I learned the material in this post from the book by Humphreys on Lie algebras and representation theory.

Recall that if {A} is any algebra (not necessarily associative), then the derivations of {A} form a Lie algebra {Der(A)}, and that if {A} is actually a Lie algebra, then there is a homomorphism {\mathrm{ad}: A \rightarrow Der(A)}. In this case, the image of {\mathrm{ad}} is said to consist of inner derivations.

Theorem 1 Any derivation of a semisimple Lie algebra {\mathfrak{g}} is inner.

 

To see this, consider {\mathrm{ad}: \mathfrak{g} \rightarrow D :=Der(\mathfrak{g})}; by semisimplicity this is an injection. Let the image be {D_i}, the inner derivations. Next, I claim that {[D, D_i] \subset D_i}. Indeed, if {\delta \in D} and {\mathrm{ad} x \in D_i}, we have

\displaystyle [\delta, \mathrm{ad} x] y = \delta( [x,y]) - [x, \delta(y)] = [\delta(x),y] = (\mathrm{ad}(\delta(x)))y.

In other words, {[\delta, \mathrm{ad} x] = \mathrm{ad}(\delta(x))}. This proves the claim.

Consider the Killing form {B_D} on {D} and the Killing form {B_{D_i}} on {D_i}. The above claim and the definition as a trace shows that {B_D|_{D_i \times D_i} = B_{D_i}}.

Now, if {D_i^{\perp}} is the orthogonal complement to {D_i} under the form {B_D}, we must have

\displaystyle D_i + D_i^{\perp} = D

but also, because of nondegeneracy of {B_{D_i}},

\displaystyle D_i \cap D_i^{\perp} = \{ 0 \}.

I claim now {D_i^{\perp} = \{0\}}. If {\delta \in D_i^{\perp}}, then for any {x \in \mathfrak{g}},

\displaystyle [\delta, \mathrm{ad}(x)] = \mathrm{ad}(\delta x) \in D_i \cap D_i^{\perp} = \{ 0 \}

because {D_i^{\perp}} is an ideal (by invariance of the Killing form again). By semisimplicity, {\delta x = 0}, and since {x} was arbitrary, we find {\delta } is the zero derivation. So {D = D_i}.

Abstract Jordan decomposition

Now assume our fields are algebraically closed.

 

Proposition 2 Let {A} be a finite-dimensional algebra. Let {\delta: A \rightarrow A} be a derivation, and regard {\delta} as an element of {End(A)} to take its Jordan decomposition {\delta = S + N}. Then {S,N} are derivations of {A} as well.

 

It is clearly enough to prove {S} is a derivation. There may be confusion caused by my using {\delta} to refer to a specific

The idea is to write {A} as a direct sum

\displaystyle A = \bigoplus_{\lambda} A_{\lambda}

where the {A_\lambda} are {\delta}-invariant subspaces with {(\delta-\lambda)} acting nilpotently on them. Then {S} acts on {A_{\lambda}} by {\lambda}.

It is enough to check that for any {a \in A_{\lambda}, b \in A_{\mu}},

\displaystyle a b \in A_{\lambda + \mu}.

For then

\displaystyle S(ab) = (\lambda + \mu)(ab) = (Sa) b + a(Sb).

Now indeed, we have a binomial-like formula

\displaystyle (\delta - \lambda - \mu)^n(ab) = \sum_{i=0}^n \binom{n}{i} (\delta - \lambda)^i a \cdot (\delta - \mu)^{n-i} b

that can be checked by induction on {n}. It shows that {ab} is annihilated by some high power of {\delta - \lambda - \mu}.

Theorem 3 (Abstract Jordan decomposition) Let {\mathfrak{g}} be a semisimple Lie algebra. Then we can write any {x \in \mathfrak{g}} uniquely as {x = s + n} where {\mathrm{ad} s} is semisimple, {\mathrm{ad} n} is nilpotent, and {s,n} commute with each other and with every element of {\mathfrak{g}} that commutes with {x}.

 

We imbed {\mathfrak{g}} as a subalgebra {D_i} of {gl(\mathfrak{g})} via the {\mathrm{ad}} mapping. Then {\mathrm{ad} x \in gl(\mathfrak{g})} splits into semisimple and nilpotent parts, {\mathrm{ad} x = S + N}. Then {S,N} are derivations of {\mathfrak{g}} by the proposition, and inner ones by Theorem 1, so we get

\displaystyle \mathrm{ad} x = \mathrm{ad} s + \mathrm{ad} n

and semisimplicity implies then {x = s + n}. Since

\displaystyle \mathrm{ad} [s,n] = [\mathrm{ad} s, \mathrm{ad} n] = 0,

we find {[s,n]=0}. The commuting properties are thus seen to follow from the corresponding ones of the normal Jordan decomposition. Uniqueness follows by the uniqueness of the Jordan decomposition in {End(\mathfrak{g})}.

Note that if we have decompositions {x = s+n, y = s' + n'}, and {x,y} commute, then

\displaystyle x+ y = (s+s') + (n + n')

is the Jordan decomposition for {x+y}.

This by itself is not all that interesting. But it turns out to be the case that semisimple elements (i.e., those whose nilpotent part is zero) in a Lie algebra act by semisimple endomorphisms in any representation. We need to talk about complete reducibility first though.

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