# Tag Info

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The divergence of a vector field $V = V^a \partial _a$ on a (pseudo-)Riemannian manifold is given by $\operatorname{div} V = V^a{}_{;a}$. In words, this is obtained by taking the trace of the total covariant derivative. In the special case of $\mathbb R^n$ with a flat Euclidean or pseudo-Euclidean metric, this yields the usual calculus formula for the ...

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Author here: It's not surprising that you don't understand why -- this chapter is just supposed to suggest ideas that will be developed fully later in the book. This particular fact follows from properties of the exponential map, developed in Chapter 5. See also the bottom of page 145.

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A simple example is boundedness: you can't deduce that a function is bounded based only on its local behavior. Said another way, the presheaf of bounded functions is not in general a sheaf.

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Instead of squares I would have taken circles. But come to think of it, we might simply check for the existenc of an equilateral triangle of side length $2$. In $(0,2)\times\Bbb R$, the points $(a,-1)$, $(a,1)$, $(\sqrt 3+a,0)$ are such a triangle (provided $0<a<2-\sqrt 3$) In $(0,1)\times \Bbb R$, there is no such triangle: Suppose $(x_1,y_1)$, ...

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Note that $Y$ vanishes on $U$. As $\varphi$ has support in $U$, $\varphi|_{M\setminus U} = 0$ and hence $\varphi Y \equiv 0$ on $M$. Therefore $\varphi Y = 0 = 0\cdot 0 = 0\cdot\varphi Y$.

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Orientation is a good example that you mentioned. A somewhat related example is the difference between a closed and an exact differential $k$-form. The former is local but the latter is global; for example the 1-form $d\theta$ on the unit circle is easily seen to be closed locally but to see that it is not exact one needs to take an integral. Stokes' ...

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It is more a matter of notation. Take $z\in H$ and $x,y\in\mathbb R^2$. Then $$\langle d_zw\, x,d_zw \,y\rangle_{w(z)}=:\frac{|dw|^2}{Im(w)^2}(x,y)=\frac{|dz|^2}{Im(z)^2}(x,y):=\langle x,y\rangle_z.$$ Added: More explicitly, write $z=z_1+iz_2$, $x=(x_1,x_2)\equiv x_1+ix_2$ and $y=(y_1,y_2)\equiv y_1+iy_2$. Then, by definition, $$\langle ... 2 As the proposition you quoted says, the connection forms have to be skew-symmetric with respect to any orthonormal frame. You computed the connection forms with respect to the coordinate frame \{\partial/\partial x, \partial/\partial y\}, which is not orthonormal for the hyperbolic metric. 2 As you say any two metrics on \mathbb{S}^1 are conformal. Let V a vector field tangent to \mathbb{S}^1 and nowhere zero. Then, if g and \tilde{g} are two Riemannian metrics it must be \tilde{g}(V,V)>0 and g(V,V)>0. If we define f=\dfrac{\tilde{g}(V,V)}{g(V,V)} one gets \tilde{g}=fg with f>0. That is, g and \tilde{g} are ... 2 This is almost tautological. "Bi-invariant" means the metric is invariant under both left and right translation. The resulting distance function satisfies, in particular,$$d(gx,gy)=d(x,y)$$Hence left translation is an isometry (as is right translation). 2 Simple connectedness is a very simple (sorry for the bad pun) example. 2 Quite a lot of topology can be done locally. The paper by Gelfand and Tsygan, Localization of Topological Invariants, has a general construction of what can come from one method of integration of local quantities. Also anything from rational homotopy, which can be built (a la Sullivan) from differential forms. Index of elliptic operators is another and ... 2 If we intend as local property a property that can be deduced from a neighborhood of a given point, than it seems to me that in your example you are using (implicitly) a global property, i.e. the fact that the manifold is compact (as far as I know this property cannot be deduced locally). In this sense as a classical example of two manifolds locally ... 2 It is not true, even if N is the Euclidean space. Just consider the curve$$\gamma: (0,2\pi)\to \mathbb{R}^2, \quad t \to (\cos t, \sin t).$$Since |\gamma'(t)|= 1 it is an isometric immersion. But it is not harmonic, because harmonic curves are geodesics. And, \gamma is not a geodesic. Edit In case \dim M=\dim N the answer is yes. See the bottom ... 2 The point is that the metric on S^n is invariant under the action of the group \mathbb{Z}_2, so it can be pushed down to the quotient. More explicitly, assume that we endow S^n with a standard round metric and let \pi \colon S^n \rightarrow \mathbb{RP}^n be the quotient map. For p \in S^n, we denote \pi(p) by [p]. Let A \colon S^n \rightarrow ... 2$$ \nabla^a \nabla_a \xi^\mu = g^{ab}g^{\mu\nu} \nabla_a \nabla_b \xi_\nu=\\ =-g^{ab}g^{\mu\nu} \nabla_a \nabla_\nu \xi_b=-g^{\mu\nu} \nabla_a \nabla_\nu \xi^a=\\ =-g^{\mu\nu} \nabla_a \nabla_\nu \xi^a +g^{\mu\nu} \nabla_\nu \nabla_a \xi^a =\\ =-g^{\mu\nu} [\nabla_a, \nabla_\nu] \xi^a=\\ =-R^{a}{}_{\sigma a\nu}\xi^\sigma=-R_{\sigma\nu}\xi^\sigma $$which is ... 2 For the path independence part: Recall that \nabla is flat iff R^{a}_{bcd}=0 everywhere on M. If the parallel transport of vectors along a path on M relative to \nabla is indeed path independent then \nabla is flat. Conversely, if \nabla is flat then, at least locally, parallel transport of vectors on M relative to \nabla is path ... 2 If you suppose that G is also compact then you have a left invariant metric on G. Let p:G\rightarrow G/H the projection and x be a point of G, you can identify the tangent space of p(x) to the orthogonal N_x of the subspace of T_xG tangent to the orbit H(x). The restriction of the invariant metric on N_x defines a metric on G/H. 2 A great circle is the same data as two antipodal points, namely the two points furthest away from the circle. That is, the space of great circles is the space of two antipodal points on S^2, which is \mathbb{RP}^2. The direction picks one of these points (say, the point that appears on the left as you follow the direction around the circle), which ... 2 You can use the construction of the complex conjugate vector space and identify a sesquilinear form with an element of V^{*} \otimes \overline{V}^{*}. An elementary tensor \varphi \otimes \psi \in V^{*} \otimes \overline{V}^{*} gives rise to a sesquilinear form g_{\varphi \otimes \psi} \colon V \times V \rightarrow \mathbb{C}  defined by$$ g(v, w) := ...

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I have studied the whole book of Rosenberg and I can asure you that it is a very good choice to start with, I have studied some parts of the books of Gilkey and Lawson-Michelson too and I can asure you they are quite advanced, it would be hard to begin with them. I guess the book of Lawson-Michelson is better for you because it covers the classical proof of ...

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If you have a directed circle, use the right-hand rule that would position your thumb in the appropriate direction. That direction will give the right point on the sphere.

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This is isomorphic to $Hom(TM,TM)$. Namely it is the endomorphism bundle of the tangent bundle. In general given two vector spaces(or bundles) we have an isomorphism between $W \otimes V^*$ and $Hom(V,W)$.

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First, let $S$ be a regular surface of $\mathbb{R}^3$with Gauss map $N$. Let $c\colon I:=(-\varepsilon,\varepsilon)\subset \mathbb{R}\longrightarrow S$ be a curve parametrized by arc length with $c(0)=p\in S$ and $c'(0)=v\in T_pS$. Then we have the following: II_p(v)=-\langle dN_p(v),v\rangle=-\langle dN_{c(0)}(c'(0)),c'(0)\rangle=-\langle (N\circ ...

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Let $G$ be a connected Lie group and $H$ a closed subgroup. The set of $G$-invariant Riemannian metrics on $G/H$ is in canonical bijection with the set of $H$-invariant scalar products on $\mathfrak{g}/\mathfrak{h}$ under the adjoint representation of $G$. If $H$ is compact, the latter is non-empty. If $H$ is trivial, this set of metrics is thus quite big ...

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The fact that it is the universal cover of a compact manifold is not essential, what you need is completeness of the Riemannian manifold $\tilde M$ (the universal cover of a compact manifold is always complete) and its simple connectivity. Assuming these two properties and the NCP property, just follow the proof of the Cartan-Hadamard theorem to conclude ...

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Yes, that's true, assuming that, when talking about a piecewise smooth curve you are talking about a map $\gamma:[a,b]\rightarrow M$ which is continuous and smooth with nonzero derivative on subintervals $(a_i,b_i)$ where $b_i = a_{i+1}$, and that $M$ is a complete Riemannian manifold (which means it's (Cauchy) complete with respect to the metric induced ...

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This post will explain why in the post of @levap , the distance preserving map $\tilde{\varphi}$ fails to be smooth. The key point is: the space $\mathbb R$ with distance $d(x,y)=|x^3-y^3|$ (which can be induced by metric tensor $g=9x^4$) is a Riemannian manifold with metric non-degenerate everywhere but at $x=0$. If you check Myers and Steenrod's ...

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I asked this question also in the forum MathOverflow even though this question is so basic. Indeed somone solved my problem almost immediately in the comments. I write the answer here because maybe can be useful to someone in a future. The answer (as I espected) was obvious and I had this problem just because tensors are a new thing for me. The object ...

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