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No. As an example (without being rigorous) consider a manifold which looks like an inifinite half cylinder parallel to the positive $z-$axis in Euclidean three space with a hemisphere attached at the bottom along an equator and smoothed out. (A bit like a hyperboloid or an one sided infinite cigar). If you look at the south (bottom) pole of the hemisphere ...

6

No, because the Ricci curvature is ($n-1$ times) the average of the sectional curvatures, so if they are all nonnegative it must be nonnegative. More precisely, the result is false regardless of the compact Einstein setting. We have, for every $p \in M$, for every unit vector $e_1 \in T_pM$, $$\operatorname{Ric}(g)(e_1)=(n-1) \sum_{j=2}^n ... 5 Yes. Here's a sketch of a proof. (The theorem references are from my Introduction to Smooth Manifolds, 2nd ed.) First, let f\colon M\to \mathbb R be a smooth exhaustion function, i.e., a function whose sublevel sets M_c = f^{-1}((-\infty,c]) are compact for all c\in\mathbb R. (Such a function exists by Prop. 2.28.) Because the sets U_c = ... 4 You're correct, but so is Topping - the curvature operator on functions is just zero! Remember that$$-R(X,Y)A = \nabla_X (\nabla_Y A) - \nabla_Y (\nabla_X A) - \nabla_{[X,Y]}A,$$so when A = f is a function you get XYf - YXf - [X,Y]f = 0. Since A(W,Z,\ldots) has all its slots filled, it is a scalar function, and thus this curvature term vanishes. 4 By taking the orientable cover if necessary, we assume that M is orientable (Note that the Betti number is not changed after taking the orientable cover). The Bochner formula for one form is$$\tag{1} \Delta_d \alpha = -\nabla^*\nabla \alpha + \text{Ric}(\alpha),$$for all \alpha. Let \alpha be a harmonic one form, then integrating the above formula ... 4 An isometry in the sense that the Riemannian metric (scalar product) is preserved is also a local isometry in the sense of geodesic distance, which is not too difficult to see since it preserves locally the length of curves (this does require some reasoning to make it strict, though). Since the Riemannian curvature tensor can be calculated in terms of the ... 4 In general, the notion of isometries for Riemannian manifolds is an infinitesimal one. A smooth map \varphi \colon (M,g) \rightarrow (N,h) between smooth Riemannian manifolds is called a local isometry if for each p \in M the differential d\varphi_p \colon (T_pM, g_p) \rightarrow (T_{\varphi(p)}N, h_{\varphi(p)}) is an isometry of inner-product spaces. ... 3 I'm a fan of Lee's Riemannian Manifolds: An Introduction to Curvature. It is definitely an introductory book; there are many deeper topics that it doesn't mention (compare to Peterson's Riemanninan Geometry). Here is an excerpt from the preface: "I have selected a set of topics that can reasonably be covered in ten to fifteen weeks, instead of making ... 3 Think back to what the Riemannian volume form is. It's a form that, when fed an oriented orthonormal frame, spits out 1. (Of course every manifold here is oriented.) What are the oriented orthonormal frames on \Sigma \subset M, where \Sigma is a (Riemannian) submanifold? Because it's a Riemannian submanifold, if (x_1, \dots, x_{d-1}) is an oriented ... 3 It is not true that \nabla_X({\rm tr}(dx^j\otimes\partial_i))=0 implies \nabla_X(dx^j\otimes\partial_i)=0; indeed this latter equation is false for most coordinate systems. Remember that you don't need to show$$\nabla_X dx^j \otimes \partial_i = -dx^j \otimes \nabla_X \partial_i,$$but only the contracted version$$(\nabla_X dx^j) (\partial_i) = -dx^j ...

3

Here is an idea of a proof: For any curve we can choose a reparametrisation such that its speed is constant, such that we actually have $L(\gamma)^2 = 2 E(\gamma)$ (this follows from going through the Cauchy-Schwarz inequality that already underlies the inequality in the text) and we only have to consider curves with constant speed. Now it is clear that if ...

3

While the tangent spaces $T_pM$ are all distinct abstract vector spaces, if $M$ has dimension $n$, then they are all isomorphic to $\mathbb{R}^n$, even though there is a priori no "canonical" isomorphism $T_p M \cong \mathbb{R}^n$. On a Riemannian manifold, parallel transport gives a way of associating to a curve $\gamma(t) \subset M$ in the manifold ...

3

$\{e_j\}$ is a basis on $V$. $\{\phi_i\}$ the complementary basis on $V^*$. That is, for all $i,j, \phi_i(e_j) = \delta_{ij}$. Now if $A\in \operatorname{End}(V)$, then $\Phi(A)(\phi_i, e_j) = \phi_i(A(e_j)) = A_{ij}$, that is, the entry in the $i^\text{th}$ row and $j^\text{th}$ column of the matrix representation of $A$ with respect to the basis ...

3


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Whenever you have a finite dimensional vector space $V$ and a metric $g$ on $V$, one can define an linear mapping $\flat: V\to V^*$, $X\mapsto X^\flat \in V^*$ given by $X^\flat (Y) = g(X, Y)$. Now let $\{e_1, \cdots, e_n\}$ be a basis of $V$, then every $X\in V$ is written as $X = X^i e_i$, or simply $X = X^i$. Let $\{e^1, \cdots, e^n\}$ be its dual basis ...

3

In general for a $(p,q)$-tensor $A$, one defines $\nabla A$ as a $(p,q+1)$ tensor by $$\nabla A(X, \cdots) = \nabla_XA(\cdots),$$ where $\nabla_X A$ is defined as (for example if $A$ is a $(2,0)$ tensor) $$\nabla_X A(Y, Z) = \nabla_X(A(Y, Z)) - A(\nabla_X Y, Z) - A(Y, \nabla _X Z).$$ So in our case $$\begin{split} \nabla^2 F(X, Y) &= (\nabla \nabla F) ... 3 Your first idea (using the uniqueness of geodesics) is good. The key formula is the derivative of the distance function: we are given a minimum x_0 of f, so we'd like to extract some information from the critical point condition Df_{x_0} = 0. The hypotheses on the manifold guarantee the distance function will be smooth away from the diagonal, so this ... 3 This is not true - you definitely need to take the variation of the metric in to account, as you correctly observe in your first few lines. You shouldn't need it to be true to get these inequalities - the extra terms in \partial_t |A|^2 due to the variation of the metric will be of the form Rc * A * A and thus can be controlled by the C |Rm||A|^2 term. ... 2 Your result is correct: one first computes \mathcal L _X \big( g (Y,Z) \big), in this one replaces X \cdot g(Y,Z) with the formula obtained from the fact that \nabla is a metric connection, then using that T(X,Y) = 0 one gets (\mathcal L _X) (Y, Z) = g(\nabla _Y X, Z) + g(Y, \nabla _Z X). The author's formula is true with one condition, though: ... 2 Actually if you calculate that at the center of a normal coordinates, writing X = \sum_i X^i \frac{\partial}{\partial x^i}, then$$\text{div} X = \sum_i \partial_i X^i.$$So using |\nabla u|^2 = \sum_j u_j^2,$$\begin{split} \text{div} \left( \frac{\nabla u}{\sqrt{|\nabla u|^2 + \epsilon^2}}\right) &= \sum_i \left( \frac{ u_i}{\sqrt{|\nabla u|^2 + ...

2

This theorem is concerned with proving an estimate, and as estimates tend to go there is a lot of wiggle room in the proof, especially since we are not concerned with the exact constants involved. Thus it is commonplace not to keep track of all the little details when we are just going to estimate them all (usually with Cauchy-Schwarz) and lump them together ...

2

The Laplacian is the trace of the Hessian, which is positive. However the Hessian is symmetric, and symmetric plus negative semidefinite implies nonpositive trace (this can be seen easily through diagonalization).

2

Just apply $\nabla_i$ and sum over $i$ to get $$\nabla_i \nabla_iR_{jkla}+\nabla_i \nabla_jR_{kila}+\nabla_i \nabla_kR_{ijla}=0.$$ Recognising $\nabla_i \nabla_i = \Delta$ (since Topping must be working in an orthonormal frame) and $R_{ijla} = -R_{jila}$ gives you your result. There are no Ricci terms because at least one of the indices $i$ we are tracing ...

2

The Hessian comparison theorem is (see e.g. this article) If the sectional curvatures of a manifold are bounded below by $M$, then the distance function $s(x) = d(p,x)$ satisfies $\nabla^2 s \le \nabla^2_M s_M$, where $s_M$ is the corresponding distance function on the space of constant curvature $M$. We know $|R_{ijkl}|<M$ and thus our sectional ...

2

Since the goal is to get an estimate $|\nabla Rm|^2 \le C/t$ which is in particular independent of spatial position, I think that what's going on here is a maximum principle argument. As written the inequality you have highlighted does not hold pointwise - you're correct that it should be $$F(x,t) \le F(x,0) + \bar C M^3 t + \int_0^t \Delta F(x,s)ds.$$ ...

2

Without the Ricci curvature bound one does not have control on the doubling constant of the space (which is the only thing that the bound is used for). This means that the spaces may contain larger and larger sets of uniformly separated points (say, distance $\ge 1$ between any two points). This precludes being Cauchy in Gromov-Hausdorff metric, since if the ...

2

Usually lower indices represent the covariant components (of a covector) and upper indices the contravariant components ( of a vector) You can see here.

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Have you heard of non-trivial covering spaces? --- fiber bundles with discrete fiber! Have you heard of the Möbius Band? --- fiber bundle over the circle with an interval as fiber! Both examples are non trivial (i.e. not product bundles).

2

A vector field at point $p$ on the differentiable manifold $M$ is smooth if in every chart $\left( {U,\varphi } \right)$ with coordinates ${x^1},...,{x^n}$ , the coefficients ${\alpha ^i}:U \to \mathbb{R}$ of the vector field in the representation ${X_p} = \sum\limits_{i = 1}^n {{\alpha ^i}\left( p \right)} \frac{\partial }{{\partial {x^i}}}{|_p}$ are ...

2

This is impossible at least if $G$ is compact and non-abelian (or a product of such group with something else). On the other hand, if $G=\mathbb R^n$ (or $\mathbb R^n/\mathbb Z^n$), then the Killing form vanishes identically and any constant or linear function satisfies the desired property. I don't have a complete answer, but let me explain why it fails for ...

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