24

j=1,k=1

proving (10.5).

10.3 Orthogonal extension of a Lie algebra.

Let us get back to the general case of a Lie algebra r acting as in¬nitesimal

orthogonal transformations on p and the map ν : r ’ §2 p given by (9.13).

Suppose that the Lie algebra r has a non-degenerate invariant symmetric bilinear

form ( , )r . We have the transpose map

ν † : §2 p ’ r

since both r and §2 p have non-degenerate symmetric bilinear forms. For y and

y in p, let us de¬ne

[y, y ]r := ’2ν † (y § y ), .

This map is an r morphism which says that

[x, [y, y ]r ] = [x · y, y ]r + [y, x · y ]r , (10.6)

where the bracket on the left denotes the Lie bracket on r. Also, we have

’2(x, ν † (y § y ))r

(x, [y, y ]r )r =

’2(ν(x), y § y )p

=

’2(ι(y)ν(x), y )p

=

(x · y, y )p .

=

166 CHAPTER 10. THE KOSTANT DIRAC OPERATOR

So we have proved

(x, [y, y ]r )r = (x · y, y )p . (10.7)

This has the following signi¬cance: Suppose that we want to make r • p into a

Lie algebra with an invariant symmetric bilinear form ( , ) such that

• r and p are orthogonal under ( , ),

• the restriction of ( , ) to r is ( , )r and the restriction of ( , ) to p is ( , )p ,

and

• [r, p] ‚ p and the bracket of an element of r with an element of p is given

by [x, y] = x · y.

Then

the r component of [y, y ] must be given by [y, y ]r .

Thus to de¬ne a Lie algebra structure on r • p we must specify the p

component of the bracket of two elements of p. This amounts to specifying

a v ∈ §3 p as we have seen, and the condition that the Jacobi identity hold

for x, y, y with x ∈ r and y, y ∈ p amounts to the condition that v ∈ §3 p be

invariant under the action of r. It then follows that if we try to de¬ne [ , ] = [ , ]v

by

[y, y ] = [y, y ]r + 2ι(y)ι(y )v

then

([z, z ], z ) = (z, [z , z ])

for any three elements of g := r • p, and the Jacobi identity is satis¬ed if at

least one of these elements belongs to r. Furthermore, for any x ∈ r we have

([[y, y ], y ], x) = ([[y, y ], y ], x)r

= ([y, y ], [y , x])p by (10.7)

= ([x, [y, y ]], y )p

= ([[x, y], y ], y )p + ([y, [x, y ]], y )p

= ([x, y], [y , y ])p + ([x, y ], [y , y])p

= (x, [y, [y , y ])r + (x, [y , [y , y]])r

or

([[y, y ], y ] + [[y , y ], y] + [[y , y], y ], x) = 0.

In other words, the r component of the Jacobi identity holds for three elements

of p.

So what remains to be checked is the p component of the Jacobi identity for

three elements of p. This is the sum

Jac(b)(y, y , y ) + [y, y ]r · y + [y , y ]r · y + [y , y]r · y .

10.4. THE VALUE OF [V 2 + ν(CASR )]0 . 167

Let us choose an“orthonormal” basis {xi }, i = 1, . . . , r of r and write

:= (xi , xi )r = ±1

[y, y ]r = i ([y, y ], xi )xi , i

i

so

[y, y ]r · y = ·y .

i ([y, y ]r , xi )xi

i

Then by (9.11) and (10.4) we see that the Jacobi identity is

ι(y)ι(y )ι(y ) v 2 + ν(Casr ) = 0

where

2

∈ U (r)

Casr := i xi

i

does not depend on the choice of basis, and ν : U (r) ’ C(p) is the extension of

the homomorphism ν : r ’ C(p). In particular, we have proved that v de¬nes

an extension of the Lie algebra structure satisfying our condition if and only if

v 2 + ν(Casr ) ∈ C (10.8)

i.e. has no component of degree four.

Suppose that this condition holds. We then have de¬ned a Lie algebra

structure on

g = r • p.

We let Pr and Pp denote projections onto the ¬rst and second components of

our decomposition. Our Lie bracket on g, denoted simply by [ , ] satis¬es

[x, x ] = [x, x ]r , x, x ∈ r (10.9)

[x, y] = x · y, x ∈ r, y ∈ p (10.10)

Pr [y, y ] = [y, y ]r = ’2ν † (y § y ) y, y ∈ p (10.11)

Pp [y, y ] = b(y, y ) = 2ι(y)ι(y )v, y, y ∈ p. (10.12)

From now on we will assume that we are over the complex numbers or that

we are over the reals and the symmetric bilinear forms are positive de¬nite.

This is not for any mathematical reasons but because the formulas become a

bit complicated if we put in all the signs. We leave the general case to the

reader.

The value of [v 2 + ν(Casr )]0 .

10.4

Condition (10.8) says that the degree four component of v 2 + ν(Casr ) vanishes.

Assume that this holds, so we have constructed a Lie algebra. We will now

compute the degree zero component of v 2 + ν(Casr ). The answer will be given

in equation (10.13) below.

168 CHAPTER 10. THE KOSTANT DIRAC OPERATOR

We can write (10.5) as

n

1

2

(v )0 = tr Pp adg (yj )Pp adg (yj )Pp

24 j=1

n

1

= tr Pp adg (yj )Pp adg (yj )

24 j=1

in view of (10.12) where adg denotes the adjoint action on all of g. On the other

hand, we have from (9.12) that

r n

1 1

2

ν(Casr )0 = tr (ad xi ) Pp = ([xi , [xi , yj ]], yj ).