Bourbaki

Exercise 1, § 8

Tag 0A6Z, permanent. Cite it as [Bourbaki, Tag 0A6Z].

Lie Groups and Lie Algebras, II, 205

(a) An element π\pi of KK is called admissible if v(π)=0v(\pi) = 0 and πp−1/p≡−1(modπoK)\pi^{p-1}/p \equiv -1 \pmod{\pi \mathfrak{o}_K}. (Note that πp−1/p∈oK\pi^{p-1}/p \in \mathfrak{o}_K since v(π)=0v(\pi) = 0.) Show that there exist fields KK satisfying the conditions of the paragraph and containing an admissible element (adjoin to Qp\mathbf{Q}_p a (p−1)(p-1)-th root of −p-p).

(b) Let π\pi be an admissible element of KK. Consider the following formal power series with coefficients in KK:

eπ(U)=1πe(πU)=∑n=1∞πn−1Un/n!, e_{\pi}(U) = \frac{1}{\pi} e(\pi U) = \sum_{n=1}^{\infty} \pi^{n-1} U^n / n!,
lπ(U)=1πl(πU)=∑n=1∞(−π)n−1Un/n. l_{\pi}(U) = \frac{1}{\pi} l(\pi U) = \sum_{n=1}^{\infty} (-\pi)^{n-1} U^n / n.

Show that these series are inverses of one another and that their coefficients belong to oK\mathfrak{o}_K (use Lemma 2).

(c) Let {U,V}\{U, V\} be a set with two elements and let H(U,V)H(U, V) be the Hausdorff series. We write

Hπ(U,V)=1πH(πU,πV). H_{\pi}(U, V) = \frac{1}{\pi} H(\pi U, \pi V).

Then Hπ(U,V)∈L^K({U,V})H_{\pi}(U, V) \in \hat{L}_K(\{U, V\}).

Show that

Hπ(U,V)=lπ(eπ(U)+eπ(V)+πeπ(U)⋅eπ(V)). H_{\pi}(U, V) = l_{\pi}(e_{\pi}(U) + e_{\pi}(V) + \pi e_{\pi}(U) \cdot e_{\pi}(V)).

Deduce that Hπ∈L^oK({U,V})H_{\pi} \in \hat{L}_{\mathfrak{o}_K}(\{U, V\}), i.e. that the coefficients of HπH_{\pi} belong to oK\mathfrak{o}_K; hence another proof of Proposition 1.

(d) Let e~π\tilde{e}_{\pi}, l~π\tilde{l}_{\pi} and H~π\tilde{H}_{\pi} denote the series obtained by reducing eπe_{\pi}, lπl_{\pi} and HπH_{\pi} modulo πoK\pi \mathfrak{o}_K; their coefficients belong to the ring oK/πoK\mathfrak{o}_K/\pi \mathfrak{o}_K.

Show that l~π(U)=U−Up\tilde{l}_{\pi}(U) = U - U^p (note that vp(n)<(n−1)θv_p(n) < (n-1)\theta if n≠1, pn \neq 1,\ p). Deduce that

e~π(U)=U+Up+⋯+Upn+⋯ \tilde{e}_{\pi}(U) = U + U^p + \cdots + U^{p^n} + \cdots

and that

H~π(U,V)=e~π(U)+e~π(V)−(e~π(U)+e~π(V))p, \tilde{H}_{\pi}(U, V) = \tilde{e}_{\pi}(U) + \tilde{e}_{\pi}(V) - (\tilde{e}_{\pi}(U) + \tilde{e}_{\pi}(V))^p,

or also

H~π(U,V)=U+V−Λp(∑n=0∞Upn,∑n=0∞Vpn), \tilde{H}_{\pi}(U, V) = U + V - \Lambda_p \left( \sum_{n=0}^{\infty} U^{p^n}, \sum_{n=0}^{\infty} V^{p^n} \right),

where Λp\Lambda_p is defined by the formula

Λp(U,V)=(U+V)p−Up−Vp,cf. Chapter I, § 1, Exercise 19. \Lambda_p(U, V) = (U + V)^p - U^p - V^p, \quad \text{cf. Chapter I, § 1, Exercise 19.}

In particular, H~π(U,V)\tilde{H}_{\pi}(U, V) belongs to L^Fp({U,V})\hat{L}_{\mathbf{F}_p}(\{U, V\}) and its homogeneous component of degree pp is −Λp(U,V)-\Lambda_p(U, V). Then

H~π(U,V)=H~π(V,U) \tilde{H}_{\pi}(U, V) = \tilde{H}_{\pi}(V, U)
H~π(U,H~π(V,W))=H~π(H~π(U,V),W)in L^Fp({U,V,W}), \tilde{H}_{\pi}(U, \tilde{H}_{\pi}(V, W)) = \tilde{H}_{\pi}(\tilde{H}_{\pi}(U, V), W) \quad \text{in } \hat{L}_{\mathbf{F}_p}(\{U, V, W\}),
where WW is a third indeterminate.

(e) Let C(U,V)=H(−U,H(−V,H(U,V)))C(U, V) = H(-U, H(-V, H(U, V))) ("Hausdorff commutator"). Then exp⁡(C(U,V))=exp⁡(−U)exp⁡(−V)exp⁡(U)exp⁡(V)\exp(C(U, V)) = \exp(-U) \exp(-V) \exp(U) \exp(V). We write

Cπ(U,V)=1πC(πU,πV). C_{\pi}(U, V) = \frac{1}{\pi} C(\pi U, \pi V).
Show that the coefficients of Cπ(U,V)C_{\pi}(U, V) belong to πoK\pi \mathfrak{o}_K (use the fact that H~π(U,V)=H~π(V,U)\tilde{H}_{\pi}(U, V) = \tilde{H}_{\pi}(V, U)).†

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