By Francis E. Burstall, Dirk Ferus, Katrin Leschke, Franz Pedit, Ulrich Pinkall
The conformal geometry of surfaces lately constructed through the authors ends up in a unified figuring out of algebraic curve concept and the geometry of surfaces at the foundation of a quaternionic-valued functionality thought. The ebook deals an hassle-free advent to the topic yet takes the reader to particularly complicated themes. Willmore surfaces within the foursphere, their Bäcklund and Darboux transforms are lined, and a brand new evidence of the category of Willmore spheres is given.
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Additional resources for Conformal Geometry of Surfaces in Quaternions
3. E(S) harmonic submanifold = E inner The energy 7. M is End(V) I S2 Fnd(EV) I XS -SX}, I YS SY}. JY E End(fff) IX (indefinite) < defined = fm := functional < dS A *dS >. with respect to variations of S are called M to Z. S is harmonic if and only if the Z-tangential component of dS vanishes: (d This condition is equivalent to * any dS)T of the F. E. : LNM 1772, pp. 31 - 38, 2002 © Springer-Verlag Berlin Heidelberg 2002 = 0. 2) = 0, (6-3) = 0. 4) fact, d(S Proof. St be Let a Q * = 4d * of S in variation fm d d Wt- E(S) the 4d = S(d dS)T * Z with = (Sd variational * dS)T.
HIE E is is the called a is equivalent to the fact 0 E F(E), projection. 1) do) and hence induces a holomorphic structure F(E) into itself, complex line bundle E. line bundle A complex line bundle E C H induces a quaternionic of H maps the on L=EH=EE)Ej CH. to the structure of a JE admits a unique extension complex structure bundle (L, J), namely right-multiplication complex quaternionic by (-i) on line bundle (L, J) C H induces a a complex quaternionic Ej. Conversely, line bundle complex The E:= Definition lift 10.
HP1 with deriva- L C H in curve Then there exist complex structures J such that We want to extend J and j = to S E a jj complex structure F(End(H)) such that SL this Q case". e. 4) dSL c L. of S is'clear: The existence bundle L' Since L' Identify not unique, L. is H = 0, R can be Q: interpreted I((S 4 + 1(SdS 4 Q+ V) E F(L), SIL complementary Ji SIP := j. := S + R is kerR, and We compute If some if and RS + SR Note that for and define 7r, RH c L c whence R2 L E) L' = It is easy to see that S is not unique.