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**Extra info for Constant Mean Curvature Surfaces, Harmonic Maps and Integrable Systems (Lectures in Mathematics. ETH Zürich)**

**Sample text**

Proof. We build a perturbation of u which is a modulation of the action of X by a function ¢ E C' (SZ, R). It gives us a map v, from St to R" such that v,(x + s$(x)X(x)) = u(x). Note that if s is sufficiently small, x 1 1 x + s4(x)X(x) is a diffeomorphism of Q. ;)X (l;))) d6 + o(s). [Il - s¢(x)d(X (x))]) +s J [- E E 0,(3=1 i=1 m +E 0=1 a a a (x) [1 + s¢(x)divX (x)] dx (x)X s(x) OU, (x, u(x), du(x)) ° (x)Xc(x)L(x, u(x), du(x))] dx + o(s) 3. Variational point of view and Noether's theorem An [u] - s E J a,p=1 in 33 aX0 (x)X' (x)H; (x) dx + o(s).

Step 1: Local existence. (x)) aM' where we denote (MA. (x)) aM :_ E M,Ao(x)kaj . 1 *
*

0 = 0. 1 We say that u: S2 S" is isotropic, if and only if 17,,Q = 0 for all a,,3 > 0 satisfying a +,3 >- 1. 1. We will apply induction on a +,3. The claim is clear for a + Q = 1, since 7701 = 7710 = (au, u) = 0. Assume now that -y E N is given, and that dap = 0 for all a,,3 such that 1