Bourbaki

Proposition 1

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

Lie Groups and Lie Algebras, chapter VI, § 3. Exponential invariants

If 2 is not a zero divisor in AA, the elements J(ep)J(e^p) for p∈P∩Cp \in P \cap C form a basis of the module of anti-invariant elements of A[P]A[P].

The weights w(p)w(p) for w∈Ww \in W and p∈P∩Cp \in P \cap C are pairwise distinct. It follows that the J(ep)J(e^p) for p∈P∩Cp \in P \cap C are linearly independent.

On the other hand, let x=∑pxpepx = \sum_p x_p e^p be an anti-invariant element of A[P]A[P]. If p0p_0 belongs to a wall, it is invariant under a reflection s∈Ws \in W and

x=∑pxpep=−s(x)=−∑pxpes(p). x = \sum_p x_p e^p = -s(x) = -\sum_p x_p e^{s(p)}.
It follows that 2xp0=02x_{p_0} = 0, so xp0=0x_{p_0} = 0. Since every element that does not belong to any wall can be written uniquely in the form w(p)w(p) with w∈Ww \in W and p∈P∩Cp \in P \cap C, we thus have
x=∑p∈P∩C∑w∈Wxw(p)ew(p).(2) x = \sum_{p \in P \cap C} \sum_{w \in W} x_{w(p)} e^{w(p)}. \tag{2}
Since w(x)=∑pxpew(p)=ε(w)∑pxpepw(x) = \sum_p x_p e^{w(p)} = \varepsilon(w) \sum_p x_p e^p, xw(p)=ε(w)xpx_{w(p)} = \varepsilon(w)x_p and we deduce from (2) that
x=∑p∈P∩CxpJ(ep), x = \sum_{p \in P \cap C} x_p J(e^p),
which completes the proof.

Consider now the element dd of the algebra A[12P]A[\frac{1}{2}P] defined by

d=∏α∈R,α>0(eα/2−e−α/2)=eρ⋅∏α∈R,α>0(1−e−α)=e−ρ⋅∏α∈R,α>0(eα−1). d = \prod_{\alpha \in R, \alpha > 0} (e^{\alpha/2} - e^{-\alpha/2}) \\ = e^{\rho} \cdot \prod_{\alpha \in R, \alpha > 0} (1 - e^{-\alpha}) \\ = e^{-\rho} \cdot \prod_{\alpha \in R, \alpha > 0} (e^{\alpha} - 1).
Since ρ∈P\rho \in P, d∈A[P]d \in A[P].

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

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Cites

Nothing.

Transcribed from the printed volume. Not translated.