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A composition series of a module is a finite chain of submodules whose nonzero successive quotient modules are simple modules. Every finite-dimensional Lie algebra representation has such a chain, viewed as a module over the universal enveloping algebra.

This is the degree-one convention for an almost commutative algebra: the exhaustive filtration is generated by a finite-dimensional first step containing $1$, the degree-zero part is $k$, and its associated graded algebra is commutative. The first step is a finite-dimensional Lie algebra under commutators. Its inclusion into $A$ extends by the universal property to a surjection from its universal enveloping algebra. The element $1$ of the first step is retained as a central Lie generator and its image is the associative identity; quotienting the first step by $k1$ before taking the bracket can lose scalar commutators. This convention is stronger than merely requiring a finitely generated commutative associated graded algebra for an arbitrary weighted filtration.

// Destination: noncommutative-algebra.bigb

For a positively graded vector space $V$, the free graded Lie algebra $\mathbb{L}(V)$ receives a linear map from $V$ and extends every graded linear map $V\to L$ uniquely to a Lie homomorphism. Over $\mathbb{Q}$, it is the Lie subalgebra generated by $V$ inside the tensor algebra $T(V)$, using $[u,v]=uv-(-1{)}^{|u||v|}vu$. Its universal enveloping algebra is $T(V)$: both have the same universal property for maps from $V$ into associative algebras. The primitives of the tensor Hopf algebra with primitive generators are exactly $\mathbb{L}(V)$.

For a ring $D$, an automorphism $\sigma$ and central $t$, adjoin $X,Y$ with $Xd=\sigma (d)X$, $Yd={\sigma }^{-1}(d)Y$, $YX=t$ and $XY=\sigma (t)$. This construction includes Weyl algebras and rank-one universal enveloping algebras.

For a graded Lie algebra over characteristic zero, the associated graded algebra of its universal enveloping algebra under the word-length filtration is its graded symmetric algebra. Ordered homogeneous monomials form a basis, with exponent at most one on each odd Lie basis element. The defining graded commutation relations reorder every word, and the Jacobi identity resolves the triple-reordering ambiguities. In particular the primitive elements of a Hopf algebra of $U(L)$ are precisely $L$: in its associated graded algebra, a primitive polynomial of symmetric length greater than one would have a nonzero mixed term under the diagonal, which is impossible in characteristic zero.

For a unitary derived representation $\rho$, its generators are anti-Hermitian. Define Hermitian observables $H_X=i\rho (X)$; then their ordinary operator commutator is $[H_X,H_Y]=i{H}_{[X,Y]}$. Thus real Lie algebra structure constants and Hermitian-operator commutation relations differ by an $i$. Products in a universal enveloping algebra require one conversion factor for every generator. In particular, the abstract bilinear Pauli-Lubanski element has representation image minus the pseudovector formed from the Hermitian physical generators.

For simply connected $Y$, rational Hurewicz homomorphism identifies ${\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q}$ with the primitive elements of a Hopf algebra of $H_{\ast }(\Omega Y;\mathbb{Q})$ and induces an isomorphism $U({\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q})\cong H_{\ast }(\Omega Y;\mathbb{Q})$ of graded Hopf algebras. The bracket on homotopy is the Samelson product. The underlying algebra theorem says that a connected graded cocommutative Hopf algebra over characteristic zero is the universal enveloping algebra of its primitive Lie algebra. These statements let a free tensor algebra of loop-space homology detect a free graded Lie algebra of homotopy.

Choose a basis $x_1,x_2,x_3$ of the Lie algebra and filter its universal enveloping algebra by word length: $F_{0}R=\mathbb{C}$ and $F_{d}R$ is spanned by products of at most $d$ basis elements. The relations $x_j{x}_i-x_i{x}_j=[x_j,x_i]$ replace a degree-two commutator by degree one, so the degree-one symbols commute.

The Poincaré-Birkhoff-Witt theorem says the ordered monomials $x_1^{a_1}{x}_2^{a_2}{x}_3^{a_3}$ are a basis. Its relevant proof can be seen by reordering an inverted neighboring pair using the displayed relation. Reordering terminates because same-length terms have fewer inversions and bracket terms have smaller word length. Disjoint swaps commute; the only overlapping three-letter reorderings differ by $[x_i,[x_j,x_k]]+[x_j,[x_k,x_i]]+[x_k,[x_i,x_j]]=0$. The Jacobi identity therefore makes the reduced ordered expression independent of the reordering path. Ordered monomials have no remaining relations and form the basis. Their degree-$d$ symbols giveFor generators $m_1,\dots ,m_q$ of the finitely generated module $M$, put $M_d={\sum }_j{F}_{d}R{m}_j$. This is a good filtration, and $\mathrm{gr}M$ is finite over the three-variable polynomial ring. Define the dimension of a filtered module as the Krull dimension of its associated graded module, equivalently the degree of the eventual cumulative polynomial ${\dim }_{\mathbb{C}}{M}_d$, or the Gelfand–Kirillov dimension of a module. Two good filtrations have bounded shifts, so the growth degree is independent of the generators. Sinceits polynomial degree is at most three. Therefore every nonzero finitely generated $R$-module has dimension at most $3$. The zero module may be assigned the usual separate empty-support convention. This bound also follows immediately from its support of a module inside affine three-space.

Put $U_0=G$ and $U_{j+1}=U_j^p$, where powers mean closed generated subgroups. The finite-quotient arguments of Question 3 pass to a powerful pro-p group: in particular $U_j$ is powerfully embedded, the power map is onto $U_{j+1}$, andFor the root assertion, apply the finite result to every finite quotient and use the finite intersection property on the closed sets of possible roots in the compact group. Each $U_j$ is finitely generated by the appropriate powers of a finite generating tuple, and $U_j/U_{j+1}$ is finite and elementary abelian. Thus the $U_j$ are open. They form an identity-neighbourhood basis: in any finite continuous quotient their images form a terminating lower p-series, so some $U_j$ is contained in the kernel. In particular ${\bigcap }_j{U}_j=1$. Collection also givesOne can see the extra index directly by induction from $[U_a,G]\subseteq U_{a+1}$: collect $[v,w^p]$ for $w\in U_b$. Its leading factor $[v,w{]}^p$ advances one layer, and the remaining iterated commutators, with their binomial coefficients, lie at least as deep. Products and limits preserve the inclusion. This is the usual strongly central property of the lower p-series.

A uniform pro-p group is a finitely generated powerful pro-p group for which all the indices $|U_j:U_{j+1}|$ equal $|G:G^p|$. The preliminary congruence gives surjective linear mapsIf $G$ is uniform, these maps are bijective. If $x\ne y$, write $y=xu$ and choose the unique layer $j$ with $u\in U_j\setminus U_{j+1}$. Bijectivity of ${\theta }_j$ and the power congruence imply $y^p(x^p{)}^{-1}\in U_{j+1}\setminus U_{j+2}$, so $x^p\ne y^p$. Thus the power map is injective. A finite-order element of a pro-p group has $p$-power order; repeated injectivity of the power map excludes every nonidentity such element. Hence a uniform group is a torsion-free group.

Conversely, suppose some ${\theta }_j$ has a nonzero kernel. Choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. At step $r\ge j+1$, if the current $x_r$ satisfies $x_r^p\in U_{r+1}$, surjectivity of ${\theta }_r$ supplies $u_r\in U_r$ withSet $x_{r+1}=x_r{u}_r$. The power congruence gives $x_{r+1}^p\in U_{r+2}$. Since the corrections tend to $1$ in the complete profinite group, the $x_r$ converge to $z$, with $z^p=1$ and $z\equiv x(\mathrm{mod}{U}_{j+1})$. In particular $z\ne 1$. A torsion-free group cannot have such an element, so every ${\theta }_j$ is injective and all layer indices are equal. This proves the root correction criterion for uniformity andFor a uniform group the power maps are continuous bijections $U_j\to U_{j+1}$, hence homeomorphisms. Their inverses define the unique roots used below. Comparison at the first differing layer also proves the power congruences in a uniform pro-p group:

The intrinsic Lie algebra of a uniform pro-p group has the same underlying set as $G$, zero equal to the group identity, and operationsHere $x^{\lambda }$ is the limit of integral powers approaching $\lambda$. The first root is taken in $U_n$ and the second in $U_{2n}$; the commutator actually lies in $U_{2n+1}$. The following faithful algebra realization verifies convergence and the Lie algebra axioms, rather than assuming them as part of the definition.

Choose a minimal tuple $a_1,\dots ,a_d$ of topological generators. The layer isomorphisms ${\theta }_j$ show that the images of $a_i^{p^j}$ are bases of $U_j/U_{j+1}$. The ordered-coordinate argument of Question 3 therefore gives a continuous bijectionExistence follows by successive digits and compactness; uniqueness follows by the first digit at which the coordinates differ, because its images are linearly independent in the corresponding power layer. Thus this map is a homeomorphism.

Here are details of the norm needed on the group algebra. In ${\mathbb{Z}}_p[[G]]$ put $b_i=a_i-1$. Its elements have unique ordered expansions ${\sum }_{\alpha }{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To verify this description, identify the ordered coordinates of $G$ with ${\mathbb{Z}}_p^d$. Elements of the completed group algebra are integral measures on this space; their values on the binomial polynomials ${\prod }_i\left(\genfrac{}{}{0ex}{}{{\lambda }_i}{{\alpha }_i}\right)$ are precisely the coefficients $c_{\alpha }$. The finite-difference identity pairs the ordered monomial $b^{\beta }$ with this binomial polynomial as ${\delta }_{\alpha \beta }$. The Mahler expansion of a continuous function therefore proves both existence and uniqueness of the measure with the specified coefficients. This also proves that the ordinary rational group algebra embeds in these expansions.

Choose $1/(p-1)<s<1$ and setThis defines the weighted group algebra of a uniform pro-p group. Its multiplication is submultiplicative in the valuation convention, $v(AB)\ge v(A)+v(B)$. Indeed, $a_i^p-1=p{b}_i+\left(\genfrac{}{}{0ex}{}{p}{2}\right)b_i^2+\cdots +b_i^p$ has valuation $1+s$; collection of ordered coordinates gives the same lower bound for $g-1$ when $g\in U_1$. Interchanging two adjacent $b_i$ in a product introduces a correction coming from $[a_i,a_j]-1$, of valuation at least $1+s>2s$. Iterating the interchange rule never lowers the total weight; the corrections have strictly increasing weight, so infinite collection converges. The same argument with $p^j$-th powers showsThe equality follows from the first nonzero coordinate digit: its leading terms are $p^j{\lambda }_i^{\prime }{b}_i$, at least one with unit ${\lambda }_i^{\prime }$, and cannot cancel in the ordered expansion. Completing the rational group algebra in this norm produces a complete normed algebra $A$ in which $G$ embeds faithfully and with its original topology.

Since $v(g-1)\ge s>1/(p-1)$, the seriesconverge on the relevant balls in $A$ and are inverse there. Their convergence follows from $v_p(r!)\le (r-1)/(p-1)$ and $v_p(r)\le (r-1)/(p-1)$; formal one-variable identities prove that they are inverse. Put $X=\log x$ and $Y=\log y$. The Baker--Campbell--Hausdorff formula in $A$ now givesbecause every term of degree at least two tends to zero after this rescaling. Similarly expansion of $\exp (-p^{n}X)\exp (-p^{n}Y)\exp (p^{n}X)\exp (p^{n}Y)$ givesThe leading term is the indicated associative-algebra commutator and all terms of degree at least three tend to zero. These expressions are the logarithms of the unique roots in the intrinsic formulas. The image $\log G$ is compact and hence closed in $A$, so both limits belong to $\log G$, and the inverse logarithm takes the limits back to $G$. Also $\log (x^{\lambda })=\lambda \log x$. Consequently $\log$ identifies the intrinsic operations with addition, scalar multiplication and commutation in $A$. The Jacobi identity follows by expanding the associative products; thus $L_G$ is indeed a Lie algebra over ${\mathbb{Z}}_p$.

The commutator roots in the intrinsic formula belong to $U_1$, so their limit gives $[L_G,L_G]\subseteq \log U_1=p{L}_G$. Iterating the power congruence gives $x^{p^n}{y}^{p^n}\equiv (xy{)}^{p^n}(\mathrm{mod}{U}_{n+1})$. Taking the unique roots shows that intrinsic addition modulo $U_1$ is the original group product modulo $U_1$. Consequently $L_G/p{L}_G$ is the original elementary abelian quotient $G/U_1$, and the images of the $a_i$ form a basis. Successively removing digits expresses every element uniquely as an additive sum ${\sum }_i{\lambda }_i{a}_i$: existence uses completeness, while a nonzero relation would have a first nonzero digit contradicting the basis in $U_j/U_{j+1}$. Hence $L_G$ is a free module of rank $d$ over the p-adic integers and is a powerful Lie lattice.

Conversely a powerful Lie algebra over p-adic integers, for odd $p$, is a finite free ${\mathbb{Z}}_p$-module $L$ with a bilinear alternating Lie bracket satisfying the Jacobi identity and $[L,L]\subseteq pL$. DefineFor completeness, integrality and convergence can be verified without guessing which individual denominators are units. In ${\mathbb{Q}}_p\otimes L$ set $M=p^{-1}L$. Then $[M,M]\subseteq M$, so its integral universal enveloping algebra has an ordered basis by the Poincaré-Birkhoff-Witt theorem. Its coefficientwise $p$-adic norm is submultiplicative, and $L=pM$ consists of elements of valuation at least one. The homogeneous degree-$r$ terms in $\log (\exp X\exp Y)$ have coefficients of valuation at least $-\lfloor (r-1)/(p-1)\rfloor$: expand the logarithm, use $v_p(k!)\le (k-1)/(p-1)$ on each nonconstant exponential factor, and $v_p(\ell )\le (\ell -1)/(p-1)$ on a product of $\ell$ such factors. Therefore evaluating degree $r$ on $X,Y\in pM$ gives valuation at leastwhich tends to infinity and is at least two when $r\ge 2$. Each homogeneous term is a rational Lie polynomial by the formal Baker--Campbell--Hausdorff formula. The ordered enveloping-algebra basis shows that its intersection with ${\mathbb{Q}}_{p}M$ is exactly $M$, so the estimate places the sum in $L$, with the non-linear part in $pL$.

Associativity follows from the identity $\exp (X\ast Y)=\exp X\exp Y$ in the completed universal enveloping algebra. Its identity is $0$, its inverse is $-X$, and $X^{\ast m}=mX$, since the bracket of $X$ with itself is zero. The same coefficient estimate for the group commutator puts it in $pL$. Conjugation in the BCH group is $\exp (\mathrm{ad}X)$; each term $\mathrm{ad}(X{)}^r/r!$ preserves $p^{j}L$, because each bracket adds a factor of $p$ and $v_p(r!)<r$. Thus the submodules $p^{j}L$ are normal subgroups for $\ast$, their intersection is zero, and their finite quotients have $p$-power order. Thus $(L,\ast )$ is a powerful pro-p group, withA module basis topologically generates this group: it generates the quotient $L/pL$, and successive power digits lift that generation through all $p^{j}L$. The BCH construction therefore gives a uniform pro-p group.

Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.

Use the degree-one convention implicit in this enveloping-algebra assertion: a degree-one almost commutative algebra has $F_{0}R=k$, a finite-dimensional $F_{1}R$ containing $1$ which generates $R$, and commutative associated graded ring. Taking the leading symbol of a commutator givesin particular $[F_{1}R,F_{1}R]\subseteq F_{1}R$. Thus $\mathfrak{g}=F_{1}R$, equipped with $[u,v]=uv-vu$, is a finite-dimensional Lie algebra. Bilinearity and antisymmetry are immediate, and expanding the three nested commutators proves the Jacobi identity.

Its inclusion into the associative algebra $R$ is a Lie algebra homomorphism. The universal enveloping algebra is $T(\mathfrak{g})$ modulo the relations $u\otimes v-v\otimes u-[u,v]$. Sending the tensor generators to their elements of $R$ kills these relations and gives an algebra homomorphism $U(\mathfrak{g})\to R$. It is surjective because $F_{1}R$ generates $R$. ThereforeThe vector $1\in F_{1}R$ is retained as a central Lie algebra generator and maps to the associative identity; its corresponding relation in the kernel is that generator minus $1_{U(\mathfrak{g})}$. One should not first replace $F_{1}R$ by $F_{1}R/k1$, since commutators of generators can be nonzero scalars.

Terminology is important here. Merely requiring a commutative finitely generated associated graded ring for an arbitrary weighted filtration is a broader definition of almost commutative algebra; it does not supply the finite-dimensional bracket-closed first step used in this argument. The degree-one assumptions make the requested enveloping presentation precise. The Noetherianity in part (b) also follows directly under the broader finite-generation convention.

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

The correspondence concerns uniform pro-p groups and powerful Lie lattices, with their natural morphisms. Put $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. A powerful pro-p group is a pro-p group whose closed commutator subgroup lies in its closed $p^{ϵ}$-power subgroup. It is uniform when it is topologically finitely generated and all successive indices in its lower p-series are equal. Equivalently, a finitely generated powerful pro-p group is uniform precisely when it is a torsion-free group. A powerful Lie algebra over p-adic integers means a Lie algebra over ${\mathbb{Z}}_p$ that is free of finite rank as a module and satisfiesHere a Lie algebra over a ring is a module with an alternating bilinear Lie bracket satisfying the Jacobi identity. Finite freeness is an essential part of the lattice convention. The zero-bracket module ${\mathbb{F}}_p$ satisfies the bracket inclusion but has torsion, and its additive group is not uniform. Thus allowing arbitrary torsion modules would make the categorical assertion false.

The basic structural facts behind the construction can be seen from the power layers. Write $P_n=G^{p^n}$ with $P_0=G$. Collection for a powerful pro-p group givesFor odd $p$, the power lifting is exactly the successive-correction argument of Question 3, followed by compactness; at $p=2$ the stronger fourth-power commutator condition makes the corresponding collection corrections work. The maps between successive layers are surjective linear maps. Uniformity makes their finite dimensions equal, so these maps are isomorphisms. Comparing the first layer in which two elements differ now shows that the power map $P_n\to P_{n+1}$ is injective, while the power-set formula makes it surjective. Thus all the roots used below exist uniquely in the indicated layer. If $g_1,\dots ,g_d$ lift a basis of $G/P_1$, the images of $g_i^{p^n}$ form a basis of $P_n/P_{n+1}$. Successively choosing their coefficients modulo $p$ gives the unique ordered expressionExistence follows by lifting one layer at a time and taking a limit; uniqueness follows by comparing the first nonzero coefficient layer. In particular $[G:P_n]=p^{nd}$ and these layers form a neighbourhood basis. These are the power-layer and coordinate facts needed in the proof, rather than an assumption that an arbitrary pro-p group has unique roots.

On the underlying set of $G$, set the additive zero to $1$, the additive negative of $x$ to $x^{-1}$, and defineThe p-adic integers act through limits of integer powers in the closed cyclic subgroup generated by $x$. In the second formula the product belongs to $P_n$; in the third the group commutator belongs to $P_{2n+ϵ}$. The symbols $p^{-n}$ and $p^{-2n}$ denote the unique group roots, not unrestricted division in a group. We next justify convergence and the Lie algebra laws together.

A useful concrete tool is a complete normed group algebra containing $G$. In its completed group algebra, put $b_i=g_i-1$. Ordered-coordinate uniqueness gives unique expansions ${\sum }_{\alpha \in {\mathbb{N}}^d}{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To see why, first work modulo $p$ and $P_n$: the ordered products of the $g_i$ with exponents below $p^n$ form the regular basis, and the change to the corresponding products of $g_i-1$ is triangular with diagonal entries one. Take inverse limits and then lift the coefficient digits successively in powers of $p$. Define the weightIt is a genuine separated, submultiplicative weight. Indeed reordering $b_j{b}_i$ produces a group-commutator correction in $P_{ϵ}$; the binomial expansion of a $p^{ϵ}$th power gives that correction weight at least $2ϵ$, so collection never decreases total weight. Also $v(pc)=1+v(c)$. Inverting $p$ and completing therefore gives a Banach algebra $B$ over ${\mathbb{Q}}_p$. For $g\in G$, ordered coordinates showwith value infinity at $g=1$. Thus the induced topology on $G$ is exactly its original topology. This construction supplies the ambient algebra without presupposing the uniform pro-p group and powerful Lie lattice correspondence. The weighted construction, including the case $p=2$, is described in Equivariant D-modules on rigid analytic spaces, Section 6.5.

Since $ϵ>1/(p-1)$, the p-adic logarithm and p-adic exponential converge on the relevant balls in $B$:The estimates $v_p(k!)\le k/(p-1)$ and $v_p(k)\le {\log }_{p}k$ show that term weights tend to infinity. Formal substitution is legitimate there and proves that the two maps are inverse. They preserve the leading weight. In particular $\log$ embeds the compact set $G$ homeomorphically into $B$, and $\log (g^{p^n})=p^n\log g$.

Put $a=\log x$ and $b=\log y$. Expanding the convergent Baker--Campbell--Hausdorff formula in $B$ givesThe first expansion has linear part $p^n(a+b)$ and all other terms have degree at least two in $p^{n}a,p^{n}b$. The second has leading term $p^{2n}(ab-ba)$ and all remaining terms have degree at least three. After division by the displayed powers of $p$, the extra factor of $p^n$ forces those errors to zero. The normalized expressions are logarithms of the unique roots already specified. Since $\log G$ is compact and closed in $B$, both limits belong to $\log G$. Scalar multiplication also remains in $\log G$ by continuity of powers. Consequently $\log G$ is a module over ${\mathbb{Z}}_p$, closed under the associative-algebra Lie bracket $ab-ba$. Bilinearity, alternation and the Jacobi identity follow by expanding products in the associative algebra $B$. Transporting these operations back gives exactly $L_G$ above.

The commutator-layer inclusion places every normalized bracket root in $P_{ϵ}$. Since $\log P_{ϵ}=p^{ϵ}\log G$ and this set is compact, passage to the limit proves $[L_G,L_G]\subseteq p^{ϵ}{L}_G$. Addition modulo $p{L}_G$ is the original multiplication modulo $P_1$: collection shows $(x^{p^n}{y}^{p^n}{)}^{p^{-n}}\equiv xy(\mathrm{mod}{P}_1)$, as the nonlinear errors lie at least one layer deeper. Thus $L_G/p{L}_G$ has dimension $d$, with basis $e_i=\log g_i$. Given $a\in \log G$, write its residue as ${\sum }_i{\lambda }_{0i}{e}_i$ and repeat on $(a-{\sum }_i{\lambda }_{0i}{e}_i)/p$. This iterative digit lifting converges because the remainders stay in the bounded compact set $\log G$. It gives $a={\sum }_i{\lambda }_i{e}_i$ with ${\lambda }_i\in {\mathbb{Z}}_p$. A nontrivial relation among the $e_i$ could be divided by the smallest power of $p$ among its coefficients, contradicting their independence modulo $p$. Hence $L_G$ is free of rank $d$, proving that it is a powerful Lie lattice.

Conversely, let $L$ be a powerful Lie lattice. In $L{\otimes }_{{\mathbb{Z}}_p}{\mathbb{Q}}_p$, defineThe formal Baker--Campbell--Hausdorff formula and its coefficient estimate are the algebraic ingredients used here. More precisely, the degree-$m$ component can be expressed as an integral Lie polynomial divided by a number of $p$-adic valuation at most $\lfloor (m-1)/(p-1)\rfloor$. One way to obtain this estimate is to bound the associative coefficients of $\log (\exp X\exp Y)$ and use an integral ordered basis of the free Lie algebra. The denominator bound $v_p(m!)+\lfloor {\log }_p{s}_p(m)\rfloor$, where $s_p(m)$ is the base-$p$ digit sum, is at most $\lfloor (m-1)/(p-1)\rfloor$. A precise primary reference for that denominator estimate is Smallest common denominators for the homogeneous components of the Baker-Campbell-Hausdorff series. A length-$m$ Lie bracket in $L$ lies in $p^{ϵ(m-1)}L$. Thus the degree-$m$ term belongs toThese exponents are nonnegative and tend to infinity; for $m\ge 2$ they are at least one. This proves integral convergence, continuity and $x\ast y\equiv x+y(\mathrm{mod}pL)$. Associativity follows from the formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ in the completed universal enveloping algebra. Substitution of three elements is justified by the same valuation estimates, so the two associative expressions agree. The identity is $0$, the inverse is $-x$, and $x^{\ast k}=kx$ because every nonlinear term involving only $x$ vanishes.

For powerfulness, formal conjugation givesThe difference $\delta =y-e^{-\mathrm{ad}x}y$ lies in $p^{ϵ}L$: its $k$th term has valuation at least $kϵ-v_p(k!)\ge ϵ$ for $k\ge 1$. The group commutator is therefore $(-y+\delta )\ast y$. Its linear term is $\delta$. Every nonlinear Lie term contains $\delta$, since it vanishes when $\delta =0$, and the extra divisibility makes all those terms lie in $p^{ϵ}L$ too. Hence $[(L,\ast ),(L,\ast )]\subseteq p^{ϵ}L$. The ideals $p^{n}L$ are open normal subgroups for $\ast$, and their finite quotients have order $p^{nd}$. The group is consequently a pro-p group. Its power sets are exactly $p^{n}L$, its lower-series indices are constantly $p^d$, and a basis of $L$ topologically generates it by the same successive digit lifting. The BCH group $(L,\ast )$ is uniform.

To check the inverse constructions, powers in the Baker--Campbell--Hausdorff formula group are ordinary multiples in $L$. Thereforeand the similarly normalized group commutator tends to $[x,y]$. The limit constructions recover the original Lie algebra exactly. In the other direction, the ambient Banach algebra construction gives $\log (xy)=\mathrm{BCH}(\log x,\log y)$, so the recovered operation is the original multiplication of $G$.

Finally a continuous group homomorphism between uniform pro-p groups preserves integer powers and powers with exponents in the p-adic integers, preserves the unique roots in the power layers, and commutes with their limits. It therefore becomes a linear Lie algebra homomorphism. Conversely a linear Lie algebra homomorphism between finite free modules is continuous and preserves every term of the convergent Baker--Campbell--Hausdorff formula, hence is a continuous group homomorphism for $\ast$. The constructions preserve identity maps and composition, and the inverse identifications are natural. They are mutually inverse functors, giving the stated category correspondence with the finite-free lattice convention.

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincaré-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.