The precise shifted L-infinity structure of a complete filtered differential graded Lie algebra

The degree of each displayed operation is one. If $a=\mathrm{\mid }x\mathrm{\mid }$, $b=\mathrm{\mid }y\mathrm{\mid }$, skewsymmetry gives $q_2(sy,sx)=(-1{)}^{b+1+ab}s[x,y]$, so $q_2(sx,sy)=(-1{)}^{(a-1)(b-1)}{q}_2(sy,sx)$. Hence $q_2$ descends to $S^2(\mathfrak{g}[1])$. The unary shifted identity is $q_1^2(sx)=s(d^{2}x)=0$.

On $(sx,sy)$ the three arity-two terms are $q_1{q}_2(sx,sy)=(-1{)}^{a+1}s[dx,y]-s[x,dy]$, $q_2(q_{1}sx,sy)=(-1{)}^{a}s[dx,y]$, and $(-1{)}^{a-1}{q}_2(sx,q_{1}sy)=s[x,dy]$. They sum to zero by the differential derivation identity.

Put $c=\mathrm{\mid }z\mathrm{\mid }$. The three arity-three unshuffle terms, with their Koszul signs, sum to $(-1{)}^{b}s([[x,y],z]-(-1{)}^{bc}[[x,z],y]+(-1{)}^{a(b+c)}[[y,z],x])$. Skewsymmetry rewrites the parenthesis as $[[x,y],z]+(-1{)}^{ab}[y,[x,z]]-[x,[y,z]]$, which is zero by graded Jacobi. In arities at least four every composition contains an operation of arity at least three and vanishes. Thus [F1] gives square zero.

The shifted filtration is complete and separated in each degree because the original one is; $q_1$ preserves it and $q_2$ adds filtration indices. All higher operations vanish. These are exactly [F2]. For $R\in F^1{\mathfrak{g}}^1$, the curvature is $C(sR)=-s(dR+[R,R]\mathrm{/}2)$ by substituting $\mathrm{\mid }R\mathrm{\mid }=1$. Thus the asserted Maurer–Cartan equation is equivalent to $C(sR)=0$.

For $\xi \in F^1{\mathfrak{g}}^0$, $q_1^{sR}(s\xi )=-s(d\xi )-s[R,\xi ]=s(-d\xi +[\xi ,R])$ by skewsymmetry. Applying this quotientwise to polynomial coefficients translates [F3] exactly to $R^{\mathrm{\prime }}=-d\xi +[\xi ,R]$. The degree of $s\xi$ is $-1$, as required for the path parameter.