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5

There is no Stokes's Theorem for general tensors. The general result is that $\displaystyle\int_M d\omega = 0$ for any compact oriented $n$-dimensional manifold $M$ without boundary and $(n-1)$-form $\omega$ on $M$. The divergence theorem you cite is for the case of $\omega = \star\eta$ where $\eta$ is the $1$-form corresponding to $X$ under the isomorphism ...


4

You do not want to give charts. Here are some hints to get you started. Compactness should follow immediately from the fact that a sphere in $\Bbb R^{2n+1}$ is compact and level sets of continuous functions are closed. In general, any time you have a set $X\subset\Bbb R^N$, say, that is defined by equations, you'd like to use those equations to define a ...


4

OK, I'll accept the challenge... The biggest difference in the pseudo-Riemannian case is that curves can have zero length, and the "Riemannian distance function" (the supremum of the lengths of curves between two points) is not a metric in the sense of metric spaces. Thus most of the results in Chapter 6 of my book don't make sense if the metric isn't ...


4

As an American president once said "it all depends on what the meaning of the word is is". If you look at the definition, it is clear that Riemannian metrics, as defined, are not distance functions. When you say that a (connected) Riemannian manifold $(M,g)$ is a metric space, you are actually saying that there exists a certain functor from the category of ...


3

All of my geometric intuition for "immersed" versus "embedded" surface is contained in my geometric intuition for "immersions" versus "embeddings". In particular, as many people have pointed out, immersions need not be injective. But, of course, even injective immersions need not be embeddings. As a very simple example, consider the map $f : (-\pi, \pi) \to ...


3

As a preamble to this response, I want to comment that we should be careful about saying "the matrix" $J$. It's true that for any $t$ you can define a linear automorphism $J_t$ of $T_{c(t)} f(\Omega)$ by your formula, but there's no obvious "natural" basis for $T_{c(t)} f(\Omega)$, and you'll need to choose one to actually write down an honest-to-goodness ...


3

"but I assume nothing like that is going on here. Its not like we are putting tangent vectors tip to tail and summing over their lengths is it? " -- in a sense it is. (1) we are integrating $\sqrt {\langle \gamma\prime, \gamma\prime\rangle}$ if $\gamma$ is a parametrical $C^1$ curve with values in $M$, i.e. if $\gamma:[a,b]\rightarrow M$ is $C^1$ you ...


3

Assume that $f\colon M\to M$ is a diffeomorphism satisfying $f^*g = g$. It's not enough just to use the fact that $\nabla$ is a metric connection; you also have to use the fact that it's torsion-free. The idea of the proof is to define $\widetilde \nabla\colon \mathfrak X(M)\times \mathfrak X(M) \to \mathfrak X(M)$ by $$ \widetilde \nabla_XY = ...


3

An oriented manifold comes equipped with oriented frames, so that the construction given may proceed. For an arbitrary manifold, this is not the case. As you suggested, one may arbitrarily choose an orientation at each point. However, this data would not necessarily depend smoothly on the point, and therefore it wouldn't produce a local frame. If by some ...


2

In general relativity, the Ricci tensor contains all the information necessary to characterize matter's effect on curvature. This comes directly from the Einstein equation: $$R_{ab} + \frac{1}{2} R g_{ab} = 8\pi T_{ab}$$ where $T_{ab}$ is the stress-energy tensor, containing all the information about the density, momentum, pressure, and shear of matter. ...


2

Your argument is essentially correct: On an $n$-manifold, the rank of the bundle $\Lambda^n T^*M$ (informally, the bundle of top forms) is $1$. In any local coordinates $(x^i)$, we the $n$-form $dx_1 \wedge \cdots \wedge dx_n$ is nonzero, and hence it locally spans $\Lambda^n T^*M$. (Note that there may be no global nonzero section of this bundle; this is ...


2

Hint: If $b(-,-)$ is a symmetric bilinear form in characteristic $\neq 2$, then $$b(x,y) = \frac{1}{2}\left(b(x+y,x+y) - b(x,x) - b(y,y)\right).$$


2

b. We compute $\nabla(\alpha)$ for an arbitrary $1$-form $\alpha(x) = \sum_{i=1}^n \alpha_i(x)\,dx_i$. We have$$\partial \hspace{.5mm} d\alpha = \partial\left(\sum_{i \neq j} D_j\alpha_i\,dx_j \wedge dx_i\right) = \partial\left(\sum_{i < j} (D_i\alpha_j - D_j\alpha_i)\,dx_i \wedge dx_j\right)$$$$ = (-1)^{3n+1}\star d\left(\sum_{i<j}(D_i\alpha_j - ...


2

If you use the standard metric on the second factor and use $(-\infty, \infty)$ instead of $(0, \infty)$ and the product metric on the product it for sure is. This is actually rather easy to see: the injectivity radius will be bounded from below and the sectional curvature will be bounded from above on the compact set $M\times [1,3]$, say, and the manifold ...


2

The space of affine connections is an affine space, not a vector space. The difference of two affine connections is a tensor $A_{ij}\,^{k}$ of type $\tbinom{1}{2}$; so the space of connections is an affine space modeled on the vector space of such tensors. Another way to say this is that Christoffel symbols do not transform tensorially but the difference of ...


2

$\newcommand{\Reals}{\mathbf{R}}$Here's a slightly different viewpoint (even for the $2$-sphere). The point $p$ in $S^{n}$ and the normalized initial velocity $v$ of the geodesic $\gamma$ may be viewed as orthogonal unit vectors in $\Reals^{n+1}$, and they span a real $2$-plane $N$ (for "normal space") through the origin. The tangent space $T_{p} S^{n}$ is ...


2

A few words.. (too long to be a comment, I think) Suppose we have a parametrization ${\bf x}(u,v) = (x^1(u,v),x^2(u,v),x^3(u,v))$ and a unit normal vector $\bf N$ all the way. For problem $1$, if you call the coefficients of the first fundamental form $$E = \langle {\bf x}_u, {\bf x}_u \rangle, \quad F = \langle {\bf x}_u,{\bf x}_v\rangle, \quad G = ...


2

For problem 2 what you did is the only correct thing. Interpreting "straight line" to mean "the solution to a linear relation between the coordinate values" then you must compute the Christoffel symbols. This is the following "meta" fact Statements about geometry as viewed in a specific coordinate system invariably requires a proof given in terms of ...


2

Suppose that $M$ is compact. For (2): It follows, by taking trace, that $\Delta f = \Delta h$, i.e., $f-h$ is harmonic and, hence, constant. Conversely, if $f-h$ is constant, Hessians are equal, of course. For (1). Again, one obtains $\Delta h= \frac{1}{f}\Delta f$. WLOG, by subtracting a constant from $h$, we can assume that $h$ is negative on the ...


2

This is defining what ways to differentiate vector fields (called the covariant derivative), which we'll denote $\nabla_X Y$ for the covariant derivative of $Y$ in the direction $X$, we'll consider valid. Property (1) is saying this is linear in $X$ over $\mathbb{R}$, and moreover that multiplication of $X$ by any function can be pulled out (you could call ...


2

1- If length were sufficient a circle of length (perimeter) $2\pi$ would be the same as a square whose side is $\frac{2\pi}{4}$. 2- a unit speed parametrization is defined as follows: if $x(t)$ is an arbitrary parametrization take $s$ to be defined by $\frac{ds}{dt}=\vert\vert x'(t)\vert\vert$. You check that $\vert\vert \frac{dx}{ds} \vert\vert=1$. $x(s)$ ...


1

If you have a Riemannian manifold then at each point on the manifold you have a matrix $\Sigma$ which can be used to represent the Riemannian metric as a quadratic form. You can define the length of a tangent vector $x$ as $$\sqrt{x^T \Sigma x}$$ In Euclidean space the matrix $\Sigma$ is just the identity matrix. Remember that from point to point the metric ...


1

A Riemannian metric is not a metric, in the same way that an alleged criminal is not (necessarily) a criminal. A Riemannian metric is a smoothly varying positive-definite bilinear form on the tangent spaces. By extension, we call any non-degenerate --- not necessarily positive-definite --- bilinear form on the tangent spaces a pseudo-Riemannian metric. In ...


1

There is a connection between the covariant derivative and affine connections: the latter generalize the former. In particular, the covariant derivative you described depends on a Riemannian metric, while you can consider affine connections of manifolds without a metric. You have surely noticed that the conditions imposed on affine connections are usual ...


1

Work over a coordinate patch $U$ around $p$ with local coordinates $\phi:U\xrightarrow{\sim\,}\Bbb R^n$ ($n$ is the dimension of $M$). Take a basis $(e_1,\dots,e_r)$ a basis of the ($r$-dimensional) vector space $P_p$. Extend it to a local basis of $P|_U$ by the formula $$\forall q\in U,\,E_i(q)=\mathrm{PT}^\nabla_{\gamma_{q}}(e_i)$$ Where ...


1

Just in order to close this matter: There are two easier existence theorems in differential geometry: a. If $E\to M$ is a fiber bundle (say, a principal bundle, for concreteness) equipped with a connection $\nabla$, then each smooth curve on $M$ lifts to a smooth horizontal curve in $E$. b. If $M$ is a compact Finsler manifold and $\nabla$ is a ...


1

In the comments there are more than one good answers to the question. However, there is some other way to think of it. Recall that $\mathcal{X}(M)$, the space of vector fields on $M$, is the Lie algebra of the Lie group $\mathrm{diff}(M)$, the group of automorphisms of $M$. The space $\mathcal{K}(M)\subset\mathcal{X}(M)$, which consists of the Killing ...


1

For a general one-parameter family of geodesics, the Lie bracket expression you wrote down has no meaning. The way to prove this, instead, is to choose smooth local coordinates $(x^i)$ on $M$, and write $\gamma$ locally as $\gamma(s,t) = (x^1(s,t),\dots,x^n(s,t))$. Then when you expand out $$ \cfrac{D}{dt} \cfrac{\partial \gamma(s,t)}{\partial ...


1

You can always set up the coordinate as $(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$, and choose the geodesic to be $\phi=0$ and $p=(1, 0, 0)$ be a point at the equatorial. Note that $X=\frac{\partial}{\partial\theta}=(\cos\theta\cos\phi, \cos\theta\sin\phi, -sin\theta)$ and $Y=\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=(-\sin\phi, ...


1

Let's look at what do Carmo actually writes for the definition of a regular surface: A subset $S \subset \mathbb{R}^3$ is a regular surface if, for each $p \in S$, there exists a neighborhood $V \subseteq \mathbb{R}^3$ and a map ${\bf x} : U \to V \cap S$ of an open set $U \subseteq \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that: ...



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