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For a point $\mathbf{x}$ on the plane, the Cauchy-Schwarz inequality gives $|d|=|\mathbf{x}\cdot \mathbf{n}|\le |\mathbf{x}|$, because $\mathbf{n}$ is a unit vector. Equality is achieved at $\mathbf{x}=d\mathbf{n}$. Thus the distance from the origin is $|d|$.

Let $\delta =\mathbf{p}\cdot \mathbf{n}-d$ and let $\mathbf{q}=\mathbf{p}-\delta \mathbf{n}$ be the perpendicular projection of the sphere's center onto the plane. Every point of the plane has form $\mathbf{x}=\mathbf{q}+\mathbf{u}$ with $\mathbf{u}\cdot \mathbf{n}=0$. The Pythagorean theorem givesIf $|\delta |=r$, the sphere equation therefore forces $\mathbf{u}=0$. There is exactly one contact point, $\mathbf{q}$; it lies on both surfaces.

Write $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The nonzero scalar triple product proves that $\mathbf{a},\mathbf{b},\mathbf{c}$ are linearly independent, so they form a positively oriented basis of ${\mathbb{R}}^3$ and define a nondegenerate tetrahedron.

For the three faces through the origin, choose the inward unit vectorsTheir signs are correct because each has positive inner product with the opposite vertex vector. Write the center as $\mathbf{p}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. It lies on the interior side of every face. Tangency of the interior sphere therefore means signed distance $r$, not $-r$, from each of those three planes. Taking inner products givesSolving proves the center formulaThis is the sphere center from three tetrahedron face distances; the interior hypothesis selects these signs rather than an exterior center.

Finally set $\mathbf{N}=\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}$. The scalar triple product identities give $\mathbf{N}\cdot \mathbf{a}=\mathbf{N}\cdot \mathbf{b}=\mathbf{N}\cdot \mathbf{c}=\tau$. In particular $\mathbf{N}\ne 0$, and the plane through the three vertices isThe second formula follows by normalizing $\mathbf{N}$ to a unit vector and applying the first distance result.

An eigenvalue can be complex even though $P$ is real, so use the complex Hermitian inner product. For a nonzero complex eigenvector $\mathbf{v}$, the real orthogonal matrix also satisfies $P^{\ast }P=P^{T}P=I$. Henceand $\boxed{|\lambda |=1}$. Taking complex conjugates in $P\mathbf{v}=\lambda \mathbf{v}$ gives $P\overline{\mathbf{v}}=\overline{\lambda }\overline{\mathbf{v}}$ because $P$ is real. The conjugate vector is nonzero, so $\boxed{{\lambda }^{\ast }=\overline{\lambda }\text{ is also an eigenvalue}}$.

For the unheaded geometrical conclusion, the third eigenvalue of $Q$ must be real: nonreal eigenvalues of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact algebraic multiplicity two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-eigenspace is a two-dimensional plane and the $-1$-eigenspace is a line. They are orthogonal: preservation of the inner product between a fixed vector and a negated vector gives $\mathbf{u}\cdot \mathbf{v}=-\mathbf{u}\cdot \mathbf{v}$. Therefore $Q$ is the orthogonal reflection in its fixed plane. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf{n}$, $Q=I-2\mathbf{n}{\mathbf{n}}^T$.

For a real orthogonal matrix, $P^{T}P=I$. ConsequentlyThus it preserves every real inner product, and in particular lengths and angles.

Put $r_t=\sqrt{p_t}-\sqrt{p}-(t/2)g\sqrt{p}$. By differentiability in quadratic mean, $\Vert r_t{\Vert }_{L^2(\mu )}=o(|t|)$ andin L2 space. The second limit follows because the first gives $\Vert \sqrt{p_t}-\sqrt{p}{\Vert }_2=O(|t|)$.

Both probability density functions integrate to one. Consequently the L2 inner product satisfiesThe Cauchy-Schwarz inequality permits passage to the limit in this inner product, givingThus every score function is a mean-zero function. This proof uses square-root differentiability in quadratic mean, without requiring differentiation under the density integral.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

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 $\mathfrak{su}(2)$.

Write $U^{\ast }={\overline{U}}^T$ for the conjugate transpose. Taking determinants in $U^{\ast }AU=A$ gives $|\det U{|}^2\det A=\det A$. Since the Hermitian matrix $A$ is invertible, every such $U$ is invertible and $|\det U|=1$. The identity belongs to $U_A$; if $U,V\in U_A$, then $(UV{)}^{\ast }A(UV)=V^{\ast }AV=A$; and multiplying $U^{\ast }AU=A$ by $(U^{\ast }{)}^{-1}$ and $U^{-1}$ gives $(U^{-1}{)}^{\ast }A{U}^{-1}=A$. Associativity comes from matrix multiplication, so this is a group.

The Hermitian form $h_A(v,w)=v^{\ast }Aw$ obeys $h_A(Uv,Uw)=h_A(v,w)$ precisely when $U\in U_A$. Thus $U_A$ is the group of linear isometries of this nondegenerate Hermitian form; positivity of $A$ is unnecessary.

For $A=I$, $U$ is a unitary matrix. Here is a direct diagonalization of a matrix proof. Over $\mathbb{C}$, its characteristic polynomial has a root, giving a unit eigenvector $v$ with $Uv=\lambda v$. Preservation of the inner product gives $|\lambda |=1$. If $w\perp v$, then $⟨Uw,Uv⟩=⟨w,v⟩=0$, so $v^{\perp }$ is invariant. The restriction is again a unitary matrix. Induction on the dimension supplies an orthonormal basis of eigenvectors. In that basis, $U$ is diagonal, with diagonal entries on the unit circle.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

Let $M$ be a closed manifold with a Riemannian metric and use $\Delta =-\mathrm{div}\mathrm{grad}$, the positive Laplace-Beltrami operator. This is the setting in which the final discrete eigenfunction expansion applies without additional boundary or noncompact spectral hypotheses. A Riemannian heat kernel is a smooth $H(t,x,y)$ for $t>0$ such thatfor every smooth initial function $f$. Here $\nu$ solves the heat equation. The limit is the statement that the initial kernel is the Dirac delta distribution on the diagonal. In the closed setting this uniquely determines the heat kernel; it is symmetric, preserves constants, is nonnegative, and satisfies the semigroup property.

A heat parametrix $P(t,x,y)$ is an approximate version with the same delta initial limit and with errorregular enough at $t=0$ to correct by a convergent integral series. One may construct it by the local Gaussian ansatzwhere $\chi =1$ near the diagonal and is supported in a convex normal neighborhood, and the smooth coefficients solve the usual radial transport equations. The leading coefficient accounts for the Riemannian volume form. Taking $q$ sufficiently large makes the residual extend continuously, with any prescribed finite number of derivatives, to $t=0$. Alternatively a smooth asymptotic summation of all transport coefficients gives a residual vanishing to every order. These local construction, extension and differentiability facts are used here as subsidiary results, as permitted.

For the correction proof take a heat parametrix whose residual is smooth up to $t=0$ on a short interval $[0,T]$. We also use its uniform integral bound ${\sup }_{0<t\le T,x}{\int }_M|P(t,x,y)|d{V}_y<\infty$, its approximate-identity limit, and the following standard differentiability property: convolution with a smooth time-dependent kernel gives a smooth kernel for $t>0$, and differentiating it yields the identity below. These properties follow from the Gaussian estimates and the delta initial limit; no conclusion about the final exact heat kernel is assumed.

Define the Volterra convolution of kernels byIt is associative wherever these integrals converge. The delta term at the upper endpoint givesSeek $H=P+P\ast Q$. Then its error vanishes exactly when $Q+R+R\ast Q=0$. The solution is the Volterra parametrix correctionThe signs matter: the first correction is $-P\ast R$.

To prove convergence, let $|R|\le C$ and $V=\mathrm{vol}(M)$. The time variables in $R^{\ast k}$ range over a simplex of volume $t^{k-1}/(k-1)!$, soThe series for $Q$ therefore converges uniformly on $[0,T]\times M\times M$. Derivatives obey analogous bounds, with polynomial factors in $k$, by the quoted residual extension and differentiation properties. Thus the series can be convolved and differentiated as above. Associativity and absolute convergence give $Q+R+R\ast Q=0$, proving $({\partial }_t+{\Delta }_x)H=0$. The correction $P\ast Q$ has integral norm $O(t)$ by the integral bound on $P$ and boundedness of $Q$. It has zero initial limit, so $H$ has precisely the required delta initial data.

For uniqueness, a smooth solution $w$ with zero initial data satisfies the heat equation energy identityIts initial norm is zero, hence $w=0$. Applying uniqueness to consecutive evolutions proves the semigroup property. Choose $s_0<T$ and, for arbitrary $t>0$, compose enough kernels $H(t/r)$ that $t/r<s_0$. The resulting kernel is independent of the subdivision by uniqueness, is smooth for positive time by the quoted differentiability property, and extends the construction to every $t>0$. This proves a parametrix determines the global heat kernel in the closed setting. The heat equation maximum principle gives nonnegativity and applying uniqueness to the constant initial function gives conservation of total mass. Self-adjointness of the positive Laplace-Beltrami operator $\Delta$ gives symmetry.

Finally let $\{{\varphi }_j\}$ be a complete complex orthonormal eigenbasis of $\Delta$, with eigenvalues ${\lambda }_j\ge 0$, repeated by multiplicity. As subsidiary analytic facts we use the compact elliptic compact elliptic spectral theorem, elliptic regularity bounds making each fixed derivative of ${\varphi }_j$ grow at most polynomially in $1+{\lambda }_j$, and a polynomial eigenvalue-counting bound. The spectral expansion of the Riemannian heat kernel converges because exponential decay makes the following series converge with every derivative when $t\ge \epsilon >0$:Indeed, evolving initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$ by direct substitution into the heat equation and uniqueness. For general $f\in L^2(M)$, completeness and the heat equation energy identity giveThe smoothly convergent kernel series represents this same operator, and equality for all smooth $f$ identifies it pointwise with the constructed heat kernel. The complex conjugate is required by the inner product; for a real eigenbasis it may be omitted. A noncompact Riemannian manifold may instead require a spectral integral, and the unqualified discrete formula should not be asserted there.

The complexification of a Lie algebra $L_0$ is $L=L_0{\otimes }_{\mathbb{R}}\mathbb{C}$, with the Lie bracket extended complex-bilinearly. Equivalently write $L=L_0\oplus i{L}_0$, whereComplex conjugation is an antilinear map and a Lie algebra automorphism whose fixed subalgebra is $L_0$.

If the solvable radical of $L_0$ is nonzero, its complexification is a nonzero solvable ideal of $L$. Conversely the solvable radical $R$ of $L$ is preserved by complex conjugation, because it is the unique largest solvable ideal. Consequentlyfor $z\in R$, both $(z+\overline{z})/2$ and $(z-\overline{z})/(2i)$ belong to $R_0$. If $R\ne 0$, $R_0\ne 0$ is a solvable ideal of $L_0$. This proves $L_0$ is semisimple if and only if $L_0{\otimes }_{\mathbb{R}}\mathbb{C}$ is semisimple. Consistently, the complex Killing form is just the complex-bilinear extension of the real one; its determinant in a real basis is unchanged by extending scalars.

Take the sl2R Lie algebra ${\mathfrak{sl}}_2(\mathbb{R})$ and the special unitary Lie algebra $\mathfrak{su}(2)$. Both complexify to ${\mathfrak{sl}}_2(\mathbb{C})$. This is immediate for the former; for the latter, the real basis $ih,e-f,i(e+f)$ consists of traceless matrices that are skew-Hermitian matrices and is also a complex basis of ${\mathfrak{sl}}_2(\mathbb{C})$.

They are not isomorphic as real Lie algebras. Their Killing forms are $B(X,Y)=4\mathrm{tr}(XY)$, but on $\mathfrak{su}(2)$ this is negative definite. On ${\mathfrak{sl}}_2(\mathbb{R})$ the basis $h,e+f,e-f$ has a diagonal Gram matrix with entries $8,8,-8$, so the signature of a quadratic form is $(2,1)$. A Lie algebra isomorphism preserves the Killing form and therefore its signature.

A real split semisimple Lie algebra has a Cartan subalgebra whose Adjoint representation of a Lie algebra is simultaneously diagonalizable over $\mathbb{R}$, so its root-space decomposition is defined over $\mathbb{R}$. The example ${\mathfrak{sl}}_2(\mathbb{R})$ is split: the diagonal Cartan subalgebra $\mathbb{R}h$ has eigenvalues $0,2,-2$ and real root spaces $\mathbb{R}e,\mathbb{R}f$. The example $\mathfrak{su}(2)$ is not split. Invariance makes every $\mathrm{ad}x$ skew-adjoint for the positive definite inner product $-B$, so its eigenvalues are purely imaginary. If it were diagonalizable over $\mathbb{R}$, all these eigenvalues would be zero and $\mathrm{ad}x=0$. The center is zero, so only $x=0$ has this property; no nonzero split Cartan subalgebra exists. Thus ${\mathfrak{sl}}_2(\mathbb{R})$ is split and $\mathfrak{su}(2)$ is not.

An orthogonal block design has treatment contrasts orthogonal to block contrasts after removing the grand mean. This is a statement about vectors on the experimental units, with the usual inner product; the two full spaces are not orthogonal because both contain the constant vector.

For a counting criterion, let $n_{ij}$ count experimental units receiving treatment $i$ in block $j$, let $r_i={\sum }_j{n}_{ij}$, let $k_j={\sum }_i{n}_{ij}$, and let $n={\sum }_i{r}_i$. The inner product of the centered indicators for treatment $i$ and block $j$ is $n_{ij}-r_i{k}_j/n$. ConsequentlyIn particular, equal-sized blocks must contain each treatment in the same proportion. Under an additive block-and-treatment model, adjustment for blocks then does not change the fitted treatment contrasts.

For one-column Gram matrices, unit Euclidean norm of the columns gives $A_S^{\ast }{A}_S=1$. The restricted isometry constant formula therefore gives ${\delta }_1=0$. For a two-column Gram matrix, put $c=⟨a_i,a_j⟩$. The difference from the identity matrix has the formup to the convention for the complex inner product. Its characteristic polynomial is ${\lambda }^2-|c{|}^2$, so the eigenvalues are $\pm |c|$ and its matrix 2-norm is $|c|$. Maximizing over pairs gives ${\delta }_2=\mu$, where $\mu$ is the mutual coherence.

Part (b), together with the minimality defining the restricted isometry constant, gives ${\delta }_s\le {\mu }_1(s-1)$. Every summand defining cumulative coherence is at most the mutual coherence, so ${\mu }_1(s-1)\le (s-1)\mu$. Thus, for normalized columns and $N\ge 2$,The upper range $s\le N$ matters because the printed definition of cumulative coherence stops at $N-1$. For $N=1$, only ${\delta }_1=0$ is needed; the maximum over pairs defining mutual coherence is otherwise empty.

The linear independence of the first $n$ elements of each orthonormal system shows that $T_n$ and $S_n$ have the same finite dimension $n$ and are closed subspaces of a Hilbert space. Apply part (b) with $U=T_n$ and $V=S_n^{\perp }$. Since $(S_n^{\perp }{)}^{\perp }=S_n$, the positive directed subspace angle cosine givesLet $Q$ be the oblique projection onto $T_n$ along $S_n^{\perp }$. Define ${\tilde{f}}_n=Qf$. Its residual $f-Qf$ is orthogonal to every ${\psi }_j$ with $1\le j\le n$, so the required measurements agree. Conversely, if $g\in T_n$ has the same measurements, then $f-g\in S_n^{\perp }$; uniqueness of the direct sum decomposition gives $g=Qf$. Thus this is finite-dimensional Hilbert sampling reconstruction.

There is also an explicit coefficient description. Use the inner product convention linear in its first entry and putThen $Cc=b$. The orthonormal systems show $\Vert P_{S_n}{\sum }_k{c}_k{\varphi }_k\Vert =\Vert Cc{\Vert }_2$ and $\Vert {\sum }_k{c}_k{\varphi }_k\Vert =\Vert c{\Vert }_2$, so the smallest singular value of $C$ is $\cos {\theta }_{T_n,S_n}>0$. Hence $c=C^{-1}b$, another direct proof of existence and uniqueness.

For $n\ge 1$, both summands of this direct sum are nonzero: the infinite orthonormal system $({\psi }_j)$ contains ${\psi }_{n+1}\in S_n^{\perp }$. The permitted oblique projection norm formula therefore applies without its degenerate exception, givingThe operator norm immediately yields the stability estimate $\Vert {\tilde{f}}_n\Vert \le \sec {\theta }_{T_n,S_n}\Vert f\Vert$. For the approximation bounds, put $e=f-P_{T_n}f$. Because $Q$ fixes $T_n$, we haveso the operator norm estimate gives $\Vert f-Qf\Vert \le \sec {\theta }_{T_n,S_n}\Vert e\Vert$. For the lower bound, $e\perp T_n$ while $P_{T_n}f-Qf\in T_n$. The Pythagorean identity givesThe unique measurement-matching reconstruction is stable and within the secant function factor of the best orthogonal projection approximation:If a zero-dimensional reconstruction is admitted, it is simply ${\tilde{f}}_0=0$ and its error equals the norm of $f$; that case is best stated directly instead of using the angle of a zero space.