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4

The curve $\gamma(t) = \operatorname{Exp}_{p}(tv + v_{0})$ is not generally a geodesic. Let $(M, g)$ be the round unit sphere in $\mathbf{R}^{3}$, and let $p = (0, 0, 1)$. In polar coordinates $(r, \theta)$ in the tangent plane $T_{p} M$, the exponential map is given by $$\operatorname{Exp}_{p}(r, \theta) = (\sin r \cos\theta, \sin r \sin\theta, \cos r). ... 3 No, the closure of X has locally finite 1-dimensional measure. Indeed, the set of non-differentiability points is contained in the cut locus of p, which is a closed set. It was proved by Hebda and Itoh that the intersection of cut locus with every compact set has finite 1-dimensional measure. See Hebda, James J. Metric structure of cut loci in ... 3 you need a metric:$$ g(\operatorname{grad}f,v)=v(f) \qquad g|_M(\operatorname{grad}f|_M,v_m)=v_m(f|_M) \\ $$for all tangent vectors (this is the definition), and if \operatorname{grad}f = V_o+V_t is the orthogonal decomposition in the ambient tangent space$$ g|_M(\operatorname{grad}f|_M,v_m)=v_m(f|_M)=v_m(f)=g(\operatorname{grad}f,v_m) ...

3

Your mistake is the following: You have $D(\sigma_m \circ \gamma)(t) = D\sigma_m \gamma'(t)$. Also $\gamma'(t) \in T_{\gamma(t)}M$ but on the other hand $D\sigma_m = -Id$ holds only on $T_mM = T_{\gamma(0)}M$. Thus you can only deduce $$D(\sigma_m \circ \gamma)(0) = -\gamma'(0).$$ For other values of $t$ the equation $D(\sigma_m \circ \gamma)(t) = ... 3$U(p,q)$is the symmetry group of the Hermitian form of type$(p,q),$defined by $$|| (z_1,\ldots,z_n)||^2 = |z_1|^2 + \cdots + |z_p|^2 - |z_{p+1}|^2 - \cdots - |z_{p+q}|^2.$$ Let's consider the case$p = n, q = 1$. Any vector in$\mathbb C^{n+1}$can have positive, zero, or negative norm, and pretty clearly this depends only on the complex line spanned by ... 2$L(\gamma)=L(\sigma)$is obvious. The difficult thing is to prove that these suprema can be written as an integral. Nevertheless, here is why one has$L(\gamma)=L(\sigma)$: Both$L(\gamma)$and$L(\sigma)$are the sup of the same set, namely the set of all sums of the form $$\sum_{k=1}^N |\gamma(t_k)-\gamma(t_{k-1})|$$ with ... 2 Let$g = dx^2 + x^2dy^2$. First observe that for$(x,y)$,$(a,b) \in (\mathbb R^+ \times \mathbb R,g)$for all$n \geq 1:$$$d((x,y),(a,b)) \leq d((1/n,y),(x,y)) + d((1/n,y),(1/n,b)) + d((1/n,b),(a,b))$$ $$\leq x + 1/n|y - b| + a.$$ It follows that $$d((x,y),(a,b)) \leq x + a.$$ Now let$p_n = (x_n,y_n)$be a cauchy sequence in$\mathbb R^+ \times ...

2

What Chris said is correct and completely to the point, but would have been too abstract for me when I was first learning. Apologies if his answer sufficed. Consider the case of $M = \mathbb{C}^n$. For each point $z \in \mathbb{C}^n$, $T_zM$ can be identified with $\mathbb{C}^n$. An almost complex structure gives you a complex structure on each tangent ...

2

Such Riemannian manifold $(M,g)$ is called "locally flat" or "locally Euclidean". Its metric is locally isometric to the standard metric on $R^n$ where $n$ is the dimension of the manifold. Without further hypothesis, this does not tell you much. Suppose however, that the metric is complete (e.g. the manifold is compact). Then $(M,g)$ is the quotient of the ...

2

First of all, you want non-positive, not non-negative curvature. In positively curved spaces $d^2$ is neither smooth nor convex (consider the sphere). The answer is negative. Consider a geodesic $\gamma:[0,1]\to M$, and on it the affine combination $$f(t) = 2d^2(\gamma(t),\gamma(0))-d^2(\gamma(t),p) \tag{1}$$ where $p$ is some point in $M$. Since $\gamma$ ...

2

First, I think this question is more suitable for www.mathoverflow.net I will consider the general setting of a Riemannian manifold $M$ (of arbitrary dimension and topology). Define the non-smoothness set $N(p)$ set of the distance function $d(p, x)$ (with fixed $p$), the cut-locus $C(p)$ of the point $p$ and the set $NU(p)$ consisting of points $q\in M$ ...

2

This question is several months old, so those who were interested probably know the answer by now. The crucial notion is that of a conjugate point, whose definition is usually given using Jacobi fields, but an equivalent definition is that $q$ is conjugate to $p$ if $\exp_p(t)=q$ and the derivative of $\exp_p$ at $t$ is not an injective linear map. It is a ...

2

Here is an explanation why people define energy in that way: For the length of a curve, we have $$L[c] = \int_I \sqrt{g_{c(t)} (\dot c(t), \dot c(t) )}\mathsf{dt}$$ as you said. However, dealing with the square root is difficult. For example, $\sqrt{t}$ is not $C^\infty$. So we would rather dealing with square of it, that is, the energy $$E[c] = ... 2 To perhaps lend some physical insight lets examine this problem in a simpler context, Physically the kinetic energy of a free particle is defined to be, E = \frac{1}{2}m v^2, where v is the magnitude of the velocity vector \vec{v} = \frac{d\vec{x}}{dt}. Thinking of this in \mathbb{R}^3 with the Euclidean metric we can say the following. The ... 2 It's the norm of the linear functional \omega_x with respect to the scalar product induced on the cotangent bundle by the Riemannian metric. E.g. in coordinates,$$|\omega_x|^2= \sum_{i,j} g^{ij}(x)\omega_i(x) \omega_j(x)$$where, in commonly used notation, (g^{ij}) is the inverse of (g_{ij}) 1 This shows you, that the connection is completely encoded in it's parallel transport and in this encoding, it's nothing than the regular derivative. Let \xi_1,...,\xi_n be a basis of E_p and s_1,...,s_n be the unique parallel sections along \gamma with s_i(0)=\xi_p. Define \tilde{S}=s\circ\gamma and let \tilde{S}(t)=\sum\sigma^j(t)s_j(t). The ... 1 The eigenfunctions u_j of the Laplacian in \Omega, corresponding to the eigenvalues \lambda_j, constitute an orthogonal (and WLOG orthonormal) basis of L^2(\Omega). Thus if v_0(x)=v(x,0), where u is a solution of the Heat Equation (with homogeneous boundary conditions) then$$ v_0=\sum_j \langle v_0,u_j\rangle u_j, $$and as ... 1 If T:V\to W is a linear transformation between two euclidean vector spaces (vector spaces with a positive definite inner product) then you define its norm square by taking the sum of squares of the entries of its matrix wrt some othonormal bases in V,W. You can check that this definition is independent of the bases chosen. An equivalent definition: ... 1 First we have to notice that \nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j}=0 for all i,j=1,2,3. Second, let \gamma(s)=(\gamma_1,\gamma_2,\gamma_3) and \gamma'(s)=(\gamma'_1,\gamma'_2,\gamma'_3), then$$\nabla_{\gamma'}\gamma'=\nabla_{\gamma'(s)}(\gamma'_1,\gamma'_2,\gamma'_3) ...

1

First of all, the comment by Semsem is valid. If the vector space is one dimensional, then $V\cap \mathbb S^n$ is not connected. That is, statement (i) is correct only if you consider vector subspaces that are not one dimensional. Let $M$ be a totally geodesic submanifold of $\mathbb S^n$. After some $O(n+1)$ on $\mathbb S^n$, we assume that the north pole ...

1

This is an old post and Jason already gave a satisfactory (negative) answer to it, but some stubborn people do not take no for an answer! One of such peole was A.D.Alexandov who asked (in 1940s or 1950s, I think) about synthetic geometric conditions on metric spaces $(M,d)$ which ensure that the distance function $d$ comes from a Riemannian metric (no a ...

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