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Identify ${\mathbb{R}}^4$ with the quaternions, and use the identification of unit quaternions with $SU(2)$ proved in the root Solution. The actionpreserves the Euclidean norm by multiplicativity of quaternion norms. It is a homomorphism $SU(2)\times SU(2)\to SO(4)$: its domain is connected, so the orthogonal image has determinant $+1$.

An element in the kernel fixes $v=1$, giving $q_L=q_R=q$. Fixing every other $v$ means $q$ commutes with all quaternions, so $q$ is real; since it has unit norm, $q=\pm 1$. Hence the kernel is the diagonal subgroup $\{(1,1),(-1,-1)\}$.

The differential acts as $v\mapsto av-vb$ for imaginary quaternions $a,b$. If it vanishes, $v=1$ gives $a=b$, and commutation with all $v$ makes $a$ real and imaginary, hence zero. The differential is injective, and both Lie algebras have dimension six. The image is therefore an open subgroup of connected $SO(4)$, hence all of $SO(4)$. Thuswhere ${\mathbb{Z}}_2$ acts diagonally, not separately on the two factors.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

For $V={\mathbb{R}}^2$ with basis $e_1,e_2$, the step-two Lie algebra has basis $e_1,e_2,[e_1,e_2]$ and its last element is central. WriteIts first level is $v=(x,y)$ and its second level isThe step-two multiplication givesConsequently the mapis an isomorphism onto the real Heisenberg group. The shift by $xy/2$ converts exponential coordinates into the matrix coordinates used for that group. In exponential coordinates the left-invariant horizontal vector fields areTaking $X,Y$ orthonormal gives exactly the Carnot-Carathéodory distance inherited from $G^2({\mathbb{R}}^2)$.

In the truncated tensor algebra, let ${\mathfrak{g}}^N(V)$ be the linear span of all iterated Lie brackets of elements of $V$ of length at most $N$, with bracket $[a,b]=a\otimes b-b\otimes a$. Brackets of length greater than $N$ vanish. The free step-N nilpotent Lie group isThe exponential and logarithm are the finite seriesIts multiplication is the tensor multiplication above, its identity is $e=(1,0,\dots ,0)$, and its inverse isThe truncated Baker--Campbell--Hausdorff formula shows that this multiplication stays in the exponential of the Lie algebra. This is the connected simply connected Lie group of the free step-$N$ nilpotent Lie algebra, rather than a discrete free nilpotent group.

Give the first layer $V$ its usual Euclidean norm. An absolutely continuous curve $\gamma$ in $G^N(V)$ is horizontal iffor an absolutely continuous control $x$ in $V$. In the truncated tensor algebra this equation is ${\dot{\gamma }}_t={\gamma }_t\otimes {\dot{x}}_t$, truncated at degree $N$. Its horizontal length is $\int |{\dot{x}}_t|dt$. The Carnot-Carathéodory distance isThe first-layer Lie brackets span the full Lie algebra, so horizontal curves reach every endpoint. Thus this infimum is finite; it is a left-invariant distance inducing the usual manifold topology. Dilations ${\delta }_{\lambda }(a{)}^{(k)}={\lambda }^k{a}^{(k)}$ give $d_{CC}({\delta }_{\lambda }g,{\delta }_{\lambda }h)=\lambda d_{CC}(g,h)$ for $\lambda >0$. Equivalently $d_{CC}(e,g)$ is the least length of a bounded variation control whose step-$N$ path signature is $g$.

For $p\ge 1$ set $N=\lfloor p\rfloor$. A weak geometric p-rough path is a continuous path $\mathbf{X}:[0,T]\to G^N(V)$, usually based at $e$, withIts increments ${\mathbf{X}}_{s,t}={\mathbf{X}}_s^{-1}\otimes {\mathbf{X}}_t$ obey ${\mathbf{X}}_{s,u}={\mathbf{X}}_{s,t}\otimes {\mathbf{X}}_{t,u}$. Equivalently their degree-$k$ tensor components have finite $p/k$-variation and satisfy the path signature algebraic constraints. This is a group-valued definition; being a limit of smooth path signatures in the same $p$-variation topology is the stronger definition of a geometric p-rough path.

For a finite-dimensional Lie algebra, the Killing form is defined using the Adjoint representation on the algebra itself, rather than the given Lie algebra representation $d$:The Jacobi identity gives ${\mathrm{ad}}_{[X,Y]}=[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]$. Using the cyclic property of the trace,The same trace identity makes the Killing form symmetric. For a basis $(t_a)$, the lowered structure constants of a Lie algebra are $c_{abc}=\kappa ([t_a,t_b],t_c)$. Antisymmetry of the Lie bracket gives $c_{abc}=-c_{bac}$, while invariance gives $c_{abc}=\kappa (t_a,[t_b,t_c])=-c_{acb}$. The two adjacent transpositions generate every permutation, proving the antisymmetry of Killing-lowered structure constants: $c_{abc}$ is totally antisymmetric, even if the Killing form is degenerate.

A semisimple Lie algebra has no nonzero solvable ideal; equivalently, its solvable radical is zero, or, in characteristic zero, its Killing form is nondegenerate. A real compact Lie algebra is the algebra of some compact Lie group, equivalently an algebra admitting a positive-definite invariant bilinear form on a Lie algebra. For a real semisimple Lie algebra the compactness criterion from the Killing form says precisely that $\kappa$ is negative-definite. A compact algebra with a nonzero abelian center has a degenerate Killing form, so the semisimplicity qualification matters.

For a semisimple Lie algebra, let ${\kappa }^{ab}$ denote the inverse matrix of ${\kappa }_{ab}=\kappa (t_a,t_b)$. Its Casimir operator in $d$ isThis contraction is independent of the chosen basis. To prove centrality directly, put $[X,t_b]=A^a{}_b{t}_a$. Invariance gives $A^{\mathsf{T}}\kappa +\kappa A=0$, hence $A{\kappa }^{-1}+{\kappa }^{-1}{A}^{\mathsf{T}}=0$. The Lie algebra representation property and the product rule for a commutator now giveIndeed, the coefficient of $d(t_c)d(t_b)$ in this expression is $(A{\kappa }^{-1}+{\kappa }^{-1}{A}^{\mathsf{T}}{)}^{cb}$. Thus $[d(X),C]=0$ for every $X$. No irreducibility of $d$ is required.

The matrix units obey $E_{ij}{E}_{pq}={\delta }_{jp}{E}_{iq}$. Consequently the general linear Lie algebra $\mathcal{M}={\mathfrak{gl}}_n(\mathbb{R})$ hasFor its Killing form, write $L_X(Z)=XZ$ and $R_X(Z)=ZX$, so that ${\mathrm{ad}}_X=L_X-R_X$. Left multiplication acts identically on each of the $n$ columns, and right multiplication identically on each of the $n$ rows. Thus $\mathrm{Tr}(L_X{L}_Y)=\mathrm{Tr}(R_X{R}_Y)=n\mathrm{tr}(XY)$. The diagonal coefficient of $L_X{R}_Y$ on $E_{ij}$ is $X_{ii}{Y}_{jj}$, giving $\mathrm{Tr}(L_X{R}_Y)=\mathrm{tr}X\mathrm{tr}Y$; the other mixed trace is identical. This proves the Killing form of the general linear Lie algebra:The scalar matrices form a nonzero central abelian ideal, so $\mathcal{M}$ is not semisimple for any $n\ge 1$. More explicitly, the radical of the Killing form is $\mathbb{R}I$, since vanishing against every $Y$ forces $nX-(\mathrm{tr}X)I=0$ by nondegeneracy of the trace pairing. For $n\ge 2$, take $H=E_{11}-E_{22}$. Then $\kappa (H,H)=4n>0$, whereas a positive-definite invariant inner product would make every adjoint map skew and hence give $\kappa (X,X)\le 0$. Therefore $\mathcal{M}$ is not compact when $n\ge 2$. For the allowed exceptional value $n=1$, the abstract abelian algebra $\mathbb{R}$ is a compact Lie algebra, being isomorphic to the algebra of $U(1)$; the particular matrix group $G{L}_1(\mathbb{R})$ is nevertheless noncompact. This distinguishes compactness of an algebra from that of a chosen group realizing it.

For the final algebra, the printed basis specifies strictly upper triangular matrices. If $i<j$ and $p<q$, the first term in $[E_{ij},E_{pq}]$ can be nonzero only when $i<j=p<q$, and then $E_{iq}$ is strictly upper triangular; the second can be nonzero only when $p<q=i<j$, and then $E_{pj}$ is strictly upper triangular. Bilinearity proves closure under the Lie bracket. To compute its own Killing form, grade $E_{ij}$ by the positive integer $j-i$. Bracketing with any basis element increases that degree whenever the result is nonzero. In a basis ordered by degree, all adjoint maps are strictly triangular, as are their products. Every such product has zero trace, so the Killing form of a nilpotent Lie algebra vanishes here. Therefore the subspace denoted by a prime in the PDF isThis is the defined radical of the Killing form, not the derived algebra; for $n=1$ both $\mathcal{L}$ and this radical are zero.

There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

The objective is a continuous faithful representation of a uniform pro-p group $G$ in some ${\mathrm{GL}}_N({\mathbb{Z}}_p)$. The proof proceeds from the intrinsic Lie lattice to a small matrix group, then extends the resulting representation to all of $G$. This last step matters: a representation defined only on an open subgroup does not yet establish linearity of the original group.

Let $L=L_G$ be the Lie algebra of a uniform pro-p group and put $\mathfrak{g}={\mathbb{Q}}_p{\otimes }_{{\mathbb{Z}}_p}L$. The uniform-group/Lie-lattice correspondence identifies the group law on $L$ with BCH. The substantial finite-dimensional Lie-theoretic input is the Ado theorem: every finite-dimensional Lie algebra over a characteristic-zero field admits a faithful finite-dimensional representation. Apply it to obtainAn Adjoint representation would not suffice here, since it kills the centre; Ado supplies injectivity even in central directions.

Choose a p-adic lattice $\Lambda$ spanning $V$. Since $L$ has finitely many lattice generators, their representing matrices have bounded denominators. There is therefore an integer $a\ge 0$ such thatOn this matrix lattice the series $\exp X={\sum }_{n\ge 0}{X}^n/n!$ and $\log (1+Y)={\sum }_{n\ge 1}(-1{)}^{n+1}{Y}^n/n$ converge and are inverse. For example $v_p(n!)\le n/(p-1)$, so the valuations of the exponential terms tend to infinity with this choice of $ϵ$; the logarithm terms do as well. They take values in $1+p^{ϵ}\mathrm{End}(\Lambda )$ and in $p^{ϵ}\mathrm{End}(\Lambda )$ respectively. The formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ holds for the convergent series.

Under the correspondence, $U=G^{p^a}$ has Lie lattice $p^{a}L$ and is an open characteristic subgroup. Definewhere ${\log }_G$ denotes the identification of $U$ with $p^{a}L$. A Lie homomorphism preserves BCH, so the exponential identity proves $\eta (uv)=\eta (u)\eta (v)$. If $\eta (u)=I$, the matrix logarithm gives $\rho ({\log }_{G}u)=0$; faithfulness of $\rho$ gives $u=1$. Thus $\eta$ is a continuous faithful integral representation of $U$.

To extend it, take the moduleValues on a transversal of $U\backslash G$ determine $f$, so $W$ is a free module over ${\mathbb{Z}}_p$ of rank $[G:U]\dim V$. These functions are continuous, since on each open coset they are given by the continuous representation $\eta$. Define $(T_{g}f)(x)=f(xg)$. The equivariance condition is preserved, and $T_g{T}_h=T_{gh}$. In a transversal basis each $T_g$ is a permutation of blocks followed by matrices from $\eta (U)$, so it belongs to $\mathrm{GL}(W)$ and depends continuously on $g$.

Faithfulness can be checked directly. If $g\notin U$, right multiplication moves the coset $U$ to a different coset, so $T_g$ moves a function supported on one block and cannot be the identity. If $g\in U$ and $T_g=1$, evaluation at $1$ gives $f(g)=f(1)$ for every $f$, whereas equivariance gives $f(g)=\eta (g)f(1)$. Since $f(1)$ is arbitrary, $\eta (g)=I$, and hence $g=1$. ThereforeCompactness of $G$ makes this continuous injection a homeomorphism onto its closed image. This proves the linearity of a uniform pro-p group for the full group, including its centre, and explains why taking a sufficiently small power lattice permits matrix integration before finite-index induction restores the whole group.

A real Lie algebra is a real vector space with a bilinear antisymmetric bracket satisfying the Jacobi identity. For matrices, the bracket is the commutator $[X,Y]=XY-YX$. The Dynamical Lie algebra isIt is found by repeatedly adjoining commutators and taking real linear spans until no new independent generators appear. Let $G$ be its connected dynamical Lie group. Under the standard assumptions of freely timed controls with unrestricted real amplitudes, the following are the relevant Lie criteria.

For pure-state controllability, $G$ must act transitively on normalized state vectors, or on their rays when global phase is disregarded. The standard finite-dimensional classification is, up to a unitary change of basis,Here $\mathfrak{sp}(n)$ is the compact symplectic Lie algebra in its defining complex $2n$-dimensional representation. Equivalently, the infinitesimal action spans the tangent directions to the whole pure-state orbit. The symplectic cases are important because pure-state reachability alone need not give complete mixed-state reachability.

For density operator controllability, the action must be transitive on every unitary orbit of a density operator, meaning on every fixed-spectrum class. The criterion is $\mathfrak{g}=\mathfrak{su}(N)$ or $\mathfrak{u}(N)$. This concerns all density operators, not merely one unusually degenerate spectrum.

For exact unitary operator controllability, the criterion is $\mathfrak{g}=\mathfrak{u}(N)$. If only the physical gate up to global phase matters, it is enough that the image after removing the scalar direction is $\mathfrak{su}(N)$, equivalently that $\mathfrak{g}+\mathbb{R}iI=\mathfrak{u}(N)$. A traceless system can implement all special-unitary gates while lacking independent control of the overall phase.

An almost commutative algebra over $\mathbb{C}$ has an exhaustive nonnegative increasing multiplicative ascending filtrationfor which the associated graded ringis a commutative finitely generated complex algebra. This includes the customary degree filtration with $F_{0}R=\mathbb{C}$, finitely many degree-one generators, and commutators lowering degree. The finite-generation clause is part of the convention used here and is necessary for the conclusion.

A finitely generated commutative complex algebra is a quotient of a polynomial ring $\mathbb{C}[X_1,\dots ,X_d]$. Repeated application of the central-variable case of the Hilbert basis theorem makes that polynomial ring Noetherian; quotients preserve the ascending chain condition by lifting ideal chains. Thus $\mathrm{gr}R$ is Noetherian.

We prove ascending filtered-graded transfer of Noetherianity directly. Give a left ideal $I\subseteq R$ the induced ascending filtration $I\cap F_{n}R$. Its associated graded module embeds into $\mathrm{gr}R$ as a homogeneous ideal, denoted $\mathrm{gr}I$. Choose finitely many homogeneous generators ${\overline{u}}_1,\dots ,{\overline{u}}_m$ of this ideal, of degrees $d_1,\dots ,d_m$, and lifts $u_j\in I\cap F_{d_j}R$. Homogeneous generators may be chosen by taking homogeneous components of any finite generating set.

For $u\in I\cap F_{n}R$, its degree-$n$ symbol can be expressed aswith terms of negative coefficient degree omitted. Lift each coefficient to $a_j\in F_{n-d_j}R$. ThenInduction on $n$, ending at $F_{-1}R=0$, expresses $u$ in the left ideal generated by the $u_j$. Hence every left ideal is finitely generated. Applying the same argument to right ideals, or to the opposite ring, provesThis degree-lowering proof terminates in finitely many steps and requires no completeness assumption on the filtered ring. Without finite generation of the graded algebra, a counterexample would be the commutative polynomial ring in infinitely many variables, whose ideals $(X_1)⊊(X_1,X_2)⊊\cdots$ do not stabilize.

For the universal enveloping algebra $U(\mathfrak{g})$, use the degree ascending filtration inherited from the tensor algebra of the finite-dimensional Lie algebra $\mathfrak{g}$. For $x,y\in \mathfrak{g}$, the defining relation is $xy-yx=[x,y]\in F_{1}U(\mathfrak{g})$, whereas the product $xy$ has filtration degree at most $2$. Thus the degree-one symbols commute in $\mathrm{gr}U(\mathfrak{g})$. Those symbols generate the graded algebra, so it is a quotient ofThis already proves the required almost-commutative property. The Poincaré-Birkhoff-Witt theorem gives the stronger identification $\mathrm{gr}U(\mathfrak{g})\cong \mathrm{Sym}(\mathfrak{g})$, but injectivity of this map is not needed for the chain-condition argument. Therefore

We consider finite-dimensional continuous complex representations of SU(2). Its maximal torus is $\mathrm{diag}(e^{i\theta },e^{-i\theta })$. The irreducible representations areThey may be realized as homogeneous polynomials of degree $n$ in two variables, with the induced linear substitution action. Their torus weights are $n,n-2,\dots ,-n$, each once. The centre $-I$ acts by $(-1{)}^n$, so precisely the even-$n$ representations descend to SO(3). In physics $n=2j$ labels integral or half-integral spin $j$.

Here is a classification argument. Average any positive definite Hermitian form over the compact group using normalized Haar measure. The averaged form is invariant, so the orthogonal complement of an invariant subspace is invariant. Repeatedly splitting gives complete reducibility. Differentiate an irreducible representation and complexify to the Lie algebra ${\mathfrak{sl}}_2$, with generators $H,E,F$ satisfying $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. Torus periodicity gives integer $H$-weights. Choose a vector $v$ of maximal weight $n$; then $Ev=0$. Induction on $k$ using the commutators givesThe chain terminates: if $F^{p}v\ne 0$ and $F^{p+1}v=0$, the displayed formula forces $n=p\ge 0$. The vectors $v,Fv,\dots ,F^{n}v$ have distinct weights, span an invariant subrepresentation, and hence span the irreducible representation. Their generator action agrees with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. Invariance under the Lie algebra is equivalent to invariance under the connected group, completing the classification. Conversely, a nonzero invariant subspace of a symmetric power contains a weight vector by projection onto torus weights. Raising it to the highest weight and then lowering generates every monomial, proving that the symmetric power is irreducible.

The character on the torus iswith the endpoints interpreted by continuity. Every representation has a unique decomposition $V\cong {⨁}_{n\ge 0}{m}_n{V}_n$. Its multiplicities can be computed from character inner products,The sine orthogonality directly verifies orthonormality of the irreducible characters. Alternatively, if $w_k$ is the dimension of the weight-$k$ subspace, the weight lists give $w_n=m_n+m_{n+2}+\cdots$, so $m_n=w_n-w_{n+2}$ for $n\ge 0$. This gives a constructive decomposition from torus weights. In particular, multiplying weight characters yields the Clebsch-Gordan decomposition $V_m\otimes V_n\cong V_{m+n}\oplus V_{m+n-2}\oplus \cdots \oplus V_{|m-n|}$. Arbitrary unitary Hilbert-space representations of this compact group similarly decompose into Hilbert direct sums of these finite-dimensional irreducibles; the finite-dimensional statement above is the standard representation-theory setting.