Graded PBW, the CE resolution, and the free Lie comparison

Choose and order a homogeneous basis of $\mathfrak{g}$. In its tensor algebra rewrite $xy$ for an inverted pair as $(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}yx+[x,y]$, and rewrite the square of an odd basis vector $x$ as $[x,x]\mathrm{/}2$. Order reductions first by word length and then by inversions. A bracket term has smaller length, and a swapped term has fewer inversions, so reduction terminates, even for an infinite basis, since every expression is finite. Disjoint reductions commute. For three distinct adjacent letters the difference of the two overlapping reductions is the signed Jacobi expression. For an odd repeated letter the overlap is $[x,[x,z]]-\frac{1}{2}[[x,x],z]=0$; the triple-odd overlap is $[x,[x,x]]=0$. These follow from graded Jacobi in characteristic zero. The overlaps with an even repeated letter have no square reduction and reduce to the same signed Jacobi expression. Induction on the reduction order therefore gives unique reduced expressions. The ordered monomials, with odd basis vectors occurring at most once, form a basis for $U(\mathfrak{g})$. Their leading terms give $\mathrm{gr}U(\mathfrak{g})=\mathrm{Sym}(\mathfrak{g})$, compatibly with the internal differential.

The displayed formulas define a differential graded left $U(\mathfrak{g})$-module. The Koszul signs in $\mathrm{\Delta }$ are those of [F4]. The coderivation identity and $Q^2=0$ follow from [F3] and the dg Lie identities [F2]. In the expansion of $D^2$, the internal/internal terms use $d^2=0$; internal/action and internal/bracket terms use the differential derivation identity; the two-action terms cancel the action of the bracket by the defining relation $xy-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}yx=[x,y]$ of $U(\mathfrak{g})$ in the Statement; and three-input terms are graded Jacobi. The coaugmentation term gives the unit summand, so $K(0)=k$.

Filter $K$ by total number of enveloping letters plus suspended letters. The internal and action terms preserve this filtration degree, and bracket terms lower it. Its associated graded is $\mathrm{Sym}(\mathfrak{g})\otimes \mathrm{Sym}(\mathfrak{g}[1])$, with differential $d_0=d_{\mathrm{i}\mathrm{n}\mathrm{t}}+\delta$, where $\delta (sx)=x$ and $\delta (x)=0$. Here $sx$ has degree $\mathrm{\mid }x\mathrm{\mid }-1$, and $d_{\mathrm{i}\mathrm{n}\mathrm{t}}(sx)=-s(dx)$. On this graded commutative algebra the degree $-1$ derivation $h$ given by $h(x)=sx$, $h(sx)=0$ satisfies $d_{\mathrm{i}\mathrm{n}\mathrm{t}}h+h{d}_{\mathrm{i}\mathrm{n}\mathrm{t}}=0$ and $\delta h+h\delta =E$, the derivation counting total letters. These identities hold on generators and hence on the whole algebra. Therefore $h\mathrm{/}w$ contracts the positive weight-$w$ part for $d_0$; weight zero is $k$. The division is permitted in characteristic zero, for both even and odd generators.

If an augmentation-zero cycle in $K$ has largest weight $w>0$, the contraction in step 2.1 writes its weight-$w$ part as a boundary in the associated graded. Lift that primitive to $K$ and subtract its differential. The new cycle has smaller largest weight. Iterating reaches weight zero, where augmentation zero means zero. This proves exactness of $K\to k$ without an unbounded spectral-sequence convergence assumption. To justify derived tensor products, filter instead by suspended length: the internal differential preserves length and the other terms lower it. Every associated quotient is $U\otimes W$ for a $k$-complex $W$. Such a module is homotopically projective because ${\mathrm{Hom}}_U(U\otimes W,E)={\mathrm{Hom}}_k(W,E)$ is acyclic for acyclic $E$, which is contractible over a field. A compatible nullhomotopy extends across successive length levels by correcting its defect in this acyclic Hom complex. Induction gives a nullhomotopy on the union. Thus $K$ is homotopically projective and the resolution computes $k{\otimes }_U^{\mathbf{L}}k=C_{\ast }(\mathfrak{g})$.

For $\mathfrak{g}=\mathrm{FreeLie}(V)$, the universal properties identify $U(\mathfrak{g})=T(V)$. Concatenation gives a dg-exact sequence $0\to T(V)\otimes V\to T(V)\to k\to 0$: every positive word has a unique last letter, and this decomposition commutes with the internal differential. The cone resolution $P=T(V)\otimes (k\oplus V[1])$, whose extra differential sends $sv$ to $v$, is homotopically projective by its two-step free-complex filtration. There is a natural $U$-module chain map $P\to K$ sending $1$ to $1\otimes 1$ and $sv$ to the length-one suspended generator. It commutes with both internal and action differentials, and induces the identity on the augmentation $k$. Since both are homotopically projective resolutions, this map is a homotopy equivalence: its acyclic homotopically projective cone has nullhomotopic identity, by the extension argument of step 3.1. Tensoring with $k$ therefore gives the asserted natural quasi-isomorphism $k\oplus V[1]\to C_{\ast }(\mathfrak{g})$.

The last map preserves coaugmentation and coproduct because every suspended generator is primitive in [F4]; its domain has no products of two suspended generators. Full linear duality over a field is exact (acyclic complexes split) and takes that coalgebra map to the augmented commutative algebra map in the Statement. Products, rather than finite-support duals, are used in each total cochain degree. Thus the assertion holds for arbitrary unbounded $V$; the finite positive case is included.