Proposition 1

Tag 0AE1, permanent. Cite it as [Bourbaki, Tag 0AE1].

Lie Groups and Lie Algebras, chapter III, § 5. Formal calculations in Lie groups

(i)

x^{[-1]} \equiv -x + B(x, x) \mod \deg 3

(ii)

x.y.x^{[-1]} \equiv y + [x, y] \mod \deg 3

(iii)

y^{[-1]}.x.y \equiv x + [x, y] \mod \deg 3

(iv)

x^{[-1]}.y^{[-1]}.x.y \equiv [x, y] \mod \deg 3

(v)

x.y.x^{[-1]}.y^{[-1]} \equiv [x, y] \mod \deg 3.

(In (i),

x^{[-1]}

of course represents the expansion of the function

x \mapsto x^{[-1]}

as an integral series about the origin; the other formulae can be interpreted analogously.) Let

g_1

and

g_2

be the homogeneous components of

x^{[-1]}

of degrees 1 and 2. Then

0 = x.x^{[-1]} \\ \equiv x + g_1(x) + B(x, g_1(x)) \mod \deg 2 \quad \text{(by (6))} \\ \equiv x + g_1(x) \mod \deg 2

and hence

g_1(x) = -x

. Then

0 = x.x^{[-1]} \\ \equiv x + (-x + g_2(x)) + B(x, -x + g_2(x)) \mod \deg 3 \\ \equiv g_2(x) - B(x, x) \mod \deg 3

and hence $g_2(x) = B(x, x)$ . This proves (i). Then, using (i),

x.y.x^{[-1]} \equiv (x + y + B(x, y)).(-x + B(x, x)) \quad \text{mod deg 3}

\equiv x + y + B(x, y) + (-x + B(x, x)) + B(x + y, -x) \quad \text{mod deg 3}

\equiv y + B(x, y) - B(y, x) \quad \text{mod deg 3}

\equiv y + [x, y] \quad \text{mod deg 3 (by (13))}

whence (ii). The proof of (iii) is analogous. Combining (i) and (iii), we obtain

x^{[-1]},y^{[-1]}.x.y \equiv (-x + B(x, x)).(x + [x, y]) \quad \text{mod deg 3}

\equiv -x + B(x, x) + x + [x, y] + B(-x, x) \quad \text{mod deg 3}

\equiv [x, y] \quad \text{mod deg 3}

whence (iv). The proof of (v) is analogous.

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

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