the complex numbers, so can consider the basis X + iY, ’X + iY, iZ and ¬nd

that

[iZ, X +iY ] = X +iY, [iZ, ’X +iY ] = ’(’X +iY ), [X +iY, ’X +iY ] = 2iZ.

These are the bracket relations for sl(2) with e = X + iY, f = ’X + iY, h =

iZ. In other words, the complexi¬cation of our three dimensional world is the

irreducible three dimensional representation of sl(2) so o(3) = sl(2) which is

simple.

To study the higher dimensional orthogonal algebras it is useful to make two

remarks:

If V is a vector space with a non-degenerate symmetric bilinear form ( , ),

we get an isomorphism of V with its dual space V — sending every u ∈ V to the

linear function u where u (v) = (v, u). This gives an identi¬cation of

End(V ) = V — V — with V — V.

Under this identi¬cation, the elements of o(V ) become identi¬ed with the anti-

symmetric two tensors, that is with elements of §2 (V ). (In terms of an or-

thonormal basis, a matrix A belongs to o(V ) if and only if it is anti-symmetric.)

Explicitly, an element u§v becomes identi¬ed with the linear transformation

Au§v where

Au§v x = (x, v)u ’ (u, x)v.

This has the following consequence. Suppose that z ∈ V with (z, z) = 0, and

let w be any element of V . Then

Aw§z z = (z, z)w ’ (z, w)z

and so U (o(V ))z = V . On the other hand, suppose that u ∈ V with (u, u) = 0.

We can ¬nd v ∈ V with (v, v) = 0 and (v, u) = 1. Now suppose in addition that

dim V ≥ 3. We can then ¬nd a z ∈ V orthogonal to the plane spanned by u

and v and with (z, z) = 1. Then

Az§v u = z,

so z ∈ U (o(V ))e and hence U (o(V ))u = V . We have proved:

1 If dim V ≥ 3, then every non-zero vector in V is cyclic, i.e the representa-

tion of o(V ) on V is irreducible.

3.3. THE ORTHOGONAL ALGEBRAS. 49

(In two dimensions this is false - the line spanned by a vector e with (e, e) = 0

is a one dimensional invariant subspace.)

We now show that

2 o(V ) is simple for dim V ≥ 5.

For this, begin by writing down the bracket relations for elements of o(V ) in

terms of their parametrization by elements of §2 V . Direct computation shows

that

[Au§v , Ax§y ] = (v, x)Au§y ’ (u, x)Av§y ’ (v, y)Au§x + (u, y)Av§x . (3.7)

Now let n = dim V ’ 2 and choose a basis

u, v, x1 , . . . , xn

of V where

(u, u) = (u, xi ) = (v, v) = (v, xi ) = 0 ∀i, (u, v) = 1, (xi , xj ) = δij .

Let g := o(V ) and write W for the subspace spanned by the xi . Set

d := Au§v

and

g’1 := {Av§x , x ∈ W }, g0 := o(W ) • Cd, g1 := {Au§x , x ∈ W }.

It then follows from (3.7) that d satis¬es (3.5). The spaces g’1 and g1 look like

copies of W with the o(W ) part of g0 acting as o(W ), hence irreducibly since

dim W ≥ 3. All our remaining axioms are easily veri¬ed. Hence o(V ) is simple

for dim V ≥ 5.

We have seen that o(3) = sl(2) is simple.

However o(4) is not simple, being isomorphic to sl(2) • sl(2): Indeed, if Z1

and Z2 are vector spaces equipped with non-degenerate anti-symmetric bilinear

forms , 1 and , 2 then Z1 —Z2 has a non-degenerate symmetric bilinear form

( , ) determined by

(u1 — u2 , v1 — v2 ) = u1 , v1 u 2 , v2 2 .

1

The algebra sl(2) acting on its basic two dimensional representation in¬nitesi-

mally preserves the antisymmetric form given by

x1 y

,1 = x1 y2 ’ x2 y1 .

x2 y2

Hence, if we take Z = Z1 = Z2 to be this two dimensional space, we see that

sl(2) • sl(2) acts as in¬nitesimal orthogonal transformations on Z — Z which is

four dimensional. But o(4) is six dimensional so the embedding of sl(2) • sl(2)

in o(4) is in fact an isomorphism since 3 + 3 = 6.

50 CHAPTER 3. THE CLASSICAL SIMPLE ALGEBRAS.

3.4 The symplectic algebras.

We consider an even dimensional space with coordinates q1 , q2 , . . . , p1 , p2 , . . . .

The polynomials have a Poisson bracket

‚f ‚g ‚f ‚g

{f, g} := ’ . (3.8)

‚pi ‚qi ‚qi ‚pi

This is clearly anti-symmetric, and direct computation will show that the Ja-

cobi identity is satis¬ed. Here is a more interesting proof of Jacobi™s identity:

Notice that if f is a constant, then {f, g} = 0 for all g. So in doing bracket

computations we can ignore constants. On the other hand, if we take g to be

successively q1 , . . . , qn , p1 , . . . , pn in (3.8) we see that the partial derivatives of f

are completely determined by how it brackets with all g, in fact with all linear

g. If we ¬x f , the map

h ’ {f, h}

is a derivation, i.e. it is linear and satis¬es

{f, h1 h2 } = {f, h1 }h2 + h1 {f, h2 }.

This follows immediately from from the de¬nition (3.8). Now Jacobi™s identity

amounts to the assertion that

{{f, g}, h} = {f, {g, h}} ’ {g, {f, h}},

i.e. that the derivation

h ’ {{f, g}, h}

is the commutator of the of the derivations

h ’ {f, h} and h ’ {g, h}.

It is enough to check this on linear polynomials h, and hence on the polynomials

qj and pk . If we take h = qj then

‚f ‚g

{f, qj } = {g, qj } =

,

‚pj ‚pj

so

‚f ‚ 2 g ‚f ‚ 2 g

{f, {g, qj }} = ’

‚pi ‚qi ‚pj ‚qi ‚pi ‚pj

‚g ‚ 2 f ‚g ‚ 2 f

{f, {f, qj }} = ’ so

‚pi ‚qi ‚pj ‚qi ‚pi ‚pj

‚

{f, {g, qj }} ’ {g, {f, qj }} = {f, g}

‚pj

= {{f, g}, qj }

as desired, with a similar computation for pk .

3.4. THE SYMPLECTIC ALGEBRAS. 51

The symplectic algebra sp(2n) is de¬ned to be the subalgebra consisting of

all homogeneous quadratic polynomials. We divide these polynomials into three

groups as follows: Let g1 consist of homogeneous polynomials in the q™s alone,

so g1 is spanned by the qi qj . Let g’1 be the quadratic polynomials in the p™s

alone, and let g0 be the mixed terms, so spanned by the qi pj . It is easy to see

that g0 ∼ gl(n) and that [g’1 , g1 ] = g0 . To check that g’1 is irreducible under

g0 , observe that [p1 qj , pk p ] = 0 if j = k or , and [p1 qj , pj p ] is a multiple of