By Biasi C., de Mattos D.

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**Extra resources for A Borsuk-Ulam Theorem for compact Lie group actions**

**Example text**

Moreover, |˜ ρ(k)| ≤ 1 . In general, in the interval −π ≤ k ≤ π, the maximum ρ˜(k) = 1 is reached only for k = 0 except if ρ(q) is non-vanishing only on a subset of the form q = a + mb m ∈ Z. with a, b ﬁxed and b > 1 , A simple example of such a situation, with a = 1, b = 2, is ρ(q) = 1 2 for q = ±1 . In a ﬁrst analysis, we exclude this situation and comment on it later. Then, the Fourier series has the properties required to prove the central limit theorem. 7): ρ(Q )Pn (Q − Q ).

The beneﬁt of considering this particular function, rather than the Fourier transform, is that the integrand is still a positive measure. The function Z(b) then is a generating function of the moments of the distribution, that is, of expectation values of monomials. Indeed, one recognizes, expanding the integrand in powers of the variables bk , the series ∞ Z(b) = =0 1 ! n bk1 bk2 . . bk xk1 xk2 . . xk . ,k =1 Expectation values can thus be obtained by diﬀerentiating the function Z(b) with respect to its arguments.

Let us point out that the proof of this convergence requires only weaker assumptions. Examples and counter-examples. (i) The distribution uniform on the segment [−1, +1] and vanishing outside, ρ(q) = 1 2 sgn(q + 1) − sgn(q − 1) , ⇒ q = 0 , q2 = 1 3 , (sgn is the sign function) is centred around zero and satisﬁes the hypotheses of the central limit theorem. Its Fourier transform is sin k 1 1 . dq eiqk = ρ˜(k) = 2 −1 k The generating function of cumulants has the expansion w(k) = ln(sin k/k) = − 61 k 2 − 4 1 180 k + O(k 6 ).