1 The groups and Lie algebrasnwallach/groups3.pdfLemma 4 If Gis a connected semisimple algebraic group over C with a ZΠgraded Lie algebra then there exists a semisimple element (i.e.

1 The groups and Lie algebras

We consider G ⊂ GL(n,C) (n × n invertible matrices) a subgroup given asthe locus of zeros of polymomial in C[xij] with xij the matrix entries of ann × n—matrix that is an element of Mn(C). G in the jargon of algebraicgroup theory is an affi ne algebraic group, i.e. a Zariski closed subgroup ofGL(n,C). Set XT equal to the transpose of X and X∗ = X̄T as usual.Let Lie(G) denote its Lie algebra which we can take to be all X ∈Mn(C)

such that etX ∈ G for all t ∈ C. Here if ‖X‖ = tr(XX∗)12 then ‖...‖ defines

a norm on the vector space Mn(C) and

‖XY ‖ ≤ ‖X‖ ‖Y ‖ , X, Y ∈Mn(C).

Thus the series

eX =∞∑m=0

Xm

m!

converges uniformly and absolutely on closed balls ‖X‖ ≤ r. This impliesthat X 7→ eX defines a complex analytic function from Mn(C) to the opensubset GL(n,C). It is standard that if X, Y ∈ Lie(G) then

[X, Y ] = XY − Y X ∈ Lie(G).

As usual, we set U(n) = {g ∈ Mn(C)|g∗g = I}. Then U(n) is a compactsubgroup of GL(n,C)(it is a closed subset of the unit sphere in Mn(C)) andso is a Lie subgroup. Its Lie algebra is precisely the set of skew-hermitiann× n matrices.If we assume that if g ∈ G then g∗ ∈ G, that is, G is a symmetric

subgroup. We set K = G ∩ U(n). Then K is also a compact Lie group andLie(K) is the set of all elements in Lie(G) that are skew-Hermitian.

Theorem 1 (cf. [GIT,Theorem 37] )Let H be a Zariski closed, reductivesubgroup of, G, a symmetric subgroup of GL(n,C) then there exists g ∈ Gsuch that gHg−1 is symmetric.

Thus we can take as a definition of reductive that there exists an imbed-ding in GL(n,C) as a symmetric subgroup.We assume that G is symmetric. A basic structural result is

Lie(G) = Lie(K)⊕ iLie(K)

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as a real vector space and the map

K × iLie(K)→ G

k,X 7→ keX

is a diffeomorphism ontoG. Clearly this implies thatK is a maximal compactsubgroup of G. One can prove that all maximal compact subgroups of G areconjugate toK. In particular, any compact subgroup of G can be conjugatedinto K.As usual we denote by Ad(g)X the action of the inner automorphism of

Lie(G) corresponding to g ∈ G on X. That is, Ad(g)X = gXg−1. We alsowrite Inn(g)x = gxg−1 for g, x ∈ G.On Lie(G) we put the bilinear form (X, Y ) = trXY . We note that (..., ...)

is G-invariant, that is,

(gXg−1, gXg−1) = tr(gXY g−1) = tr(XY ) = (X, Y ).

Then since G is invariant under g 7−→ g∗ (hence so is Lie(G)) the form isnondegenerate.

Lemma 2 If H ⊂ G is Zariski closed and reductive then (..., ...)|Lie(H) isnon-degerate.

Proof. We have observed above that there exists g ∈ G so that gHg−1 isinvariant under h 7−→ h∗. Hence the form is nondegenerate on gLie(H)g−1.The G—invariance of the form implies the result.We note that the converse is also true. Here are examples of reductive

groups1. GL(n,C).2. SL(n,C) = {g ∈ GL(n,C)| det(g) = 1}.3. O(n,C) = {g ∈ GL(n,C)|gIgT = I}.4. SO(n,C) = O(n,C) ∩ SL(n,C).5. Let

J =

[0 I−I 0

]with I the n× n identity matrix. Set Sp(n,C) = {g ∈ GL(n,C)|gJgT = J}.Exercise. These five groups and arbitrary finite products of them are

symmetric subgroups.

These groups and their coverings and quotients give the full set of classicalgroups.

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2 Gradings of Lie algebras

Let g be a Lie algebra over C and let S be a commutative semi-group (writtenadditively) then an S—grading of g is a direct sum decomposition

g =⊕s∈S

gs

with gs possibly 0 and[gs, gt] ⊂ gs+t

for s, t ∈ S.

2.1 Z—gradingsThe main examples that are of interest to us are when S = Z or S = Z/nZwith n a stricly positive integer. We will only be looking at finite dimensionalLie algebras.We have the following characterization of a Z—grading.

Lemma 3 A Z—grading of g is equivalent to having an algebraic homomor-phism φ : C× → Aut(g) via φ(z)x = znx for x ∈ gn.

Proof. We leave it to the reader to check that the formula in the statementdefines an algebraic homomorphism of C× to Aut(g). If φ is an algebraichomomorphism of C× to Aut(g) then g is the direct sum of spaces gχ withφ(z)x = χ(z)x for x ∈ gχ with χ an algebraic homomorphism of C× to itself(i.e. an algebraic character). The algebraic characters of C× are exactly theones given by χn(z) = zn. If we set gn = gχn we have our grade.Exercises. 1.When is a Z/nZ grading a Z—grading?2. Let k > 0 be an integer and let g =

⊕s∈Z gs be a Z—grading of g. Let

for n ∈ Z, [n] be its class mod k. Then defining g[n] =⊕

r≡nmod k gr we havea Z/nZ grading.

2.2 Z/nZ gradingsWe now consider the case when we have a Z/nZ-grading. Then if ζ is ann—th root of unity we can define θ : g→ g by θ|g[m] = ζmI. Then θ defines anautomorphism of g of order n. If θ is an automorphism of order n and if ζ isa primitive n—th root of unity then we can define g[m] = {x ∈ g|θx = ζmx}and this is a Z/nZ grading.

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2.3 Relationship with groups 1. Z—grades.We now consider the case when G is a connected reductive algebraic groupover C and g = Lie(G) and g is Z—graded

g =⊕j∈Z

gj

then let φ : C× → Aut(g) be the algebraic homomorphism

φ(z)v = zjv, v ∈ gj.

Let g = c ⊕ [g, g] with c the center of g and [g, g] the commutator algebrawhich is semi-simple. We observe that

φ(C×)c = c

andφ(C×)[g, g] ⊂ [g, g].

Let C be the center of G and let G′ be the derived group then G = CG′.G′ is connected and Lie(G′) = [g, g] inherits a grade from g. Ad(G) = Ad(G′)and since all of the group actions that Vinberg studies are by subgroups ofthe adjoint group of the Lie algebra we will emphasize G′ and g′ = [g, g].For the rest of this subsection we will assume that g = [g, g]. The linear

map D : g→ g given byD|gj = jI

defines a derivation of g. All derivaions of g are inner so D = ad(x) withx ∈ g. We have proved

Lemma 4 If G is a connected semisimple algebraic group over C with a Z—graded Lie algebra then there exists a semisimple element (i.e. diagonalizable)x ∈ Lie(G) such that ad(x) has integral eigenvalues and the eigenspaces ofad(x) give the grade.

2.4 Relationship with groups 2. Z/nZ—grades.We now assume that G is a connected reductive algebraic group with g =Lie(G) having a Z/nZ—grade,

g =⊕

[j]∈Z/nZ

g[j].

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Let ζ be a primitive n—th root of unity and

θ|g[j] = ζjI.

As aboveg = c⊕ [g, g]

and θc = c and θ[g, g] =[g, g]. The same non-interaction between the gradesof the center and the derived algebras prevails.Assumeing that G is semi-simple (i.e. c = 0) then Ad : G → Ad(G) is

a covering homomorphism and Ad(G) is the identity component of Aut(G).We have

Lemma 5 Let G be a connected semi-simple Lie group with a Z/nZ—gradedLie algebra g = Lie(G). If θ is as above and τ(g) = θgθ−1 for g ∈ Ad(G)then identifying g with ad(g), dτ = θ.

2.5 The affi ne action associated with a grade.

Let G be a connected reductive algebraic group with g = Lie(G) having a Zor a Z/mZ grade with m ∈ Z>0. So we have a graded decomposition

g =⊕s∈S

gs

with S = Z or Z/mZ. In both cases we will write [j] for the correspondingelement of S. Let h denote the Lie algebra g[0] and let V = g[1]. Let H be theconnected Lie subgroup of G with Lie algebra h. Then the pair (Ad(H)|V , V )of a group and module for the group is what we will call the affi ne actionassociated with the grade.

Lemma 6 The group Ad(H)|V is a connected reductive algebraic subroup ofGL(V ) that depends on g and not the choice of G.

Proof. We note that V = c∩g[1]⊕ [g, g]∩g[1]. Relative to this decompositionthe elements of H act trivially on the first summand and preserve the second.We may therefore assume that g = [g, g]. If S = Z let x ∈ g be suchthat adx gives the grade. If S = Z/mZ then let θ be as in the previoussubsection. In the first case we note that H is the identity component of{h ∈ H|Ad(h)x = x} and in the second H is the identity component of

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{h ∈ H|Ad(h)θ = θAd(h)} in bothe cases this defines Z—closed subgroup ofG and since x and θ are semi-simple H is reductive. In both cases Ad(H)|V isthe connected subgroup of GL(V ) with lie algebra {ady|y ∈ h} so it dependsonly on g.This leads us to the main objects of our study. We use the term Z-closed

for Zariski closed.

Definition 7 A Vinberg pair will be a pair (L, V ) with L a connected, re-ductive Z-closed subgroup of GL(V ) such that there exists a graded reductiveLie algebra

g =⊕s∈S

gs

with S = Z or Z/mZ such that V = g[1] and L = Ad(H)|V with H as in theprevious Lemma.

Definition 8 A direct sum of Vinberg pairs is a pair (L, V ) with L = L1 ×· · · × Lr and V = V1 ⊕ V2 ⊕ ... ⊕ Vr with (Li, Vi) a Vinberg pair and L actsby the block diagonal action.

Definition 9 A morphism of direct sums of Vinberg pairs (L, V ) and (L′, V ′)is a pair, (φ, T ) of an algebraic group homomorphism φ : L→ L′ and a linearmap T : V → V ′ such that Tgv = φ(g)Tv for g ∈ L and v ∈ V .

Exercise. When is a direct sum of Vinberg pairs isomorphic with aVinberg pair?

2.6 Decomposition into simple constituents.

In the last section we showed how one can assign a reductive algebraic group,H, and an algebraic H—module V to a reductive graded Lie algebra. In thissection we will describe a decomposition of into what we will call simpleconstituents.Let G and g be as in the previous section. Let

g = c⊕n⊕j=1

g(j)

be a decomposition of g into the direct sum of its center, c, and its simpleideals

g(1), g(2), ..., g(n).

6

If the grade on g is a Z—grade then we have seen that each of the summandsinherits the grade. So in this case a Z—graded simple constituent will just bea Z—graded simple Lie algebra. We have

Lemma 10 Notation as above. If g is Z—graded then g = c⊕⊕n

j=1 g(j) is a

decomposition of g into a direct sum of Z—graded Lie algebras. Furthermorethe corresponding affi ne group action is given as follows

V = c[1] ⊕ g(1)[1] ⊕ ...⊕ g

(n)[1] = c[1] ⊕ V (1) ⊕ ...⊕ V (n)

and if we take G(i) toe be the connected subgroup of G with Lie algebra g(i)

and H(i) the connected subroup of G(i) with Lie algebra g(i)[0] then

Ad(H)|V = {Ic[1]} × Ad(H(1))|V (1) × · · · × Ad(H(n))|V (n) .

We now consider the case when the grade is a Z/mZ—grade. The situationis almost the same except for one complication.In this case we define θ : g → g by θ|g[j] = ζjI with ζ a primitive m—the

root of unity. Then θc = c and θ permutes the simple fractors. In this casethe permutation breaks up into disjoint cycles. We concider one such cycleof length d and relabel so that the cycle is

1→ 2→ ...→ d→ 1.

We note that θ preserves g̃ =⊕d

j=1 g(j) and we analyze its action on this

space. If x ∈⊕d

j=1 g(j) then x =

∑xi and relative to this decomposition we

haveθx =

∑θxi

And θ : g(j) → g(j+1) for j = 1, ..., d − 1 and θ : g(d) → g(1) is a Lie algebraisomorphism. We denote the η—eigenspace for θ by sub—η. Thus g̃1 = g̃[0]

consists of the elements such that θxi = xi+1 for i = 1, ..., d−1 and θxd = x1.So x can be written

d−1∑i=0

θix1

with the additional condition θdx1 = x1. Noting that [g(i), g(j)] = 0 if i 6= j

we see that if we set θ(1) equal to θd then the map φ1 : g(1)1 → g̃1 given by

φ1(x1) =

d−1∑i=0

θix1

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is a Lie algebra isomorphism. Furthermore if η is an m—th root of unityand if x ∈ g̃η then we have θxi = ηxi+1, i = 1, ..., d − 1 and θxd = ηx1 soxi = η−iθix1 and as before θ

dx1 = ηdx1. We see that the map φη : g(1)

ηd→ g̃η

given by

φη(x1) =

d−1∑i=0

η−iθix1

defines a linear isomorphism that it addition satisfies

φη([x1, y1]) = [φ1(x1), φη(y1)]

for x1 ∈ g(1)1 and y1 ∈ g(1)

ηd.

We can now prove the analogue of Lemma 10.

Proposition 11 Let (L, V ) be a Vinberg pair. Then (L, V ) is isomorphicwith a direct sum of a vector space ({I}, Vo) and Vinberg spaces (Lj, Vj) withcorresponing graded Lie algebra simple.

The above argument suggests the following let g be a Lie algebra over Cand let g{d} be the direct sum of d copies of g. Let θ be an automorphism ofg and define

θ{d}(x1, ..., xd) = (θxd, x1, ..., xd−1)

then θd{d} = (θ, ..., θ).

Lemma 12 The map (x1, ..., xd) 7−→∑θi−1xi defines an isomorphism of

the graded Lie algebra (θd{d}, (g(1)){d}) onto (θ, g̃).

Exercise. Prove this lemma.

2.7 Maximal compact groups.

In this subsection we take G to be semi-simple, connected and such that g =Lie(G) is Z/mZ—graded. We assume that there exists, τ , an automorphismof G such that dτ = θ. We assume as, we may, that G is a Z—closed subgroupof GL(n,C) that is invariant under adjoint. We set K = G ∩ U(n).We note that τ(K) is a maximal compact subgroup of G (this is true for

any continuous isomorphism). Thus there exists g ∈ G such that gτ(K)g−1 =K. Hence, if we define τ ′(x) = gτ(x)g−1 then τ ′ is an automorphism of G and

8

if θ′(X) = gθ(X)g−1 then θ′ is an element of orderm such that τ ′(eX) = eθ′X .

We will therefor assume that τ(K) = K and θ(Lie(K)) = Lie(K).If θm = 1 with m <∞, we also fix the inner product

〈X, Y 〉 =1

m

m−1∑l=0

tr(θlX(θlY )∗)

and the symmetric bilinear form

(X, Y ) =1

m

m−1∑k=0

tr(θlX(θlY )).

We note that, since X ∈ Lie(K) is skew Hermitian, if X 6= 0, X ∈ Lie(K)then (X,X) < 0. This implies that each term in the above sum for X =Y 6= 0 in Lie(K) is negative. Thus (..., ...) is negative definite on Lie(K) andtherefore positive definite on iLie(K). In particular, (..., ...) is non-degenerateon Lie(G). We also note that since θAd(g)θ−1 = Ad(τg) for g ∈ G the form(..., ...) is both G and θ—invariant Similarly, we have an inner product 〈..., ...〉that is K and θ—invariant.We will denote Lie(G) by g. Let for ζ ∈ C, gζ = {X ∈ g|θX = ζX}.

Then we note that if the order of θ is m <∞ thena) gζ = 0 if ζm 6= 1.b) Set Gτ = {g ∈ G|τ(g) = g} then Lie(Gτ ) = g1.c) If θ is arbitrary but semisimple then (gζ , gµ) = 0 inless ζµ = 1 and

the corresponding pairing of gζ with gζ−1 is perfect. Thus in particular,dim gζ = dim gζ−1 .

Definition 13 A Vinberg triple (G, V, τ) consists of a connected semi-simplealgebraic group G, as above, and automorphism, τ , of G such that θ = dτ issemi-simple as an automorphism of g = Lie(G), V is the θ—eigenspace for anontrivial eigenvalue of θ.

There is also a notion of a θ—space which is a pair (H, V ) of a connectedalgebraic groupH and an algebraicH—module V such that there is a Vinbergtriple (G, V, τ) such that H is the identity component of Gτ . We note thatif (L, V ) is a Vinberg pair then if L is non-trivial (L, V ) is a θ—space and if(H,V ) is a θ—space with H the identity component of Gτ and (G, V, τ) is aVinberg triple then (Ad(H)|V , V ) is a Vinberg pair

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Lemma 14 If (H,V ) is a θ—space with θ of finite order. Then there existsa Vinberg triple (G, V, τ) so that if g = Lie(G) then

g =⊕j

gζj

with dτ |gζj

= ζjI with ζ a primitive m—th root of unity, V = gζ and H is theidentity component of Gτ .

Proof. By the definition of θ—space there (G̃, τ , V ) so that H is the identitycomponent of G̃τ and V = Lie(G̃)ζ with ζ is a root of unity. Let m bethe order of ζ then ζ is a primitive m—th root of unity. We may assumethat G̃ is contained in GL(n,C) and is invariant under Hermitian adjoint.Then g = ⊕jLie(G̃)ζj is reductive Lie subalgebra. As above we may assumethat τ normalizes K̃ and that there is an inner product on Lie(G̃) so thatdτ and Ad(K̃) act unitarily. Let K be the connected subgroup of K̃ with

Lie(K) = g ∩ Lie(K̃).Since(Lie(G̃)µ

)∗= Lie(G̃)−µ and

Lie(K) = {x ∈ g|x∗ = −x}

we have spanC(Lie(K)) = g. We assert that K is closed. Indeed, let U be itsclosure. Then (τ |U)m = I. Thus spanCLie(U) = ⊕jLie(G̃)ζj∩spanCLie(U)since the right hand side is contained in g and the left hand side containsLie(K) we conclude that Lie(U) = Lie(K) so K = U (since both groups areconnected). Now set G equal to the Zariski closure of U in G̃. Connectednessimplies that the identity component of Gτ is H. This completes the proof.

Definition 15 A Vindberg space is a Vindberg triple (G, τ, V ) such that ifg = Lie(G) then g =

⊕j gζj with either ζ a primitive m—th root of unity or

ζ is of infinite order and V = gζ.

2.8 A lemma of Richardson

In this section we will describe some ideas of Richardson.

Lemma 16 Let G ⊂ GL(n,C) be algebraic and assume that H is a Zariskiclosed connected subgroup of G. Assume that V ⊂ Cn is a H—invariant sub-space such that if v ∈ V then Lie(G)v∩V = Lie(H)v. If x ∈ V then Gx∩Vis a disjoint union of a finite number of orbits of H all of which are of thesame dimension and are open in Gx ∩ V

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Let u ∈ Gx ∩ V . ThenTu(Gx ∩ V ) ⊂ Lie(G)u ∩ V = Lie(H)u ∩ V.

Also, sinceHu ⊂ Gu ∩ V = Gx ∩ V

we see that dim (Gx ∩ V ) = dim(Hu). This implies that the orbits of H inGx∩V all have the same dimension and are open in Gx∩V . Hence they areclosed and since they are irreducible they are the irreducible components ofthe variety Gx ∩ V .Lemma 17 Let (G, V, τ) be a Vinberg space and set H = Gτ . If x ∈ V then[g, x] ∩ V = [Lie(H), x].

Proof. If y ∈ g then y =∑m−1

k=0 yζk (as usual). If [y, x] ∈ V then [y, x] =[y, x]ζ = [y1, x] ∈ Ad(Lie(H))x. Since it is clear that [g, x]∩ V ⊃ [Lie(H), x]the Lemma follows.

2.9 Basic properties of Vinberg spaces 1.Infinite order.

Let (G, V, τ) be a Vinberg space relative to θ = dτ , Lie(G) = g = ⊕gζk withζ of infinite order (all but a finite number of the gζk = 0) and V = gζ . Weassume that K is as in the previous section and is τ—invariant. As aboveH = Gτ . We can think of g having a Z—grading setting gk = gζk .

Lemma 18 If x ∈ V then ad(x) is nilpotent.

Proof. If y ∈ gζk then ad(x)ry ∈ gζk+r . Since only a finite number of gζk arenon-zero the result follows.Note in the next lemma we won’t assume that ζ is of infinite order.

Lemma 19 If (G, V, τ) is a Vinberg space then under H the identity com-ponent of Gτ the number of nilpotent orbits in V is finite.

Proof. We replace G with Ad(G) and look upon G ⊂ GL(g) a Zariski closedsubgroup. We note that G acting on g has a finite number of nilpotent orbits.The result now follows from Lemmas 17 and 16.We note that this result implies

Proposition 20 If (G, V, τ) is a Vinberg space such that every element x ∈V is nilpotent then under H the identity component of Gτ then there are onlya finite number of H-orbits and there exists v ∈ V such that Hv is Zariskiopen in V .

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2.10 Basic properties of Vinberg spaces 2.Finite order

Let (G, V, τ) be a Vinberg space and assume that τ is of order m and relativeto θ = dτ , Lie(G) = g = ⊕m−1

k=0 gζk with ζ a primitive m—th root of unity andV = gζ . We assume that K is as in the previous section and is τ—invariant.As above H = Gτ

If x ∈ V we set x = xs + xn equal to its Jordan decomposition in g.That is, xs has the property that it is a diagonalizable element of Mn(C)(i.e.semisimple), xn is nilpotent and [xs, xn] = 0. This decomposition is unique.We note that if x ∈ g is semisimple then θx is also semisimple similarly if xis nilpotent θx is nilpotent. Thus θ(xs) = θ(x)s and θ(xn) = θ(x)n. Since,x ∈ gζ we have θx = ζx so xs, xn ∈ V .

Theorem 21 If x ∈ V then Hx is closed if and only if x = xs.

Proof. We first prove that if x ∈ V then Hx contains xs. If xn = 0 there isnothing to prove. Otherwise, set e = xn and let f, h ∈ gxs(that is commutewith xs) be such that

[e, f ] = h, [h, f ] = −2f, [h, e] = 2e

Then h =∑m−1

k=0 hζk with θhζk = ζkhζk . Thus θ[h, e] = 2ζx on the one handand

θ[h, e] =m−1∑k=0

ζk+1[hζk , e].

This implies that [hζk , e] = 0 unless zk = 1. Similarly, [hζk , xs] = 0 for all k.Thus there exists u = h1 ∈ Lie(H) such that [u, xs] = 0 and [u, xn] = 2xn.Hence

Ad(e−tu)x = xs + e−2txn.

This implies our assertion and therefore proves that if x isn’t semisimple ing then Hx isn’t closed. So if Hx is closed then x is semisimple in g.We now show that if x is semisimple then Hx is closed. We note that Hx

contains a closed orbit, Hu with u ∈ V . We note that u ∈ Gx = Gx sincex is semisimple in g. Lemmas 17 and 16. above prove that dimAd(H)x =dimAd(H)u. Since Ad(H)u is closed, irreducible and contained in Ad(H)xwhich is irreducible we see that Ad(H)u = Ad(H)x.

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3 The structure of a Vinberg pair

In this section we delve into the more delicate aspects of Vinbergs results.

3.1 Cartan Subspaces

Definition 22 Let (G, V, τ) be a Vinberg space. A Cartan subspace of V isa subspace that consists of semi-simple commuting elements and is maximalsubject to this property. The rank of V is the maximum of the dimensions ofCartan subspaces.

Remark 23 The rank of (G, V, τ) is 0 if and only if every element of V isnilpotent.

For the rest of this section we will assume that (G, V, τ) is a Vinbergspace with θ = dτ of order m. We will also asume that G is connected.Fix a a Cartan subspace of V . Let T ⊂ G be the intersection of all Zariski

closed sugroups, U , of G such that Lie(U) ⊃ a.

Lemma 24 T is an algebraic torus that is contained in the identity compo-nent of the center of the centralizer of a in G.

Proof. Consider the Z—closure, T1, of exp(a). Then T1 is abelian and itis contained in every Z—closed subgroup that whose Lie algebra contains a.Thus since the Lie algebra of a Lie group is the same as the Lie algebra of itsidentity component we see that T1 = T and T is connected and abeleian. LetG1 be the centralzer of a in G then G1 is reductive furthermore the center Cof G1 has the property that Lie(C) ⊃ a. Hence T is contained in the identitycomponent of the center of G1. Hence, T consists of semi-simple elementsand is connected so it is a torus.We will set T = Ta.

Proposition 25 dimTa = ϕ(m) dim a with ϕ(m) the Euler Totient function(the number of 0 < j < m with gcd(j,m) = 1).

Proof. Set T = Ta. Then τ : T → T is an algebraic automorphism. Thecharacter group, T̂ , of T is a free abelian group of rank equal to dimT . Thusif we identify a character with its differential we find that Lie(T )∗ has abasis λ1, ..., λd such that θ = dτ has an integral matrix. Let e1, ..., ed be the

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dual basis to λ1, ..., λd then θei =∑ajiej with aij ∈ Z. The characteristic

polynomial, f(x), of θ|Lie(T ) is

f(x) =d∑

k=0

ckxk

with cj ∈ Z. But since θm = 1 we have a factorization over Q[ζ]

f(x) =m−1∏k=0

(x− ζk)dk .

By definition of Ta and the maximality of a, d1 = dim a. Let for 0 < j < mand gcd(j,m) = 1, σ : Q[ζ]→Q[ζ] be the element of the Galois group Q[ζ]over Q defined by σ(ζ) = ζj. Thus

f(σ(x)) =d∑

k=0

ckσ(x)k = σ(d∑

k=0

ckxk)

also

σ(f(x)) =m−1∏k=0

(σ(x)− ζkj)dk

and

f(σ(x)) =m−1∏k=0

(σ(x)− ζk)dk .

This implies that dj = d1. Hence, dimLie(T ) ∩ gζj = dim a. This yields thelower bound dimT ≥ ϕ(m) dim a.To prove the upper bound we show that if

b =∑

gcd(m,j)=1

Lie(T ) ∩ gζj

then exp(b) is Zariski closed in T . We may assume that T ⊂ (C×)n. We note

that if h is the m—th cyclotomic polynomial that b = kerh(θ|Lie(T )). Let

h(x) =

ϕ(m)−1∑j=0

rjxj.

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As is standard, aj ∈ Z. We consider the regular homomorphism of T to T.

β(z) =

φ(m)−1∏j=0

τ(z)rj .

If z = exp(u) with u ∈ b then

β(z) = exp(

ϕ(m)−1∑j=0

rjθju) = 1.

Thus ker dβ is equal to b. Hence the identity compoent of ker β is an algebraicsubtorus of T whose Lie algebra contains a. This implies that T = exp b andso dimT ≤ ϕ(m) dim a.

3.2 Restricted roots

We continue with the notation of the previous section.Let a be a Cartan subspace of V and let Σ(a) be the set elements λ ∈ a∗,

λ 6= 0 such that there exists x ∈ g such that x 6= 0 and [h, x] = λ(h)x for allh ∈ a. Then Σ(a) is called the restricted root system of the triple. If λ ∈ a∗set

gλ = {x ∈ g|[h, x] = λ(h)x, h ∈ a}.Then

g = g0⊕ ⊕

λ∈Σ(a)

gλ.

Assume that Σ(a) = ∅. The space a consists of semi-simple elementsthus a is contained in the center of g. The maximality in the definition of aCartan subspace implies that a is the unique Cartan subspace. Also, sinceevery semi-simple element in V is contained in some Cartan subspace we seethat every semi-simple element in V is in a. Lemma 24 implies that T = Ta iscentral. Setting G′ = G/T and g′ = Lie(G′) then (G′, V/a, τ |G′) is a Vinbergtriple of rank 0. We eliminate this complication by adhering to the followingdefinition:

Definition 26 A Vinberg triple will be called semi-simple if G is semi-simple(that is, Lie(G) has 0 center). We will abreviate this to SS Vinberg triple.

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Lemma 27 Assume that (G, V, τ) is a semi-simple Vinberg triple. If a is anon-zero Cartan subspace of V then the set Σ(a) spans the dual space of a.

Proof. If h ∈ a satisfies Σ(a)(h) = {0} then since h is semi-simple [h, g] = 0.Thus h = 0.For the rest of this section we will assume that the rank of (G, V, τ) is

semi-simple and not 0. Fix for the moment a Cartan subspace, a. Then anelement, h, of a is called regular if λ(h) 6= 0 for all λ ∈ Σ(a). We set a′ equalto the set of regular elements of a.If S ⊂ V is a set of commuting semi-simple elements then we set GS =

{g ∈ G|gs = s, s ∈ S}. Then GS is a reductive alebraic group and

(GS, Lie(GS) ∩ V, τ |GS)

is a Vinberg triple. We note the Lie(GS) = {x ∈ g|[x, s] = 0, s ∈ S} = Cg(S).

Lemma 28 If x ∈ Cg(a)ζ then xs ∈ a.

Proof. We have seen that xs ∈ V . Furthermore, if [x, h] = 0 then [xs, h] = 0.The lemma follows.

Lemma 29 Let h ∈ a′ then Cg(h)ζ = Cg(a)ζ.

Proof. This follows since the restricted root decomposition implies Cg(a) =g0 = Cg(h).

Proposition 30 Let H be the identity component of Gτ . Set

r(a) = {x ∈ Cg(a)ζ |xs ∈ a′}.

Then Ad(H)r(a) has non-empty Zariski interior in V .

Proof. We first prove that the map

Φ : H × r(a)→ V

(h, x) 7→ Ad(h)x

has surjective differential for all (h, x) ∈ H × r(a). Let (..., ...) be the formdefined in section 2.2 then both G and θ leave it invariant. This implies

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that it induces a perfect pairing between V = gζ and gζ−1 . We note that ifx, v ∈ r(a) and h ∈ H then if X ∈ Lie(H) then

dΦh,x(Xh, v) = Ad(h)([X, x] + v).

If this differential is not surjective then there would exist u ∈ gζ−1 such that

(u, [X, x] + v) = 0

for all v ∈ r(a) and X ∈ Lie(H) = g1. this implies that

0 = (u, [X, x]) = ([x, u], X)

so [u, x] is orthogonal to g1. So since [u, x] ∈ g1 this implies that [u, x] = 0.As before, ker adx ⊂ ker adxs which u ∈ Cg(xs) = Cg(a). Hence u ∈ Cg(a)ζ−1and is orthogonal to Cg(a)ζ . Thus u = 0. This implies that Φ has open imagein the Euclidian topology. Thus the image of Φ is Zariski dense and so hasnon-zero Zariski interior.

Corollary 31 If a and b are Cartan subspaces of V then there exists h ∈ Hsuch that ha = b.

Proof. The sets Ad(H)r(a) and Ad(H)r(b) both have non-empty Zariskyinterior. This implies that they must have non-trivial intersection. Thusthere is h ∈ H, x ∈ r(a) such that Ad(h)x ∈ r(b). Thus Ad(h)xs ∈ b. SoAd(h)a ⊂ b. The maximality implies equality.

Corollary 32 Let a be a Cartan subspace of V . If x ∈ V and Ad(H)v isclosed in V then there exists h ∈ H such that Ad(h)x ∈ a.

Proof. If b is maximal in V subject to the conditions that pairs of elementsin b commute, every element in b is semisimple and x ∈ b. Then b is a Cartansubspace of V so the previous corollary implies that there is an h ∈ H suchthat Ad(h)b = a.

Corollary 33 Let a be a Cartan subspace of V . Then Ad(H)r(a) is Zariskiopen in V .

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Proof. Let for X ∈ V

det(tI − adX) =

p∑j=0

tjDp−j(X)

with p = dim g. Since each Dp−j is H invariant Dp−j(X) = Dp−j(Xs). Thereexists h ∈ H and x ∈ a such that Xs = Ad(h)x. For x ∈ a we have

det(tI − adX) = tdim g0∏

λ∈Σ(a)

(t− λ(x))dim gλ .

Set u = dim g0. Thus if x ∈ a then Dp−j = 0 for j < u and

Dp−u(x) = (−1)p−u∏

λ∈Σ(a)

λ(x)dim gλ .

This implies that Ad(H)r(a) = {x ∈ V |Dp−u(x) 6= 0}.

3.3 The Weyl group

If G is a Lie group, H is a closed subgroup and a is a subspace of Lie(G)then the normalizer of a in H is the subgroup

NH(a) = {g ∈ H|Ad(g)a ⊂ a}

and the centralizer of a is the subgroup

CH(a) = {g ∈ H|Ad(g)x = x, x ∈ a}.

The Weyl group of a is the group

WH(a) = NH(a)/CH(a)

which we will think of as a group of linear maps of a to a.

Lemma 34 If adx is diagonalizable for every element in a and [x, y] =0, x, y ∈ a then Lie(NH(a)) = Lie(CH(a)).

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Proof. We note that

Lie(NH(a)) = {x ∈ Lie(H)|ad(x)a ⊂ a}

andLie(CH(a)) = {x ∈ Lie(H)|ad(x)a = 0}.

If x ∈ Lie(NH(a)) and y ∈ a then [x, y] ∈ a so (ady)2 x = 0. This impliesthat [y, x] = 0 since ady is diagonalizable. Hence y ∈ Lie(CH(a)).

Corollary 35 If a satisfies the conditions of the previous lemma and if G isa linear algebraic group and H is a Z—closed subgroup then WH(a) is finite.

Proof. The previous lemma implies that CH(a) contains the identity com-ponent of NH(a) since an algebraic group has a finite number of connectedcomponents the result follows.We now return to the situation of the previous section.

Corollary 36 Let a be a Cartan subspace of V then WH(a) is a finite groupwhich we call the Weyl group of the Vinberg space.

We note that since any two Cartan spaces of V are conjugate by H theWeyl group is, up to isomorphism, independent of the choice of a.

Proposition 37 If x, y ∈ a, a Cartan subspace of V then Ad(H)x = Ad(H)yif and only if WH(a)x = y.

Proof. We assume x, y ∈ a and that there exists h ∈ H with hx = y.We consider the Vinberg triple (CG(y), Lie(CG(y))∩ V, τ |CG(y)). Then a andAd(h)a are Cartan subspaces of Lie(CG(y)) ∩ V . Hence there exists u inthe identity component of CG(y)τ such that Ad(u)Ad(h)a = a. We haveAd(uh)x = Ad(u)y = y. Thus uh ∈ NH(a) and Ad(uh)x = y.We now come to Vinberg’s generalization of the Chevalley restriction

theorem. Our proof is a bit simpler than the original.

Theorem 38 Assume that G is semi-simple. The restriction map resV/a :

O(V )H → O(a)WH is an isomorphism of algebras.

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Proof. Set W = WH(a). We first observe1. If x, y ∈ a and Wx ∩ Wy = ∅ then there exists f ∈ O(V )H with

f(x) = 0 and f(y) = 1.Indeed, if x, y ∈ a then Ad(H)x and Ad(H)y are Z-closed in a. The

previous proposition implies that Ad(H)x ∩ Ad(H)y = ∅. Theorem 36 in[GIT] asserts the existance of f .Set B = resV/a(O(V )H) and let q(B) be the quotient field of B. Let F

be the field of rational functions on a. Then F is the quotent field of O(a).We assert that2. F is a normal extension of q(B).Consider,

det(tI − adX) =

p∑j=0

tjDp−j(X)

for X ∈ V (see the previous sub-section) then Dp−j ∈ O(V )H so h(t) =∑pj=0 t

jresV/aDp−j(X) is in q(B)[t]. The roots of this polynomial are 0 andthe elements of Σ(a). Since the span of Σ(a) is a∗ we see that F is thesplitting field of h(t) and thus a normal extension of q(B). This implies that

q(B) = {f ∈ F |σf = f, σ ∈ Gal(F/q(B))}.

Also,3. If σ ∈ Gal(F/q(B)) then σ(a∗) = a∗ hence σO(a) = O(a).Let U = {σ ∈ Aut(O(a))|σ|B = I}.Then U = Gal(F/q(B))|O(a). We now

come to the crux of the argument. Let σ ∈ U then we set for x ∈ a, f ∈ O(a)

δσ,x(f) = σf(x).

This defines a homomorphism of O(a) to C. The Nullstellensatz implies thatthere exists x1 ∈ a such that δσ,x(f) = f(x1) for all f ∈ O(a). By thedefinition of U we see that f(x1) = f(x) for all f ∈ B. 1. implies that thereexists s ∈ W so that x1 = sx. Hence if f ∈ O(a)W then σf = f . Thisimplies that O(a)W ⊂ B. Since the converse is obvious the result follows.

3.4 The GIT quotient

Let (H,V ) be a Vinberg pair corresponding to a graded Lie algebra, g =⊕m−1k=0 gζk with ζ a primitive m—th root of unity. Then O(V )H is a Noetherian

algebra and the variety specmaxO(V )H is an affi ne variety denoted V//H

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(the categoric or GIT quotient). Since O(V )H is a subalgebra of O(V ) wehave a morphism Φ : V → V//H Let a be a Cartan subspace of V and letl = dim a = rank(g).

Theorem 39 Each fiber of Φ is connected, of codimension l and has a finitenumber of H—orbits. In paricular, each fiber has a unique closed orbit and aunique open orbit.

Proof. We note that the fiber Φ−1(Φ(0)) is the set of nilpotent elementsin V which decomposes into a finite number of H—orbits (Theorem 19). Wenote that Φ(x) = Φ(y) if and only if Hxs = Hys. Indeed, consider theset Fx = Φ−1(Φ(x)). We note that if y ∈ Fx then f(x) = f(y) for allf ∈ O(V )H . This implies that f(xs) = f(ys) for all f ∈ O(V )H . Thisimplies that Hxs = Hys. We also have x = xs + ys with [xs, ys] = 0. Thusthe set of elements in Fx is precisely the collection of elements in V of theform h(xs + z) with z nilpotent and in Cg(xs)∩V . Indeed, if y ∈ Fx we haveseen that ys = hxs for some h ∈ H. Thus

h−1y = xs + h−1ys = xs + z

implying the assertion. But (Hxs , Cg(xs)∩V ) is a Vindberg pair. This impliesthat the set of such z breaks up into a finite number of Hxs orbits.Let V ′′ be the set of points v ∈ V such that dimAd(H)v is of maximal

dimension. Then V ′′ is Zariski open and dense in V . This implies that themaximal dimension is the generic dimension. Since each fibre of Φ containsan open orbit and the generic dimension of the fiber of Φ is the minimaldimension (see, for example, Shafarevich, Basic Algebraic Geometry, p.60,Theorem 1.3.7) we must have all fibers have the same dimension. We notethat the dimension of V//H is the same as that of a/W which is l. Thus theminimal dimension of a fiber is dimV − l.

3.5 Complex reflections

We assume that (G, V, τ) is a semi-simple Vinberg Space with θ = dτ andθm = 1. Let a be a Cartan subspace of V and let Σ(a) be the root system ofa. Let λ ∈ Σ(a) and set g[λ] equal to the

g0 ⊕⊕

µ∈Σ(a)∩Cλ

gµ.

We now collect some properties of g[λ].

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Lemma 40 The subalgebra g[λ] is invariant under θ. In fact, θgµ = gζ−1µ.

Proof. The last assertion implies the first. Let y ∈ gµ and let x ∈ a. Then

[x, θy] = θ[θ−1x, y] = ζ−1µ(x)θy.

Lemma 41 g[λ] is reductive.

Proof. Let B denote the Killing form of g. Then we have B(gµ, gν) = 0unless, µ+ ν = 0. This implies that B is nondegenerate on g[λ].

Lemma 42 Let y ∈ gµ and write y =∑m−1

j=0 yζj relative to the grade on gcorrespondng to θ then if x ∈ a and µ(x) 6= 0 then1. (ad(x)

µ(x))my1 = y1.

2. yζj = (ad(x)µ(x)

)jy1, j = 0, ...,m− 1.

Proof. λ(x)y =∑

j ad(x)yζj . Since x ∈ gζ , ad(x)yζj = λ(x)yζj+1 . Thisimplies the result.We consider, [g[λ], g[λ]] which is a semi-simple Lie algebra that is Z/mZ

graded. We also note that a∩[g[λ], g[λ]] 6= 0. Also, the restriced root systemof g[λ] on a consists of multiples of λ. Thus we see that Lemma 27 implesthat

dim([g[λ], g[λ]] ∩ a

)= 1.

We have proved

Lemma 43 There exists a unique element of Hλ ∈ [g[λ], g[λ]] ∩ a such thatλ(Hλ) = 1.

Proposition 44 Let (G, V, τ) be a rank one semi-simple Vinberg space thenits Weyl group is a non-trivial cyclic group.

Proof. Let (L, V ) be the corresponing Vinberg pair. Lemma 11 implies thatV is isomorphic with a direct sum of simple Vinberg pairs. Since the rank isone only one of the pairs has positive rank, which is one. Thus we can assumethat g = Lie(G) is simple and that θ = dτ has order m. Let a be a Cartansubspace of V . Then since the rank is 1, Σ(a) ⊂ Cλ with λ a restricted root.This implies that the Weyl group is isomorphic with a subgroup of C×. Thus

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it is a cyclic group (see the appendix to this subsection). Assume that thegroup is trivial. We will derive a contradiction. Since O(a)W = O(a) we sethat O(V )H = C[u] with u ∈ V ∗. Let Hλ ∈ a be defined by λ(Hλ) = 1.Since Hλ is semi-simple and a = CHλ we may assume that u(Hλ) = 1.Wealso note that keru is H-invariant thus complete reducibility implies that

V = Cv ⊕ keru

with Ad(H)v ⊂ Cv. We also note that since u generates the H—invariantskeru is precisely the set of nilpotent elements in V . Let Ad(h)v = χ(h)vbe the action of H on Cv. Let v = vs + vn be its Jordan decomposition. Ifh ∈ H,

Ad(h)vs = (Ad(h)v)s = χ(h)vs.

Since u(v) 6= 0, vs 6= 0. So we may assume v = vs. There exists h ∈ H suchthat Ad(h)v = cHλ thus

χ(h)v = cHλ.

We therefore see that we may assme that v = Hλ.Since the Weyl group istrivial we see that χ is the trivial character. We are finally ready for thecontradiction. Let y ∈ gλ and let y =

∑m−1k=0 yζk(ζ a primitive m—th root of

unity) relative to the grade. Lemma 42 implies that

yζk = ad(Hλ)ky1

and ad(Hλ)my1 = y1. But Ad(H) fixes Hλ. Hence y = 0. This is the desired

contradiction.

Definition 45 If X is a complex vector space then a complex reflection is alinear transformation s : X → X such that ker(s− I) has codimension 1 andit has a non-trivial root of unity as an eigenvalue.

Proposition 46 Let (G, V, τ) be a semi-simple Vinberg space and let a bea Cartan subspace of V with Σ(a) the corresponding root system and let Hbe the identity component of Gτ . Let λ ∈ Σ(a) then there exists a complexreflection sλ ∈ WH(a) = W such that ker(sλ− I) = kerλ and the order of sλis the order of the Weyl group of g[λ].

Proof. We note that if y ∈ g[λ] and if x ∈ a and λ(x) = 0 then [x, y] = 0.Let G[λ] be the connected group correseponding to g[λ] and let H [λ] be the

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identity component of H ∩G[λ]. Then we know that WH[λ]([g[λ], g[λ]]∩ a) is acyclic group. Let s be a generator and let l be its order. Let h ∈ H [λ] ⊂ Hbe a representative. Then Ad(h)Hλ = µHλ with µ a primative l—th root ofunity. Also, Ad(h)| kerλ = I. Thus Ad(h)|a is a complex reflection of order l.

Lemma 47 We continue the notation in the previous proposition. If λ ∈Σ(a) and x ∈ a then

sλ(x) = x+ (µ− 1)λ(x)Hλ

with µ a primative l—th root of unity.Proof. a = CHλ ⊕ kerλ. If x ∈ a then x = λ(x)Hλ + (x − λ(x)Hλ) is itsecpression relative to the direct sum decomposition. Thus

sλ(x) = µλ(x)Hλ + x− λ(x)Hλ.

We note that the reflection sλ depends on the choice of primitive l—rootof unity but otherwise is the same for any multiple of λ that is a root.

Proposition 48 We maintain the notation of the Proposition above. Let W̃be the subgroup of W generated by the reflections sλ for λ ∈ Σ(a). Then W̃is a normal subgroup of W . If g is simple then W̃ acts irreducibly on a.

Proof. If s ∈ W and λ ∈ Σ(a) then sλ = λ ◦ s−1 ∈ Σ(a). This anthe previous Lemma implies that W̃ is normal. Suppose that g is simpleof W̃ is and a = A ⊕ B with W̃A = A and W̃B = B. If a ∈ A thensλ(a) = (l− 1)λ(a)Hλ + a. Thus if λ(a) 6= 0 then Hλ ∈ A. This implies thatif λ|A 6= 0 then λ|B = 0 and vice-versa. Set ΣA = {λ ∈ Σ(a)|λ|A 6= 0} andΣB = {λ ∈ Σ(a)|λ|B 6= 0}. Then Σ(a) = ΣA ∪ ΣB a disjoint union (sinceΣ(a) spans the dual space of a). We also note that the above observationimpleis that if λ ∈ ΣA and µ ∈ ΣB then cλ+ dµ ∈ Σ(a) only if cd = 0. Let

ZA =∑λ∈ΣA

gλ, ZB =∑λ∈ΣB

gλ

and let gA (resp. gB) be the Lie subalgebra of g generated by ZA (resp. ZB).Noting that [ZA, ZB] = 0 we see that gA and gB are ideals in g. Thus one ofthem must be 0. Hence A or B is 0.We finally come to the crux of the matter.

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3.5.1 Appendix

The purpose of this appendix is to prove two technical Lemmas needed forthe subsection above and to recall a standard result with a reference. Thefirst uses an argument that was posted by Andrea Petracci on the internet.

Lemma 49 If G is a finite subgroup of the multiplicative group of a fieldthen G is cyclic.

Proof. We note that this result follows from the following observation:Let G be a finite group with n elements. If for every d|n we have∣∣{x ∈ G|xd = 1}

∣∣ ≤ d , then G is cyclic.Indeed, if F is a field then the number of solutions of a polynomial of

degree d is at most d.We now prove the observation. Fix d|n and consider the set Gd consisting

of the elements of G with order d. Suppose that Gd 6= ∅ and let y ∈ Gd.Then

〈y〉 = {yj|j ∈ Z} ⊂ {x ∈ G|xd = 1}.But the subgroup 〈y〉 has cardinality d so 〈y〉 = {x ∈ G|xd = 1} . ThereforeGd is the set of generators of the cyclic group 〈y〉 of order d. We have shownthat |Gd| = ϕ(d) with ϕ(d) Euler’s totient function (the number of positiveintegers less than d relatively prime to d).We have proved that Gd is empty or has cardinality ϕ(d) , for every d|n

. So we haven = |G| =

∑d|n

Gd ≤∑d|n

ϕ(d) = n.

Therefore |Gd| = ϕ(d) for every d dividing n . In particular Gd 6= ∅ . Thisproves that G is cyclic.The next result is due to Vinberg.

Lemma 50 Let V be a finite dimensional vector space over C and assumethat G is a finite subgroup of GL(V ) such that (V − {0})/G is simply con-nected in the quotient topology. G is generated by its elements that havenon-zero fixed points.

Proof. Let H be the subgroup of G generated by those elements with non-zero fixed points. Then H is a normal subgroup of G. Hence

(V − {0})/G = ((V − {0})/H)/(G/H).

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So (V − {0})/H is a covering space of (V − {0})/G with the group of decktransformations G/H. But (V −{0})/G is simply connected so we must haveG/ = H.The standard result is

Proposition 51 Let G be reductive and connected. Assume that x ∈ Lie(G)is semi-simple. Then CG(x) is connected.

Proof. Let T be the Zariski closure of {etx|t ∈ C} in G. Then T is analgebraic torus. The result now follows from Corollary 11.12 in [B] page 152.

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