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By S. Waner

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Xn) = 0. Show that Xi is identically zero in any coordinate system. 2. Give and example of a contravariant vector field that is not covariant. Justify your claim. 3. Verify the following claim If V and W are contravariant (or covariant) vector fields on M, and if å is a real number, then V+W and åV are again contravariant (or covariant) vector fields on M. 4. 7: If Ci is covariant and Vj is contravariant, then Ck Vk is a scalar. 5. Let ˙: Sn’E1 be the scalar field defined by ˙(p1, p2, . . , pn+1) = pn+1.

Thus, we can think of covariant tangent fields as nothing more than 1forms. Proof Here is the one-to-one correspondence. Let F be the family of 1-forms on M (or U) and let C be the family of covariant vector fields on M (or U). Define ∞: C’F by ∞(Ci)(Vj) = Ck Vk . In the homework, we see that Ck Vk is indeed a scalar by checking the transformation rule: C—k V—k = ClVl. The linearity property of ∞ now follows from the distributive laws of arithmetic. We now define the inverse 30 §: F’C by (§(F))i = F(∂/∂xi).

An en ·e1 + 0 V·e2 = a1 e1 ·e2 + ... + an en ·e2 ... V·en = a1 e1 ·en + ... ei]g* *. ei], 62 as required.  For reasons that will become clear later, let us now digress to look at some partial derivatives of the fundamental matrix [g* *] in terms of ambient coordinates. ∂ ∂  ∂ys ∂ys [g ] = p  q r ∂xp qr ∂x ∂x ∂x ∂2 ys ∂ys ∂2 ys ∂ys = p q r + r p q ∂x ∂x ∂x ∂x ∂x ∂x or, using “comma notation”, gqr,p = ys,pq ys,r + ys,rp ys,q Look now at what happens to the indices q, r, and p if we permute them (they're just letters, after all) cyclically in the above formula (that is, p→q→r), we get two more formulas.

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