By David Bachman

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If V2 = z, 1, xy (x,y,z) and ω is the differential 2-form, x 2 y dx ∧ dy − xz dy ∧ dz, then ω(V1 , V2 ) = x 2 y dx ∧ dy − xz dy ∧ dz( 2y, 0, x = x2y (x,y,z) , z, 1, xy (x,y,z) ) 2y z 0 1 − xz = 2x 2 y 2 − x 2 z, 01 −x xy which is a function from R3 to R. Notice that V2 contains the vector 3, 1, 2 (1,2,3) . So, from the previous example we would expect that 2x 2 y 2 − x 2 z equals 5 at the point (1, 2, 3), which is indeed the case. 1. Let ω be the differential 2-form on R3 given by ω = xyz dx ∧ dy + x 2 z dy ∧ dz − y dx ∧ dz.

5 What about surfaces? 31 1 (ai ) But lim a→0 φ1 (ai+1 )−φ is precisely the definition of the derivative of φ1 a dφ1 1 at ai , da (ai ). Hence, we have lim a→0 F (ai ) = f (φ1 (ai ))| dφ da (ai )|. 1. Check that −1 1 f (φ1 (a))| dφ da |da = π 0 1 f (φ1 (a))| dφ da |da. 2. 1 Recall that dφ da is a vector, based at the point φ(a), tangent to M. If we think 1 of a as a time parameter, then the length of dφ da tells us how fast φ1 (a) is moving 1 along M. How can we generalize the integral, −1 1 f (φ1 (a))| dφ da |da?

Dx ∧ dy + dx ∧ dz. 3. 3dx ∧ dy + dy ∧ dx + dx ∧ dz. 4. dx ∧ dy + 3dz ∧ dy + 4dx ∧ dz. 18. Find a 2-form which is not the product of 1-forms. In doing this exercise, you may guess that, in fact, all 2-forms on Tp R3 can be written as a product of 1-forms. Let’s see a proof of this fact that relies heavily on the geometric interpretations we have developed. Recall the correspondence introduced above between vectors and 1-forms. If α = a1 dx + a2 dy + a3 dz then we let α = a1 , a2 , a3 . If V is a vector, then we let V −1 be the corresponding 1-form.

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