By Kalantari I., Welch L.

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**Example text**

But in order to do so, we need to integrate over a two dimensional region. So let U be some open subset of M and let ψ : U → π −1 U ⊂ O be a map satisfying π ◦ ψ = id. So ψ assigns a frame to each point of U in a differentiable manner. Let C be a curve on M and suppose that C lies in U . Then the surface determines a ribbon along this curve, namely the choice of frames from which e1 is tangent to the curve (and pointing in the positive direction). So we have a map R : C → O coming from the geometry of the surface, and (with now necessarily different notation from the preceding section) R∗ Θ12 = kds is the geodesic curvature of the ribbon as studied above.

Dt t If ω is a linear differential form, then we may compute i(Y )ω which is a function whose value at any point is obtained by evaluating the linear function ω(x) on the tangent vector Y (x). Thus i(φ∗t Y )φ∗t ω(x) = dφ∗t ω(φt x), dφ−t Y (φt x) = {i(Y )ω}(φt x). In other words, φ∗t {i(Y )ω} = i(φ∗t Y )φ∗t ω. We have verified this when ω is a differential form of degree one. e. a function, since then both sides are zero. But then, by the derivation property, we conclude that it is true for forms of all degrees.

If Z = YC is the restriction of a vector field Y to C we can define its “derivative” Z , also a vector field along C by YC (t) := ∇C (t) Y. 6) If g is a smooth function defined in a neighborhood of the image of C, and h is the pull back of g to I via C, so h(t) = g(C(t)) then the chain rule says that h (t) = d g(C(t)) = C (t)g, dt the derivative of g with respect to the tangent vector C (t). 2) implies that (hZ) = h Z + hZ . 6) hold. Indeed, to prove uniqueness, it is enough to prove uniqueness in a coordinate neighborhood, where Z j (t)(∂i )C .

### A blend of methods of recursion theory and topology: A П 0^1 tree of shadow points by Kalantari I., Welch L.

by Donald

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