By Isaac Chavel

The fundamental objectives of the booklet are: (i) to introduce the topic to these attracted to studying it, (ii) to coherently current a few uncomplicated thoughts and effects, presently utilized in the topic, to these operating in it, and (iii) to give a few of the effects which are appealing of their personal correct, and which lend themselves to a presentation now not overburdened with technical equipment.

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Additional resources for Eigenvalues in Riemannian Geometry

Sample text

Then P(u) is afundarnental domain of T, that is, no two distinct points of P(u) __ represent the same point of T, and the image of the closed parallelopiped P(u), under the covering, is all of T. Note that v01 T = V O P(u) ~ for every u E r. We simply refer to P(u) as a copy of T. Also note that if A: R" + R" is a linear transformation for which r = A(Z'), then r*= (A*)-'(E"),where A* is the adjoint of A. Therefore, if T* is the torus determined by r*then we have vol T = ldet A1 = (vol T*)-'; 3.

R, be pairwise disjoint normal domains in M , whose boundaries, when intersecting d M , do so transversally. Given an eigenvalue problem on M , consider, for each r = 1, . , m, the eigenvalue problem on R, obtained by requiring vanishing Dirichlet data on dR, n M and by leaving the 18 1. The Laplacian original data on aR, n aM unchanged. Arange all the eigenvalues of in an increasing sequence a,, . ,R, 0I v1 I I v2 * * a with each eigenvalue repeated according to its multiplicity, and let the eigenvalues of M be given as in (79).

From (44) we would conclude that t , E (0,n/2&). But 1(6) I A(n/2fi) = nK implies that { ( V c , - v's,)s"-'}(t) = {n(6) - n K } 1: Vs: is nonnegative on (0, 6). Therefore at t = t , we have V ( t , ) 2 0-a tradiction. Thus for K > 0, 6 > n/2<~, we have p(6) > 1(6). con- THEOREM4. If K = 0, that is, if M K= R", then there exist positive constants c D ,cNsuch that A(6) = Ci/62, p(6) = Ci/62 for all 6 > 0. PROOF:For K = 0 we have S,(t) = t, C,(t) = 1, so the dzerential equation under study is y" + ( n -t 1) ~ 1(1 + n - 2) y = 0.