Whitehead's lemma (Lie algebra)

Without mentioning cohomology groups, one can state Whitehead's first lemma as follows: Let ${\displaystyle {\mathfrak {g}}}$ be a finite-dimensional, semisimple Lie algebra over a field of characteristic zero, V a finite-dimensional module over it, and ${\displaystyle f\colon {\mathfrak {g}}\to V}$ a linear map such that

${\displaystyle f([x,y])=xf(y)-yf(x)}$.

Then there exists a vector ${\displaystyle v\in V}$ such that ${\displaystyle f(x)=xv}$ for all ${\displaystyle x\in {\mathfrak {g}}}$. In terms of Lie algebra cohomology, this is, by definition, equivalent to the fact that ${\displaystyle H^{1}({\mathfrak {g}},V)=0}$ 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 ${\displaystyle H^{2}({\mathfrak {g}},V)=0}$.

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 ${\displaystyle V}$ be irreducible under the ${\displaystyle {\mathfrak {g}}}$-action and let ${\displaystyle {\mathfrak {g}}}$ act nontrivially, so ${\displaystyle {\mathfrak {g}}\cdot V\neq 0}$. Then ${\displaystyle H^{q}({\mathfrak {g}},V)=0}$ for all ${\displaystyle q\geq 0}$.[3]

As above, let ${\displaystyle {\mathfrak {g}}}$ be a finite-dimensional semisimple Lie algebra over a field of characteristic zero and ${\displaystyle \pi$ :{\mathfrak {g}}\to {\mathfrak {gl}}(V)} a finite-dimensional representation (which is semisimple but the proof does not use that fact).

Let ${\displaystyle {\mathfrak {g}}=\operatorname {ker} (\pi )\oplus {\mathfrak {g}}_{1}}$ where ${\displaystyle {\mathfrak {g}}_{1}}$ is an ideal of ${\displaystyle {\mathfrak {g}}}$. Then, since ${\displaystyle {\mathfrak {g}}_{1}}$ is semisimple, the trace form ${\displaystyle (x,y)\mapsto \operatorname {tr} (\pi (x)\pi (y))}$, relative to ${\displaystyle \pi }$, is nondegenerate on ${\displaystyle {\mathfrak {g}}_{1}}$. Let ${\displaystyle e_{i}}$ be a basis of ${\displaystyle {\mathfrak {g}}_{1}}$ and ${\displaystyle e^{i}}$ the dual basis with respect to this trace form. Then define the Casimir element ${\displaystyle c}$ by

${\displaystyle c=\sum _{i}e_{i}e^{i},}$

which is an element of the universal enveloping algebra of ${\displaystyle {\mathfrak {g}}_{1}}$. Via ${\displaystyle \pi }$, it acts on V as a linear endomorphism (namely, ${\displaystyle \pi (c)=\sum _{i}\pi (e_{i})\circ \pi (e^{i}):V\to V}$.) The key property is that it commutes with ${\displaystyle \pi ({\mathfrak {g}})}$ in the sense ${\displaystyle \pi (x)\pi (c)=\pi (c)\pi (x)}$ for each element ${\displaystyle x\in {\mathfrak {g}}}$. Also, ${\displaystyle \operatorname {tr} (\pi (c))=\sum \operatorname {tr} (\pi (e_{i})\pi (e^{i}))=\dim {\mathfrak {g}}_{1}.}$

Now, by Fitting's lemma, we have the vector space decomposition ${\displaystyle V=V_{0}\oplus V_{1}}$ such that ${\displaystyle \pi (c):V_{i}\to V_{i}}$ is a (well-defined) nilpotent endomorphism for ${\displaystyle i=0}$ and is an automorphism for ${\displaystyle i=1}$. Since ${\displaystyle \pi (c)}$ commutes with ${\displaystyle \pi ({\mathfrak {g}})}$, each ${\displaystyle V_{i}}$ is a ${\displaystyle {\mathfrak {g}}}$-submodule. Hence, it is enough to prove the lemma separately for ${\displaystyle V=V_{0}}$ and ${\displaystyle V=V_{1}}$.

First, suppose ${\displaystyle \pi (c)}$ is a nilpotent endomorphism. Then, by the early observation, ${\displaystyle \dim({\mathfrak {g}}/\operatorname {ker} (\pi ))=\operatorname {tr} (\pi (c))=0}$; that is, ${\displaystyle \pi }$ is a trivial representation. Since ${\displaystyle {\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]}$, the condition on ${\displaystyle f}$ implies that ${\displaystyle f(x)=0}$ for each ${\displaystyle x\in {\mathfrak {g}}}$; i.e., the zero vector ${\displaystyle v=0}$ satisfies the requirement.

Second, suppose ${\displaystyle \pi (c)}$ is an automorphism. For notational simplicity, we will drop ${\displaystyle \pi }$ and write ${\displaystyle xv=\pi (x)v}$. Also let ${\displaystyle (\cdot ,\cdot )}$ denote the trace form used earlier. Let ${\displaystyle w=\sum e_{i}f(e^{i})}$, which is a vector in ${\displaystyle V}$. Then

${\displaystyle xw=\sum _{i}e_{i}xf(e^{i})+\sum _{i}[x,e_{i}]f(e^{i}).}$

Now,

${\displaystyle [x,e_{i}]=\sum _{j}([x,e_{i}],e^{j})e_{j}=-\sum _{j}([x,e^{j}],e_{i})e_{j}}$

and, since ${\displaystyle [x,e^{j}]=\sum _{i}([x,e^{j}],e_{i})e^{i}}$, the second term of the expansion of ${\displaystyle xw}$ is

${\displaystyle -\sum _{j}e_{j}f([x,e^{j}])=-\sum _{i}e_{i}(xf(e^{i})-e^{i}f(x)).}$

Thus,

${\displaystyle xw=\sum _{i}e_{i}e^{i}f(x)=cf(x).}$

Since ${\displaystyle c}$ is invertible and ${\displaystyle c^{-1}}$ commutes with ${\displaystyle x}$, the vector ${\displaystyle v=c^{-1}w}$ has the required property. ${\displaystyle \square }$

• Jacobson, Nathan (1979). Lie algebras (Republication of the 1962 original ed.). Dover Publications. ISBN 978-0-486-13679-0. OCLC 867771145.