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A Lie group is a group which is a finite-dimensional smooth Hausdorff second-countable manifold, with smooth multiplication and inversion. For the special unitary group, consider the real vector spaceThe diagonal entries of a skew-Hermitian matrix supply $n$ real parameters, and the entries above the diagonal supply $2\left(\genfrac{}{}{0ex}{}{n}{2}\right)$; the trace-zero condition removes one real parameter. Thus ${\dim }_{\mathbb{R}}\mathfrak{su}(n)=n^2-1$.

The matrix facts used to construct charts are these: the matrix exponential is analytic, its derivative at zero is the identity, and it has an analytic inverse matrix logarithm on a sufficiently small neighborhood of the identity. On those neighborhoods $\log (e^X)=X$, $e^{\log U}=U$, and $\det (e^X)=e^{\mathrm{tr}X}$. For a unitary matrix in that neighborhood, the spectral theorem writes $U=W\mathrm{diag}(e^{i{\theta }_j})W^{\ast }$ with principal angles close to zero, and $\log U=W\mathrm{diag}(i{\theta }_j)W^{\ast }$ is skew-Hermitian.

Choose a norm ball of radius $\epsilon <\pi /n$ small enough to lie in the exponential's inverse domain. If $U$ is unitary, $\det U=1$ and $\Vert \log U\Vert <\epsilon$ in operator norm, thenso its trace is zero. Conversely $X\in \mathfrak{su}(n)$ gives $(e^X{)}^{\ast }{e}^X=I$ and $\det (e^X)=1$. Thus the logarithm identifies an open neighborhood $V$ of $I$ in $SU(n)$, with its matrix subspace topology, with the open ball $B_{\epsilon }\subset \mathfrak{su}(n)$. Choosing a real linear identification $\mathfrak{su}(n)\cong {\mathbb{R}}^{n^2-1}$ turns this into a chart.

For every $A\in SU(n)$ take the translated chartTheir overlap maps are $X\mapsto \log (B^{-1}A{e}^X)$ wherever both charts are defined, so they are real analytic. They cover the group; the matrix topology is Hausdorff and second-countable. This constructs the real manifold directly, rather than merely counting equations. Matrix multiplication is polynomial in real/imaginary entries and inversion here is $U\mapsto U^{\ast }$; in the logarithm charts both are smooth. The logarithm charts for the special unitary group therefore proveFor $n=1$ it is the one-point group of dimension zero.

For $n=2$, let the first column be $(a,b)$, of unit norm. Every unit vector orthogonal to it is a scalar of modulus one times $(-\bar{b},\bar{a})$. The determinant of the resulting matrix is that scalar, so determinant one fixes it to $1$. Henceis bijective, with inverse given by taking the first column. The forward map is real polynomial in the four coordinates; composing with the local logarithm charts verifies smoothness into the constructed manifold. The inverse is a smooth matrix-coordinate projection and, in the usual embedded sphere charts, is smooth into $S^3$. Thus $\boxed{SU(2)\cong S^3\text{ by a diffeomorphism}}$, the explicit realization of SU(2) as the three-sphere.

Use logarithms to base two, so all entropies are in bits. For a discrete probability distribution $p$ and a density operator $\rho$ with eigenvalues ${\lambda }_j$, the Shannon entropy and Von Neumann entropy are respectivelyUse $0\log 0=0$. The second definition is basis-independent by the spectral theorem; for a density operator diagonal with probabilities $p_x$, it agrees with the first.

The classical relative entropy, or Kullback-Leibler divergence, of $p$ from $q$ isIt is $+\infty$ if some $p_x>0$ has $q_x=0$. The quantum relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and is $+\infty$ otherwise. Matrix logarithms are taken on the corresponding positive supports. In a common eigenbasis this reduces to classical relative entropy; both divergences are nonnegative and are generally asymmetric.

To prove Fano's inequality, let the finite alphabet size be $m\ge 2$ and let $E$ indicate an incorrect guess. Since $E$ is determined by $(X,Y)$, the chain rule for conditional entropy first givesExpanding in the opposite order givesConditioning reduces entropy, so $H(E\mid Y)\le H(E)=h_2(p_e)$, the binary entropy function. When $E=0$, knowledge of $Y$ determines $X=f(Y)$ and the conditional entropy is zero. When $E=1$, $X$ can lie in at most the $m-1$ alphabet values other than $f(Y)$. Entropy on $m-1$ values is at most ${\log }_2(m-1)$: its deficit from that number is the nonnegative relative entropy from the uniform distribution. Averaging the two cases therefore givesCombining the two expansions provesThis proof of Fano's inequality via an error indicator works for any alphabet-valued guessing rule, not just an optimal one. If $m=1$, there is no guessing uncertainty and the claim is simply $H(X\mid Y)=0$; that case is treated separately rather than writing $\log 0$.

By the spectral theorem, write $V=U\mathrm{diag}(v_1,\dots ,v_p)U^T$, where $U$ is orthogonal and each $v_j>0$. Choose the whitening transformationA linear image of a multivariate normal vector is multivariate normal, and $LV{L}^T=I$. Thus $Z_i=L(Y_i-\mu )$ are independent $N_p(0,I)$ random vectors. Their whole joint distribution is independent of $\mu ,V$.

Define $\bar{Z}=n^{-1}{\sum }_i{Z}_i$ and $S_Z=n^{-1}{\sum }_i(Z_i-\bar{Z})(Z_i-\bar{Z}{)}^T$. These satisfyThereforeThe right side is a function of independent standard multivariate normal observations only, so its distribution depends on $n,p$ but not on $\mu ,V$. This is the same affine invariance of Hotelling's statistic that permits multivariate mean tests with unknown covariance. The proof concerns the jointly transformed mean and covariance; whitening the mean alone would not establish the conclusion. Inverses exist almost surely under $n>p$. If $V$ is singular, no square matrix can whiten it to $I$, so nonsingularity of the population covariance is necessary as well.

A compatible almost complex structure is a smooth endomorphism $J:TX\to TX$ satisfying $J^2=-I$, preserving $\omega$, and making $g_J(u,v)=\omega (u,Jv)$ a positive-definite inner product. Preservation and $J^2=-I$ make this bilinear form symmetric.

To construct one, choose any Riemannian metric $h$ and define $A$ by $h(Au,v)=\omega (u,v)$. The endomorphism $A$ is skew-adjoint and invertible. Consequently $-A^2$ is positive definite and self-adjoint. Let $B=(-A^2{)}^{1/2}$ be its positive square root and setThe spectral theorem gives $B$ fibrewise, and the positive-square-root operation is smooth on positive-definite matrices, so this defines a smooth endomorphism. Since $A$ commutes with $B$, $J^2=-I$. Also $J^{\ast }=-J$, so $J^{\ast }J=I$. Commutation then givesThis is the metric construction of a compatible almost complex structure, proving nonemptiness.

If we start with $h=g_J$ for an already compatible $J$, then $h(Ju,v)=\omega (u,v)$, so $A=J$, $B=I$, and the construction recovers $J$. For two compatible structures, linearly interpolate their metrics,and apply the construction to $h_t$. This is a continuous smooth-in-$t$ path with endpoints $J_0,J_1$. Thus the space is nonempty and path connected. Interpolating instead to a single fixed auxiliary metric actually gives the contractibility of compatible almost complex structures.

Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

A spatial finite difference method and time discretization for a linear evolution problem give a recurrence $U^{n+1}=B_{h,k}{U}^n$. To obtain numerical stability on $0\le nk\le T$, one needs a bound $\Vert B_{h,k}^n\Vert \le C_T$ uniform in the permitted grid sizes and time steps. More generally a growth estimate $\Vert B_{h,k}^n\Vert \le C{e}^{cnk}$ is sufficient. The finite-time stability versus power boundedness distinction matters: growth per step of $1+O(k)$ can be compatible with the latter finite-time estimate, whereas unrestricted geometric growth at a fixed Courant number is not.

The eigenvalue method diagonalizes $B_{h,k}$ into modes. If $B_{h,k}$ is a normal matrix, the spectral theorem gives $\Vert B_{h,k}^n{\Vert }_2={\max }_j|{\mu }_j{|}^n$. Thus $|{\mu }_j|\le 1$ gives contraction, and $|{\mu }_j|\le e^{ck}$ gives the preceding growth estimate. A diagonalizable matrix $B=V\Lambda V^{-1}$ has $\Vert B^n{\Vert }_2\le \Vert V{\Vert }_2\Vert V^{-1}{\Vert }_2{\max }_j|{\mu }_j{|}^n$, so its condition number must remain bounded uniformly as the mesh changes. A stable-looking list of eigenvalues alone is insufficient.

Indeed,Both eigenvalues are one, but the Jordan block gives growth proportional to $n\sim T/k$, violating uniform finite-time numerical stability. A nonnormal Fourier amplification matrix can similarly exhibit large amplification even with its eigenvalues inside the unit disk. Eigenvector conditioning and the full powers of the amplification operator are essential.

For constant-coefficient schemes on a periodic or infinite grid, a Fourier transform diagonalizes scalar translation-invariant stencils. This yields von Neumann stability analysis using the Fourier amplification symbol $G(\theta )$. The discrete Parseval identity turns a bound on $G(\theta {)}^n$ into a discrete L2 norm estimate. For systems, the symbol is a matrix; a uniform power bound for matrix Fourier symbols is required, rather than a bound on its spectral radius alone.

For the heat equation, a central finite difference in space and the explicit Euler method in time giveThe scalar condition $|G|\le 1$ for every frequency is exactly $0\le r\le 1/2$. With zero Dirichlet boundary conditions, the spatial matrix is symmetric and diagonalized by sine modes, leading to the same uniform sufficient restriction. By contrast, the Backward Euler diffusion scheme has $G(\theta )=[1+4r{\sin }^2(\theta /2){]}^{-1}$ and is stable for every $r\ge 0$.

For the advection equation $u_t+a{u}_x=0$, $a>0$, centered space with the explicit Euler method gives $G=1-i\nu \sin \theta$, $\nu =ak/h$, and $|G{|}^2=1+{\nu }^2{\sin }^2\theta$. At fixed positive Courant number, repeated powers are unbounded as the mesh is refined. The upwind finite difference scheme instead haswhich gives contraction precisely for $0\le \nu \le 1$. This comparison shows why the spatial stencil and time method must be analyzed together.

Boundary treatment can destroy the normality or mode decomposition available on a periodic grid, so a finite-interval scheme needs its own uniform operator estimate. Multilevel time schemes likewise require the root condition for a multistep method and uniform bounds for the associated amplification matrices; simple scalar root tests do not establish these bounds for arbitrary systems. Once consistency of a numerical method, a well-posed linear evolution problem and the required numerical stability estimate hold, the Lax equivalence theorem gives numerical convergence. Eigenvalues provide a powerful route to that estimate when the accompanying eigenvector and boundary analysis is justified.

For a positive operator $A$, the spectral theorem writes $A={\sum }_j{a}_j|j⟩⟨j|$ with $a_j\ge 0$. Hence its unique positive square root of an operator is $\sqrt{A}={\sum }_j\sqrt{a_j}|j⟩⟨j|$. Conjugating this spectral decomposition shows $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$ for a unitary operator $U$.

Let $\Delta ={\rho }_1-{\rho }_2$ and ${\rho }_i^{\prime }=U{\rho }_i{U}^{†}$. Then ${\Delta }^{\prime }=U\Delta U^{†}$ and $({\Delta }^{\prime }{)}^{†}{\Delta }^{\prime }=U{\Delta }^{†}\Delta U^{†}$. The square-root covariance and the cyclic property of the trace giveFor the quantum fidelity, covariance first gives $({\rho }_1^{\prime }{)}^{1/2}=U{\rho }_1^{1/2}{U}^{†}$. ThereforeApply the same square-root identity and the cyclic property of the trace to obtainThus unitary invariance of trace distance and fidelity givesThe same $U$ acts on both quantum states: these quantities depend on the relative distinguishability of the quantum states, not on a choice of basis.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

Because $A$ is a real symmetric matrix, the spectral theorem gives $\Vert A{\Vert }_2=\rho (A)$. For $\beta \rho <1$, the Neumann series satisfiesLet $m={\sum }_i{X}_i(0)$ be the initial infected count. Since its entries are indicator random variables, $\Vert X(0){\Vert }_2=\sqrt{m}$, while $\Vert 1{\Vert }_2=\sqrt{n}$. The Cauchy-Schwarz inequality and the operator norm bound yieldThus a sufficient asymptotic condition for a small outbreak isFor example, it holds when $m=o(n)$ and $\beta \rho \le 1-\eta$ for a fixed $\eta >0$. For any fixed $\delta >0$, the Markov inequality then givesThis is the spectral condition for a small discrete SIR outbreak. A uniform gap and few initial infections are needed for this conclusion; the finite-population condition $\beta \rho <1$ alone does not control an asymptotically vanishing gap or an initially macroscopic infected set.

For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

Two finite-dimensional Hermitian matrices are related by unitary conjugation if and only if they have the same multiset of eigenvalues. Necessity follows because conjugation preserves the characteristic polynomial; sufficiency follows by mapping orthonormal eigenbases in the spectral theorem. Consequentlyand no member of the first pair is equivalent to a member of the second. For an explicit example, the Hadamard gate $H$ gives $H{\rho }_0{H}^{†}={\rho }_2$ and $H{\rho }_3{H}^{†}={\rho }_1$. Multiplying $H$ by a scalar phase does not change either conjugation.

Let $x_{\ast }=A^{-1}b$ and $e^{(k)}=x^{(k)}-x_{\ast }$. The Richardson iteration givesBy the spectral theorem, a real symmetric positive definite matrix has an orthonormal eigenbasis with positive eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$. The error in the $i$th eigenvector direction is multiplied by $(1-\tau {\lambda }_i{)}^k$. Thus convergence for every initial vector occurs exactly when $|1-\tau {\lambda }_i|<1$ for every $i$, or $0<\tau {\lambda }_i<2$. Since $\rho (A)={\lambda }_{\max }$,Necessity also follows from an initial error in an offending eigenvector direction. At $\tau =0$ it remains unchanged; at $\tau =2/\rho (A)$ the largest-eigenvalue error alternates sign; outside the interval some error does not decay. Hence neither endpoint can be included.

A unitary family with the required Weyl relations isTranslations preserve the $L^2$ norm and the multiplier has modulus one. Multiplication of two such operators leaves the phase $x_1{y}_2-y_1{x}_2$ after extracting $W(x_1+x_2,y_1+y_2)$, proving the relation. Translation and modulation are strongly continuous, first on compactly supported smooth functions by direct estimates, then on all of $L^2$ by density and unitarity.

If a bounded operator $A$ commutes with all $W$, it commutes with translations and every modulation $e^{2iyt}$. The spectral theorem for the multiplication coordinate therefore makes it commute with all bounded multiplication operators and interval projections. One can see its form directly: on a finite interval $I$, set $h_I=A{1}_I$. For bounded $\varphi$ supported in $I$, commutation gives $A\varphi =\varphi h_I$. Testing characteristic functions shows $|h_I|\le \Vert A\Vert$ almost everywhere. On increasing intervals these functions agree on overlaps and yield a single $h\in L^{\infty }$ with $A=M_h$.

Commutation with translations says $h(t+x)=h(t)$ almost everywhere for every $x$. Convolving with a smooth approximate identity produces smooth functions invariant under every translation, hence constants. Passing to the local $L^1$ limit makes $h$ constant almost everywhere. Therefore $A$ is a scalar operator, and the unitary family acts irreducibly.

Principal component analysis replaces correlated measurements by uncorrelated linear compounds ordered by decreasing variance. Keeping the leading compounds gives a low-dimensional summary that preserves as much variance as possible under orthonormal projection. It is an unsupervised description of variation, rather than a guarantee that the high-variance directions are best for classification or prediction.

Write $X_c=X-\mu$. For a unit loading vector $q$, the variance of $q^T{X}_c$ is $q^T\Sigma q$. The spectral theorem gives an orthonormal eigenbasis $q_1,\dots ,q_p$ with eigenvalues ${\lambda }_1\ge \cdots \ge {\lambda }_p\ge 0$. If $q={\sum }_j{b}_j{q}_j$ and ${\sum }_j{b}_j^2=1$, thenEquality is attained at $q_1$. Requiring the next loading to be orthogonal to the first excludes its eigendirection, so the same argument gives $q_2$ and variance ${\lambda }_2$, and continues inductively. Repeated eigenvalues permit any orthonormal basis of the tied eigenspace. These are the population principal components.

With $Q=(q_1,\dots ,q_p)$, set $Y=Q^T{X}_c$. ThenIn particular the components are uncorrelated. They need not be independent unless, for example, $X$ is multivariate normal. FinallyThe trace equality follows from $Q^{T}Q=I$ and cyclicity of trace, so the rotation preserves total variance. The explained variance of a principal component is its eigenvalue divided by this total.

The SU(3) Lie algebra is the compact real algebra of traceless skew-Hermitian matrices; its complexification is ${\mathfrak{sl}}_3(\mathbb{C})$, a special linear Lie algebra. In an antihermitian Lie algebra representation, the two commuting Cartan operators $H_1,H_2$ are Hermitian matrices, and the root operators satisfy $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA weight diagram plots the pairs $\mu$, recording the weight multiplicity $\dim V_{\mu }$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the roots of a root systemThe raising operators $E_{+}^a$ move a nonzero weight vector by $+{\alpha }_a$ and the lowering operators move it by $-{\alpha }_a$. These six root directions form the A2 root system. Each root and its negative generate an SU(2) representation ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt{3}{H}_2/2$ constrain the two simple-root strings.

Choose a linear functional positive on the three positive roots. A maximal weight for this ordering is killed by every raising operator. In an Irreducible Lie algebra representation it generates the entire representation by lowering, and its weight space has dimension one. The two SU(2) representation strings show that its Dynkin labels are nonnegative integers:Conversely, each such pair specifies exactly one finite-dimensional irreducible highest-weight representation. Write it as $R(p,q)$. The fundamental weights in these Cartesian coordinates are ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe dimension follows from the three positive-root factors in the Weyl dimension formula: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every weight negated. The SU(3) highest-weight coordinates here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.

The weight diagram is invariant under the Weyl group, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the highest weight. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda +$ root lattice. A point in a different weight lattice coset is not a weight merely because it lies inside the polygon.

The shell multiplicities in an SU(3) weight diagram can be stated geometrically. The outer boundary has weight multiplicity one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual weight multiplicity by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min (p,q)+1$; when $p=q$ it is just the central point. A triangular weight diagram has multiplicity one everywhere. Thus $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, place a copy of the second weight diagram at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining dominant weight, identify its irreducible highest-weight representation, subtract that representation's whole diagram with its weight multiplicities, and repeat. This highest-weight character subtraction must exhaust every weight, not merely the corners. Unlike the Clebsch-Gordan decomposition for SU2, an SU(3) product may contain repeated copies of an irreducible representation.

The two three-dimensional Fundamental representations of sl3 are $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, the largest weight is $2{\omega }_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $3\otimes \overline{3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the octet and singlet in a fundamental SU3 tensor product and the two-quark product areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation:The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.