Exponential and logarithm in a noncommutative complete filtered algebra

The terms $X^j$ and $N^j$ belong to $F^{j}E$. Therefore all displayed series have finite reductions modulo every $F^{n}E$ and define unique limits. Multiplication respects these limits because the filtration is multiplicative. In particular any one-variable formal series with rational coefficients and zero-order constant defined in $k$ can be evaluated at a positive-order element: each coefficient calculation becomes finite in every quotient.

In $\mathbb{Q}[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$, define $\mathrm{\ell }(t)={\sum }_{j\ge 1}(-1{)}^{j-1}{t}^j\mathrm{/}j$. Termwise differentiation gives ${\mathrm{\ell }}^{\mathrm{\prime }}(t)=(1+t{)}^{-1}$. The formal chain rule, obtained by expanding products coefficient by coefficient, gives $h^{\mathrm{\prime }}=h\mathrm{/}(1+t)$ for $h=\exp \mathrm{\ell }(t)$ and $h(0)=1$. From $(1+t)h^{\mathrm{\prime }}=h$ the coefficient recursion gives $h=1+t$: its coefficient of $t$ is $1$ and every higher coefficient is zero. Also the formal derivative of $\mathrm{\ell }(\exp (t)-1)$ is $1$ and its constant coefficient is zero, so $\mathrm{\ell }(\exp (t)-1)=t$. Each assertion is an identity of scalar coefficient series.

Induction on $r$ proves $({\mathrm{ad}}_X{)}^r(T)={\sum }_{j=0}^r(-1{)}^j\left(\genfrac{}{}{0px}{}{r}{j}\right)X^{r-j}T{X}^j$: the induction step multiplies once on the left and subtracts once on the right, and Pascal's identity collects the overlapping terms. Every summand has filtration order at least $r$, even if $T$ has order zero, so the resulting exponential series converges.

Substitute $N$ and $X$ into the two scalar identities modulo $F^{n}E$ and pass to the inverse limit, using step 1.1. This proves $\exp (\log (1+N))=1+N$ and $\log (\exp X)=X$. The substitution is sound in the noncommutative algebra because every identity involves powers of one fixed element only. The coefficient of total degree $r>0$ in $\exp (X)\exp (-X)$ is $X^r{\sum }_{j=0}^r(-1{)}^j\mathrm{/}(j!(r-j)!)=0$; the degree-zero coefficient is $1$. Thus the exponential inverse is $\exp (-X)$ and both inverse bijections have the stated domains.

Expanding $\exp (X)T\exp (-X)$ modulo $F^{n}E$ and collecting the terms containing a total of $r$ copies of $X$ gives ${\sum }_{j=0}^r(-1{)}^j{X}^{r-j}T{X}^j\mathrm{/}((r-j)!j!)=({\mathrm{ad}}_X{)}^r(T)\mathrm{/}r!$. Passing to limits gives the conjugation identity. If $E$ is graded and $\mathrm{\mid }X\mathrm{\mid }=0$, its graded commutator with $T$ is exactly the ordinary commutator used here, so the same proof applies in every fixed homogeneous degree. In the degreewise-complete variant, perform the scalar substitutions and exponential inverses entirely in the complete unital algebra $E^0$. Every term $X^{r-j}T{X}^j$ has degree $d$, so the conjugation coefficient sums and their limit lie in the complete component $E^d$. Multiplicativity gives continuity of the left and right $E^0$ actions on $E^d$, justifying those limits. This proves the graded variant without allowing a sum of endomorphisms in varying degrees.