By Fuchs L.

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1. A path is a continuous mapping : [a, b] → Rn . We call (a) the initial point and (b) the final point. The image of the path, ([a, b]), is called the arc1 of . If ([a, b]) ⊂ Ω , we say that is a path in Ω . 1. The line segment joining two points x, y ∈ Rn is the arc [x, y] := ([0, 1]), where : [0, 1] → Rn denotes the path (t) = x + t(y − x) = ty + (1 − t)x. 2. Let j : [0, 2π ] → R2 be given by j (t) := (cos( jt), sin( jt)). Then for every j ∈ Z \ {0}, the arc j ([0, 2π ]) is the unit circle x2 + y2 = 1 in R2 .

J=1 After identifying vectors with linear forms, it is quite natural to identify a vector field on a set U with a mapping that associates to each point of U a linear form. 1. Let U ⊂ Rn be an open set. A differential form of degree 1 on U, or simply a 1-form, is a mapping ! : U ⊂ Rn → L (Rn , R) = (Rn )∗ . , a vector x ∈ U, and an integer j ∈ {1, . . (x)(e j ) ∈ R by f j (x). (x) ∈ (Rn )∗ , for any h = (h1 , . . h j = j=1 n ∑ j=1 f j (x)h j = n ∑ f j (x)dx j j=1 (h). = n ∑ f j · dx j j=1 and call f j the component functions of !.

Moreover, we have ˜ : U −→ V and G : V −→ Rm with ˜ and G differentiable (respectively of class Cq ) on their respective domains. Now the chain rule gives the conclusion. The same argument clearly proves the following further variation of the chain rule. 2. Let α : [a, b] −→ R be differentiable at each point of [a, b] and let ˇ : [c, d] −→ Rm be a path that has a derivative at each point of [c, d] with α ([a, b]) ⊂ [c, d]. Then ˇ ◦ α : [a, b] −→ Rm has a derivative at each point of [a, b] and (ˇ ◦ α ) (t) = ˇ (α (t))α (t), for every t ∈ [a, b].

### Gesammelte mathematische Werke by Fuchs L.

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