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## Hot answers tagged differential-geometry

5

A smooth function $\lambda$ such that $d(\lambda \omega) =0$ is called an integrating factor for $\omega$. If $\omega$ is a nonvanishing $1$-form, then $\omega$ has a nonvanishing integrating factor in a neighborhood of each point if and only if $\omega\wedge d\omega=0$. The necessity of this condition follows from the Poincaré lemma: If ...

5

From $\| \gamma'(s) \| = 1$, it follows that $$0 = \frac{d}{ds} \langle \gamma'(s),\gamma'(s) \rangle = 2\langle \gamma''(s),\gamma'(s)\rangle.$$ So $\gamma'$ and $\gamma''$ are perpendicular. Assume that $\gamma''(s) \neq 0$. Thus $N(s) := \gamma''(s)/\|\gamma''(s)\|$ locally defines a unit normal field to $\gamma$. Since $\gamma$ parametrizes a level set ...

3

Let me write $\alpha = (\alpha_1, \cdots, \alpha_k)$ and $I =(i_1, \cdots, i_k)$. Then using $$w = \sum_I w_I dx^{i_1}\wedge \cdots \wedge dx^{i_k} = \sum_\alpha w_\alpha dy^{\alpha_1}\wedge \cdots \wedge dy^{\alpha_k}$$ and $dy^{\alpha_j} = \frac{\partial y^{\alpha_j}}{\partial x^i} dx^i$, we have $$w_I = \sum_{\alpha} w_\alpha \frac{\partial ... 2 This is not a complete answer, since this is worth working through on your own. Instead, here is the main idea for all of these questions: if you have a U \subset B with a trivialization of \xi|_U, you now have a way of constructing C^0 sections (C^\infty sections if B is smooth) by taking "constant" sections times a cut-off function. This is why ... 2 For a unit speed curve, 1 = (x')^2 + (y')^2. Differentiating and dividing by 2 gives 0 = x'' x' + y'' y'. Geometrically, the velocity (x', y') and acceleration (x'', y'') are orthogonal. Since the velocity is non-vanishing by hypothesis, there exists a real \kappa such that (x'', y'') = \kappa(-y', x'); taking magnitudes shows |\kappa| is in ... 2 I see two serious errors off the bat. First, in your formula for the metric, you have \sin\phi where you should have \sin\theta. But that was a typo you corrected in the next line. More substantively, your formula for E should have 1/r^2. Then everything works as it should. EDIT: Oops, one more. You need d of everything once you have the ... 1 The angle \alpha:=\angle APB is equal to \angle AQB where Q is a point on the circle passing through A, B and P; let M be its center. We know that 2\alpha=\angle AMB and that the first coordinate of M is 2. So the smaller the second coordinate y_M of M the bigger \alpha becomes. The smallest value of y_M such that P is an ... 1 The Inverse Function Theorem tells you that a smooth map with invertible derivative is an open map. I'm skeptical about your definition, by the way; I've never seen a definition thst says "differentiable" rather than C^1. The latter is absolutely needed for applications of the Inverse Function Theorem. 1 A hyperbolic metric on n-dimensional smooth manifold M is a Riemannian metric g in M which is locally isometric to the standard Riemannian metric on the hyperbolic n-space H^n. Locally isometric here means that every point in M admits a neighborhood U so that the restriction of g to U is isometric to an open subset of the hyperbolic n-space. Such metric g ... 1 The principle symbol arise naturally when you take the Fourier transform, where the symbol appears the top order multiplier. So the choice of including i is immaterial since then i^{m} is a constant. The important thing is the property of \sigma(D) (like whether it is elliptic, hyperbolic, invertible, etc), and that would not be changed by multiplying ... 1 Part (a) says that given a topology on A, there's at most one differentiable structure on A that makes (A, i) a submanifold of M. It's possible, once you have proved this, that for several different topologies on A, there are corresponding differentiable structures that make (A, i) a submanifold. Part (b) specifies a topology (the relative ... 1 Solving the Frenet-Serret equations with given functions \kappa and \tau will always give you a solution, namely a curve with the said curvature and torsion functions. The condition for this to be well-defined is that the curvature should be positive. In the case you mentioned one will therefore obtain a solution so long as s is less than \pi/2. A ... 1 There are several equivalent ways to show a set is an embedded submanifold, and I think each allows a proof of that statement in a straightforward manner. One such approach would look as follows (with some of the details left to you): (For the sake of completeness, here is the claim: M\subset\mathbb{R}^n is assumed to be a k-dimensional embedded ... 1 For a curve, ds = \sqrt{dx^2+dy^2}. Also, \tan \theta=\frac{dy}{dx}. So, we have \begin{eqnarray} \frac{ds}{d\theta}&=&\frac{ds}{dx}.\frac{dx}{d\theta}=\frac{\frac{ds}{dx}}{\frac{d\theta}{dx}}\\ &=&\frac{\sqrt{1+\left(\frac{dy}{dx}\right)^2}}{\frac{d\left(\tan^{-1}\frac{dy}{dx}\right)}{dx}} ... 1 Pez, if I understand correctly, you want to know how each polygon vertex v = (a_i,b_i) moves as you move each edge along the edge normal by \epsilon? First, compute the normals N_1, N_2 of the two edges adjacent to v. Then the inflated vertex position v' is given by$$v' = v + \frac{N_1+N_2}{\|N_1+N_2\|} \epsilon \sec \frac{\psi}{2}, where ...

1

It's useful to think of differential forms as antisymmetric multilinear mappings, i.e a $p$-form eats $p$ arbitrary vector fields and gives you an ordinary function of $n$ variables (where $n$ is the dimension of your manifold). The interior product is a mapping from a $p$ form $\omega$ to a $(p-1)$ form, since you've fixed one argument of the $p$ form to ...

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