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6

The differential is an isometry of vector spaces. The local map between Manifolds is a radial isometry, but in general it will fail to map geodesic distance spheres from the base point to isometric distance spheres. Simply because these need not admit an isometry. So a path orthogonal to geodesic rays starting in the base points will be mapped onto a path ...


6

Examples in dimension $3$ follow from hyperbolization theorems for 3-manifolds combined with Mostow rigidity. Suppose that you have a compact 3-manifold $M$ with connected boundary $\partial M$ of genus $\ge 2$, such that $M$ is irreducible, atoroidal, and acylindrical and $\partial M$ is incompressible --- in brief this says $M$ has no "badly embedded" ...


6

Since there is some ambiguity concerning what are your hypotheses, it is not guaranteed that your proof is exhaustive. It is known that there is a $C^1$-isometric embedding of the flat torus inside $\mathbb{R}^3$ (this is related to Nash embedding theorem). However, for the reason you presented and because of Gauss' theorema egregium, there is no ...


5

The figure below, taken from an earlier MSE answer, shows a (green) shortest path on a locally flat metric. Is it "a Euclidean straight line"? Depends on what that phrase means...       Only after unfolding the surface to a plane is it a straight segment in $\mathbb{R}^3$.


5

Here's a specific example of such a manifold, known as the Thurston tripus or knotted Y. It is described in Example 3.3.12 of Thurston's Three-Dimensional Topology and Geometry, Volume 1. The following picture shows a hyperbolic polyhedron in $\mathbb{H}^3$ under the Poincaré ball model. It has the combinatorial structure of a truncated tetrahedron. ...


5

First, note that the transformation rule defines an action of $GL_+(2, \Bbb R)$ (the group of $2 \times 2$ real matrices with positive determinant) on the upper half-plane, namely via $$\pmatrix{a & b\\c & d} \cdot z := \frac{a z + b}{c z + d}.$$ For any such transformation, if we scale each of $a, b, c, d$ by $\lambda \in \Bbb R - \{0\}$, the ...


4

This somehow 'depends'. If, as usual, in Riemannian Geometry, you are looking at a smooth ($C^\infty$) manifold with smooth metric, then the answer is, for sure, yes. Only if you are reducing the differentiability assumptions on the manifold and the metric you may run into trouble at a certain point. What you are looking at (e.g. in local coordinates) is an ...


3

NB: in the following, it is assumed that $X, Y, Z$ and $R$ are continuous. If $C \subset TM$ is the compact subset from which $X$, $Y$ and $Z$ are taken, then $C \times C \times C$ is compact (Tychonoff's theorem), and the map $R(X, Y)Z: C \times C \times C \to TM \tag{1}$ is continuous; this follows by virtue of the fact that in any local coordinate ...


3

A natural approach is to show that the map $f(x,v)= (dL_{x^{-1}})_x(v)$ is smooth: The map $F:G\times G \to G$ given by $F(x,y)=xy$ satisfies $dF((x,u),(y,v))=(dL_{x})_y(v)+(dR_y)_x(u)$. Embed $TG$ in $T(G\times G)\cong TG \times TG$ via $(x,v)\mapsto ((x^{-1},0),(x,v))$. The composition $TG \hookrightarrow T(G \times G) \to TG$ is smooth and gives the ...


3

I think it is just the composition of smooth maps. Let's try to state this formaly. Consider the "double-tangent bundle" (this is not an official notation, I'm not sure it have a standard name) $T^2G=\bigsqcup_{x\in G}T_x G \times T_x G$ and $\pi : T^2G \rightarrow G$ the canonical projection. Then, we can define on $T^2G$ the map (with $\mathfrak{g}=T_eG$ ...


3

The sectional curvature of a Riemannian manifold $(M, g)$ of real dimension $n > 2$ is not a scalar function on $M$, but a scalar function $K$ on the Grassmannian bundle of $2$-planes in the tangent bundle. To say $(M, g)$ "has constant curvature" means $K$ is a constant function, i.e., the sectional curvature is the same for all $2$-planes (at all ...


3

$$ \sum_j g^{ij} g_{kj} = \delta^i_k $$ Now, $\delta^i_i = 1$, with no summation, so $$ \sum_i \sum_j g^{ij} g_{ij} = \sum_i \delta_i^i = \sum_i 1 = n. $$ You may be confusing whether or not you are using summation convention.


3

Note: $$\sum_{i,j=1}^n g^{ij} g_{ij} = \text{tr} (g^{-1}g) = \text{tr} (I_n) = n.$$


3

Exponential maps are in general not isometries (otherwise every Riemannian manifold would be flat), and the term "linear" does not make sense since general Riemannian manifolds do not have a linear structure.


3

Besides Hopf-Rinow you can use a general fact. Namely that the flow of a vector field defined on a compact manifold is complete i.e. defined for all time. Then you should know that the geodesic flow is defined by a vector field (some times called spray) defined on the unit tangent bundle of the compact manifold. Such unit tangent bundle is also compact hence ...


3

This seems to be false, even if one assumes that $\Sigma$ is isometrically embedded. Counterexample: Let $M$ be the $2$-sphere and $\Sigma$ some great circle in $M$; the volume form $\Omega$ of $\Sigma$ is some parallel one-form on $\Sigma$. $\Omega$ can't possibly extend even locally to some neighborhood of $M$, because $M$ doesn't admit even locally ...


3

$\newcommand{\Reals}{\mathbf{R}}$To be invariant about it, if $v$ is a tangent vector at $e$, then the one-parameter subgroup $x(t) = \exp(tv)$ is a curve through $e$ with $x'(0) = v$, and $$ dL_{g}(v) = \frac{d}{dt}\bigg|_{t=0} L_{g}\exp(tv), $$ just as you say. To get more detailed and/or explicit information, you need to invoke specifics about the group ...


3

Your answer is almost correct. You have that straight lines are the critical points of the arc-lenght functional. These lines minimize arc-lenght if they are timelike or lightlike, but they maximize arc-lenght if spacelike. This is sort of expected, since the connection of this Lorentz plane is the same as the connection from $\Bbb R^2$ and the Christoffel ...


2

Yes, it is. Let $\Phi$ be an isotopy. That is $\Phi:I\times M\to M$ is smooth, such that $\Phi(0,p)=p$ and $\Phi(1,p)=\phi(p)$. Let $\overline{\Phi}:I\times \overline{M}\to\overline{M}$ denote the lift of $\Phi$ equal to the identity at time $0$. By compactness, there is some $\lambda>0$ such that for every $(t,p)\in I\times M$ we have ...


2

First of all you need to make sense of "averaging along all planes containing $U$". This can be done: Let $U^\perp \subset T_pM$. Then all planes containing $U$ are in one-to-one correspondence with the unit sphere $S^{n-2} \subset U^\perp$. Thus the average $A$ is given by $$A = \frac{1}{Vol(S^{n-2})} \int_{S^{n-2}} \ \mathrm R(U, V, U, V) d\sigma = ...


2

As noted in the comment, it is easier to work with complex coordinate. It will be even easier if we break down the general transformation to simpler ones: Note that when $a, c\neq 0$, $$\begin{split} \frac{az + b}{cz+d} &= \frac ac \frac{cz + \frac{cb}{a}}{cz+ d}\\ & = \frac ac - \frac{d - \frac{cb}{a}}{cz + d} \\ &= \frac ac - \frac 1a ...


2

If $g$ is the hyperbolic metric of curvature $-1$ on $\mathcal{H}\colon = \{ \mathcal{Im} z > 0\}$, $g = \frac{d x^2+ d y^2}{y^2}$, then it's easy to show that $g$ is invariant under all the transformations $z \mapsto \frac{az + b}{c z +d}$, where $a,b,c,d$ real and $ad - b c >0$. We want uniqueness of invariant metrics. Let $g'$ be any other ...


2

The implication is a consequence of Ricci identity. The notations $\nabla_k$ is standard: Let, in general, that $A$ is a $(p, q)$-tensor, one can define a $(p, q+1)$ tensor $\nabla A$. In your specific example, there are two tensors: The function, $\omega$, which is a $(0,0)$ tensor, and It's exterior derivative $d\omega$, which is a one form (that is, a ...


2

If you just want to prove the equation in the last line of your post then it is enough to compute $$R(\partial_i,\partial_j) \partial_k \, .$$ Notice that the last term (which bother you) does not appear since the Lie bracket $[\partial_i,\partial_j] = 0$.


2

Your claim is too general, not to mention that you should clarify what you mean by "Euclidean straight line". As Joseph O'Rourke showed in his answer, there are spaces which admit a locally flat metric but do not contain any line. On the other hand, there are spaces which can be embedded in a Euclidean space where some geodesics are straight lines (in the ...


2

To say $g = \frac{dx^2+dy^2}{y^2}$ is to say $g = \frac{1}{y^2} (dx \otimes dx+dy \otimes dy)$. Then $\gamma'(t) = \frac{d\gamma^1}{dt}\frac{\partial}{\partial x}\bigg{|}_{(t,t)}+\frac{d\gamma^2}{dt}\frac{\partial}{\partial y}\bigg{|}_{(t,t)}$ hence $$g(\gamma'(t),\gamma'(t)) = \left(\frac{1}{y^2}(dx \otimes dx+dy \otimes ...


2

Assume that there is a family of immersions $\varphi_t : M^n \to \mathbb R^{n+1}$ so that $$\frac{\partial }{\partial t} \varphi_t = f_t\ \nu_t,$$ where $f_t$ are smooth functions on $M$. Let $A_t$ denote the area given by $\varphi_t$. Then one has $$\frac{\partial}{\partial t} A_t = -\int_M f_t H d\mu = -\langle f_t, H\rangle_t$$ Note that $\langle f , ...


2

For a curve $(x,y) = \bigl(x(t),y(t)\bigr)$ in $\mathbb{H}$, the parallel transport of a vector $\textbf{v}$ is determined by the following differential equation: $$ \frac{d\textbf{v}}{dt} \;=\; \frac{1}{y}\frac{dy}{dt}\,\textbf{v} \,-\, \frac{1}{y} \frac{dx}{dt}\,\textbf{v}^\perp $$ where $\textbf{v}^\perp$ is the vector obtained by rotating $\textbf{v}$ ...


1

The matrix representation of $g$ at the point $(x_1,x_2,x_3)$ is$$\left(\begin{array}{ccc}x_1^2&0&0\\0&1&0\\0&0&1\end{array}\right).$$It is clearly symmetric at any point, and positive definite for $x_1\neq0$. At the point $(1,0,0)$ the metric $g$ coincides with the standard metric, and so the norm of any tangent vector with respect ...


1

I am not sure where the problem is but let me present my solution. In order to find the transport of $(0,1)$ at point $a(t)$ we need to find parallel vector field $V$ along $a$ with $V(0)=(0,1)$ and compute $V(t)$. Let: $$V(t)=V_1(t) \partial_1(a(t)) + V_2(t) \partial_2(a(t))$$ Than: $$D_tV=\sum_{k}(V_k'+\sum_{i,j}\Gamma^k_{ij}a_i'V_j)\partial_k$$ It is ...



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