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By P. Baird

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Suppose ~ *h has eigenvalues v1 , ••• , v with respect to g, where v1 > ••• > v > 0. e. if T. is the eigenspace of "·• then a subset of x 1 , ••• X J J forms an orthonormal basis of T.. The energy density of the map . 1) "k/2 • H P E ~(0 2 T• M), then let P .. 1) P(X. ) denote the components of P with 1 J respect to X 1 , ••• , Xm. Then 47 eO~)gij -u~*h) .. s1J.. )o .. 1 1J q ( 1 - 2 oiq) 2 Now - e(0) ~ IJ. 1 IJ q therefore s1J.. 2) o .. •. ,l'n=l = 0. m Example 3. 3. 2 A: M - 1f 0 is a harmonic morphism, as in Section 1.

Define f: sm- 1 - R by f(z) = coss. The level hypersurfaces are (m-2)-dimen- sional small spheres. Each focal variety is a single point r- 1 (1) and f-1(-1). (iii) on Hm- 1 there are three distinct cases. (a) For each z E H m-1 , write z = (coshs, sinhs y) ,yES m 1 - and s E [O,co). Let f: Hm-1- R be defined by f(z) = coshs. The level hypersurfaces are (m-1)- dimensional spheres. There is one focal variety which is a point. 34 Let z E Hm- 1 be written as z = (coshs x, sinhs), x E Hm-2, s E [O,oc) • pefine f : H m-1 - R by f(z) = sinhs.

0 is conformal. _0 :(M,g)- (N,h), then s 0 = 0 if and only if dim M = 2 and 0 is conformal. 3 (N,h) is a harmonic map of the 2-sphere into a Riemannian manifold N, then 0 is conformal. Proof Choose local isothermal coordinates z on s 2 , so that g = p(z) dz dz. Then a tensor T E ~(0 2 T* M) has a type decomposition of the form T = T ( 2 ' 0 ) + T ( 1 • 1 ) + T(0, 2 ), where T(i,j) is spanned by dzi dzj locally. A calculation shows that (0*h)(1, 1 ) =e(0)g. Thus ~ 0 =s/ 2 •0) + s~0,2), where sJ 0 • 2) = s/2,0).

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