Nonlinear Equations - UFRJ

76 [CH. 6: EXPONENTIAL SUMS AND SPARSE POLYNOMIAL SYSTEMS
2. The map m : M → A ⊂ (R n ) ∗ is volume preserving, in the
sense that for any measurable U ⊆ A,
Proof. We compute explicitly
Vol(m −1 (U)) = π n Vol(U)
m(x) =
∑a∈A ac 2a re(x)
ae
∑
a∈A c 2a re(x)
ae
where we assimilate a to a 1 dq 1 + · · · + a n dq n .
Every vertex of A is in the closure of the image of m. Indeed,
let a ∈ (R n ) ∗ be a vertex of A and let p ∈ R n be a vector such that
ap ≥ a ′ p for all a ′ ≠ a. In that case, m(e tp ) → a when t → ∞.
Also, it is clear from the formula above that the image of m is a
subset of A.
The will prove that the image of m is a convex set as follows:
f(x) = −m(x) = − 1 2
log K(x, x) is a convex function. Its Legendre
transform is
f ∗ (α) = αx + m(x)
Therefore, the domain of f ∗ is {−m(x) : x ∈ R n } which is convex
(Proposition 6.6).
Now, we consider the map ˆm from M to A × R n ⊂ C n /2πZ n ,
given by
ˆm(x + √ −1y) = m(x) + √ −1y.
The canonical symplectic form in C n is η = dx 1 ∧dy 1 +· · ·+dx n ∧
dy n . We compute its pull-back ˆm ∗ η:
Differentiating,
ˆm ∗ η = η(D ˆmu, D ˆmv)
D ˆm(x + √ −1y) : ẋ + √ −1ẏ ↦→ D 2 ( 1 2 log K(x, x))ẋ + √ −1ẏ