Without mentioning cohomology groups, one can state Whitehead's first lemma as follows: Let be a finite-dimensional, semisimple Lie algebra over a field of characteristic zero, V a finite-dimensional module over it, and a linear map such that
- .
Then there exists a vector such that for all .
In terms of Lie algebra cohomology, this is, by definition, equivalent to the fact that for every such representation. The proof uses a Casimir element (see the proof below).[2]
Similarly, Whitehead's second lemma states that under the conditions of the first lemma, also .
Another related statement, which is also attributed to Whitehead, describes Lie algebra cohomology in arbitrary order: Given the same conditions as in the previous two statements, but further let be irreducible under the -action and let act nontrivially, so . Then for all .[3]
As above, let be a finite-dimensional semisimple Lie algebra over a field of characteristic zero and :{\mathfrak {g}}\to {\mathfrak {gl}}(V)}
a finite-dimensional representation (which is semisimple but the proof does not use that fact).
Let where is an ideal of . Then, since is semisimple, the trace form , relative to , is nondegenerate on . Let be a basis of and the dual basis with respect to this trace form. Then define the Casimir element by
which is an element of the universal enveloping algebra of . Via , it acts on V as a linear endomorphism (namely, .) The key property is that it commutes with in the sense for each element . Also,
Now, by Fitting's lemma, we have the vector space decomposition such that is a (well-defined) nilpotent endomorphism for and is an automorphism for . Since commutes with , each is a -submodule. Hence, it is enough to prove the lemma separately for and .
First, suppose is a nilpotent endomorphism. Then, by the early observation, ; that is, is a trivial representation. Since , the condition on implies that for each ; i.e., the zero vector satisfies the requirement.
Second, suppose is an automorphism. For notational simplicity, we will drop and write . Also let denote the trace form used earlier. Let , which is a vector in . Then
Now,
and, since , the second term of the expansion of is
Thus,
Since is invertible and commutes with , the vector has the required property.