# Tagged Questions

A branch of differential geometry dealing with Riemannian manifolds. *Riemannian manifolds* are smooth manifolds with an inner product smoothly attached to the tangent space of each point. Usually, Riemannian geometry focuses on the notions of distance, curvature, and shape. Consider using this tag ...

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### Gross-Zagier formulae outside of number theory

(Edit: I have asked this question on MO.) The Gross-Zagier formula and various variations of it form the starting point in most of the existing results towards the Birch and Swinnerton-Dyer ...
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### PDEs on manifold: what changes from Euclidean case?

I know some PDE theory for nice open domains in $\mathbb{R}^n.$ I want to know what the changes are when I switch to other domains like manifolds. For example, do things like Poincare's inequality ...
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### Why is the Laplacian important in Riemannian geometry?

As I've learned more Riemannian geometry, many of my teachers have said that studying the Laplacian (and its eigenvalues) is very important. But I must admit, I've never fully understood why. ...
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### Review on Riemannian Geometry

I'm currently reading through Griffiths and Harris Principles of Algebraic Geometry, and the only subject in the foundational material section that I am not completely comfortable with is riemannian ...
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### Yarn-like functions

When wrapping yarn around a ball you cannot make sharp turns or the yarn will fall off. If we think of the yarn as a curve on the surface of the sphere, we would say it must have curvature less than ...
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### The space of Riemannian metrics on a given manifold.

For a finite-dimensional smooth (Hausdorff, second-countable) manifold $M$, consider the set $$\mathcal{Met}(M) = \{ g : g \text{ is a Riemannian metric on }M \}.$$ I'd like to know about the typical ...
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### Is the Nash Embedding Theorem a special case of the Whitney Embedding Theorem?

The Whitney Embedding Theorem states that every smooth manifold can be embedded in Euclidean space. The Nash Embedding Theorem states that every Riemannian manifold can be embedded in Euclidean space....
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### Are there spaces that 'look the same' at every point, but are not homogeneous?

A metric space is homogeneous if for any two points there is a global isometry that maps one into the other. It is locally homogeneous if any two points have isometric neighborhoods, i.e. the space '...
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### On the variation of a Kähler metric on a surface by pullback of the complex structure

Let $\Sigma$ be a compact, connected, oriented surface, and let $\rho\in\Omega^2(\Sigma)$ be a fixed volume form. Then any (almost) complex structure $J\in\Omega^0(M;\operatorname{End}TM)$ compatible ...
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### Existence of a Riemannian metric inducing a given distance.

Let $M$ be a smooth, finite-dimensional manifold. Suppose $M$ is also a metric space, with a given distance function $d: M \times M \rightarrow \mathbb{R}_{+}$, which is compatible with the original (...
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### Why is $\langle \operatorname{grad} f, X\rangle_g$ independent of the metric on a Riemannian manifold?

Let $(M,g)$ be a Riemannian manifold and let $f \in C^{\infty}(M)$. Let $X$ be a smooth vector field on $M$. In smooth local coordinates $(x^i)$ on $M$, we can write $g = g_{ij} dx^i \otimes dx^j$ as ...
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### Why are we interested in closed geodesics?

There's a lot of work about the existence, number and other properties of closed geodesics on a Riemannian manifold (belonging to some specific class of manifolds). In the case of geodesics ...
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### Geometric meaning of symmetric connection

If $(M, g)$ is Riemannian manifold, there is unique connection $\nabla$, called Levi-Civita connection, satisfying the following: 1) Compatibility with Riemannian metric, i.e. $\nabla(g)$=0 2) ...
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### Non-ellipticity of Yang-Mills equations

Let $D=\text{d}+A$ be a metric connection on a vector bundle with curvature $F=F_D$. How does one prove that the Yang-Mills equations $$\frac{\partial}{\partial x^i}F_{ij}+[A_i,F_{ij}]=0$$ from ...
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### How can I understand the three-dimensional space forms?

Here is what I know: A space form is defined as a manifold admitting a Riemannian manifold of constant sectional curvature A classical result of Cartan states that a manifold is a space form if and ...
Let $(M, g_{ab})$ be a spacetime and define a new metric, $\tilde{g}_{ab}$, on $M$ by $\tilde{g}_{ab} = \Omega^2 g_{ab}$, where $\Omega$ is a smooth, positive function. Let $\nabla_a$ denote the ...