By Francis E. Burstall, Dirk Ferus, Katrin Leschke, Franz Pedit, Ulrich Pinkall (auth.)

The conformal geometry of surfaces lately constructed by way of the authors ends up in a unified knowing of algebraic curve idea and the geometry of surfaces at the foundation of a quaternionic-valued functionality conception. The ebook bargains an effortless creation to the topic yet takes the reader to fairly complicated themes. Willmore surfaces within the foursphere, their Bäcklund and Darboux transforms are coated, and a brand new evidence of the class of Willmore spheres is given.

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**Additional resources for Conformal Geometry of Surfaces in S 4 and Quaternions**

**Example text**

A surface in S', if and only if S > A get < 0,,0 we curve = A < 0, V) L in 0,0 >. > = 0. 2 In the half-space Poincar6 model of the hyperbolic space, geodesics are orthogonally intersect the boundary. e. those 2-spheres in RP' that orthogonally intersect the separating isotropic S3. 2. In particular, it induces an isometry of the standard Riemannian metric of RP' which fixes S3. Given a 2-sphere S E End(EV), S2 -I, that intersects S3 in a point 1, we use affine coordinates, as in Example 4, with 1 =:F vH and w or euclidean circles that = .

DN 2 2 But H(NdN - *dN). 3 The Willmore Condition in Affine Coordinates We use the notations of the previous Proposition 12, and in addition abbre- viate v, = dR+R*dR. Note that V Proposition 13. -dR +. *dRR = The Willmore -dR -- integrand A A *A > = 16 For f : M -4 R, this is * dR = -v. given by 1 1 < R - JRdR - *dR12 is the classical = 4 (IHI2 - K integrand 1 < A A *A >= 4 (Ih 12 - K)Idfl2. - K-L)JdfJ2. 3 The Willmore Condition in Affine Coordinates 45 Proof. < A A *A > 8 traceR(-A' 1 4 Now see We Proposition now 1V) 4 Re( - (*A)) = IV12 = 2 16 4 express the and * A = JdR + R * dR12 16 jRdR - *dR12.

AAQ=O Therefore < dS A *dS > 4 < = (*Q - *A) (A A Q) - > =-4<*QAQ> -4< *AAA> 4 < = and similarly S, A *Q > +4 < A A *A >, for < dS A SdS >. Lemma 7. Let V be B E Q End(V) a vector space, L C V quaternionic a quaternionic line, such that S2 SB T, _j image B -BS, = C L. Then 2 traceR B with equality if and only if BIL = < 0, 0- may assume B 54 0. Then L L. Let 0 E L\101, and Proof. We SL = So Then 'X2 =-1, and BS = OA, = -SB BO can be applied to A = or OIL. -PA. B20 = _JpJ2 0, traceR B2 IL = -41p 12.