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An $R$-point means a unital algebra homomorphism over a field $\alpha :k_q[x,y]\to R$, as in points of a noncommutative algebra. It is specified by $X=\alpha (x)$ and $Y=\alpha (y)$ satisfying $YX=qXY$. For $R=\mathbb{C}$ and $q$ not a root of unity, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

For matrix-valued points of a noncommutative algebra, put $H={\mathbb{C}}^n$ and let $H_{\lambda }(X)=\ker (X-\lambda I{)}^n$ be a generalized eigenspace. From $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ is nilpotent on $V_X$, although $X$ is invertible there.

Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ generalized eigenspaces is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero generalized eigenspaces all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero generalized eigenspaces:The zero generalized eigenspace of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero eigenvalues, hence are nilpotent matrices. This proves the spectral decomposition of a quantum-plane point:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.

The Jordan–Chevalley decomposition is characterized byOver $\mathbb{C}$, a semisimple linear operator is a diagonalizable one. These conditions determine the two parts uniquely. Concretely, on the generalized eigenspace for an eigenvalue $\lambda$, the semisimple part is $\lambda I$ and the nilpotent part is $X-\lambda I$; both are polynomials in $X$. To see uniqueness, any commuting candidate parts preserve each generalized eigenspace; diagonalize the proposed semisimple part there, and nilpotence then forces its eigenvalue to be the sole eigenvalue $\lambda$ of $X$ on that space.

For the displayed matrix, $\det (tI-X)=(t-1{)}^2$ andSince $I$ is semisimple and commutes with $N$, the required parts are

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

A finite-dimensional complex Lie algebra is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form is the symmetric bilinear formThe cyclic trace identity and ${\mathrm{ad}}_{[z,x]}=[{\mathrm{ad}}_z,{\mathrm{ad}}_x]$ giveThus the Killing form is an invariant bilinear form on a Lie algebra. In particular its radical $R=\{x:B_L(x,L)=0\}$ is an ideal.

We supply the trace argument needed for nondegeneracy rather than assuming the Cartan criterion for semisimplicity. The matrix form of the Cartan solvability criterion says: if $A\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ and $\mathrm{tr}(uv)=0$ for every $u\in [A,A]$, $v\in A$, then $A$ is solvable. To prove this direction, fix $u\in [A,A]$. Let $V={⨁}_{\lambda }{V}_{\lambda }$ be its generalized eigenspace decomposition, and define $s$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, the semisimple part of ${\mathrm{ad}}_u$ has eigenvalue $\lambda -\mu$, whereas ${\mathrm{ad}}_s$ acts by $\overline{\lambda }-\overline{\mu }=\overline{\lambda -\mu }$. Polynomial interpolation on the finitely many eigenvalues, with all derivatives through the sizes of the nilpotent blocks set to zero, therefore givesHere the same difference always has the same conjugate, so the interpolation is consistent; the prescribed zero derivatives remove every nilpotent block. Since ${\mathrm{ad}}_u(A)\subseteq [A,A]$, it follows that $[s,A]\subseteq [A,A]$.

Write $u={\sum }_j[a_j,b_j]$. By cyclicity and the trace hypothesis,On the other hand, $\mathrm{tr}(us)={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$. Thus all eigenvalues of $u$ vanish, and every element of $[A,A]$ is nilpotent. The Engel theorem makes $[A,A]$ nilpotent as a Lie algebra, hence solvable; $A/[A,A]$ is abelian, so $A$ is solvable. This is the Conjugate-spectrum proof of Cartan solvability.

Apply this to the Killing radical $R$. For $x,y\in R$, the adjoint actions preserve $R$ and act as zero on $L/R$, because $R$ is an ideal. Computing traces in a basis adapted to $R$ givesThe matrix Lie algebra ${\mathrm{ad}}_R(R)$ consequently satisfies the trace criterion and is solvable. Its kernel is $Z(R)$, an abelian ideal; a central extension of a solvable algebra is solvable. Thus $R$ is solvable. This proves the reusable assertion that the Killing radical is a solvable ideal. Semisimplicity forces $R=0$, and hence $B_L$ is nondegenerate.

A Cartan subalgebra is a nilpotent subalgebra $H$ which is self-normalizing:For a complex semisimple algebra this is equivalently a maximal toral subalgebra. The nilpotent, self-normalizing definition permits a proof of the restricted nondegeneracy without first assuming the toral characterization.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the adjoint action of $H$:For completeness, the stability underlying this decomposition follows directly from nilpotence of $H$. For fixed $h$, every ${\mathrm{ad}}_h$ on $H$ is nilpotent. If $T=\rho (h)$ and $S=\rho (y)$ in a finite-dimensional representation, then $({\mathrm{ad}}_T{)}^{r}S=0$ for sufficiently large $r$. The identityshows that $S$ preserves each generalized eigenspace of $T$. Starting with a basis of $H$, refine these primary decompositions successively; all summands remain $H$-invariant. Each resulting summand has only one eigenvalue for each basis element. The Lie theorem triangularizes the action on that summand, so those eigenvalues extend to a single linear character $\lambda$ on all of $H$. This proves the displayed decomposition and nilpotence of all shifted operators there.

Since $H$ is nilpotent, $H\subseteq L^0$. The space $L^0$ is a subalgebra: repeated use of the derivation rule for ${\mathrm{ad}}_h$ shows that the bracket of two generalized zero-eigenvectors is another such vector. If $L^0/H\ne 0$, the adjoint action of $H$ on this quotient consists entirely of nilpotent maps. The Engel theorem supplies a nonzero coset $x+H$ with $[H,x]\subseteq H$, contradicting self-normalization. Therefore the zero generalized weight space of a Cartan subalgebra is exactly $H$.

Finally, $B_L(H,L^{\lambda })=0$ when $\lambda \ne 0$. Choose $h\in H$ with $\lambda (h)\ne 0$ and put $T={\mathrm{ad}}_h$. On $H$, a power $T^m$ vanishes. On $L^{\lambda }$, $T$ is invertible. For $u\in H$ and $v\in L^{\lambda }$, write $v=T^{m}w$; invariance givesIf $u\in H$ is orthogonal to $H$, it is now orthogonal to every summand of $L$, so nondegeneracy of $B_L$ implies $u=0$. The restriction $B_L{|}_{H\times H}$ is nondegenerate. This establishes the nondegeneracy of the Killing form on a Cartan subalgebra for every Cartan subalgebra, without requiring a chosen root basis.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

Let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be a finite-dimensional Lie algebra representation. The form denoted $B_V$ isIt is the Trace form of a Lie algebra representation; the Killing form $B$ without a subscript is specifically the case $V=\mathfrak{g}$, $\rho =\mathrm{ad}$. The distinction matters: the form of the trivial representation cannot detect whether the algebra is semisimple.

Linearity of $\rho$ and trace proves bilinearity, and cyclicity of the trace gives symmetry. With $A=\rho (X)$, $D=\rho (Y)$ and $C=\rho (Z)$, the representation identity givesThus it is an invariant bilinear form on a Lie algebra, and in particular the Adjoint representation preserves the Killing form.

The Cartan solvability criterion has two useful formulations. For a finite-dimensional complex Lie algebra $\mathfrak{g}$,Its matrix version says that a Lie subalgebra $L\subseteq \mathfrak{gl}(V)$ is solvable precisely when $\mathrm{tr}(xy)=0$ for all $x\in [L,L]$ and $y\in L$. We prove the matrix version first, with ordinary trace in $V$.

If $L$ is solvable, the Lie theorem makes all its matrices upper triangular. Their commutators are strictly upper triangular, so multiplying such a matrix by an upper triangular one still has zero diagonal and hence zero trace. This proves the easy direction.

Conversely assume the trace-orthogonality condition and fix $x\in [L,L]$. We show that all eigenvalues of $x$ vanish. Use its Additive Jordan decomposition $x=x_s+x_n$ with $[x_s,x_n]=0$. On the generalized eigenspace $V_{\lambda }$ of $x$, define an auxiliary endomorphism $y$ to be $\overline{\lambda }I$. This $y$ need not be in $L$; we only need control of its commutator with $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}{x}_s$ acts by $\lambda -\mu$ and $\mathrm{ad}y$ by $\overline{\lambda }-\overline{\mu }$. Choose a polynomial $q$ with $q(0)=0$ and $q(\lambda -\mu )=\overline{\lambda -\mu }$ at the finitely many distinct differences. Polynomial interpolation supplies it because equal differences have equal conjugates. Therefore $\mathrm{ad}y=q(\mathrm{ad}{x}_s)$.

The adjoint compatibility of additive Jordan decomposition identifies $\mathrm{ad}{x}_s$ as the semisimple part of $\mathrm{ad}x$. Elementary Jordan–Chevalley decomposition gives $\mathrm{ad}{x}_s=p(\mathrm{ad}x)$ for a polynomial $p$ with $p(0)=0$. Hence $\mathrm{ad}y$ is a polynomial in $\mathrm{ad}x$ with zero constant term. Since $\mathrm{ad}x$ maps $L$ into $[L,L]$ and preserves that Lie algebra ideal, we obtainNo assumption that $x_s$, $x_n$ or $y$ belongs to $L$ was made.

Write $x={\sum }_i[a_i,b_i]$ with $a_i,b_i\in L$. Cyclicity and the assumed orthogonality now giveOn the other hand the nilpotent parts have zero trace on each generalized eigenspace, soThus all $\lambda$ are zero and $x$ is a nilpotent endomorphism. This is the Conjugate-spectrum proof of Cartan solvability.

The permitted Engel theorem, in the form needed here, states: a finite-dimensional Lie subalgebra of endomorphisms in which every element is nilpotent has a nonzero vector annihilated by all its elements, and iteration on quotients makes every element simultaneously strictly upper triangular. Apply it to $[L,L]$. The strictly upper triangular algebra is nilpotent, so $[L,L]$ is a nilpotent Lie algebra and therefore solvable. Since $L/[L,L]$ is abelian, the derived series of a Lie algebra of $L$ terminates too. This proves the matrix criterion.

Apply it to $L=\mathrm{ad}\mathfrak{g}$. The condition on $B$ is exactly the matrix condition on $L$. Thus $\mathrm{ad}\mathfrak{g}$ is solvable. The kernel of $\mathrm{ad}$ is the center of a Lie algebra, which is abelian, so $\mathfrak{g}$ is solvable as well: once the derived series maps to zero it is central, and its next term vanishes. Conversely a solvable $\mathfrak{g}$ has solvable adjoint image, proving the abstract Cartan solvability criterion in both directions.

A finite-dimensional Lie algebra is a semisimple Lie algebra when it has no nonzero solvable Lie algebra ideal, equivalently its solvable radical is zero. Let $R=\mathrm{rad}B=\{x:B(x,\mathfrak{g})=0\}$. Invariance makes $R$ an ideal. For $x,y\in R$, $\mathrm{ad}x$ induces zero on $\mathfrak{g}/R$, and the block trace gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$, where $B_R$ is the adjoint Killing form of $R$ itself. The Cartan solvability criterion therefore makes $R$ solvable. If $\mathfrak{g}$ is semisimple, $R=0$.

Conversely suppose $B$ is a nondegenerate bilinear form. If a nonzero solvable ideal existed, its last nonzero derived term $A$ would be an abelian ideal of $\mathfrak{g}$. For $a\in A$ and $x\in \mathfrak{g}$, the operator $\mathrm{ad}a\mathrm{ad}x$ has image in $A$ and is zero on $A$, so its square and its trace are zero. Thus $B(A,\mathfrak{g})=0$, contradicting nondegeneracy. This is Abelian ideals lie in the radical of the Killing form. We conclude the Cartan criterion for semisimplicity:

Work over $\mathbb{C}$. Decompose $V$ as the direct sum of the generalized eigenspaces $V_{\lambda }$ of $X$. Define $X_s$ to be $\lambda I$ on $V_{\lambda }$ and set $X_n=X-X_s$. Then $X_s$ is a diagonalisable endomorphism, $X_n$ is a nilpotent endomorphism, and they commute. For uniqueness, any commuting decomposition $X=S+N$ has $S$ and $N$ commuting with $X$, hence preserving each $V_{\lambda }$. Decompose $V_{\lambda }$ further into eigenspaces of $S$. On a nonzero such space with eigenvalue $\sigma$, the operator $X=\sigma I+N$ has only the eigenvalue $\sigma$, so $\sigma =\lambda$. Since $S$ is diagonalizable, $S=\lambda I$ throughout $V_{\lambda }$. Thus $S=X_s$ and $N=X_n$. This is the Additive Jordan decomposition.

The Chinese remainder theorem for the pairwise coprime polynomials $(t-\lambda {)}^{r_{\lambda }}$ also gives a polynomial $p$ with $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^{r_{\lambda }})$, where $r_{\lambda }$ is the largest Jordan block size. Consequently $X_s=p(X)$ and $X_n=X-p(X)$. This polynomial description shows that both parts preserve every $X$-invariant subspace.

On $\mathrm{Hom}(V_{\lambda },V_{\mu })$ the operator $\mathrm{ad}{X}_s$ is the scalar $\mu -\lambda$, so it is diagonalizable. The nilpotence of commutation by a nilpotent endomorphism makes $\mathrm{ad}{X}_n$ nilpotent. They commute, since $[\mathrm{ad}{X}_s,\mathrm{ad}{X}_n]=\mathrm{ad}[X_s,X_n]=0$. Uniqueness of the Additive Jordan decomposition therefore gives the adjoint compatibility of additive Jordan decomposition:Now let $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ be a complex semisimple Lie algebra and $X\in \mathfrak{g}$. Since the semisimple part of $\mathrm{ad}X$ is a polynomial in $\mathrm{ad}X$, it preserves $\mathfrak{g}$. Thus $[X_s,\mathfrak{g}]\subseteq \mathfrak{g}$. By the Weyl complete reducibility theorem, the Adjoint representation of $\mathfrak{g}$ on $\mathrm{End}(V)$ has a decomposition $\mathrm{End}(V)=\mathfrak{g}\oplus M$ into invariant subspaces. Write $X_s=y+z$ with $y\in \mathfrak{g}$ and $z\in M$. For $a\in \mathfrak{g}$, the vector $[z,a]=[X_s,a]-[y,a]$ lies in $\mathfrak{g}$, and invariance of $M$ puts it in $M$ as well. Hence $[z,a]=0$.

Decompose $V={⨁}_j{U}_j$ into Irreducible Lie algebra representations. Each $U_j$ is preserved by $X_s=p(X)$ and by $y$, hence by $z$. The Schur lemma makes $z{|}_{U_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every representing element of $\mathfrak{g}$ has zero trace on every $U_j$. Also $\mathrm{tr}(X_s{|}_{U_j})=\mathrm{tr}(X{|}_{U_j})$, because $X_n$ is nilpotent there. It follows that $c_j\dim U_j=\mathrm{tr}(z{|}_{U_j})=0$. In characteristic zero, $c_j=0$, so $z=0$. Therefore $X_s\in \mathfrak{g}$ and $X_n=X-X_s\in \mathfrak{g}$. This proves that semisimple matrix Lie algebras are closed under additive Jordan decomposition, including representations with repeated isomorphic irreducible summands.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

The centre manifold theorem gives, near an equilibrium of a sufficiently smooth system, a local invariant manifold tangent to the generalized eigenspaces with zero-real-part eigenvalues, expressible as a graph of a function over them. Its reduced dynamics govern local equilibria and their stability in the presence of strictly stable transverse directions. The manifold need not be unique or analytic; finite Taylor jets can nevertheless be determined from invariance. Treating a parameter as a variable with $\dot{\mu }=0$ gives the extended centre manifold for a parameter.

At $r=1$ the linearized $(x,y)$ block has eigenvalues $0,-2$, and the $z$ direction has eigenvalue $-1$. Thus the origin is a nonhyperbolic equilibrium, with stable dimension two and non-extended centre dimension one. The extended centre dimension is two. The transformed equations areThe graph of a function satisfies $V(0,0)=Z(0,0)=0$ and $DV(0,0)=DZ(0,0)=0$, with invariance equations $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$. Symmetry permits $V$ odd and $Z$ even in $u$. Giving $u$ weight one and $\mu$ weight two, comparison yieldsHere $O_{\mathrm{w}}(j)$ means terms of weighted order at least $j$ under the stated scaling. If $a<1$, the pitchfork bifurcation is supercritical: stable branches $u\sim \pm \sqrt{\mu /(1-a)}$ appear for $\mu >0$ as the origin loses stability. If $a>1$, it is a subcritical pitchfork bifurcation: unstable branches occur for $\mu <0$, while the origin is stable there.

For $a=1$, assign $\mu$ weight four. The leading cubic terms in $\dot{u}$ cancel. The invariance equations now give $V=u^3/2+O_{\mathrm{w}}(5)$ and $Z=u^2+0u^4+O_{\mathrm{w}}(6)$: $Z_u\dot{u}$ first contributes at weight six and $v^2$ also first contributes there. Substitution then givesThe derivative along the nonzero branches is $-2\mu >0$, so they are unstable. The origin is stable for $\mu <0$, unstable for $\mu >0$, and nonlinearly unstable at $\mu =0$ because the leading term is $+u^5$. This is a degenerate subcritical pitchfork bifurcation.

View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

Write $D=\mathrm{ad}y$. The generalized eigenspace decomposition of the linear map $D$ isfor any sufficiently large $N$. Because $D$ is a derivation, the generalized-eigenspace bracket lemma givesConsequently $L_{0,y}$ is a Lie subalgebra.

The set $I(K)=\{x\in L:[x,K]\subseteq K\}$ is the normalizer of a Lie subalgebra $K$. Certainly $L_{0,y}\subseteq I(L_{0,y})$. Conversely, if $x\in I(L_{0,y})$, then $y\in L_{0,y}$ givesOn the direct sum of the nonzero generalized eigenspaces, $D$ is invertible. Hence the nonzero-eigenvalue component of $x$ vanishes, and

Now let $K$ be a Lie subalgebra containing $L_{0,y}$. Since $y\in K$, the subspace $K$ is $D$-invariant. The generalized zero eigenspace of the induced map on $L/K$ is the image of $L_{0,y}$, hence is zero. If $x\in I(K)$, then $Dx=[y,x]\in K$, so $x+K$ lies in that zero eigenspace. Thus $x\in K$ and

A nilpotent Lie algebra is one whose lower central serieseventually reaches zero. Suppose $L$ is nilpotent and $K⊊L$. Choose the least $r\ge 2$ for which ${\gamma }_r(L)\subseteq K$. Then ${\gamma }_{r-1}(L)⊈K$, and anysatisfies $[x,K]\subseteq [L,{\gamma }_{r-1}(L)]={\gamma }_r(L)\subseteq K$. Therefore $x\in I(K)\setminus K$, proving the normalizer condition for a nilpotent Lie algebra

It remains to prove the converse needed here. The Engel lemma states that if a finite-dimensional Lie algebra of linear maps consists of nilpotent maps, then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra $H$, induction applied to the action of $H$ on $L/H$ produces $x\notin H$ with $[H,x]\subseteq H$. Thus $H$ is an ideal of codimension one. Induction also gives a nonzero common kernelThe ideal property makes $W$ invariant under $L$; a nilpotent representative of a basis of $L/H$ has a nonzero kernel on $W$, yielding a vector killed by all of $L$.

Apply the lemma to the Adjoint representation. It produces a nonzero element of the center of a Lie algebra. Induction on $\dim L$, followed by passage to the quotient by this center, proves Engel theorem: if every $\mathrm{ad}y$ is nilpotent, then $L$ is nilpotent. The hypothesis $L_{0,y}=L$ says exactly that every $\mathrm{ad}y$ is nilpotent, so

For finite matrices satisfying $YX=qXY$, with $q$ not a root of unity, $Y$ sends the generalized eigenspace of $X$ at $\lambda$ into the one at $q^{-1}\lambda$. Finiteness of the spectrum makes $Y$ nilpotent on the sum of the nonzero $X$ generalized eigenspaces. Exchanging $X,Y$ gives the corresponding statement for $X$. Splitting the zero generalized eigenspace of $X$ by that of $Y$ gives the displayed invariant direct sum; both operators are nilpotent on $U$.

Let $U\subseteq W\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ be vector subspaces and $M=\{T:[T,W]\subseteq U\}$. If $\alpha \in M$ and $\mathrm{tr}(\alpha \beta )=0$ for all $\beta \in M$, then $\alpha$ is a nilpotent endomorphism.

To prove it, let $\beta$ act by $\overline{\lambda }$ on the generalized eigenspace of $\alpha$ of eigenvalue $\lambda$. Polynomial interpolation on eigenvalue differences, together with adjoint compatibility of additive Jordan decomposition, expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Since $\mathrm{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta \in M$. Taking the matrix trace on each generalized eigenspace yields $0={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$, so every eigenvalue is zero. This is the linear-algebra step behind the Cartan solvability criterion; $U,W$ need not be Lie subalgebras.