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Use $D_j=-i{\partial }_{x_j}$, so the Fourier transform of $D^{\alpha }u$ is ${\lambda }^{\alpha }\hat{u}$. Distinguish the full degree-$N$ polynomial $P(\lambda )$ from its homogeneous top-degree part $p_N(\lambda )$, the principal symbol. The elliptic differential operator condition isIt does not require the full polynomial to be nonzero at small frequencies. Compactness of the unit sphere gives $c_0={\min }_{|\theta |=1}|p_N(\theta )|>0$. Homogeneity and the lower-degree remainder implyFor sufficiently large $|\lambda |$ the second term is at most half the first, so the high-frequency lower bound for an elliptic polynomial isHere $⟨\lambda ⟩$ is the Japanese bracket. The order-zero case just means a nonzero constant and is immediate.

For real $s$, the Sobolev space definition with the current Fourier normalization isChanging the harmless constant in the norm gives the same space. A distribution on $X$ belongs to the Local Sobolev space $H_{\mathrm{loc}}^s(X)$ when $\chi u$, extended by zero outside $X$, belongs to $H^s({\mathbb{R}}^n)$ for every test function $\chi \in C_c^{\infty }(X)$.

A compactly supported distribution of finite order of a distribution $m$ has a smooth Fourier transform satisfying $|\hat{u}(\lambda )|\le C(1+|\lambda |{)}^m$, by applying the finite-order estimate to a fixed cutoff times the exponential. Thus its weighted squared transform is bounded by $C⟨\lambda {⟩}^{2(s+m)}$. This is integrable precisely in the sufficient range $2(s+m)<-n$, and provesThis is the negative Sobolev regularity of a compactly supported distribution; the strict inequality is important, and is not a claim that this sufficient threshold is optimal for each distribution.

To establish elliptic regularity without assuming the answer as an a priori smoothness hypothesis, first obtain a global constant-coefficient gain. For a compactly supported distribution $w$ with $P(D)w\in H^q$, the polynomial lower bound at high frequency givesOn $|\lambda |\le R$, the transform of $w$ is smooth and bounded. Therefore $P(D)w\in H^q$ implies $w\in H^{q+N}$. Possible low-frequency zeros of $P$ are harmless; we never divide by them.

Two elementary mapping facts supply the variable-coefficient argument. Distributional derivatives of order $j$ map $H^r$ into $H^{r-j}$. Sobolev multiplication by a smooth cutoff is bounded on $H^r$ for every real $r$, including negative ones. Indeed, for a compactly supported smooth $a$, the Peetre weight inequalityand $\widehat{aw}=(2\pi {)}^{-n}\hat{a}\ast \hat{w}$ reduce the bound to Young's convolution inequality with the integrable kernel $⟨\zeta {⟩}^{|r|}|\hat{a}(\zeta )|$. Smooth coefficients only need this property on compact subsets, where they can be multiplied by another cutoff.

Write the lower-order part as $B={\sum }_{|\alpha |<N}{f}_{\alpha }{D}^{\alpha }$, with $N\ge 1$, and suppose provisionally that $u\in H_{\mathrm{loc}}^r$ on a relatively compact neighborhood. For a cutoff function $\chi$ supported there,The bracket is the operator commutator. Its order is at most $N-1$: in the product rule, every surviving term has at least one derivative falling on $\chi$. Choose a second cutoff function equal to one near $\mathrm{supp}\chi$ when estimating the products. The derivative and smooth-multiplication bounds then giveThe global gain just proved applies to the compactly supported distribution $\chi u$, and yieldsThis is the cutoff bootstrap for local elliptic regularity; it works for real indices, not just nonnegative integers.

There is always a legitimate starting index. Around any fixed point choose ${\chi }_0=1$ on a smaller neighborhood. The compactly supported distribution ${\chi }_{0}u$ has the negative Sobolev regularity proved above, so $u\in H_{\mathrm{loc}}^{r_0}$ on that neighborhood for some finite $r_0$. The index need not be uniform over all of $X$. Repeating the one-step gain finitely many times reaches $s+N$, or the target is already reached if $r_0\ge s+N$. Since the point was arbitrary,For $N=0$ the result follows directly by dividing by the nonzero constant.

Finally, if $Lu=0$, its forcing belongs to every Local Sobolev space. The gain consequently gives every local Sobolev order for $u$. For each integer $j\ge 0$, choose an order larger than $j+n/2$ and apply the Sobolev embedding theorem to a localized solution. It has a $C^j$ representative; the representatives agree because they represent the same distribution. Thus every distributional solution of $Lu=0$ is smooth on $X$.

One version of the Klainerman-Sobolev inequality, for a sufficiently decaying smooth function $\psi$ on ${\mathbb{R}}^{1+3}$, isHere $I$ denotes a word of length $|I|$ in the following eleven commutation vector fields for the wave equation:The first four are spacetime translation vector fields; the next three are spatial rotation vector fields; the next three are Lorentz boost vector fields; and $S$ is the scaling vector field. The vector field method for wave equations uses these vector fields because their commutators with the d'Alembert operator obeyIn particular the Klainerman-Sobolev inequality impliesThe displayed L2 norms are spatial norms at fixed time; the spacetime vector fields can contain time derivatives. No wave equation assumption is needed for the Klainerman-Sobolev inequality itself.

We establish a quantitative almost global existence for wave equations estimate. Take $0<\epsilon \le 1$, fix an integer $m\ge 6$, and use the commutation vector fields for the wave equation from the preceding part. Put $\partial =({\partial }_t,\nabla )$ and define the commuted wave energyAll these L2 norms are finite on any smooth existence interval by finite propagation speed. At $t=0$, the polynomial coefficients of the vector fields are bounded on the fixed compact support of the Cauchy data. Whenever a higher time derivative occurs, use ${\varphi }_{tt}=\Delta \varphi -{\varphi }_t^2$ and its differentiated versions to express it in terms of initial spatial derivatives. Every term contains at least one factor of $\epsilon$; consequentlyfor a constant depending only on finitely many derivatives and the support radius of ${\varphi }_0,{\varphi }_1$.

The commutators $[Z,{\partial }_{\mu }]$ are constant linear combinations of translations. Together with $[\square ,S]=2\square$ and the Leibniz rule, this shows that each commuted source is a finite linear combination of productsThis statement includes the extra copies of the original source produced by the scaling vector field. In each product put the factor with fewer commutations in the Lp norm $L^{\infty }$ and the other in the L2 norm. The lower order is at most $\lfloor m/2\rfloor$. Applying the Klainerman-Sobolev inequality to ${\partial }_{\mu }{Z}^J\varphi$ costs at most two additional commutations; commuting those past ${\partial }_{\mu }$ introduces only lower-order translations. Since $\lfloor m/2\rfloor +2\le m$,The inhomogeneous wave energy estimate now givesLet $T_{\epsilon }={\epsilon }^{-N}$. Use a bootstrap argument with $A_m(t)\le 2C_0\epsilon$ up to the smaller of $T_{\epsilon }$ and the maximal existence time. The energy estimate improves this toFor each fixed $N$,Choose ${\epsilon }_N\le 1$ so that $4C_m{C}_0\epsilon \log (1+{\epsilon }^{-N})\le 1/2$ for every $0<\epsilon \le {\epsilon }_N$. Then $A_m(t)\le (3/2)C_0\epsilon$, a strict improvement. A continuity argument closes the bootstrap argument.

Finally, the translation terms in $A_m$ control ordinary spatial Sobolev norms of $\partial \varphi$. The missing L2 norm of $\varphi$ satisfiesThus the full local-existence Sobolev norms remain bounded on this finite interval. The smooth continuation criterion for semilinear wave equations extends the solution past any finite endpoint before $T_{\epsilon }$. To see smooth persistence explicitly, the tame Sobolev product estimate gives $\Vert {\varphi }_t^2{\Vert }_{H^k}\le C_k\Vert {\varphi }_t{\Vert }_{\infty }\Vert {\varphi }_t{\Vert }_{H^k}$. Ordinary differentiated wave energy estimates therefore bound each higher derivative energy by its initial value times $\exp (C_k{\int }_0^t\Vert {\varphi }_t(s){\Vert }_{\infty }ds)$. This is finite on the interval already controlled by the base commuted wave energy; no separate ${\epsilon }_N$ is needed for each derivative order. ThereforeFor $\epsilon =0$ the zero solution is global. The same energy estimate in fact permits an exponential lower bound for the lifespan, which is stronger than any fixed inverse power.

For $a\in R$, write $m_a$ for multiplication by $a$ in ${\mathrm{End}}_k(R)$. The algebraic differential operator ring is defined by the filtrationThe order of an algebraic differential operator is the least $j$ for which $\theta \in D_k^j(R)$; the zero operator may be assigned order $-\infty$. In degree zero, commuting with all multiplication maps means $\theta (a)=a\theta (1)$, so $D_k^0(R)=\{m_a:a\in R\}\cong R$. The commutator identity $[UV,m_a]=U[V,m_a]+[U,m_a]V$ proves by induction that $D_k^r{D}_k^s\subseteq D_k^{r+s}$, so the union is a ring under addition and composition.

The derivations of an algebra formEvery such derivation has $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, so has differential order at most one. Conversely, if $\theta \in D_k^1(R)$, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]$ is a multiplication map. Evaluating at $1$ shows it is $m_{\delta (a)}$; evaluating at $b$ gives the Leibniz rule. Hence the first-order algebraic differential operator decomposition isIt is a direct sum as $k$-vector spaces: a multiplication map that is a derivation vanishes at $1$, and is zero.

The complex Weyl algebra has generators $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ and relationsTake $n\ge 1$. If a nonzero unital module $M$ had finite dimension of a vector space over $\mathbb{C}$ $m$, its representing matrices would satisfy $[{\partial }_1,x_1]=I_m$. Taking the matrix trace gives $0=m$, impossible in characteristic zero. The Weyl algebra has no nonzero finite-dimensional modules. For $n=0$ the algebra is $\mathbb{C}$ and this assertion has the obvious exception.

For a nonzero finitely generated Weyl algebra module, use the Bernstein filtrationThe ordered monomials form a vector space basis, soTo justify the ordered basis, the relations move all ${\partial }_i$ to the right and give spanning. Independence follows in the action on $\mathbb{C}[x_1,\dots ,x_n]$: apply a putative zero operator ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x){\partial }^{\mathbf{b}}$ coefficientwise to the formal exponential $e^{t\cdot x}$, obtaining ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x)t^{\mathbf{b}}=0$. Thus every coefficient vanishes.

Choose a finite-dimensional nonzero generating subspace $M_0\subseteq M$ and set $M_j=F_j{A}_n{M}_0$. The associated graded module is finitely generated over the polynomial associated graded ring, so the Hilbert-Serre theorem implies ${\dim }_{\mathbb{C}}{M}_j$ is eventually a polynomial. DefineThis Bernstein growth dimension of a Weyl algebra module equals the Gelfand–Kirillov dimension of a module, not its ordinary vector-space dimension. If a different finite generating subspace is used, each lies in some fixed filtration step of the other. The corresponding $M_j$ are sandwiched between shifted filtrations, so the degree is unchanged.

We prove Bernstein inequality for Weyl algebra modules using an injectivity estimate. The faithful finite-step action of a Weyl algebra claim is that the mapis injective for every $j$. Induct on $j$. For $j=0$, a scalar annihilating $M_0\ne 0$ is zero. Suppose $a\in F_j{A}_n$ annihilates $M_j$. For any generator $z=x_i$ or ${\partial }_i$, the commutator $[a,z]$ belongs to $F_{j-1}{A}_n$ and annihilates $M_{j-1}$: both $a(zv)$ and $z(av)$ vanish. The induction hypothesis gives $[a,z]=0$ for every generator. In the ordered basis,In characteristic zero these identities force $a$ to be a scalar. Since it annihilates $M_0$, it is zero. This proves injectivity.

ConsequentlyThe left side grows as a positive constant times $j^{2n}$; the right side has growth degree $2d(M)$. Comparing powers provesThe polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with the usual multiplication and differentiation actions, has $d(M)=n$, showing the bound is sharp.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.

Introduce a parameter $t$ to keep track of total degree. The power series for the matrix exponential givesSet $S=X+Y$ and $C=\frac{1}{2}[X,Y]$, where the commutator is $[X,Y]=XY-YX$. Expanding a second matrix exponential givesSincethe two expressions agree through total degree two. More explicitly, applying the local matrix logarithm to the first expansion gives $\log (e^{tX}{e}^{tY})=t(X+Y)+\frac{1}{2}{t}^2[X,Y]+O(t^3)$. This proves the displayed order of the Baker--Campbell--Hausdorff formula. The Lie algebra is closed under the commutator, so its displayed exponent lies in the same Lie algebra. The expansion is a formal power series identity, or a convergent identity sufficiently near the identity matrix; this argument does not assert unrestricted global convergence.

The next two terms of the Baker--Campbell--Hausdorff formula are the two cubic commutators:Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\frac{1}{24}[Y,[X,[X,Y]]]$.

For the Pauli matrices, use the Pauli matrix multiplication lawPut $N=\mathbf{n}\cdot \mathbf{\tau }$. The antisymmetric term disappears when contracted with $n_a{n}_b$, so $N^2=|\mathbf{n}{|}^{2}I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the matrix exponential into even and odd terms of its power series yieldsThe Pauli matrices are Hermitian matrices, and the components of $\mathbf{n}$ are real, so $N^{†}=N$. For real $\alpha$, the resulting matrix $U$ satisfiesAlso $\mathrm{tr}N=0$ and $N^2=I$, so its two eigenvalues are $1,-1$. The eigenvalues of $U$ are therefore $e^{i\alpha },e^{-i\alpha }$, whose product is one. Thus $U$ is a unitary matrix with determinant one: it belongs to the SU(2) group.

Multiplying the two exact Pauli matrix exponentials, and using ${\tau }_1{\tau }_2=i{\tau }_3$, givesIn particular, the ${\tau }_3$ term has a minus sign. Its quadratic Taylor expansion isTaking $X=i\alpha {\tau }_1$, $Y=i\beta {\tau }_2$, the Pauli matrix commutator identity gives $[X,Y]=-2i\alpha \beta {\tau }_3$. The quadratic exponent in the Baker--Campbell--Hausdorff formula is therefore $Z=i\alpha {\tau }_1+i\beta {\tau }_2-i\alpha \beta {\tau }_3$. In $e^Z=I+Z+\frac{1}{2}{Z}^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator ${\tau }_1{\tau }_2+{\tau }_2{\tau }_1=0$ gives $Z^2=-({\alpha }^2+{\beta }^2)I+O(3)$, reproducing the exact product's quadratic expansion.

Solving the massive subsidiary conditions. In light-cone coordinates, use $d{s}^2=-2d{X}^{+}d{X}^{-}+{\delta }_{IJ}d{X}^{I}d{X}^J$ and write $d=D-2$. The divergence condition readsThus, when ${\partial }_{-}$ is invertible,First apply this with $n=-$, then with $n=I$, and finally with $n=+$; symmetry supplies the mixed components already determined. The trace condition becomesThe independent components are $h_{--}$, $h_{-I}$ and the trace-free part of $h_{IJ}$. Under transverse rotations they form a scalar representation, a vector representation, and a symmetric traceless rank-two tensor. ThereforeThis is the light-cone decomposition of a massive spin-two field for $D\ge 3$. In $D=2$, the trace and divergence give $h_{+-}=h_{++}=0$ and ${\partial }_{+}{h}_{--}=0$. The massive wave equation then forces $h_{--}=0$, so there are no polarizations, consistent with the zero value of the printed count. The massive particle little group is $SO(D-1)=SO(d+1)$, and its symmetric traceless square branches asThe mixed components with the extra direction give the vector; one independent trace combination gives the scalar. These are exactly the polarizations of a massive spin-two field. The remaining independent components retain the massive Klein-Gordon equation.

Transverse bosonic modes and mass levels. The variables ${\alpha }_k^I$ are Fourier amplitudes of the physical transverse open-string mode expansion. Classically, reality requires ${\alpha }_{-k}^I=({\alpha }_k^I{)}^{\ast }$. The symplectic term in the action fixes their quantum commutators:With string tension convention ${\alpha }^{\prime }=(2\pi T{)}^{-1}$, the zero mode of the constraint givesThe longitudinal nonzero modes have already been removed in light-cone gauge in string theory. Quantum normal ordering introduces the string intercept $a$, giving the open bosonic string mass spectrumEach bosonic occupation number is a nonnegative integer, so $N$ is a nonnegative integer weighted by oscillator frequency.

Suppressing the common momentum label, the lowest light-cone levels of an open bosonic string areThe oscillator vacuum is annihilated by every positive ${\alpha }_k^I$. At level one there are only $d$ vector polarizations. For a Lorentz-consistent vector, these are the transverse polarizations of a massless particle, transforming under the rotation part of its massless particle little group. A massive vector would need $D-1$ polarizations, including a scalar under $SO(d)$ that is absent here. Thus the first bosonic vector level must be massless, fixing $a=1$.

At level two the commuting creation operators give a symmetric square. Its scalar trace and symmetric traceless square, together with the mode-two vector, are the massive-spin-two decomposition above. In the consistent bosonic theory they form one massive spin-two field with $m^2=1/{\alpha }^{\prime }$. The covariant equations describe its propagation while eliminating the redundant components. For $D=26$ the transverse counts are $300+24=324$, the symmetric traceless rank-two tensor dimension of $SO(25)$.

Half-integer fermionic modes. The Neveu–Schwarz sector has antiperiodic worldsheet Majorana fermions. Its Neveu–Schwarz fermionic oscillators obeyThe oscillator vacuum satisfies ${\alpha }_k^I|0;p⟩=0$ for $k>0$ and $b_r^I|0;p⟩=0$ for $r>0$. A negative fermion mode is a fermionic creation operator for a transverse worldsheet excitation. The Neveu–Schwarz level operator and mass condition areThe smallest positive frequency is $1/2$, so the only first-excited states areThe same vector-polarization argument requires them to be massless in Lorentz-consistent quantization, fixing $a_{\mathrm{NS}}=1/2$.

At $N=1$, ${\alpha }_{-1}^I|0;p⟩$ is a vector, while $b_{-1/2}^I{b}_{-1/2}^J|0;p⟩$ is the exterior square: interchanging the indices changes the sign and equal indices give zero. Together they branch from an antisymmetric tensor:Thus the Neveu–Schwarz level-one massive tensor has $(D-1)(D-2)/2$ polarizations and mass squared $1/(2{\alpha }^{\prime })$. It differs from spin two because the two-fermion tensor is antisymmetric and has neither the symmetric trace-free representation nor its scalar trace. At $D=10$, the count is $8+28=36$, compared with $44$ for a massive spin-two field. The specified states are before the GSO projection; the usual tachyon-removing GSO projection also removes this integer level.

Write $O=\int d[\bar{\psi },\psi ]e^{-\sum \bar{\psi }\psi }|\psi ⟩⟨\psi |$. For bosons the measure is ${\prod }_{n}d\mathrm{Re}{\psi }_{n}d\mathrm{Im}{\psi }_n/\pi$ over $\mathbb{C}$ in each mode, with ${\bar{\psi }}_n={\psi }_n^{\ast }$. For fermions it is an ordered Berezin integral over independent ${\bar{\psi }}_n,{\psi }_n$; choose $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$, so $\int d\bar{\psi }d\psi e^{-\bar{\psi }\psi }=1$.

For one bosonic mode, $a|\psi ⟩=\psi |\psi ⟩$ and $⟨\psi |a={\partial }_{\bar{\psi }}⟨\psi |$. Integration by parts in the Gaussian measure gives $aO=Oa$; the conjugate argument gives $a^{†}O=O{a}^{†}$. Boundary terms vanish because of the Gaussian weight.

For one fermionic mode, put $n=a^{†}a$ and move Grassmann coefficients to the left. The weighted projector isIts ordinary commutators are $[a,W]=-a+\psi I$ and $[a^{†},W]=a^{†}-\bar{\psi }I$. Their Berezin integrals vanish, so again $[a,O]=[a^{†},O]=0$. Equivalently the sole surviving coefficient in $\int W$ is $-\bar{\psi }\psi I$, giving $I$ directly.

The modes factorize. In the irreducible Fock space representation, commuting with every creation and annihilation operator makes $O$ a scalar multiple of the identity. Its vacuum matrix element is the normalized Gaussian integral, equal to one. Thus the coherent-state resolution of identity isFor infinitely many modes, this argument first uses a finite-mode regulator.

Every commutator $[A,B]=AB-BA$ has zero trace, by the cyclic identity. Hence its linear span $W$ is contained in $\ker \mathrm{Tr}$. For the reverse inclusion, the matrix units satisfyThe off-diagonal matrix units and the differences $E_{ii}-E_{nn}$ for $i<n$ span all traceless matrices: a traceless diagonal part is ${\sum }_{i<n}{a}_{ii}(E_{ii}-E_{nn})$. Thus every element of the kernel is a linear combination of commutators, provingThis establishes that traceless matrices are spanned by commutators. For $n=1$, both vector subspaces are zero. Notice that the argument needs a span of commutators; it does not assume that every such sum has already been written as a single commutator.

Retain the order of the noncommuting position operator and momentum operator. Using their commutator $[\hat{x},\hat{p}]=i\hslash$, expansion givesThus the two factorizations of the quantum harmonic oscillator are

Use the canonical commutation relations $[{\hat{x}}_i,{\hat{x}}_j]=[{\hat{p}}_i,{\hat{p}}_j]=0$ and $[{\hat{x}}_i,{\hat{p}}_j]=i\hslash {\delta }_{ij}I$. For compactness write products without hats in the calculation, on a common domain such as smooth compactly supported wavefunctions. Expanding the commutator givesThe middle two commutators vanish because all the relevant indices differ. The first is $-i\hslash x_2{p}_1$, since $[p_3,x_3]=-i\hslash I$; the last is $i\hslash x_1{p}_2$, since $[x_3,p_3]=i\hslash I$. ThusThis is one of the orbital angular momentum commutation relations.

The product rule for a commutator gives $[A,B^2]=[A,B]B+B[A,B]$. Using the permitted cyclic angular momentum commutation relations,The first two expressions cancel, even though the factors need not commute. HenceThe squared angular momentum operator commutes with each of its components by the same cyclic calculation.