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6

To begin with, nothing has to be smooth. There are deep theories concerning manifolds which are only $C^k$ for some $k$, or topological manifolds, which don't even have to be differentiable. But when considering the category of smooth manifolds and smooth maps, certain constructions have to be smooth in order for the resulting objects to fit into this ...


5

Not true. Try lines of longitude on a sphere ($\theta=\text{constant}$). Then $\omega = d\phi$ and if you compute $(d\delta+\delta d)\omega$, you don't get $0$.


2

No, not in general. For example, the image of a circle in a plane under a homeomorphism of athe plane can be a curve of positive area. Indeed, here you can see the construction of a simple curve of positive area. It can be made closed by adding an arc, and then, by the Schoenflies theorem, it is the image of a circle under a homeomorphism of $\mathbb R^2$. ...


2

One way to guarantee this is to pick a Riemannian metric on $M$, then pick the sets $U_i$ so that each is a very small locally convex ball in $M$. From this it follows that each set $U_{i_1} \cap \cdots \cap U_{i_k}$ is actually a convex set in a ball, and so is contractible. Added: To address the question in the comment, it is proved in almost any ...


2

For any point $x$ other than $x_0$, suppose $\gamma$ is the geodesic connecting $x_0$ and $x$ whose length is $r$. Let $V=\frac{\partial}{\partial r}$, and $w$ be any vector that is orthogonal to $v=V(x)$. Parallel transport $w$ along $\gamma$ to get a vector field (along $\gamma$) $W$. Now we have $$\frac{d}{dr}(Wf)=VW(f)=\operatorname{Hess}(f)(V,W)$$ ...


2

First of all, you sure put a lot of effort in solving this problem, as well as in making your post as clear as possible. I thank you for that, it was interesting and fun to read. It seems that your approach is the right one to begin with, but at some point you made things more complicated than they should be. Namely, the embedded submanifold $N$ can be ...


1

A vector field $X$ along a curve $\alpha$ is parallel if $$\nabla_TX=0$$ This equation means that the vector field $X$ does change along $\alpha$. Geometrically, all values of $X$ along $\alpha$ seems to be parallel.


1

Geodesics are only locally length-minimizing, i.e., only for $t$ sufficiently close to $1$ will $\gamma$ necessarily be the geodesic of shortest length joining $\gamma(1)$ and $\gamma(t)$. All that can be said in general is that the distance in question is at most $\left|1-t \right| \, |X_p|$. To elaborate, there may be multiple geodesics joining ...


1

Take $M = \mathbb{R}^2$ with its standard metric. With respect to the standard coordinates $(x,y)$ each affine connection on $M$ is written as $$ \nabla = \mathrm{d} + A $$ where $A$ is a 2 by 2 matrix of $1$-forms on $M$. Remarks: $\nabla$ is compatible with the metric if and only if $A$ is skew-symmetric, i.e. $A \in \mathfrak{o}(2)$ $\nabla$ is the ...


1

The pullback of a Riemannian metric is treated like the general pullback for tensor fields. In other words, $f^*\mathrm{g}(X_q,Y_q)=\mathrm{g}(f_*X_q,f_*Y_q)$. If we are working in a specific chart, we can write out the metric as $$\mathrm{g}=\sum_{i,j}g_{ij}\mathrm{d}x^i\otimes\mathrm{d}x^j.$$ For the ease of calculation, we often just write this out as a ...


1

First of all, I assume that you are asking about forms which are only locally Lipschitz, otherwise the answer will be different (the function $x^2$ on the real line is only locally Lipschitz). Then the simplest thing to do is to write your form in local coordinates and require all the component functions to be locally Lipschitz. There are alternative ...


1

I'm turning my comment into an answer because it got too long. You may want to look at this question and answer, in which I give a way of constructing a connection. The many choices involved in the construction should make it clear that many connections exist. When you say "projecting the usual directional derivative onto the tangent space", I assume you ...


1

First, let's compute the mentioned example in detail: Example Find the Killing fields $Z$ of the standard metric $\bar{g} := dx^2 + dy^2$ on the plane $\mathbb{R}^2$. If we write $$Z = f \partial_x + g \partial_y,$$ for some functions $f,g$ of $(x, y)$, then $$g(Z, \cdot) = f \,dx + g \,dy$$ and so the bilinear form $(X, Y) \mapsto g(\nabla_X Z, Y)$ is ...


1

No. In the second line, you've incorrectly used the summation convention. The upper $a$ in the first line is a dummy index, already contracted with a lower $a$, so you can't introduce another $a$ into the formula.


1

Depending on which class of paths you consider, you will get different path spaces, which are smooth infinite-dimensional manifolds, modelled on differen model spaces. Let's assume you choose smooth paths. Then this is a manifold modelled on the Fr├ęchet space $C^\infty_0(S^1, \mathbb{R}^n$. Here the zero indicates that we mean the space of paths that are ...



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