# Tag Info

6

No, it's not true. Two connections $\nabla_1$ and $\nabla_2$ have the same geodesics if and only if their difference tensor $D(X,Y) = (\nabla_1)_X Y - (\nabla_2)_X Y$ is antisymmetric, meaning $D(X,Y) = - D(Y,X)$ for all $X,Y$. Let $\overline\nabla$ denote the Euclidean connection on $\mathbb R^3$, and choose a smooth bump function $\psi$ supported in a ...

5

Connections are not tensors, but that does not mean they are not coordinate-independent objects! A linear connection is a map sending two vector fields $X,Y$ to another vector field $\nabla_X Y$ which satisfies the rules $\nabla_{fX+Z} Y = f \nabla_X Y + \nabla_Z Y$ and $\nabla_X (fY + Z) = f\nabla_X Y + (\nabla_X f)Y + \nabla_X Z$. This is an abstract ...

5

Consider $\mathbb{R}\hookrightarrow\mathbb{R}^3\to\mathbb{R}^2$ where the projection map is given by projection onto the first two coordinates. I'm going to define connections in terms of one-forms. Since one-forms annihilate a two-dimensional subspace, the kernel of the one-form will give the horizontal subbundle of the tangent bundle of the total space. ...

5

This seems to be false in general. Define the following vector fields on $\mathbb R^3$: $$X = \frac{\partial}{\partial x} - y\frac{\partial}{\partial z}, \qquad Y = \frac{\partial}{\partial y} + x\frac{\partial}{\partial z}, \qquad Z = \frac{\partial}{\partial z},$$ and let $g$ be the metric for which $\{X,Y,Z\}$ is an orthonormal basis. Then $Z$ is a ...

5

When you look at a circle, you are seeing its extrinsic curvature, which is also what your link is calculating. That is a property of how the circle is imbedded into another manifold, not a property of the circle as a manifold itself. The curvature being referred to here is the intrinsic curvature, which is defined strictly in terms of the manifold itself, ...

4

Following Élie Cartan, you want to think of your differential system $dR - R\omega = 0$ on $M\times SO(n)$. This $\mathfrak{so}(n)$-valued $1$-form is integrable, as you said, because of vanishing curvature. The integral manifolds of this differential system will locally be the graphs of functions $R\colon U\to SO(n)$.

4

Replace $u$, in the Wiki param, with $u + \pi/2$, and swap $x$ and $y$, and you'll more or less have the book's version. As for "proving" the parameterization, that's hard to say: a parameterization isn't a statement that requires proof. But to see how such a parameterization is developed in the first place, the answers here should prove useful.

4

Project $S^2$ to the plane, then compose with an embedding $\Bbb R^2 \to S^2$. This is a submersion on each of the hemispheres. You can do a similar construction for any smooth manifold $M$; pick a smooth map $M \to \Bbb R^n$ with surjective differential at some point. (That we can do this is an exercise.) Now compose with a smooth embedding $\Bbb R^n ... 4 As mentioned in a previous answer, connections are quite intrinsic. I will take the slightly more pedestrian view that a connection is a collection of maps from type$(k,l)$tensor fields to type$(k,l+1)$fields that is linear, satisfies a Leibniz rule, commutes with contraction, agrees with the differential for smooth functions and is usually assumed to be ... 4 If your interpretation of "smooth submanifold" is smooth embedded submanifold (meaning that it has the subspace topology), then these two definitions are equivalent. The proof of the existence of a holomorphic slice works just like the smooth case, but using the holomorphic version of the inverse function theorem instead of the smooth one. Here's a sketch of ... 3 Not always. Since the canonical bundle is a complex line bundle, it's classified topologically by its first Chern class$c_1 \in H^2(X, \mathbb{Z})$, which is also the first Chern class of the cotangent bundle of$X$. There's no reason that this should be zero in general. For example, if$X$is a closed Riemann surface of genus$g$, then$c_1$vanishes iff ... 3 In contrast to what is written in the question and some of the comments, a connection is independent of coordinates. The intrinsic way to define connections is the following. For simplicity, we treat only connections on the tangent bundle, even though a similar definition can be applied for any vector bundle. Let$M$be a smooth manifold. Let$TM$denote ... 3 General idea: Take the set$U=U_{\alpha_0\ldots\alpha_p}$. By performing$\delta$twice on some form on$U$, you will get the set$V=U_{\alpha_0\ldots\alpha_p\beta_0\beta_1}$twice, and the two forms on$V$will cancel each other out. This is due to the$(-1)^p$in the definition of$\delta$. To get the right feel, you may calculate$\delta$explicitly for ... 3 The isometry group of a Riemannian manifold is a finite-dimensional Lie group. A Killing field defines a 1-parameter subgroup$\Bbb R \to \text{Isom}(M)$by flowing; and conversely, taking the derivative of a 1-parameter subgroup$\Bbb R \to \text{Isom}(M)$at$e$gives a Killing field. By exponentiating, one sees that 1-parameter subgroups of a Lie group ... 3 Okay, this is more of a collection of rambling ideas, but it is way to long for a comment, so I will have to post it as an answer. To give a short answer, I have never seen something completely like this but I have some observations to be made, that hopefully may help. First of all considering one of your side notes, you can can use the Hausdorff-distance ... 2 By definition, we have$$x_c=\frac{\int_Vxdxdydz}{\int_Vdxdydz},$$or equivalently,$$x_c\int_Vdxdydz=\int_Vxdxdydz.$$Hence,$$\int_V(x-x_c)dxdydz=\int_Vxdxdydz-x_c\int_Vdxdydz=0.$$ 2 Here's a fairly detailed sketch: Let$M$be an$n$-manifold. Suppose, contrapositively, that$M$is orientable, and fix a maximal oriented atlas, i.e., a maximal atlas for which all the transition maps have Jacobian with positive determinant. Lemma 1: If$(U, \phi)$is a chart and$U$is connected, then$(U, \phi)$is either compatible with the ... 2 Shouldn't the rank be constant on a neighborhood of that point to make this conclusion? It is sufficient to check at that point if the rank is maximal, but otherwise the rank can "jump" up suddenly, meaning you would need more coordinates. It can't jump down suddenly, because the determinant function is smooth. It takes some time for a nonzero continuous ... 2 A way to explicitly construct such kind of vector field is to use the stereographic projection to transplant a constant vector field like$\frac{\partial}{\partial x}$from the$(x,y)$plane. Using the projection the sphere is parameterized as $$r(x,y)=(\frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}, \frac{-1+x^2+y^2}{1+x^2+y^2})$$ Then ... 2 Using$x$and$y$as parameters makes perfect sense. This means you are using the diffeomorphism$\Phi(x,y) = (x,y,\sin x+\sin y)$to cover the whole surface with a coordinate patch. Pushing forward the standard vector fields$\frac{\partial}{\partial x}$and$\frac{\partial}{\partial y}$from$\mathbb{R}^2$to the surface gives a basis of the tangent space, ... 2 To be a manifold, a space$X$has to be covered in coordinate maps$\varphi: A \subseteq \Bbb R^n \to X$. This is part of the definition. In particular, consider$\varphi$to be a coordinate map onto a neighborhood of your point$x$. If$X$is embedded into$R^m$by some other map$\psi$, then$\psi \circ \varphi$is a smooth map that could be considered a ... 2 If$M$is a closed codimension-1 submanifold of$\mathbb R^n$, smoothly embedded, then the first method is in fact completely general, in the sense that if you have such a manifold, you can compute the signed distance to$M$and call this$f$;$0$is then a regular value, and$f^{-1}(0) = M$. For things like the (open) Mobius band, however, that's not ... 2 Of course there is. Compose the restriction of the projection$(x,y,z)\in\mathbb R^3\mapsto (x,y)\in\mathbb R^2$to the sphere$S^2$with the map$(x,y)\mapsto(x/2,y/2)\in\mathbb R^2$with the map$(x,y)\in(x,y,\sqrt{1-x^2-y^2})\in S^2$. 2 In the equation $$\int_{\partial \Omega} \alpha = \int_{\Omega} d\alpha$$ it is assumed that$\alpha$is defined on$\Omega$. If$i\colon\partial\Omega\to\Omega$is the inclusion, it induces a restriction map$i^*$on differential forms in the opposite direction. Really the$\alpha$on the left-hand side is$i^*\alpha$. We don't usually bother with ... 1 There's a reason$\dfrac{\omega^n}{n!}$shows up all over complex geometry as the induced volume form. You are correct and there is an error in whatever you're reading. 1 To prove this you need the useful lemma that if$U$is a neighborhood of$0$in$\Bbb R^n$and$g\colon U\to\Bbb R$is smooth with$g(0)=0$, then there are smooth functions$g_i\colon U\to\Bbb R$so that $$g(x) = \sum_{i=1}^n x_ig_i(x).$$ Moreover,$g_i(0) = \dfrac{\partial g}{\partial x_i}(0)$. Now use the properties of a derivation. 1 I'll just type in the answer of egreg (thanks!) to get it over with. This fits the context very well. In the paper$\alpha_i$is a multi-index,$\alpha_i = (a_1,a_2,\dots,a_d)$, with$a_{j} \in \mathbb{N}$and$\alpha_i!=a_1! \cdot a_2! \cdot \ldots \cdot a_d!$. This seems clear looking at equations A.6 and A.7. See ... 1 Theorem: For any closed set$C\in \mathbb{R^n}$there exists a smooth function$f:\mathbb{R^n} \to \mathbb{R}$s.t$f^{-1}(0)=C$. so by using this theorem it is clear thatfor any closed parametrized curve that property which you are asking is true. Otherwise in genral it is not true...for an instance consider$f : (0,1) \to \mathbb{R^2}f(x)= ...

1

Eric, even in Guillemin & Pollack you'll find a more general version. Look at Exercise 14 on p. 56. It removes the compactness hypothesis on $Z$. There are also interesting questions to ask along the lines of this: If $f\colon\Bbb R^n\to\Bbb R^n$ (replace with manifolds if you wish) is a local diffeomorphism at each point, what condition(s) are ...

1

I'm assuming you are working with the definitions given in Do Carmo's "Curves and Surfaces" book and that $V \subseteq \mathbb{R}^3$ is an open set. First, let us write precisely the domains and ranges of the maps involved: The map $\mathbf{x}_1 \colon U_1 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_1 \subseteq \mathbb{R}^2$ is an ...

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