In mathematics, specifically the theory of Lie algebras, Lie's theorem states that, over an algebraically closed field of characteristic zero, if π : g → g l ( V ) {\displaystyle \pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)} is a finite-dimensional representation of a solvable Lie algebra, then there is a flag V = V 0 ⊃ V 1 ⊃ ⋯ ⊃ V n = 0 {\displaystyle V=V_{0}\supset V_{1}\supset \cdots \supset V_{n}=0} of invariant subspaces of π ( g ) {\displaystyle \pi ({\mathfrak {g}})} with codim ⁡ V i = i {\displaystyle \operatorname {codim} V_{i}=i}, meaning that π ( X ) ( V i ) ⊆ V i {\displaystyle \pi (X)(V_{i})\subseteq V_{i}} for each X ∈ g {\displaystyle X\in {\mathfrak {g}}} and i.

Put in another way, the theorem says there is a basis for V such that all linear transformations in π ( g ) {\displaystyle \pi ({\mathfrak {g}})} are represented by upper triangular matrices. This is a generalization of the result of Frobenius that commuting matrices are simultaneously upper triangularizable, as commuting matrices generate an abelian Lie algebra, which is a fortiori solvable.

A consequence of Lie's theorem is that any finite dimensional solvable Lie algebra over a field of characteristic 0 has a nilpotent derived algebra (see #Consequences). Also, to each flag in a finite-dimensional vector space V, there correspond a Borel subalgebra (that consist of linear transformations stabilizing the flag); thus, the theorem says that π ( g ) {\displaystyle \pi ({\mathfrak {g}})} is contained in some Borel subalgebra of g l ( V ) {\displaystyle {\mathfrak {gl}}(V)}.

Counter-example

For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p (see the proof below), but can fail for representations of dimension p. An example is given by the 3-dimensional nilpotent Lie algebra spanned by 1, x, and d/dx acting on the p-dimensional vector space k[x]/(xp), which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the p-dimensional representation (considered as an abelian Lie algebra) gives a solvable Lie algebra whose derived algebra is not nilpotent.

Proof

The proof is by induction on the dimension of g {\displaystyle {\mathfrak {g}}} and consists of several steps. (Note: the structure of the proof is very similar to that for Engel's theorem.) The basic case is trivial and we assume the dimension of g {\displaystyle {\mathfrak {g}}} is positive. We also assume V is not zero. For simplicity, we write X ⋅ v = π ( X ) ( v ) {\displaystyle X\cdot v=\pi (X)(v)}.

Step 1: Observe that the theorem is equivalent to the statement:

  • There exists a vector in V that is an eigenvector for each linear transformation in π ( g ) {\displaystyle \pi ({\mathfrak {g}})}.

Indeed, the theorem says in particular that a nonzero vector spanning V n − 1 {\displaystyle V_{n-1}} is a common eigenvector for all the linear transformations in π ( g ) {\displaystyle \pi ({\mathfrak {g}})}. Conversely, if v is a common eigenvector, take V n − 1 {\displaystyle V_{n-1}} to be its span and then π ( g ) {\displaystyle \pi ({\mathfrak {g}})} admits a common eigenvector in the quotient V / V n − 1 {\displaystyle V/V_{n-1}}; repeat the argument.

Step 2: Find an ideal h {\displaystyle {\mathfrak {h}}} of codimension one in g {\displaystyle {\mathfrak {g}}}.

Let D g = [ g , g ] {\displaystyle D{\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]} be the derived algebra. Since g {\displaystyle {\mathfrak {g}}} is solvable and has positive dimension, D g ≠ g {\displaystyle D{\mathfrak {g}}\neq {\mathfrak {g}}} and so the quotient g / D g {\displaystyle {\mathfrak {g}}/D{\mathfrak {g}}} is a nonzero abelian Lie algebra, which certainly contains an ideal of codimension one and by the ideal correspondence, it corresponds to an ideal of codimension one in g {\displaystyle {\mathfrak {g}}}.

Step 3: There exists some linear functional λ {\displaystyle \lambda } in h ∗ {\displaystyle {\mathfrak {h}}^{*}} such that

V λ = { v ∈ V | X ⋅ v = λ ( X ) v , X ∈ h } {\displaystyle V_{\lambda }=\{v\in V|X\cdot v=\lambda (X)v,X\in {\mathfrak {h}}\}}

is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).

Step 4: V λ {\displaystyle V_{\lambda }} is a g {\displaystyle {\mathfrak {g}}}-invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let Y ∈ g {\displaystyle Y\in {\mathfrak {g}}}, v ∈ V λ {\displaystyle v\in V_{\lambda }}, then we need to prove Y ⋅ v ∈ V λ {\displaystyle Y\cdot v\in V_{\lambda }}. If v = 0 {\displaystyle v=0} then it's obvious, so assume v ≠ 0 {\displaystyle v\neq 0} and set recursively v 0 = v , v i + 1 = Y ⋅ v i {\displaystyle v_{0}=v,\,v_{i+1}=Y\cdot v_{i}}. Let U = span ⁡ { v i | i ≥ 0 } {\displaystyle U=\operatorname {span} \{v_{i}|i\geq 0\}} and ℓ ∈ N 0 {\displaystyle \ell \in \mathbb {N} _{0}} be the largest such that v 0 , … , v ℓ {\displaystyle v_{0},\ldots ,v_{\ell }} are linearly independent. Then we'll prove that they generate U and thus α = ( v 0 , … , v ℓ ) {\displaystyle \alpha =(v_{0},\ldots ,v_{\ell })} is a basis of U. Indeed, assume by contradiction that it's not the case and let m ∈ N 0 {\displaystyle m\in \mathbb {N} _{0}} be the smallest such that v m ∉ ⟨ v 0 , … , v ℓ ⟩ {\displaystyle v_{m}\notin \langle v_{0},\ldots ,v_{\ell }\rangle }, then obviously m ≥ ℓ + 1 {\displaystyle m\geq \ell +1}. Since v 0 , … , v ℓ + 1 {\displaystyle v_{0},\ldots ,v_{\ell +1}} are linearly dependent, v ℓ + 1 {\displaystyle v_{\ell +1}} is a linear combination of v 0 , … , v ℓ {\displaystyle v_{0},\ldots ,v_{\ell }}. Applying the map Y m − ℓ − 1 {\displaystyle Y^{m-\ell -1}} it follows that v m {\displaystyle v_{m}} is a linear combination of v m − ℓ − 1 , … , v m − 1 {\displaystyle v_{m-\ell -1},\ldots ,v_{m-1}}. Since by the minimality of m each of these vectors is a linear combination of v 0 , … , v ℓ {\displaystyle v_{0},\ldots ,v_{\ell }}, so is v m {\displaystyle v_{m}}, and we get the desired contradiction. We'll prove by induction that for every n ∈ N 0 {\displaystyle n\in \mathbb {N} _{0}} and X ∈ h {\displaystyle X\in {\mathfrak {h}}} there exist elements a 0 , n , X , … , a n , n , X {\displaystyle a_{0,n,X},\ldots ,a_{n,n,X}} of the base field such that a n , n , X = λ ( X ) {\displaystyle a_{n,n,X}=\lambda (X)} and

X ⋅ v n = ∑ i = 0 n a i , n , X v i . {\displaystyle X\cdot v_{n}=\sum _{i=0}^{n}a_{i,n,X}v_{i}.}

The n = 0 {\displaystyle n=0} case is straightforward since X ⋅ v 0 = λ ( X ) v 0 {\displaystyle X\cdot v_{0}=\lambda (X)v_{0}}. Now assume that we have proved the claim for some n ∈ N 0 {\displaystyle n\in \mathbb {N} _{0}} and all elements of h {\displaystyle {\mathfrak {h}}} and let X ∈ h {\displaystyle X\in {\mathfrak {h}}}. Since h {\displaystyle {\mathfrak {h}}} is an ideal, it's [ X , Y ] ∈ h {\displaystyle [X,Y]\in {\mathfrak {h}}}, and thus

X ⋅ v n + 1 = Y ⋅ ( X ⋅ v n ) + [ X , Y ] ⋅ v n = Y ⋅ ∑ i = 0 n a i , n , X v i + ∑ i = 0 n a i , n , [ X , Y ] v i = a 0 , n , [ X , Y ] v 0 + ∑ i = 1 n ( a i − 1 , n , X + a i , n , [ X , Y ] ) v i + λ ( X ) v n + 1 , {\displaystyle X\cdot v_{n+1}=Y\cdot (X\cdot v_{n})+[X,Y]\cdot v_{n}=Y\cdot \sum _{i=0}^{n}a_{i,n,X}v_{i}+\sum _{i=0}^{n}a_{i,n,[X,Y]}v_{i}=a_{0,n,[X,Y]}v_{0}+\sum _{i=1}^{n}(a_{i-1,n,X}+a_{i,n,[X,Y]})v_{i}+\lambda (X)v_{n+1},}

and the induction step follows. This implies that for every X ∈ h {\displaystyle X\in {\mathfrak {h}}} the subspace U is an invariant subspace of X and the matrix of the restricted map π ( X ) | U {\displaystyle \pi (X)|_{U}} in the basis α {\displaystyle \alpha } is upper triangular with diagonal elements equal to λ ( X ) {\displaystyle \lambda (X)}, hence tr ⁡ ( π ( X ) | U ) = dim ⁡ ( U ) λ ( X ) {\displaystyle \operatorname {tr} (\pi (X)|_{U})=\dim(U)\lambda (X)}. Applying this with [ X , Y ] ∈ h {\displaystyle [X,Y]\in {\mathfrak {h}}} instead of X gives tr ⁡ ( π ( [ X , Y ] ) | U ) = dim ⁡ ( U ) λ ( [ X , Y ] ) {\displaystyle \operatorname {tr} (\pi ([X,Y])|_{U})=\dim(U)\lambda ([X,Y])}. On the other hand, U is also obviously an invariant subspace of Y, and so

tr ⁡ ( π ( [ X , Y ] ) | U ) = tr ⁡ ( [ π ( X ) , π ( Y ) ] | U ] ) = tr ⁡ ( [ π ( X ) | U , π ( Y ) | U ] ) = 0 {\displaystyle \operatorname {tr} (\pi ([X,Y])|_{U})=\operatorname {tr} ([\pi (X),\pi (Y)]|_{U}])=\operatorname {tr} ([\pi (X)|_{U},\pi (Y)|_{U}])=0}

since commutators have zero trace, and thus dim ⁡ ( U ) λ ( [ X , Y ] ) = 0 {\displaystyle \dim(U)\lambda ([X,Y])=0}. Since dim ⁡ ( U ) > 0 {\displaystyle \dim(U)>0} is invertible (because of the assumption on the characteristic of the base field), λ ( [ X , Y ] ) = 0 {\displaystyle \lambda ([X,Y])=0} and

X ⋅ ( Y ⋅ v ) = Y ⋅ ( X ⋅ v ) + [ X , Y ] ⋅ v = Y ⋅ ( λ ( X ) v ) + λ ( [ X , Y ] ) v = λ ( X ) ( Y ⋅ v ) , {\displaystyle X\cdot (Y\cdot v)=Y\cdot (X\cdot v)+[X,Y]\cdot v=Y\cdot (\lambda (X)v)+\lambda ([X,Y])v=\lambda (X)(Y\cdot v),}

and so Y ⋅ v ∈ V λ {\displaystyle Y\cdot v\in V_{\lambda }}.

Step 5: Finish up the proof by finding a common eigenvector.

Write g = h + L {\displaystyle {\mathfrak {g}}={\mathfrak {h}}+L} where L is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in V λ {\displaystyle V_{\lambda }} for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of h {\displaystyle {\mathfrak {h}}}, the proof is complete. ◻ {\displaystyle \square }

Consequences

The theorem applies in particular to the adjoint representation ad : g → g l ( g ) {\displaystyle \operatorname {ad} :{\mathfrak {g}}\to {\mathfrak {gl}}({\mathfrak {g}})} of a (finite-dimensional) solvable Lie algebra g {\displaystyle {\mathfrak {g}}} over an algebraically closed field of characteristic zero; thus, one can choose a basis on g {\displaystyle {\mathfrak {g}}} with respect to which ad ⁡ ( g ) {\displaystyle \operatorname {ad} ({\mathfrak {g}})} consists of upper triangular matrices. It follows easily that for each x , y ∈ g {\displaystyle x,y\in {\mathfrak {g}}}, ad ⁡ ( [ x , y ] ) = [ ad ⁡ ( x ) , ad ⁡ ( y ) ] {\displaystyle \operatorname {ad} ([x,y])=[\operatorname {ad} (x),\operatorname {ad} (y)]} has diagonal consisting of zeros; i.e., ad ⁡ ( [ x , y ] ) {\displaystyle \operatorname {ad} ([x,y])} is a strictly upper triangular matrix. This implies that [ g , g ] {\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]} is a nilpotent Lie algebra. Moreover, if the base field is not algebraically closed then solvability and nilpotency of a Lie algebra is unaffected by extending the base field to its algebraic closure. Hence, one concludes the statement (the other implication is obvious):

A finite-dimensional Lie algebra g {\displaystyle {\mathfrak {g}}} over a field of characteristic zero is solvable if and only if the derived algebra D g = [ g , g ] {\displaystyle D{\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]} is nilpotent.

Lie's theorem also establishes one direction in Cartan's criterion for solvability:

IfVis a finite-dimensional vector space over a field of characteristic zero and g ⊆ g l ( V ) {\displaystyle {\mathfrak {g}}\subseteq {\mathfrak {gl}}(V)} a Lie subalgebra, then g {\displaystyle {\mathfrak {g}}} is solvable if and only if tr ⁡ ( X Y ) = 0 {\displaystyle \operatorname {tr} (XY)=0} for every X ∈ g {\displaystyle X\in {\mathfrak {g}}} and Y ∈ [ g , g ] {\displaystyle Y\in [{\mathfrak {g}},{\mathfrak {g}}]}.

Indeed, as above, after extending the base field, the implication ⇒ {\displaystyle \Rightarrow } is seen easily. (The converse is more difficult to prove.)

Lie's theorem (for various V) is equivalent to the statement:

For a solvable Lie algebra g {\displaystyle {\mathfrak {g}}} over an algebraically closed field of characteristic zero, each finite-dimensional simple g {\displaystyle {\mathfrak {g}}}-module (i.e., irreducible as a representation) has dimension one.

Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional g {\displaystyle {\mathfrak {g}}}-module V, let V 1 {\displaystyle V_{1}} be a maximal g {\displaystyle {\mathfrak {g}}}-submodule (which exists by finiteness of the dimension). Then, by maximality, V / V 1 {\displaystyle V/V_{1}} is simple; thus, is one-dimensional. The induction now finishes the proof.

The statement says in particular that a finite-dimensional simple module over an abelian Lie algebra is one-dimensional; this fact remains true over any base field since in this case every vector subspace is a Lie subalgebra.

Here is another quite useful application:

Let g {\displaystyle {\mathfrak {g}}} be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical rad ⁡ ( g ) {\displaystyle \operatorname {rad} ({\mathfrak {g}})}. Then each finite-dimensional simple representation π : g → g l ( V ) {\displaystyle \pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)} is the tensor product of a simple representation of g / rad ⁡ ( g ) {\displaystyle {\mathfrak {g}}/\operatorname {rad} ({\mathfrak {g}})} with a one-dimensional representation of g {\displaystyle {\mathfrak {g}}} (i.e., a linear functional vanishing on Lie brackets).

By Lie's theorem, we can find a linear functional λ {\displaystyle \lambda } of rad ⁡ ( g ) {\displaystyle \operatorname {rad} ({\mathfrak {g}})} so that there is the weight space V λ {\displaystyle V_{\lambda }} of rad ⁡ ( g ) {\displaystyle \operatorname {rad} ({\mathfrak {g}})}. By Step 4 of the proof of Lie's theorem, V λ {\displaystyle V_{\lambda }} is also a g {\displaystyle {\mathfrak {g}}}-module; so V = V λ {\displaystyle V=V_{\lambda }}. In particular, for each X ∈ rad ⁡ ( g ) {\displaystyle X\in \operatorname {rad} ({\mathfrak {g}})}, tr ⁡ ( π ( X ) ) = dim ⁡ ( V ) λ ( X ) {\displaystyle \operatorname {tr} (\pi (X))=\dim(V)\lambda (X)}. Extend λ {\displaystyle \lambda } to a linear functional on g {\displaystyle {\mathfrak {g}}} that vanishes on [ g , g ] {\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}; λ {\displaystyle \lambda } is then a one-dimensional representation of g {\displaystyle {\mathfrak {g}}}. Now, ( π , V ) ≃ ( π , V ) ⊗ ( − λ ) ⊗ λ {\displaystyle (\pi ,V)\simeq (\pi ,V)\otimes (-\lambda )\otimes \lambda }. Since π {\displaystyle \pi } coincides with λ {\displaystyle \lambda } on rad ⁡ ( g ) {\displaystyle \operatorname {rad} ({\mathfrak {g}})}, we have that V ⊗ ( − λ ) {\displaystyle V\otimes (-\lambda )} is trivial on rad ⁡ ( g ) {\displaystyle \operatorname {rad} ({\mathfrak {g}})} and thus is the restriction of a (simple) representation of g / rad ⁡ ( g ) {\displaystyle {\mathfrak {g}}/\operatorname {rad} ({\mathfrak {g}})}. ◻ {\displaystyle \square }

See also

Sources