research31 min studyLesson 501 in Mathematics

Ideals, solvability, and Levi decomposition

Separate the solvable radical from a semisimple complement without confusing a vector-space splitting with a direct product.

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What you’ll understand

  • Compute derived and lower-central series in a small Lie algebra.
  • Identify a solvable radical and a Levi complement.
  • Explain why a semidirect decomposition need not split as commuting ideals.

Continuous symmetry has an algebraic structure beyond its dimension or a matrix realization. Ideals record pieces stable under every internal bracket, and the solvable radical collects the part that repeatedly loses complexity under commutation. Levi decomposition separates that radical from a semisimple complement. Throughout this lesson the Lie algebras are finite dimensional over the complex numbers, a field of characteristic zero; these hypotheses are essential to the structural theorem being used.

An ideal is invariant under every adjoint action

A subspace I is an ideal when [g,I] is contained in I. This is stronger than being a subalgebra, which only requires [I,I] to stay in I. The ideal condition makes a quotient bracket on g/I well defined. It is also exactly invariance of I under the adjoint representation. An invariant subspace in some other representation is a different object, so a preserved coordinate line in the defining matrix representation does not by itself identify an ideal of the matrix Lie algebra.

The derived series begins with g and repeatedly brackets the current space with itself. Solvability means that this series eventually becomes zero. The lower-central series instead brackets the current space with all of g; its eventual vanishing defines nilpotence. Every nilpotent Lie algebra is solvable, but the converse fails. The distinction concerns how brackets are nested, not whether selected matrices in a realization happen to be nilpotent. A Lie algebra can contain semisimple and nilpotent operators while its derived series still terminates.

g(0)=g,g(k+1)=[g(k),g(k)],γ1g=g,γk+1g=[g,γkg]\mathfrak g^{(0)}=\mathfrak g,\quad \mathfrak g^{(k+1)}=[\mathfrak g^{(k)},\mathfrak g^{(k)}],\qquad \gamma_1\mathfrak g=\mathfrak g,\quad \gamma_{k+1}\mathfrak g=[\mathfrak g,\gamma_k\mathfrak g]
Derived-series termination defines solvability; lower-central-series termination defines nilpotence. The bracket of two subspaces means the linear span of all their brackets.↔ Scroll wide equations horizontally.

The radical is canonical; a complementary Levi subalgebra is not

The radical r is the largest solvable ideal. In finite dimension, sums of solvable ideals remain solvable, so such a largest ideal exists. The quotient g/r is semisimple, meaning it has no nonzero solvable ideal. Levi's theorem in characteristic zero states that a semisimple subalgebra s maps isomorphically onto this quotient and complements r as a vector space. The theorem is an existence result about a subalgebra, not a claim that any arbitrary linear complement closes under brackets.

The bracket between s and r can be nonzero, producing a semidirect product. The complement s need not be an ideal and is generally not canonical, whereas r is intrinsic. Characteristic-zero conjugacy results relate Levi complements by suitable inner automorphisms, but do not select a preferred one without additional structure. This lesson uses the existence and semidirect interpretation, rather than attempting the cohomological proof. In positive characteristic or infinite dimension the same statement cannot simply be quoted unchanged.

g=s⋉r,[(x,u),(y,v)]=([x,y], x⋅v−y⋅u+[u,v]r)\mathfrak g=\mathfrak s\ltimes\mathfrak r,\qquad [ (x,u),(y,v) ]=([x,y],\,x\cdot v-y\cdot u+[u,v]_{\mathfrak r})
Over ℂ in finite dimension, a Levi complement s is semisimple and acts on the radical by derivations. The displayed decomposition is a vector-space direct sum with a potentially nontrivial cross-action.↔ Scroll wide equations horizontally.

Reductivity is a special kind of splitting

A finite-dimensional complex Lie algebra is reductive when its radical equals its center. In that case the radical commutes with a Levi complement and the decomposition becomes a direct product of a central abelian algebra and a semisimple algebra. The algebra gl(n,C) is the basic example. Semisimple means the radical is zero, so its center is also zero; reductive permits a central part. These definitions should not be inferred merely from diagonalizability of one selected matrix or complete reducibility in one selected representation.

For a genuine semidirect example, let sl(2,C) act on its standard two-dimensional module and let that module be an abelian ideal. The resulting algebra has a two-dimensional radical and a nontrivial cross-action. Its semisimple quotient is sl(2,C), yet the whole algebra is not reductive because the radical is not central. This example explains why a quotient classification does not reconstruct the original bracket: one must also know the action on the radical and, before splitting, the extension data.

Decompose gl(2,C) explicitly

Find the radical and a Levi complement of gl(2,C), and decompose A=[[3,2],[1,1]].

  1. The trace-zero matrices form sl(2,C), a semisimple ideal. Scalar matrices form a central abelian ideal CI, so they are contained in the radical.

  2. The quotient by CI is isomorphic to sl(2,C) through X↦X−tr(X)I/2. Since that quotient is semisimple, the central ideal is exactly the radical.

  3. For A, the trace is 4, giving A=2I+[[1,2],[1,−1]]. The summands commute because the first is scalar; the second belongs to the trace-zero Levi ideal.

The result

Here the Levi decomposition is a direct product gl(2,C)=sl(2,C)⊕CI. Commuting summands are a special feature of this reductive example.

Explore the model

Interactive lab

Interactive lab · matrix

Inspect a nontrivial affine action

Reset, set a=2 and b=1. The selected matrix [[a,b],[0,1]] acts on (x,1) as (2x+1,1); predict that conjugating the translation generator E₁₂ rescales it by 2. Compare a fresh run with Reset, a=1, b=1, where that factor is 1. The diagram shows the defining linear action; the conjugation is checked algebraically. This affine family illustrates a solvable semidirect action, not a Levi-factor finder.

Inspect a nontrivial affine actionThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn action on an ideal can be nontrivial even when the underlying vector space splits.

Check your understanding

What does a Levi decomposition generally guarantee?

Try it yourself

Let g have basis h,e and bracket [h,e]=e over ℂ. Compute its derived series and lower-central series, then identify its radical and Levi factor.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

After the first bracket, compare [Ce,Ce] with [g,Ce].

Compare with the solution

The derived series is g, Ce, 0, so g is solvable. The lower-central series is g, Ce, Ce, and continues without vanishing, so g is not nilpotent. Its radical is all of g and its semisimple quotient is zero; a Levi complement is the zero algebra. Solvability does not imply nilpotence, and the existence of a Levi decomposition does not require a nonzero Levi summand.

Take it with you

  • Ideals are adjoint-invariant subspaces, not arbitrary invariant subspaces in another module.
  • The solvable radical is canonical while a Levi complement usually is not.
  • Semidirect and commuting direct-product splittings encode different bracket data.

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