By B.H. Gross, B. Mazur

Ebook by means of Gross, B.H.

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**Sample text**

3 is the case corresponding to q = 1. For a holonomy invariant l/ we have dl/ = 0 and 0: = 0 is a suitable choice, hence the De Rham class in question vanishes. If however Ex r F ---t B arises from a representation of r by isometries of a Riemannian metric gF on F, then this will turn F into a Riemannian foliation. Consider again an arbitrary foliation on (M, g), with induced metric gQ on Q. 18) (8(Y)gQ)(X,X') = YgQ(X, X') - gQ(7f[Y, X], X') - gQ(X,7f[Y,X']), where X, X' E r LJ... Therefore (8(Y)gQ)(X,X') = Yg(X,X') - g([Y,X],X' ) - g(X, [Y,X']) = g(V'WY,X ' ) + g(X, V'W,Y).

FOT Y E V(F) and X, X' E 48 5 TRANSVERSAL RIEMANNIAN GEOMETRY An infinitesimal automorphism Y is transversally metric, if 8(Y)gQ = o. If this holds, Y = 1f(Y) is called a transversal Killing field (Molino [Mo7,8]). For the point foliation with L = 0 this is the usual definition of a Killing vector field. 22). Thus ~8(Y)gQ is the symmetric part of the 2-linear form V'w on Q. A transversally oriented Riemannian foliation has a canonical holonomy invariant transversal volume v. We state the following fact.

15 PROPOSITION. a is symmetric. Proof. a(X, Y) - a(Y, X) = -\7 X7r(Y) + \7Y7r(X) +7r[X, Y] = -T\7(X, Y), which vanishes. D In the sequel we will use the symbol a for the restriction of this form to L. 16) for U, V E fL. a is the second fundamental form of the leaves of F in (M, g). The equation a = 0 (along L) holds if and only if each leaf of F is a totally geodesic submanifold of (M,g). To give an interpretation of the second fundamental form, consider Y E f Ll.. which is an infinitesimal automorphism of F.

### Arithmetic on Elliptic Curves with Complex Multiplication by B.H. Gross, B. Mazur

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