Bourbaki

Proposition 14

Tag 06CK, permanent. Cite it as [Bourbaki, Tag 06CK].

Algebra, chapter IV, § 4. Formal power series

— We have l(e(X))=e(l(X))=Xl(e(X)) = e(l(X)) = X.

The series l and e have no constant term and their terms of degree 1 are equal to X. By Prop. 10 of IV, p. 37 it suffices to prove the formula l(e(X))=Xl(e(X)) = X. By the formulae (36) and (37) and Cor. 3 of IV, p. 33 we have

D(l(e(X)))=(1+e(X))−1D(e(X))=(eX)−1eX=1 D(l(e(X))) = (1 + e(X))^{-1} D(e(X)) = (e^X)^{-1} e^X = 1

whence I(e(X))=XI(e(X)) = X.

Let K be a Q-algebra, then the elements of K[[I]] without constant term form a commutative group E\mathscr{E} under addition. The elements of K[[I]] with constant term 1 form a commutative group A\mathbf{A} under multiplication (IV, p. 30). For each f∈Ef \in \mathscr{E}, we can define the elements e∘fe \circ f and I∘fI \circ f of E\mathscr{E}, and by Prop. 14 above, the mappings f↦l∘ff \mapsto l \circ f and f↦e∘ff \mapsto e \circ f are mutually inverse permutations of E\mathscr{E}; clearly they are continuous. Since exp⁡X=e(X)+1\exp X = e(X) + 1, we see that the exponential mapping f↦exp⁡f=e∘f+1f \mapsto \exp f = e \circ f + 1 is a continuous bijection of E\mathscr{E} onto M\mathscr{M}. By formula (4) of IV, p. 29 and Prop. 13, we have exp⁡(f+g)=(exp⁡f)(exp⁡g)\exp(f+g) = (\exp f)(\exp g) for f,g∈Ef, g \in \mathscr{E}. Thus the exponential is an isomorphism of the topological group E\mathscr{E} onto the topological group A\mathbf{A}.

The inverse isomorphism of M\mathscr{M} onto E\mathscr{E} is called the logarithm and is written g↦log⁡gg \mapsto \log g. We thus have log⁡g=l(g−1)\log g = l(g-1) for gg in M\mathscr{M}, and in particular,

(38)

log⁡(1+X)=l(X). \log(1+X) = l(X).

Since the logarithm is a homomorphism of M\mathscr{M} into E\mathscr{E}, the formula (1+X)(1+Y)=1+(X+Y+XY)(1+X)(1+Y) = 1 + (X+Y+XY) implies

(39)

l(X)+l(Y)=l(X+Y+XY). l(X) + l(Y) = l(X+Y+XY).

Let (uλ)λ∈L(u_\lambda)_{\lambda \in L} be a summable family of elements of B\mathscr{B}, then the family (exp⁡uλ)λ∈L(\exp u_{\lambda})_{\lambda \in L} is multipliable and we have

(40)

exp⁡(∑λ∈Luλ)=∏λ∈Lexp⁡uλ. \exp \left( \sum_{\lambda \in L} u_{\lambda} \right) = \prod_{\lambda \in L} \exp u_{\lambda}.

Similarly, if (fλ)λ∈L(f_\lambda)_{\lambda \in L} is a multipliable family of elements of A, the family (log⁡fλ)λ∈L(\log f_\lambda)_{\lambda \in L} is summable and we have

(41)

log⁡(∏λ∈Lfλ)=∑λ∈Llog⁡fλ. \log \left( \prod_{\lambda \in L} f_\lambda \right) = \sum_{\lambda \in L} \log f_\lambda .

Let g∈Ag \in A, and let D be a continuous derivation of K[[I]]K[[I]]. We have log⁡g=l(g−1)\log g = l(g - 1), hence by Cor. 3 of IV, p. 33 and (37) we have

(42)

Dlog⁡g=D(g)/g. D \log g = D(g)/g .

The expression D(g)/gD(g)/g is called the logarithmic derivative of g (relative to D).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed edition, 2003, Springer. Not translated.