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Use a real Lie algebra convention $[T_a,T_b]=f_{ab}{}^c{T}_c$. Its Adjoint representation of a Lie algebra acts on $\mathfrak{g}$ itself:Linearity is immediate. The Jacobi identity givesso $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, the defining Lie algebra representation condition. In the basis $T_b$, its matrix entries areThus the adjoint generators are the structure constants arranged as matrices. The representation has kernel equal to the center of a Lie algebra; it need not be faithful for an arbitrary algebra.

The Adjoint representation of a Lie group is ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$. It satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$, preserves the identity, and sends inverses to inverse matrices. Differentiating $e^{tX}Y{e}^{-tX}$ gives $[X,e^{tX}Y{e}^{-tX}]$, soConsequently the positive adjoint exponentials furnish the representation on elements $g=e^X$ and their products, and the intrinsic conjugation action defines it globally.

The inverse-conjugation formula in the PDF needs a minus adjoint exponent with this standard definition. Its first-order term is $Y-[X,Y]$, whereas $e^{+{\mathrm{ad}}_X}Y$ has first-order term $Y+[X,Y]$. For example $X=T_1,Y=T_2$ in a nonabelian algebra with $[T_1,T_2]\ne 0$ already distinguishes the two. The inverse conjugation and adjoint antirepresentations identity explains the group-order issue as well: $F(g)={\mathrm{Ad}}_{g^{-1}}$ satisfies $F(gh)=F(h)F(g)$. Retaining the printed inverse conjugation as a left action without this order reversal would not give an ordinary group representation.

The Killing form is the symmetric bilinear formTo prove degeneracy for a non-semisimple Lie algebra, take its nonzero solvable radical and the last nonzero member $\mathfrak{a}$ of its derived series of a Lie algebra. This is a nonzero abelian ideal of a Lie algebra. For $x\in \mathfrak{a}$, ${\mathrm{ad}}_x$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and kills $\mathfrak{a}$. Every ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Hence ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps the full space into $\mathfrak{a}$ and has zero restriction there, so its trace is zero. Thus $B(x,y)=0$ for every $y$. This is the abelian ideals lie in the radical of the Killing form argument, and establishes a nonzero kernel and vanishing determinant.

For a compact real semisimple Lie algebra, the adjoint action is unitary in an invariant positive inner product. Its infinitesimal generators are skew-Hermitian, givingStrictness follows because the adjoint kernel is the zero center. This explains the compactness criterion from the Killing form: “strictly negative” means negative-definite, not that every entry of its matrix is negative.

In the first three-generator example, take columns to be images of the basis $(X,Y,H)$. Direct use of the brackets givesTaking traces of products yields the Killing form for cyclic three-generator bracketsIt is nondegenerate, hence semisimple by the preceding degeneracy result, but it has mixed signature and is not compact. An explicit realization iswhich span the real traceless two-by-two matrices and satisfy exactly these brackets.

Changing the sign of $[X,H]$ changes the first and third adjoint matrices towhile ${\mathrm{ad}}_Y$ is unchanged. Nowthe negative-definite compact case. The basis $X=-i{\sigma }_1$, $Y=-i{\sigma }_2$, $H=-i{\sigma }_3$ realizes it as the real special unitary Lie algebra $\mathfrak{su}(2)$.

A quadratic form on a finite-dimensional real vector space is a function $q(v)=B(v,v)$ for a symmetric bilinear form $B$. Conversely $B(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$ recovers that bilinear form. Two quadratic forms are congruent if a linear isomorphism $T$ satisfies $q_2(v)=q_1(Tv)$; in coordinates this is matrix congruence, $A_2=P^T{A}_{1}P$, with $P$ invertible.

Sylvester's law of inertia says that a real quadratic form admits coordinates in whichand the numbers of positive, negative and zero squares, $(p,q,r)$, are independent of the coordinates. Here $p+q+r=\dim V$; $p$ and $q$ are the maximal dimensions of subspaces on which the quadratic form is positive definite and negative definite, while $r$ is the dimension of its radical of a bilinear form. These numbers are the inertia of a bilinear form. The rank of a quadratic form is $p+q$, and the signature of a quadratic form, in the signed-difference convention, is $p-q$.

Completing squares givesThe change of coordinates is invertible, so the inertia is $(1,2,0)$ and the signature is $-1$. If signature is recorded as an ordered pair instead, that pair is $(1,2)$.

A finite-dimensional Lie algebra over the complex numbers $L$ is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form isThe cyclic property of the trace makes this bilinear form symmetric and gives its invariance of a bilinear form on a Lie algebra:It follows that $R=\{x:B_L(x,L)=0\}$ is an ideal of a Lie algebra. For $x\in R$, $\mathrm{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the matrix trace splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The Cartan solvability criterion implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: the Killing form is nondegenerate.

For completeness, the trace step in the Cartan solvability criterion is precisely the mechanism of the previous solution. For a complex matrix Lie algebra $\mathfrak{a}$ with $\mathrm{tr}(xy)=0$ for $x\in [\mathfrak{a},\mathfrak{a}]$, $y\in \mathfrak{a}$, set $U=[\mathfrak{a},\mathfrak{a}]$ and $W=\mathfrak{a}$. If $\beta \in M$ and $x=[a,b]$, then $\mathrm{tr}(x\beta )=\mathrm{tr}(a[b,\beta ])=0$, since $[b,\beta ]\in U$. Linearity and the trace orthogonality nilpotence lemma show that every member of $U$ is nilpotent. The Engel theorem makes $U$ nilpotent and hence $\mathfrak{a}$ solvable. Apply this to $\mathfrak{a}=\mathrm{ad}R$; the kernel of this Adjoint representation of a Lie algebra is the abelian center of $R$, so $R$ is solvable as claimed.

For an arbitrary complex Lie algebra, a Cartan subalgebra $H$ means a nilpotent Lie algebra that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the action of $H$ on $L$. Its zero generalized weight space isWe have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne 0$, the Engel theorem gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.

For a nonzero generalized weight $\lambda$, choose $h_0\in H$ with $\lambda (h_0)\ne 0$. The operator $\mathrm{ad}{h}_0$ is invertible on $L^{\lambda }$ and nilpotent on $H$. For $h\in H$, $z\in L^{\lambda }$, write $z=(\mathrm{ad}{h}_0{)}^{m}w$ with $m$ large enough that $(\mathrm{ad}{h}_0{)}^{m}h=0$. Invariance of the Killing form givesOn the other hand, $H$ is solvable, so the Lie theorem triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\mathrm{ad}[a,b]$ is strictly upper triangular, while $\mathrm{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus {⨁}_{\lambda \ne 0}{L}^{\lambda }$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.

Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus $H$ is a maximal abelian subalgebra, indeed $C_L(H)=H$.

A separated dual pair consists of real vector spaces $E,F$ and a bilinear form $⟨x,y⟩$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.

The weak topology $\sigma (E,F)$ is the coarsest topology making all maps $x\mapsto ⟨x,y⟩$, $y\in F$, continuous. It makes $E$ a Hausdorff locally convex space, with topology generated by the seminorms $p_y(x)=|⟨x,y⟩|$. A neighbourhood base at zero consists of finite intersectionsSeparation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.

We determine the continuous dual of a weak topology. Suppose a linear functional $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell (x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell (x)=0$. Hence $\ell$ factors through the image ofExtend the resulting linear functional on $T(E)$ to ${\mathbb{R}}^m$. It has the form $z\mapsto {\sum }_j{a}_j{z}_j$, giving $\ell (x)=⟨x,{\sum }_j{a}_j{y}_j⟩$. Conversely every such evaluation is continuous by the definition of the weak topology. Since the pairing separates $F$, this identification is injective, andFor the remaining parts every topological assertion about $E^{\prime }$ or its subsets uses the stipulated weak-star topology $\sigma (E^{\prime },E)$.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

The finite element method replaces an infinite-dimensional variational problem by a finite-dimensional one, usually using functions that are polynomial on small mesh elements. Ritz method and Galerkin method describe how the discrete equations are selected; neither intrinsically requires piecewise polynomials, though those spaces make assembly local and sparse.

For a concrete realization of the two-point problem, choose homogeneous Dirichlet boundary conditions at both endpoints of $(0,1)$. The differential expression alone is not a complete boundary value problem, so this boundary choice is an explicit assumption. Let $p\in W^{1,\infty }$ with $p\ge p_0>0$, $q\in L^{\infty }$ with $q\ge 0$, and $f\in L^2$. Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_2$. By the Poincare inequality this is equivalent to its full first-order Sobolev norm. Multiplying by a test function and applying integration by parts givesThe endpoint terms vanish by the essential boundary condition. The form is a symmetric bilinear form, bounded byand it is a coercive bilinear form:The load is a bounded linear functional. The Lax-Milgram theorem gives a unique weak solution and a bound $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\ast }}/p_0$. Under the stated coefficient regularity it solves the differential equation in the usual weak sense.

The Ritz method minimizes the energyover $V$, or over a conforming finite-dimensional subspace $V_h\subset V$. Its first variation is $a(v,w)-\ell (w)$; symmetry and coercivity make the functional strictly convex. In fact, if $u$ is the weak solution,Thus the energy minimizer is exactly the weak solution. The Galerkin method instead directly asks for $u_h\in V_h$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. For this symmetric coercive problem, Ritz-Galerkin equivalence for a symmetric coercive form shows that the discrete minimizer and Galerkin solution coincide. For a nonsymmetric problem, a Galerkin formulation can still apply but the same quadratic minimization generally cannot represent its full bilinear form; appropriate coercivity or inf-sup conditions must then justify the chosen spaces.

Subtract the continuous and discrete weak equations to obtain Galerkin orthogonality $a(u-u_h,v_h)=0$. Since $a$ defines the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity givesIn a general reference norm, the Céa lemma yieldsThe proof uses $a(e,e)=a(e,u-v_h)$ together with coercivity and continuity. Dense approximation spaces therefore imply convergence; the estimate reduces a numerical error problem to an approximation problem.

For implementation, partition the interval by nodes $0=x_0<\cdots <x_N=1$. Choose continuous piecewise-linear functions and the interior piecewise-linear hat functions ${\varphi }_j$ as a basis, with the two boundary coefficients prescribed. Write $u_h={\sum }_j{U}_j{\varphi }_j$. The coefficients satisfyThe stiffness matrix is symmetric positive definite: $U^{T}KU=a(u_h,u_h)>0$ for nonzero interior coefficients. The local supports make it tridiagonal in this one-dimensional linear-element case.

On an element of length $h_e$, for constant element coefficients $p_e,q_e$, the two-node matrix and constant-load vector areFor variable coefficients, integrate $p,q,f$ against the same shape functions rather than silently replacing them by constants. Element contributions are added at shared nodes, and boundary values are eliminated or lifted into the right side. A sparse direct solve or a suitable iterative method gives the nodal coefficients. Numerical quadrature should preserve the required accuracy and, with positive coefficients and weights, the positive energy structure.

A finite element interpolation estimate gives ${\inf }_{v_h}\Vert u-v_h{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ for these elements, where $h$ is the largest element size and the solution has the stated regularity. The Céa lemma then gives first-order energy error. If the dual elliptic problem has $H^2$ regularity, the Aubin–Nitsche duality argument gives second-order $L^2$ error: for $e=u-u_h$, solve $a(v,z)=(e,v{)}_{L^2}$, then use $a(e,z)=a(e,z-I_{h}z)$ to gain one further factor of $h$. Higher-order elements improve rates when the solution is sufficiently smooth.

Nonzero Dirichlet data are handled by a boundary lifting and an affine trial space. Neumann boundary conditions enter naturally through the integrated boundary term; Robin terms modify both the form and the load. With pure Neumann conditions and $q=0$, constants lie in the kernel: existence needs the load/flux compatibility condition, uniqueness needs a mean constraint, and the preceding Dirichlet coercivity argument cannot simply be reused. Conforming approximation, boundary conditions and coercivity are the ingredients connecting the finite-element construction to a justified Ritz or Galerkin solution.

A rigorous Mumford–Shah functional permits nonsmooth image signals and free discontinuities. One classical admissible class consists of relatively closed countably rectifiable sets $K\subset \Omega$ with finite Hausdorff measure ${\mathcal{H}}^1(K)$, and $u\in W^{1,2}(\Omega \setminus K)\cap L^{\infty }(\Omega )$ with finite energy. No exterior boundary values are prescribed. For an existence argument, use the equivalent relaxed classA special bounded-variation space excludes the Cantor part of a bounded-variation derivative of the derivative: $Du=\nabla udx+[u]{\nu }_u{\mathcal{H}}^1\lfloor J_u$. The jump set of a bounded-variation function $J_u$ is the relaxed image edge set. Clipping to $[-M,M]$ decreases squared fidelity, does not increase the gradient term, and does not create jumps, so this bound loses no minimizers.

Take a minimizing sequence and compare with a constant image signal. Its $L^2$ gradient norms and jump lengths are bounded. Alsoso the sequence is bounded in $BV$. The SBV compactness theorem for bounded values, superlinear gradient growth and bounded jump measure yields an $L^1$ limit in $SBV$, weak convergence of gradients in $L^2$, and lower semicontinuity of both the Dirichlet term and the jump measure. The uniform value bound upgrades convergence to $L^2$, so fidelity converges. This proves existence of a relaxed minimizer. Essential closedness of Mumford–Shah jump sets then supplies a relatively closed representative $K=\overline{J_u}\cap \Omega$, without added length, and $u\in W^{1,2}(\Omega \setminus K)$. This completes the outline for the classical pair problem. Arbitrary Hausdorff convergence of image edge sets alone is not an adequate substitute for these compactness and regularity results. No uniqueness is claimed for segmentation.

As $\alpha \to \infty$ with $\beta$ fixed, bounded energy forces $\nabla u\to 0$ in $L^2$. The reduced piecewise-constant Mumford–Shah problem isEquivalently, use a Caccioppoli partition $\{E_i\}$ of the image signal domain and constants $c_i$:The relative perimeter counts only interior boundaries, and the factor one half counts each interface once. Adjacent equal-valued regions can be merged, removing unnecessary boundaries.

For fixed $K$, let $E_i$ be its positive-area regions. Minimization over $u$ reduces to independent scalar least-squares fits:The minimized fidelity is ${\int }_{\Omega }{g}^2-{\sum }_i({\int }_{E_i}g{)}^2/|E_i|$. Thus region means in piecewise-constant segmentation give the optimal grey values for a fixed segmentation.

For a fixed full spatial function $u$, the image edge set must contain its jumps; any extra curve only adds length. The optimal choice is its essential jump set, with a relatively closed representative when appropriate. There is no independent relocation of boundaries while that full function is held fixed. A different common alternating step fixes the values $c_i$ but allows the labels $E_i$ to move. It minimizes the fidelity-plus-perimeter partition functional above. Without the perimeter term each point takes its nearest grey value; with it, interface length is penalized. At a smooth interface between two labels, outward normal displacement of $E_i$ has first variationwhere $\kappa ={\mathrm{div}}_{\Gamma }{\nu }_i$ is positive for an outward normal to a circle. The stationary segmentation interface curvature balance isThis is the geometric interpretation of optimizing boundaries with fixed grey levels, and distinguishes it from fixing the whole spatial image signal.

As $\beta \to \infty$ with $\alpha$ fixed, a constant competitor bounds the minimum independently of $\beta$, forcing ${\mathcal{H}}^1(J_u)\to 0$. compactness in the relaxed formulation leaves no jump or Cantor part of a bounded-variation derivative, so the limit is in $W^{1,2}(\Omega )$ on the connected rectangle. The reduced edge-free Mumford–Shah limit isA set of zero length can be omitted; this does not impose a zero image signal or a Dirichlet boundary value. Comparison with any fixed $H^1$ competitor and lower semicontinuity justify the limit minimization.

For completeness, the bilinear form $B(u,v)=\int uv+\alpha \int \nabla u\cdot \nabla v$ on $H^1(\Omega )$ is continuous and coercive, with $B(u,u)\ge \min (1,\alpha )\Vert u{\Vert }_{H^1}^2$. The right-hand side $\int gv$ is bounded because $g\in L^2$ on the bounded rectangle. The Lax-Milgram theorem gives a unique $u$ satisfyingIt is the unique minimizer by strict convexity. Formally,with the Neumann condition understood through this weak formulation. Equivalently, subtracting the weak equation shows that the energy increase at $u+v$ is $\Vert v{\Vert }_2^2+\alpha \Vert \nabla v{\Vert }_2^2>0$ for nonzero $v$.

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.