Choose some z ∈ z such that ±i (z ) = 0 and ±i (z ) = ±j (z ) for any i = j.

This is possible since the ±i are all linearly independent. Bracketing the above

equation by z gives

±( z )ai xi ’ ±i (z )bi yi = 0

by the relations (8.4) and (8.5). Repeated bracketing by z and using the van

der Monde (or induction) argument shows that ai = 0, bi = 0 and hence that

z = 0.

We have proved that the elements xi , yj , zk in m are linearly independent.

The element

[xi1 , [xi2 , [· · · [xit’1 , xit ] · · · ]]]

is an eigenvector of zi with eigenvalue

ci1 i + · · · + cit i .

For any pair of elements µ and » of z— (or of h— ) recall that

µ »

8.2. THE FIRST FIVE RELATIONS. 141

denotes the fact that » ’ µ = ki ±i where the ki are all non-negative integers.

For any » ∈ z— let m» denote the set of all m ∈ m satisfying

[z, m] = »(z)m ∀z ∈ z.

Then we have shown that the subalgebra x of m generated by x1 , . . . , x is

contained in

m+ := m» .

0»

Similarly, the subalgebra y of m generated by the yi lies in

m’ := m» .

»0

In particular, the vector space sum

y+z+x

is direct since z ‚ m0 . We claim that this is in fact all of m. First of all, observe

that it is a subalgebra. Indeed, [yi , xj ] = ’δij zi lies in this subspace, and hence

[yi , [xj1 , [· · · [xjt ’1 , xjt ] · · · ] ∈ x for t ≥ 2.

Thus the subspace y + z + x is closed under ad yi and hence under any product

of these operators. Similarly for ad xi . Since these generate the algebra m we

see that y + z + x = m and hence

x = m+ and y = m’ .

We have shown that

m = m’ • z • m+

where z is an abelian subalgebra of dimension , where the subalgebra m+ is

generated by x1 , . . . , x , where the subalgebra m’ is generated by y1 , . . . , y , and

where the 3 elements x1 , . . . , x , y1 . . . , y , z1 , . . . , z are linearly independent.

There is a further property of m which we want to use in the next section in

the proof of Serre™s theorem. For all i = j between 1 and de¬ne the elements

xij and yij by

xij := (ad xi )’cji +1 (xj ), yij := (ad yi )’cji +1 (yj ).

Conditions (8.6) and (8.7) amount to setting these elements, and hence the ideal

that they generate equal to zero. We claim that for all k and all i = j between

1 and we have

ad xk (yij ) = 0 (8.9)

and

ad yk (xij ) = 0. (8.10)

142 CHAPTER 8. SERRE™S THEOREM.

By symmetry, it is enough to prove the ¬rst of these equations. If k = i then

[xk , yi ] = 0 by (8.3) and hence

ad xk (yij ) = (ad yi )’cji +1 [xk , yj ] = (ad yi )’cji +1 δkj hj

by (8.2) and (8.3). If k = j this is zero. If k = j we can write this as

(ad yi )’cji (ad yi )hj = (ad yi )’cji cij yi .

If cij = 0 there is nothing to prove. If cij = 0 then cji = 0 and in fact is strictly

negative since the angles between all elements of a base are obtuse. But then

(ad yi )’cji yi = 0.

It remains to consider the case where k = i. The algebra generated by xi , y, zi

is isomorphic to sl(2) with [xi , yi ] = zi , [zi , xi ] = 2xi , [zi , yi ] = ’2yi . We have

a decomposition of m into weight spaces for all of z, in particular into weight

spaces for this little sl(2). Now [xi , yj ] = 0 (from (8.3)) so yj is a maximal weight

vector for this sl(2) with weight ’cji and (8.9) is just a standard property of a

maximal weight module for sl(2) with non-negative integer maximal weight.

8.3 Proof of Serre™s theorem.

Let k be the ideal of m generated by the xij and yij as de¬ned above. We wish

to show that

g := m/k

is a semi-simple Lie algebra with Cartan subalgebra h = z/k and root system

¦. For this purpose, let i now denote the ideal in m+ generated by the xij and

j be the ideal in m’ generated by the yij so that

i + j ‚ k.

We claim that j is an ideal of m. Indeed, each yij is a weight vector for

z, and [z, m’ ] ‚ m’ , hence [z, j] ‚ j. On the other hand, we know that

[xk , m’ ] ‚ m’ + z and [xk , yij ] = 0 by (8.9). So (ad xk )j ‚ j by Jacobi. Since

the xk generate m+ Jacobi then implies that [m+ , j] ‚ j as well, hence j is an

ideal of m. Similarly, i is an ideal of m. Hence i + j is an ideal of m, and since

it contains the generators of k, it must coincide with k, i.e.

k = i + j.

In particular, z©k = {0} and so z projects isomorphically onto an -dimensional

abelian subalgebra of g = m/k. Furthermore, since j © m+ = {0} and i © m’ =

{0} we have

g = n’ • h • n+ (8.11)

as a vector space where

n’ = m’ /j, and n+ = m+ /i,

8.3. PROOF OF SERRE™S THEOREM. 143

and n+ is a a sum of weight spaces of h, summed over » 0 while n’ is a sum

of weight spaces of h with » 0. We have to see which weight spaces survive

the passage to the quotient. The sl(2) generated by xi , yi , zi is not sent into zero

by the projection of m onto g since zi is not sent into zero. Since sl(2) is simple,

this means that the projection map is an isomorphism when restricted to this

sl(2). Let us denote the images of xi , yi , zi by ei , fi , hi . Thus g is generated by

the 3 elements

e1 , . . . , e , f1 , . . . , f , h1 , . . . , h

and all the axioms (8.1)-(8.7) are satis¬ed.

We must show that g is ¬nite dimensional, semi-simple, and has ¦ as its

root system.

First observe that ad ei acts nilpotently on each of the generators of the

algebra g, and hence acts locally nilpotently on all of g. Similarly for ad fi .

Hence the automorphism

„i := (exp ad ei )(ad ’ fi )(exp ad ei )

is well de¬ned on all of g. So if si denotes the re¬‚ection in the Weyl group W

corresponding to i, we have

„i (g» ) = gsi » .

Notice that each of the m» is ¬nite dimensional, since the dimension of m» for

» 0 is at most the number of ways to write » as a sum of successive ±i , each

such sum corresponding to the element [xi1 , [xi2 , [· · · , xit ] · · · ]. (In particular

mk± = {0} for k > 1.) Similarly for » 0. So it follows that each of the g» is

¬nite dimensional, that

∀w ∈ W

dim gw» = dim g»