) for all x,y,zG. If π:e(x)x, then by using induction on the length of an expression in F, we see that π extends to a central homomorphism of F onto G.

(a) (π,F) covers all central extensions of G. To see this, let (E,π) be any central extension of G. Choose e(x)E such that πe(x)=x. Since π is central, the e(x)'s satisfy the condition on the e(x)'s. Hence, there is a homomorphism φ:FE such that φe(x)=e(x), and
thus πφ=π.

(b) If E=𝒟F and π also denotes the restriction of π to E, then (π,E) covers all central extensions uniquely. By (iii), we have that (π,E) covers all central extensions. If (π,E) is a central extension of G and if πφ=π=πφ, then φ(x)φ(x)-1 center of E. Thus, ψ:xφ(x)φ(x)-1 is a homomorphism of E into an Abelian group. Since E=𝒟E, ψ is trivial and φ=φ.

Remark: Part (a) shows that if G is any group then there is a central extension covering all others.

(iv') If (π,E) is a central extension of G which covers all others and if E=𝒟E, then (π,E) is a u.c.e.

(v) If π:EF and ψ:FG are central extensions, then so is ψπ:EG, provided E=𝒟E.


If akerψπ, let φ be the map φ:x(a,x)=axa-1x-1,xE. Now φ(x) center of E, since π(a,x)=(πa,πx)=1 because πa center of F. Now φ is a homomorphism, so φ is trivial, and a center of E.

(vi) Exercise: In (v), (π,E) is a u.c.e. of F if and only if (ψπ,E) is a u.c.e. of G.

Definition: A group G is said to be centrally closed if (id,G) is a u.c.e. of G.

(vii) Corollary: If (π,E) is a u.c.e. of G, then E is centrally closed.

(viii) If E is centrally closed, then every central extension ψ:FE of E splits; i.e, there exists a homomorphism φ:EF such that ψφ=id.

(ix) (π,E) is a u.c.e. of G if and only if every diagram of the form

E E π π G ρ G

can be uniquely completed, where (π,E) is a central extension of G and ρ is a homomorphism.


One direction is immediate by taking G=G and ρ=id. Conversely, suppose the diagram is given and (ψ,E) is a u.c.e. Let H be the subgroup of G×E, H={(x,e)|ρx=πe}. If ψ:HG is given by ψ(x,e)=x, then ψ is central. Since (π,E) is a u.c.e. of G, there is a unique homomorphism θ:EH such that ψθ=π. Now the homomorphism φ:EE, φ=ψθ, where ψ:HE is given by ψ(x,e)=e, satisfies ρπ=ρψθ=πψθ=πφ. If ρπ=πφ, then let θ:EH be given by θ(e)=(π(e),φ(e)). Since ψθ=π, we have θ=θ and φ=ψθ=ψθ=φ, proving the uniqueness of φ.

Definition: A linear (respectively projective) representation of a group G is a homomorphism of G into some GL(V) (respectively PGL(V)).

Since GL(V) is a central extension of PGL(V) we have the following result:

(x) Corollary: If (π,E) is a u.c.e. of G, then every projective representation of G can be lifted uniquely to a linear representation of E.

(xi) Topological situation: If G is a topological group one can replace the condition G=𝒟G by G is connected, the condition (π,E) is a u.c.e. by (π,E) is a universal covering group in the topological sense, and the condition G is centrally closed by G is simply connected in the above discussion and obtain similar results.

Definition: If (π,E) is a u.c.e of the group G, then we call kerπ the Schur multiplier of G.

If we write kerπ=M(G) to indicate the dependence on G, then a homomorphism φ:GG leads to a corresponding one M(φ):M(G)M(G) by (ix). Thus M is a functor from the category of groups G such that G=𝒟G to the category of Abelian groups with the following property: if φ is onto, then so is M(φ).

Remark: Schur used different definitions (and terminology) since he considered only finite groups but did not require that G=𝒟G. If G=𝒟G our definitions are equivalent to his. One of Schur's results, which we shall not use, is that if G is finite then so is M(G).

Theorem 10: Let Σ be an indecomposable root system and k a field such that |k|>4 and, if rankΣ=1, then |k|9. If G is the corresponding universal Chevalley group (abstractly defined by the relations (A),(B),(B'),(C) of §6), if G is the group defined by the relations (A),(B),(B') (we use (B') only if rankΣ=1), and if π is the natural homomorphism from G to G, then (π,G) is a u.c.e. of G.

Remark: There are exceptions to the conclusion. E.g. SL2(4) and SL2(9) are such. Indeed SL2(4)PSL2(5) and SL2(5) is a central extension of PSL2(5). For SL2(9) see Schur. It can be shown that the number of couples (Σ,k) for which the conclusion fails is finite.


Since |k|>4, G=𝒟G, G=𝒟G, and u.c.e. exist for both G and G. The conclusion becomes G is centrally closed, by the above remarks. We need only show that every central extension (ψ,E) of G splits, i.e., there exists θ:GE so that ψθ=idG; i.e., the relations defining G can be lifted to E.

We may assume E=𝒟E; but then (πψ,E) is a central extension of G by (v). We need only show

(1) If (ψ,E) is a central extension of G, then the relations (A),(B),(B') can be lifted to E.

Let C=kerψ, a central subgroup of E. We have:

(2) A commutator (x,y) with x,yE depends only on
the classes mod C to which x and y belong.

Choose ak* so that c=a2-10. Then in G (hα(a),xα(t))=xα(ct) for all αΣ,tK. We define φxα(t)E (αΣ,tk) so that ψφxα(t)=xα(t) and so that

(3) (φhα(a),φxα(t))=φxα(ct) and then φh's (and later φw's) in terms of the φx's by the same formulas which define the h's and the x's. in terms of the x's. Note that this choice is not circular because of (2). We shall show that the relations (A),(B),(B') hold with φx's in place of x's.

(4) φhφxα(t) (φh)-1 = φ(hxα(t)h-1) for all hH, αΣ, tk.

Set hxα(t)h-1=xα(dt) with dk*. Conjugating (3) by φh, we get ( φhα(a), φxα(dt) ) = φhφxα(ct) φ(h)-1, and the left side equals φxα(cdt)= φ(hxα(ct)h-1) by (3). Similarly we have

(4') φnφxα(t)(φn)-1= φ(nxα(t)n-1) for all nN, αΣ, tk.

(5) If α and β are roots, α+β0, and α+β is not a root, then φxα(t) and φxβ(u) commute for all t,uk. Set φxα(t) φxβ(u) φxα(t)-1 = f(t,u)φ xβ(u), t,uk, f(t,u)C. We must show f(t,u)=1. Clearly, from the definitions we have

(6) f is additive in both positions.

(5a) Assume αβ. If (α,β)=0, then f(t,u)=f(tv2,u) (v0) by Lemma 20(c) and by (4) with h=hα(v). If (α,β)>0, then f(t,u)=f(tvd,u) where d=4-α,ββ,α by Lemma 20(c) and by (4) with h=hα(v2)hβ(v-β,α). In both cases, f(t(1-vd),u)=1 by (6) for some d=1,2, or 3. Choose v so vd-10. Then we get f1.

(5b) Assume α=β and rankΣ>1. If there is a root γ so that α,γ=1, set h=hγ(v) in the preceding argument and obtain (*) f(t,u)=f(tv,uv). Choose v so that v-v20 and 1-v+v20. By (*) and (6), f(t(v-v2),u)= f(t,u/(v-v2))= f(t,u/v)f(t,u/(1-v))= f(vt,u)f((1-v)t,u)= f(t,u), whence f(t(1-v+v2),u)=1 and f1. If there is no such γ, then Σ is of type Cn and α is a long root. In this case, however, α=β+2γ with β and γ roots. Thus, ( φxβ(t), φxγ(u) ) = gφxβ+γ (±tu)φ xβ+2γ (±tu2) with gC, by Lemma 33. Since φxα(v) commutes with all factors but the last by (5a), it also commutes with the last.

(5c) Assume α=β and rankΣ=1. At least we have f(t,u)=f(tv2,uv2), t,uk, vk*, using h=hα(v) in the argument above. We may also assume that |k| is not a prime. If it were, then xα(t) and xα(u) would be powers of xα(1) and (5) would be immediate. Referring to the proof of (5b), we see it will suffice to be able to choose v so that v,1-v are squares and v-v20, 1-v+v20. If k is finite of characteristic 2, this is possible since all elements of k are squares. Otherwise, set v=(2w/(1+w2))2. Then 1-v=((1-w2)/(1+w2))2, and we need only choose w so that 1+w20, v-v20, and 1-v+v20. Since at most 13 values of w are to be avoided and |k|25 in the present case, this too is possible. T his completes the proof of (5).

(7) φ preserves the relations (A). The element x= φxα(tc-1) φxα(uc-1) (φxα((t+u)c-1))-1 is in C, and hence the transform of x by hα(a) is x itself. However, by (3), (4), (5) this transform is also xφxα(t) φxα(u) (φxα(t+u))-1.

(8) φ preserves the relations (B). We have φxα(t) φxβ(u) φxα(t)-1 = f(t,u) φxiα+jβ (cijtiuj) φxβ(u), where f(t,u)C. One proves f=1 by induction on n, the number of roots of the form iα+jβ i,j*. If n=0, this is just (5). If n>0, the inductive hypothesis and (7) imply f satisfies (6), and then the argument in (5a) may be used.

(9) φ preserves the relations (B'). This follows from (4').

This completes the proof of the theorem.

Exercise: Assume is the original Lie algebra with coefficients transferred by means of a Chevalley basis to a field k whose characteristic does not divide any Nα,β0. Also assume Σ is indecomposable of rank>1. Prove:

(a) The relations [Xα,Xβ]=Nα,βXα+β,α+β0, form a defining set for . Hint: define Hα=[Xα,X-α] and show that the relations of Theorem 1 hold.
(b) =𝒟, the derived algebra of .
(c) Every central extension of splits. Hint: parallel the proof of Theorem 10.

Corollary 1: The relations (A),(B),(B') can be lifted to any central extension of G.

Corollary 2:

(a) G is centrally closed. Each of its central extensions splits. Its Schur multiplier is trivial. It yields the u.c.e. of all the Chevalley groups of the given type, and covers linearly all of the projective representations of these groups.
(b) If k is finite or more generally an algebraic extension of a finite field, then (a) holds with G replaced by G.


This follows from various of the generalities at the beginning of this section.

E.g., if k is finite, |k|>4 and SL2(9) is excluded,
then SL2(k), Spn(k), and Spinn(k) all have trivial Schur multipliers, and the natural central extensions SLnPSLn, SpnPSpn, SpinnSOn are all universal.

Corollary 3: Assume G,G, and π are as above. If k* is infinite and divisible (uk*,n implies there exists vk* with vn=u), then the Schur multiplier of G; i.e., C=kerπ, is also divisible.


Elements of the form f(t,u)= hα(t) hα(u) hα(tu)-1 in G αΣ generate C. We have f(t,vw2)= f(t,v)f(t,w2) by Lemma 39(a). By induction, we get f(t,w2n)= f(t,w2)n for arbitrary n. Since for uk* we can find wk* such that u=w2n, the proof is complete.

Corollary 3a: If k* is infinite and divisible by a set of primes including 2, then C is also divisible by these primes.

Corollary 3b: If k* is infinite and divisible, then any central extension of G by a kernel which is a reduced group (no divisible subgroups other than 1) is trivial; i.e. it splits.


Let (ψ,E) be a central extension of G with kerψ reduced. Since (π,G) is u.c.e. we have φ:GE so that ψφ=π. Since C=kerπ is divisible, so is φCkerψ. Hence φC=1 and kerφkerπ. Thus, there is a homomorphism θ:GE so that θπ=φ. Therefore, ψθπ=π on G and ψθ=1 on G.

Corollary 3c: If k* is infinite and divisible, then any finite dimensional projective representation of G can be lifted uniquely to a linear representation.


Assume σ:GPGL(V). Since G=𝒟G, we have σ:GPSL(V). Let ρ:SL(V)PSL(V) be the natural projection. Since dimV is finite, we have kerρ is finite and thus kerρ is reduced. Consider the central extension (ψ,E) of G where E= { (x,y)| σx=ρy,xG, ySL(V) } G×SL(V) and ψ(x,y)=x, (x,y)E. Now kerψ=1×kerρ is reduced, so by Corollary 3b, we have θ:GE with ψθ=1 on G. If ψ(x,y)=y, (x,y)E, then σ=ψθ:GSL(V)GL(V) with ρσ=ρψθ=σψθ=σ.

Example: Corollary 3c says, for example, that every finite dimensional representation of SLn() can be lifted to a linear one. (The novelty is that the representation is not assumed to be continuous.)

Theorem 11: If Σ is an indecomposable root system, if chark=p0, and if G=𝒟G (i.e. we exclude |k|=2, Σ of type A1,B2, or G2 and |k|=3, Σ of type A1), then (π,G) uniquely covers all central extensions of G for which the kernel has no p-torsion.


By Theorem 10, we could assume |k|4 or |k|=9. However, the proof does not use this assumption or Theorem 10. If (ψ,E) is a central extension of G such that C=kerψ has no p-torsion, then we wish to show (A), (B), and (B') can be lifted to E.

(1) Assume C is divisible by p. Choose φxα(t)E so that ψφxα(t)=xα(t) and (φxα(t))p=1, αΣ, tk. We claim relations (A), (B) and (B') hold on the φx's.

(1a) If α,β are roots, α+β not a root, and α+β0, then φxα(t) and φxβ(u) commute, t,uk. We have φxα(t) φxβ(u) φxα(t)-1 = fφxα(u) with fC. Taking pth powers, we get 1=fp which implies f=1 since C has no p-torsion.

(1b) The relations (A) hold. Taking pth powers of φxα(t) φxα(u) =fφα(t+u), fC, we get f=1 as before.

(1c) Exercise: Relations (B) and (B') also hold.

(2) General case.

(2a) C can be embedded in a group C which is divisible by p and has no p-torsion. We have a homomorphism θ of a free Abelian group F onto C. Now FQ is a divisible group, and we can identify F with FFQ. Hence C=F/kerθFQ/kerθ=D, say, and D is a divisible group. Moreover, since C has no p-torsion, CDp=1 where Dp is the p-component of D. Thus, C projects faithfully into D/Dp=C which is divisible and has no p-torsion.

(2b) Conclusion of proof. Form E=EC, the direct product of E and C with C amalgamated, and define ψ:EG by ψ(ec)=ψ(e), eE, cC.

Now (ψ,E) satisfies the assumptions of (1) so the relations (A), (B), (B') can be lifted to E. However, by Lemma 32', the lifted group is its own derived group and hence contained in E.

Corollary 1: Every projective representation of G in a field of characteristic p can be lifted to a linear one.

Corollary 2: The Schur multiplier of G is a p-group.


These are easy exercises.

Since the kernel of the map π:GG above turns out to be the Schur multiplier of G, its structure for k arbitrary is of some interest. The result is:

Theorem 12: (Moore, Matsumoto) Assume Σ is an indecomposable root system and k a field with |k|>4. If G is the universal Chevalley group based on Σ and k, if G is the group defined by (A), (B), (B'), and if π is the natural map from G to G with C=kerπ, the Schur multiplier of G, then C is isomorphic to the abstract group A generated by the symbols f(t,u) (t,uk*) subject to the relations:

(a) f(t,u)f(tu,v)= f(t,uv)f(u,v), f(1,u)= f(u,1)=1
(b) f(t,u)f(t,-u-1)= f(t,-1)
(c) f(t,u)= f(u-1,t)
(d) f(t,u)= f(t,-tu)
(e) f(t,u)= f(t,(1-t)u)
and in the case Σ is not of type Cn (n1)(C1=A1) the additional relation:
(ab') f is bimultiplicative.
In this case relations (a)-(e) may be replaced by (ab') and
(c') f is skew
(d') f(t,-t)=1
(e') f(t,1-t)=1.
The isomorphism is given by φ:f(t,u) hα(t) hα(u) hα(tu)-1, α is a fixed long root.

Remark: These relations are satisfied by the norm residue symbol in class field theory, which is a significant aspect of Moore's work.

Partial Proof.

(1) If 𝔥α is the group generated (in G') by all hα(t), α a fixed long root, then C𝔥α. We know that hα(t) hα(u) hα(tu)-1, αΣ, tk*, form a generating set for C. Using the Weyl group we can narrow the situation to at most two roots α,β with α long, β short, and (α,β)>0. Hence, β,α=1 and (hα(t),hβ(u))= hβ(tu) hβ(t)-1 hβ(u)-1 = hα(t) hα(uα,β) hα(tuα,β)-1 by Lemma 37(f). This shows a will suffice.

(2) φ is a mapping onto C. This follows from (1).

(3) φ is a homomorphism. We must show that the relations hold if f(t,u) is replaced by hα(t)hα(u)hα(tu)-1. The relations (a) are obvious. A special case (u=1) of (e) has been shown in Lemma 39(d). The other relations (b), (c), (d) follow from the commutator relations connecting the h's and the w's.

(3') Assume Σ is not of type Cn. In this case there is a root γ so that α,γ=1. Thus f(t,v)= hγ(u) f(t,v) hγ(u)-1 = hα(tu) hα(u)-1 hα(uv) hα(tuv)-1 = f(t,u)-1 f(t,uv) or f(t,uv)= f(t,u) f(t,v) . By relation (c), f(uv,t)= f(u,t) f(v,t).

(4) φ is an isomorphism. This is done by constructing an explicit model for G.

Now let G be a connected topological group. A covering of G is a couple (π,E) such that E is a connected topological group and π is a homomorphism of E onto G which maps a neighborhood of 1 in E homeomorphically onto a neighborhood of 1 in G; i.e., which is a local isomorphism. A covering is universal if it covers all other covering groups. If (id,G) is a universal covering, we say that G is simply connected.


(a) A covering (π,E) of a connected group is necessarily central as was noted at the beginning of this section.
(b) If a universal covering exists, then it is unique and each of its coverings of other covering groups is unique. This follows from the fact that a connected group is generated by any neighborhood of 1.
(c) If G is a Lie group, then a universal covering for G exists and simple connectedness is equivalent to the property that every continuous loop can be shrunk to a point (See Chevalley, Lie Groups or Cohn, Lie Groups.)

Theorem 13: If G is a universal Chevalley group over viewed as a Lie group, then G is simply connected.

Before proving Theorem 13, we shall first state a lemma whose proof we leave as an exercise.

Lemma 41: If t1,t2,,tn are complex numbers such that |t1|<ε, i=1,2,,n and i=1nti=0, there exist ti and tj such that |ti+tj|<ε.

Proof of Theorem 13.

Let (π,E) be a covering of G. Locally π is invertible so we may set φ=π-1 on some neighborhood of 1 in G. We shall show that φ can be extended to a homomorphism of G onto E, i.e., (id,G) covers (π,E). It suffices to show that φ can be extended to all of G so that the relations (A), (B), (B'), (C) hold on the φx's.

Consider the relations

(A) φxα(t) φxα(u)= φxα(t+u) αΣ.

Since φ is locally an isomorphism, there is ε>0 such that (A) holds for |t|<ε, |u|<ε. If tk, t=Σti, |ti|<ε, then set φxα(t)= φxα(ti). Using induction and Lemma 41, we see that φxα(t) is well defined. Clearly, (A) then holds for all t,uk. Alternatively, we could note that 𝔛α is topologically equivalent to and hence simply connected. Thus, φ extends to a homomorphism of 𝔛α into E and (A) holds. Clearly the extension of φ to 𝔛α is unique.

To obtain the relations (B), let α,β be roots α±β, let S be the set of roots of the form iα+jβ (i,j+), and let 𝔛S be the corresponding unipotent subgroup of G. Topologically, 𝔛S is equivalent to n for some n, and is hence simply connected. As before φ can be extended to a homomorphism of 𝔛S into E, and the relations (B) hold. This extension is consistent with those above, by the uniqueness of the latter.

We now consider hα(t)= xα(t) x-α(-t-1) xα(t). wα(-1)= xα(t-1) x-α (1-t-1) xα(-1) xα(t-1)wα(-1) where xy=y-1xy. Hence, if t is near 1 in , then φhα(t) is near 1 in E. Thus, φhα(t) is multiplicative near 1 and hence Abelian everywhere (recall that ψ is central). We then have φhα(u)= φhα(t) φhα(u) φhα(t)-1 = φhα(t2u) φhα(t2)-1 by Lemma 37(f). Since has square roots, we have φhα is multiplicative, i.e., (C) holds.

Examples: SLn(), Spn(), and Spinn() are simply connected. These cases can also be proved by induction on n. (See Chevalley, Lie Groups, Chapter II.)


(a) If is replaced by in the preceding discussion, then relations (A), (B) can be lifted exactly as before. Also φhα is still multiplicative if one of the two arguments is positive. Further kerπ is generated by φhα(-1)2, α a fixed long root, and, if type Cn (n1) is excluded, then hα(-1)4=1 or wα(-1)8=1. Prove all of this.

(b) Moore has constructed a universal covering of G and has determined the fundamental group in case k is a p-adic field, using appropriate modifications of the definitions (here G is totally disconnected.)

(c) Let G be a Chevalley group over k,G the corresponding universal group, and π:GG the natural homomorphism. If k is algebraically closed and if only appropriate coverings are allowed, then (π,G) is a universal covering of G in the sense of algebraic groups.

We close this section with a result in which the coefficients may come from any ring (associative with 1). The development is based in part on a letter from J. Milnor. Let R be the ring, and let GL(R) be the group of infinite matrices which are equal to the identity everywhere except for a finite invertible block in the upper left hand corner. Thus, GLn(R)GL(R), n=1,2,. Let E(R) be the subgroup of GL(R) generated by the elementary matrices 1+tEij (tR,ij,i,j=1,2,), where Eij is the usual matrix unit. For example, if R is a field, then E(R)=SL(R), a simple group whose double coset decomposition involves the infinite symmetric group. Indeed, if R is a Euclidean domain, then E(R)=SL(R).

Lemma 42:

(a) E(R)=𝒟GL(R)
(b) E(R)=𝒟E(R)


The relation (1+tEik,1+Ekj)=1+tEij shows (b) and hence also E(R)𝒟GL(R). If x,yGLn(R), then xyx-1y-1E(R) because in GL2n(R) we have

(1) [ xyx-1y-10 01 ] = [ x0 0x-1 ] [ y0 0y-1 ] [ (yx)-10 0yx ]
(2) [x00x-1]= [1x01] [101-x-11] [1-101] [101-x1]
(3) [1x01]= i=1n j=n+12n (1+xijEij), if x=(xij).

We call K1(R)=GL(R)/E(R) the Whitehead group of R. This concept is used in topology. The case in which R=[G] is of particular interest.

Example: If R is a Euclidean domain, then K1(R)=R*, the group of units. (See Milnor, Whitehead Torsion.)

By Lemma 42 and (iv), E(R) has a u.c.e. (π,U(R)). Set K2(R)=kerπ. This notation is partly motived by the following exact sequence.

1K2(R) U(R)GL(R) K1(R)1.

K2 is a functor from rings to Abelian groups with the following property: if RR is onto, then so is the associated map K2(R)K2(R).

Remark: K2 is known to the lecturer in the following cases:

(a) If R is a finite field (or an algebraic extension of a finite field), then K2=1.
(b) If R is any field, see Theorem 12.
(c) If R=, then |K2|=2.
Here (a) follows from Theorem 9 and the next theorem, and a proof of (c) will be sketched after the remarks following the corollaries to the next theorem.

Theorem 14: Let U(R) be the abstract group generated by the symbols xij(t) (tR,ij,i,j=1,2,) subject to the relations

(A) xij(t) is additive in t.
(B) ( xik(t), xj(u) ) = { xij(tu) ifk=,ij. 1 ifk,ij.
If π:U(R)E(R) is the homomorphism given by xij(t)1+tEij, then (π,U(R)) is a u.c.e. for E(R).


(a) π is central. If xkerπ, choose n large enough so that x is a product of xij's with i,j<n. Let Fn be the subgroup of U(R) generated by the xkn's (kn,k=1,2,). Now by (A) and (B), any element of Pn can be expressed as kxkn(tk). Since in E(R) this form is unique, πPn is an isomorphism. Also by (A) and (B), xij(t) Pnxij (t)-1 Pn if i,j<n. Thus, xPnx-1Pn. If yPn, then π(x,y)=1, and since πPn is an isomorphism we have (x,y)=1. In particular, x commutes with all xkn(t). Similarly, x commutes with all xnk(t) and hence with all xij(t)=(xin(t),xnj(1)). Thus, x is in the center of U(R).

(b) π is universal. From (B), it follows that U(R)=𝒟U(R). Hence it suffices to show it covers all central extensions. Let (ψ,A) be a central extension of E(R) and let C be the center of A. We must show that we can lift the relations (A) and (B) to A. Fix i,j,ij and choose pi,j. Choose yij(t)ψ-1xij(t) so that (*) (yip(t),ypj(1))=yij(t). We will prove that the y's satisfy the equations (A) and (B).

(b1) If ij,k, then yik(t) and yj(u) commute. Choose qi,j,k, and write yj(u)= c ( yq(u), yqj(1) ) , cC. Since yik(t) commutes up to an element of C with yq(u) and yqj(1), it commutes with yj(u). Hence

(b2) {yij(t)}, i,j fixed, is Abelian.

(b3) The relations (A) hold. The proof is exactly the same as that of statement (7) in the proof of Theorem 10.

(b4) yij(t) in (*) is independent of the choice of p. If qp,i,j, set w= yqp(1) ypq(-1) yqp(1). Transforming (*) by w and using (b1) we get (*) with q in place of p.

(b5) The relations (B) hold. We will use:

(**) If, a,b,c are elements of a group such that a commutes with c and such that (b,c) commutes with (a,b) and c, then (a,(b,c))=((a,b),c). Since a commutes with c,(a,(b,c))=((a,b),(b,c),c). The other conditions insure ((a,b),(b,c)c)=((a,b),c). Now assume i,j,k are distinct. Choose qi,j,k, so that ( yik(t), ykj(u) ) = ( yik(t), ( ykq(u), yqj(1) ) ) = ( ( yik(t), ykq(u) ) , yqj(1) ) = ( yiq(tu), yqj(1) ) =yij(tu) by (*), (**), and (b4).

This completes the proof of the theorem.

Let Un(R) denote the subgroup of U(R) generated by yij(t) with i,jn.

Corollary 1: If n5, then Un(R) is centrally closed.

Corollary 2: If R is a finite field and n5 then SLn(R) is centrally closed.


This follows from Corollary 1 and the equations En(R)= SLn(R)= Un(R).

Remarks: (a) It follows that if R is a finite field and if SLn(R) is not centrally closed, then either |R|=9, n=2 or |R|4 and n4. The exact set of exceptions is: SL2(4), SL2(9), SL3(2), SL3(4), SL4(2).

* Exercise: Prove this.

(b) The argument above can be phrased in terms of roots, etc. As such, it carries over very easily to the case in which all roots have one length. The only other exception is D4(2).

(c) By a more complicated extension of the argument, it can also be shown that the universal Chevalley group of type Bn or Cn over a finite field (or an algebraic extension of a finite field) is centrally closed if n is large enough. Hence, only a finite number of universal Chevalley groups with Σ indecomposable and k finite fail to be centrally closed.

Now we sketch a proof that K2() is a group of order 2. The notation U,Un, above will be used. The proof depends on the following result:

(1) For n3, SLn() is generated by symbols xij(i,j=1,2,,n;ij) subject to the relations

(B) ( xik, xj ) = { xij ifk=, ij, 1 ifk, ij.
(C) If wij= xij xij-1 xij, hij= wij2, then hij2=1.

Identifying xij with the usual xij(1) and using xij(t)= xij(1)t, we see that the relations (B) here imply those of Theorem 14. Since the last relation may be written hij(-1)2= hij(1) and ±1 are the only units of , we have SLn() defined by the usual relations (A), (B), (C) of §6. Perhaps there are other rings, e.g., the p-adic integers, for which this result holds. For the proof of (1) see W. Magnus, Acta Math. 64 (1934), which gives the reference to Nielsen, who proved the key case n=3 (it takes some work to cast Nielsen's result into the above form). The case n=2, with (B) replaced by (B') w12x12 w12-1 =x21-1, is simpler and is proved in an appendix to Kurosh, Theory of Groups.

Now let xij,wij,hij refer to elements of Un().

(2) If Cn is the kernel of πn:Un()SLn() and n3, then Cn is generated by h122m and (h122)2=1.

As usual, we only require (C) when i,j=1,2, and h122Cn Setting h122=1 amounts to dividing by h122, which thus equals Cn. The relation h23h12h23-1 =h12-1, which may be deduced from (B) as in the proof of Lemma 37, then yields h122=h12-2.

(3) h1221 if n3.

Assume not. There is a natural map Un()Un(), xijxij(1). This maps h122 onto h12(-1)2, which (see Remark (a) after the proof of Theorem 13) generates the kernel of Un()SLn(). Thus SLn() is centrally closed, hence simply connected. Since SLn() can be contracted to SOn (by the polar decomposition, which will be proved in the next section) which is not simply connected since SpinnSOn is a nontrivial covering, we have a contradiction.

It now follows from (2), (3) and Theorem 14 that |K2()|=2.

By Corollary 1 above the same conclusion holds with SL() replaced by any SLn() with n5.

Exercise: Let SAn be SLnx translations of the underlying space, kn with k again a field. I.e., SAn is the group of all (n+1)×(n+1) matrices of the form [xy01] where xSLn, ykn. SAn is generated by xij(t), tk, ij, i=1,2,,n, j=1,2,,n+1. Prove:

(1) If the relation
(C) hij(t) is multiplicative.
is added to the relations (A) and (B) of Theorem 14, a complete set of relations for SAn is obtained.
(2) If k is finite, (C) may be omitted.
(3) If n is large enough, the group defined by (A) and (B) is a u.c.e. for SAn.
(4) Other analogues of results for SLn.
We remark that SA2() is the universal covering group of the inhomogeneous Lorentz group, hence is of interest in quantum mechanics.

Notes and References

This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson. This work was partially supported by Contract ARO-D-336-8230-31-43033.

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