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For a complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra consisting of elements whose Adjoint representation matrices are diagonalizable. Equivalently it is a nilpotent self-normalizing Lie subalgebra. The root-space decomposition iswhere the roots are the nonzero weights of that Adjoint representation.

We use these properties of the Killing form $B$: it is nondegenerate on $\mathfrak{g}$ and on $\mathfrak{h}$, it is invariant, distinct root spaces are orthogonal unless their roots sum to zero, and $B$ pairs ${\mathfrak{g}}_{\alpha }$ nondegenerately with ${\mathfrak{g}}_{-\alpha }$. Define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$, and choose $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)=1$. Since weights add, $[e,f]\in \mathfrak{h}$, and invariance givesThe nonisotropic root lemma shows $\alpha (t_{\alpha })\ne 0$. Here is its short proof: if this number were zero, the span of $e,f,t_{\alpha }$ would be a Solvable Lie algebra with $t_{\alpha }$ central. Apply the Lie theorem to its action on $\mathfrak{g}$. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}e,\mathrm{ad}f]$ would be strictly upper triangular and hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$ makes it diagonalizable. It would therefore be zero, putting $t_{\alpha }$ in the zero center of a Lie algebra of $\mathfrak{g}$, contrary to its definition. SetThenso their span is the sl2 subalgebra associated with a root.

The abstract reduced crystallographic root system axioms are as follows. In a finite-dimensional real inner-product space $E$, the set $R$ is finite, consists of nonzero vectors and spans $E$; for every $\alpha \in R$, one has $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; the root reflection $s_{\alpha }(\beta )=\beta -2(\beta ,\alpha )\alpha /(\alpha ,\alpha )$ preserves $R$; and $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for every pair of roots. We verify the positive-definite real form as well as these axioms, rather than assuming the complex Killing form is already positive.

First $R$ is finite and has no zero element by its definition. The roots span ${\mathfrak{h}}^{\ast }$ over $\mathbb{C}$: any $h\in \mathfrak{h}$ annihilated by all roots commutes with the whole root-space decomposition and is central, hence zero. Therefore the $t_{\alpha }$, and also the $H_{\alpha }$, span $\mathfrak{h}$ over $\mathbb{C}$. For every root $\beta$, the classification of finite-dimensional sl2 representations applied to the Adjoint representation of the root subalgebra gives $\beta (H_{\alpha })\in \mathbb{Z}$. Put ${\mathfrak{h}}_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\{H_{\alpha }:\alpha \in R\}$. Every root takes real values on this space, andThe strict inequality follows because the roots span ${\mathfrak{h}}^{\ast }$. This also shows that the complexification of ${\mathfrak{h}}_{\mathbb{R}}$ injects into $\mathfrak{h}$: an equality $h_1+i{h}_2=0$ with real $h_j$ would contradict positivity of $B(h_1,h_1)$ and $B(h_2,h_2)$. Since its complex span is all of $\mathfrak{h}$, it is a real form. Moreover $B(H_{\alpha },H_{\alpha })=4/\alpha (t_{\alpha })>0$, so $t_{\alpha }$ is a real scalar multiple of $H_{\alpha }$. The dual inner product thus makes $E={\mathrm{span}}_{\mathbb{R}}R$ a Euclidean space, as in the Euclidean subspace of a Cartan subalgebra.

To prove reducedness without assuming it, consider the root-subalgebra modulewith absent root spaces understood to be zero. Its $H_{\alpha }$ weights are even, so every nontrivial Irreducible Lie algebra representation in it has even positive highest weight and one-dimensional weight-zero space. The action of $X_{\alpha }$ on its weight-zero space $\mathfrak{h}$ has image exactly $\mathbb{C}{X}_{\alpha }$, of dimension one. Therefore there is precisely one nontrivial irreducible summand, the already embedded adjoint ${\mathfrak{sl}}_2$ module of highest weight $2$. Thus $\dim {\mathfrak{g}}_{\alpha }=1$ and $2\alpha \notin R$. If $\beta =c\alpha$ is any root on the same real line, the integral numbers $\beta (H_{\alpha })=2c$ and $\alpha (H_{\beta })=2/c$ have product $4$. Hence $c$ is one of $\pm \frac{1}{2},\pm 1,\pm 2$; the half and double cases are excluded by applying the preceding argument to the appropriate root. This proves the root-space reducedness lemma and $R\cap \mathbb{R}\alpha =\{\pm \alpha \}$.

For a root $\beta$ not parallel to $\alpha$, the sum of root spaces ${⨁}_{j\in \mathbb{Z}}{\mathfrak{g}}_{\beta +j\alpha }$ is stable under the root ${\mathfrak{sl}}_2$. Its integer $H_{\alpha }$ weights are symmetric under sign in each irreducible summand. Therefore the weight $-\beta (H_{\alpha })$ also occurs, at the root $\beta -\beta (H_{\alpha })\alpha$. The Killing form normalization givesso this root is exactly $s_{\alpha }(\beta )$. For $\beta =\pm \alpha$ the reflection just swaps the two roots. This proves reflection invariance and the Cartan integer condition. All axioms of the reduced crystallographic root system have now been verified.

The form attached to $V$ is $B_V(x,y)={\mathrm{tr}}_V(xy)$. More generally, for a Lie algebra representation $\rho$, the Trace form of a Lie algebra representation isThe unqualified Killing form is the special case of the Adjoint representation,The distinction matters: a Trace form of a Lie algebra representation can be degenerate even when $\mathfrak{g}$ is semisimple, for example on the trivial Lie algebra representation.

The Trace form of a Lie algebra representation is bilinear and symmetric, because $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. It is an invariant bilinear form on a Lie algebra:This follows by expanding both commutators and cyclically permuting factors under the matrix trace. Equivalently,Its radical of a bilinear form is an ideal of a Lie algebra, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The Killing form is also preserved by every automorphism of a Lie algebra, because the corresponding adjoint operators are conjugate. On a complex finite-dimensional Lie algebra, the Cartan criterion for semisimplicity says that the Killing form is nondegenerate exactly when the Lie algebra is semisimple. The Cartan solvability criterion says that $\mathfrak{g}$ is solvable exactly when $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.

We next construct the sl2 subalgebra associated with a root. Use the root-space decompositionFor $u\in {\mathfrak{g}}_{\beta }$, $v\in {\mathfrak{g}}_{\gamma }$, invariance of the Killing form givesThus $B({\mathfrak{g}}_{\beta },{\mathfrak{g}}_{\gamma })=0$ unless $\beta +\gamma =0$, and $B({\mathfrak{g}}_{\beta },\mathfrak{h})=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak{g}$ now implies that $-\alpha$ is a root and that $B$ pairs ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ nondegenerately.

Nondegeneracy of $B{|}_{\mathfrak{h}}$ defines a unique $t_{\alpha }\in \mathfrak{h}$ byChoose $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$. Their Lie bracket lies in the zero root space, namely $\mathfrak{h}$, andTherefore $[x,y]=t_{\alpha }$.

The essential nonisotropic root lemma is that $\alpha (t_{\alpha })\ne 0$. Suppose instead that it vanished. Then $[t_{\alpha },x]=[t_{\alpha },y]=0$, so $⟨x,y,t_{\alpha }⟩$ would be a Solvable Lie algebra with derived algebra $\mathbb{C}{t}_{\alpha }$. Apply the Lie theorem to its action on $\mathfrak{g}$ by the Adjoint representation. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}x,\mathrm{ad}y]$ is strictly upper triangular in a suitable basis, hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$, so the root-space decomposition makes $\mathrm{ad}{t}_{\alpha }$ diagonalizable. A diagonalizable nilpotent linear map is zero. Thus $t_{\alpha }$ is central in $\mathfrak{g}$. The center of a Lie algebra of a semisimple Lie algebra is zero; equivalently a central element lies in the radical of the Killing form. This forces $t_{\alpha }=0$, contradicting $\alpha \ne 0$.

Writing $c=\alpha (t_{\alpha })$, defineThe root-space decomposition and $[x,y]=t_{\alpha }$ giveThe three vectors are linearly independent because they lie in the distinct summands ${\mathfrak{g}}_{\alpha }$, $\mathfrak{h}$, and ${\mathfrak{g}}_{-\alpha }$. Their span is therefore a copy of the sl2 Lie algebra.

The weight lattice consists of the functionals integral on all coroots. With $H_{\alpha }$ the coroot above, the weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$ for the simple roots ${\alpha }_j$. Here $P$ lies in the real span of the roots, viewed inside ${\mathfrak{h}}^{\ast }$.

The classification of finite-dimensional sl2 representations says that every finite-dimensional complex sl2 Lie algebra representation is a direct sum of irreducibles $L_m$, $m\in {\mathbb{Z}}_{\ge 0}$, on which the standard $H$ has eigenvalues $m,m-2,\dots ,-m$. Restrict any finite-dimensional Lie algebra representation of $\mathfrak{g}$ to each sl2 subalgebra associated with a root. If $v$ has weight $\lambda$, then $H_{\alpha }v=\lambda (H_{\alpha })v$, so $\lambda (H_{\alpha })$ is an integer. Thus every weight lies in $P$. The same restrictions show that the commuting simple coroots act diagonalizably, justifying the simultaneous weight-space decomposition.

For $\mathfrak{sp}(4)$, the roots are $\pm 2{\epsilon }_1$, $\pm 2{\epsilon }_2$, and $\pm {\epsilon }_1\pm {\epsilon }_2$. Work on ${\mathbb{C}}^4$ with the alternating bilinear form having matrixThe symplectic Lie algebra isUsing the matrix units $E_{ij}$, take the Cartan subalgebraDefine ${\epsilon }_i(a_1{h}_1+a_2{h}_2)=a_i$. A regular diagonal element of $\mathfrak{h}$ has centralizer precisely $\mathfrak{h}$, and every element of $\mathfrak{h}$ acts diagonalizably. Thus it is a Cartan subalgebra. The requested Cartan decomposition is the root-space decompositionChoose positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $2{\epsilon }_1$. The symplectic root sl2 triple are given explicitly byFor the negative root spaces, use the corresponding $Y_{\alpha }$. These eight root vectors, together with $h_1,h_2$, form a basis: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the matrix unit identityverifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ for every row. The diagonal differences verify $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$ and $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$. Thus each row supplies a basis of the required sl2 subalgebra associated with a root.