# Lie's theorem

The set of Lie , named after Sophus Lie , is a mathematical theorem from the theory of Lie algebras. It ensures the existence of a common eigenvector for all elements of a resolvable Lie algebra over a the null space different, finite - vector space and from this it follows that such a Lie algebra to a partial algebra of upper triangular isomorphic is. ${\ displaystyle \ mathbb {C}}$

## Designations

This article is a finite - vector space with dimension . The field of complex numbers can be replaced by any algebraically closed field of characteristic 0, which is omitted in the following formulations. A flag in is an ascending chain of subspaces with for everyone . ${\ displaystyle V}$${\ displaystyle \ mathbb {C}}$${\ displaystyle \ mathrm {dim} \, V \, = n> 0}$${\ displaystyle \ mathbb {C}}$${\ displaystyle V}$${\ displaystyle V_ {0} = \ {0 \} \ subset V_ {1} \ subset V_ {2} \ subset \ ldots \ subset V_ {n} = V}$${\ displaystyle \ mathrm {dim} \, V_ {i} \, = i}$${\ displaystyle i = 0, \ ldots, n}$

Furthermore denote the general linear Lie algebra over . For a Lie algebra, let the so-called derived Lie algebras be defined recursively by and , where the latter stands for the subspace generated by all products . A Lie algebra is called solvable if there is one with . A related term is the Nile potency . One defines and recursively and calls a Lie algebra nilpotent if there is one with . Since it appears that the solvability follows from the Nile power, the converse generally does not hold. ${\ displaystyle {\ mathfrak {gl}} (V)}$${\ displaystyle V}$${\ displaystyle L}$${\ displaystyle L ^ {(i)}}$${\ displaystyle L ^ {(0)}: = L}$${\ displaystyle L ^ {(i + 1)}: = [L ^ {(i)}, L ^ {(i)}]}$${\ displaystyle [x, y], x, y \ in L ^ {(i)}}$${\ displaystyle i \ in \ mathbb {N}}$${\ displaystyle L ^ {(i)} = \ {0 \}}$${\ displaystyle L ^ {0}: = L}$${\ displaystyle L ^ {i + 1}: = [L ^ {i}, L]}$${\ displaystyle i}$${\ displaystyle L ^ {i} = \ {0 \}}$${\ displaystyle L ^ {(i)} \ subset L ^ {i}}$

## First formulation of the sentence

• Let it be a solvable Lie algebra. Then there is a common eigenvector for.${\ displaystyle L \ subset {\ mathfrak {gl}} (V)}$${\ displaystyle L}$

More precisely, this means that there is a vector out such that for each there is a multiple of , in particular for each there is an invariant subspace, that is, it is mapped by in itself. ${\ displaystyle v \ neq 0}$${\ displaystyle V}$${\ displaystyle xv}$${\ displaystyle x \ in L}$${\ displaystyle v}$${\ displaystyle \ mathbb {C} \ cdot v}$${\ displaystyle x \ in L}$${\ displaystyle x}$

## Second formulation of the sentence

• Let it be a solvable Lie algebra. Then there is an -invariant flag.${\ displaystyle L \ subset {\ mathfrak {gl}} (V)}$${\ displaystyle L}$

This means precisely that a flag is with all and . This formulation intensifies the first one, because apparently every vector is a common eigenvector. ${\ displaystyle V_ {0} = \ {0 \} \ subset V_ {1} \ subset V_ {2} \ subset \ ldots \ subset V_ {n} = V}$${\ displaystyle xV_ {i} \ subset V_ {i}}$${\ displaystyle x \ in L}$${\ displaystyle i = 0, \ ldots, n}$${\ displaystyle V_ {1} \ setminus \ {0 \}}$

Conversely, the flag of the second formulation is constructed as follows using induction from the first. The induction start is . If one has to be constructed already, so ${\ displaystyle V_ {0} = \ {0 \}}$${\ displaystyle V_ {i}}$${\ displaystyle 0 \ leq i

${\ displaystyle \ varphi: L \ rightarrow {\ mathfrak {gl}} (V / V_ {i}), \ quad \ varphi (x) (v + V_ {i}): = xv + V_ {i}}$

because of the -invariance of a well-defined representation of on the quotient space , and this is because of not the null space. Then, as a homomorphic image, a solvable Lie algebra is solvable again and based on the first formulation there is a common eigenvector . If one sets now , one easily calculates that this subspace is -invariant, with which the inductive construction is finished. ${\ displaystyle L}$${\ displaystyle V_ {i}}$${\ displaystyle L}$ ${\ displaystyle V / V_ {i}}$${\ displaystyle i ${\ displaystyle \ varphi (L) \ subset {\ mathfrak {gl}} (V / V_ {i})}$ ${\ displaystyle w + V_ {i} \ in V / V_ {i}}$${\ displaystyle V_ {i + 1} = V_ {i} + \ mathbb {C} \ cdot w}$${\ displaystyle L}$

## Inferences

As a first important conclusion one can relate solvable Lie algebras to upper triangular matrices. If a flag is as in the second formulation, a base of can be constructed by means of a basic supplementary sentence , so that there is a base of for each . If you represent the elements with respect to this basis as matrices , you get upper triangular matrices because of . We therefore have: ${\ displaystyle V_ {0} = \ {0 \} \ subset V_ {1} \ subset V_ {2} \ subset \ ldots \ subset V_ {n} = V}$${\ displaystyle v_ {1}, \ ldots, v_ {n}}$${\ displaystyle V}$${\ displaystyle v_ {1}, \ ldots, v_ {i}}$${\ displaystyle i}$${\ displaystyle V_ {i}}$${\ displaystyle x \ in L}$${\ displaystyle xV_ {i} \ subset V_ {i}}$

• A solvable Lie algebra is isomorphic to a subalgebra of the Lie algebra of the upper triangular matrices.${\ displaystyle L \ subset {\ mathfrak {gl}} (V)}$

As mentioned above, nilpotent Lie algebras are solvable, but the reverse is generally not true. Therefore, the following statement is surprising at first glance:

• If there is a solvable Lie algebra, then is nilpotent.${\ displaystyle L \ subset {\ mathfrak {gl}} (V)}$${\ displaystyle L ^ {(1)} = [L, L]}$

According to the first inference, is isomorphic to a subalgebra of the upper triangular matrices. Since a commutator of two upper triangular matrices is a strict upper triangular matrix, i.e. the diagonal elements are all 0, is isomorphic to a subalgebra of the nilpotent Lie algebra of the strict upper triangular matrices and is therefore nilpotent itself. ${\ displaystyle L}$${\ displaystyle L ^ {(1)} = [L, L]}$

As a further consequence one can construct a flag of ideals in a solvable Lie algebra:

• If there is a solvable Lie algebra, there is a flag of ideals.${\ displaystyle L \ subset {\ mathfrak {gl}} (V)}$${\ displaystyle L_ {0} = \ {0 \} \ subset L_ {1} \ subset \ ldots \ subset L}$

As a proof, note that the image of the adjoint representation is a solvable Lie algebra in . According to Lie's theorem above, there is an invariant flag there, and the invariant subspaces of are ideals. Such a chain of ideals is called a Holder series of Lie algebra. ${\ displaystyle {\ mathfrak {gl}} (L)}$${\ displaystyle L}$

If there is an irreducible representation, then with is also solvable, so it has a common eigenvector according to Lie's theorem above . The one-dimensional subspace spanned by is then invariant, so it has to agree because of the irreducibility , and that is the claim. ${\ displaystyle \ pi: L \ rightarrow {\ mathfrak {gl}} (V)}$${\ displaystyle L}$${\ displaystyle \ pi (L) \ subset {\ mathfrak {gl}} (V)}$${\ displaystyle v \ not = 0}$${\ displaystyle v}$${\ displaystyle V}$

## Other basic bodies

The set of Lie is also true Fax foils vector spaces over algebraically closed bodies of characteristic zero. In contrast, there are counterexamples for vector spaces over the real numbers or over bodies with positive characteristics.

## literature

• Joachim Hilgert, Karl-Hermann Neeb: Lie groups and Lie algebras , Vieweg (1999), ISBN 3-528-06432-3 , Chapter II, §2: Nilpotents and solvable Lie algebras
• James E. Humphreys: Introduction to Lie Algebras and Representation Theory , Springer-Verlag (1972), ISBN 0-387-90052-7 , Chapter 4.1: Lie's Theorem