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By Lucien Guillou, Alexis Marin

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Additional info for A la recherche de la topologie perdue: I Du côté de chez Rohlin, II Le côté de Casson

Example text

If X is not spacelike, then we can define ||X|| = ||X||2 = gijXiXj . In the exercise set you will show that null need not imply zero. Note Since “X, X‘ is a scalar field, so is ||X|| is a scalar field, if it exists, and satisfies ||˙X|| = |˙|·||X|| for every contravariant vector field X and every scalar field ˙. The expected inequality ||X + Y|| ≤ ||X|| + ||Y|| need not hold. ) Arc Length One of the things we can do with a metric is the following. A path C given by xi = xi(t) is non-null if ||dxi/dt||2 ≠ 0.

Let ˙: Sn’E1 be the scalar field defined by ˙(p1, p2, . . , pn+1) = pn+1. (a) Express ˙ as a function of the xi and as a function of the x–j (the charts for stereographic projection). (b) Calculate Ci = ∂˙/∂xi and C—j = ∂˙/∂x–j. (c) Verify that Ci and C—j transform according to the covariant vector transformation rules. 6. Is it true that the quantities xi themselves form a contravariant vector field? Prove or give a counterexample. 7. 7 are inverse functions. 33 8. 8(d). That is, obtained from the dot product with some contrravariant field.

61 First, let us restrict to M is embedded in Es with the metric inherited from the embedding. The projection of dX/dt along M will be called the covariant derivative of X (with respect to t), and written DX/dt. To compute it, we need to do a little work. First, some linear algebra. 1 (Projection onto the Tangent Space) Let M ¯ Es be an n-manifold with metric g inherited from the embedding, and let V be a vector in Es,. (∂/∂xj) as usual. Proof We can represent V as a sum, V = πV + V⊥, where V⊥ is the component of V normal to Tm .

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