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12

Your $M$ is not compact, and $\omega$ is not compactly supported. So you have not contradicted Stokes's theorem.


10

Congratulation! You found something which leads to idea of closed but not exact forms and cohomology groups. To see the problem take $M= \overline{B^2(1)} \setminus B^2(\epsilon)$. Than $\partial M = S^1(1) - S^1(\epsilon)$. Now the Stokes theorems says: $$ \int_{S^1(1)} \omega - \int_{S^1(\epsilon)} \omega = \int_M d\omega $$ You might think that integral ...


4

For (nice) space curves with a Frenet frame, by convention, curvature $\kappa$ is always positive. There are two ways in which "signed curvature" is used to refer to curves. (1) For regular plane curves, we can decide that the unit tangent vector $T$ and principal normal vector $N$ will always make a right-handed basis. Then $T'(s)=\kappa(s)N(s)$, and ...


3

Ok, I undestood what is going on here. It actually a triviality, but a very confusing one. Everything is clear if one have in mind that $\tilde{P}$ is actually a multilinear function. Recall that the differential of a linear transformation is actually the same linear transformation. For example $$T:\mathbb{R^n}\rightarrow \mathbb{R}$$ then the differential ...


3

No curve between two points on a Riemannian manifold ever maximizes arc length. Given any curve $\gamma$ from $p$ to $q$, you can always find a longer curve, e.g. by leaving $\gamma$ for a little while and taking a detour, and then coming back and rejoining $\gamma$ where you left off. The dashed curve in your drawing is a saddle point for the distance ...


3

It suffices to take $\alpha = f\, dx^{i_1}\wedge\dots\wedge dx^{i_p}$ and $\beta=g\,dx^{j_1}\wedge \dots\wedge dx^{j_q}$. Then $\alpha\wedge\beta = fg\, dx^{i_1}\wedge\dots\wedge dx^{i_p}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $d\alpha\wedge\beta$ and $\alpha\wedge ...


3

When changing the basis with respect to which a bilinear form is represented, we have$$A'=P^tAP,$$where $A$ and $A'$ are the representing matrices and $P$ is the transition matrix. When one of the matrices represents $g$ with respect to an orthonormal basis, we have$$A'=P^tIP,$$ and thus$$\det A'=\det P^t\cdot\det P=(\det P)^2.$$The last sentence in the ...


2

The vector fields $Y_i$ do necessarily span an integrable distribution $E \subseteq T(U \times \overline{U})$. $E$ is necessarily a smooth subbundle of $T(U \times \overline{U})$, so it suffices to show $E$ is closed under the Lie bracket. Note that $[X_i, \overline{X}_j] = 0$ for every $i$, $j$ because $X_i$ only differentiates with respect to the first $n$ ...


2

First of all, note that in the formula you give us, $$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2\dots i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}, $$ there is a summation involved! In fact, there are $n-p$ sums, since the indices are being contracted. I won't use Einstein's convention in ...


2

Sorry to resurrect, but we leave a $($detailed$)$ proof here that $TM$ has the structure of an oriented $2n$-manifold, even if the $n$-manifold $M$ is non-orientable. Let $\{(U_\alpha, \phi_\alpha)\}_{\alpha \in A}$ be a smooth atlas of $M$, and let $V_\alpha = \phi_\alpha(U_\alpha) \subset \mathbb{R}^n$. Then $(\phi_\alpha)_*: TU_\alpha \to TV_\alpha = ...


2

Let $\overline{X}_i$, $\overline{U}$ be as in the setup from the linked question. The distribution $E$ is integrable by the linked question, so by the Frobenius Theorem, there exists an injective immersed submanifold $M \subseteq U \times \overline{U}$ near $(p, q)$ such that $T_{(a, b)}M = E_{(a, b)}$ for all $(a, b) \in M$. Let $\pi:U \times \overline{U} ...


2

I assume that $ f: \mathbb{R}^{4} \times \mathbb{R}^{4} \to \mathbb{R} $, as mentioned by anon. Now, a tensor can be defined as a multi-linear mapping. Hence, if $ f $ were a tensor, it would have to satisfy $$ \forall \mathbf{x},\mathbf{y} \in \mathbb{R}^{4}, ~ \forall \lambda \in \mathbb{R}: \quad f(\lambda \cdot \mathbf{x},\mathbf{y}) = \lambda \cdot ...


2

Let $E \stackrel{\pi}{\to} M$ be the vector bundle, then given a smooth section $s \in \Gamma(E)$, what does the connection output when applied to $s$? We need an element of $\Gamma(T^* M \otimes E)$, which is a section of the tensor product bundle $T^* M \otimes E$. This is given by the following data: for any point $x \in M$ in the base space, $(Ds)(x) = ...


1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$If you have a surface embedded in $\Reals^{3}$, you can (as you note) use the "ambient" Euclidean inner product to take dot products of tangent vectors. However, that's mild overkill; in order to take dot products of tangent vectors to a surface, all you really need is the first fundamental form. If ...


1

This is actually easier. Let $C = \text{supp}(\omega)$ and $U = \Sigma \setminus C$. Then $\omega |_U = 0$ and so $d(\omega |_U) = 0$. But as $d\omega)|_U = d(\omega|_U)$, we have $d\omega = 0$ on $U$ and so $\text{supp}\omega \subset C$.


1

The fact that $\nabla \dot{x}(s)\in T_{x(s)}L$ implies that $\nabla\dot{x}(s)=\tilde{\nabla}\dot{x}(s)$, where $\tilde{\nabla}$ is the Levi-Civita connection of $L$. Now, if $\gamma$ is a geodesic in $L$, then$$\nabla\dot{\gamma}=\tilde{\nabla}\dot{\gamma}=0,$$and hence $\gamma$ is a geodesic in $\mathbb{R}^n$ too.


1

Not if this is true on only one chart. For example, consider the 1-form metrically dual to the north-south vector field on the round sphere. Now multiply it by a bump function supported away from a small neighborhood of the north pole to produce a second 1-form. The two 1-forms agree away from that neighborhood. In particular, they agree on a chart covering ...


1

The tangent space of $M=SO(n)$ at any point $q\in M$ is: $$T_q(M)=\{xq|x+x^T=0\}$$ Let $df=0$ on $T_q(M)$ we got $Tr(Dxq)=Tr(qDx)=0$ for all skew-symmetric $x$. Let $A=qD$ and let $x_{ij}=-x_{ji}=1$ for fixed $i,j$, and all other entries of $x$ be $0$, $Tr(Ax)=A_{ji}-A_{ij}=0$, we conclude that $A=qD$ is symmetric. Now $A_{ij}=\sum ...


1

From your definition, $$ (\nabla_XT)(\lambda, Y ) = \nabla_X[T(\lambda, Y )] - T(\nabla_X \lambda, Y ) - T(\lambda, \nabla_X Y ) . $$ To calculate $\nabla_i T^j_{\ k}$, note (We use the simplified notation $\partial_i = \frac{\partial}{\partial x^i}$) $$\nabla_i T^j_{\ k} = (\nabla_{\partial _j} T)(dx^j , \partial_k) = \nabla_{\partial_i}[T(dx^j , ...


1

A useful fact: Let $X,Y,$ be vector fields on $U$, and define vector fields on $U\times U$ by $\overline{X}=(X,0),\quad \overline{Y}=(0,Y).$ Since $\overline{X}$ depends only on the first coordinate and $\overline{Y}$ depends only on the second, we have by definition $$\mathcal{L}_XY=0,$$in other words$$[X,Y]=0.$$And to the question: We ...


1

Every manifold has at every point $p\in M$ a tangent space $T_pM$. In fact, $T_pM=\text{Im}(d_a\varphi)$ for any parametrization $\varphi:V\to \mathbb R^m$ defined on an open set $V\subset\mathbb R^d$, $d=\dim(M)$, with $a\in V$, $p=\varphi(a)$ and $\varphi(V)$ an open nbhd of $p$ in $M$. In other words $T_pM$ is the linear space generated by the columns of ...


1

There is nothing that you can simplify in this, except if some further simplification comes from the explicit form of $F$ and $\alpha$. Also the action of the Hodge operator on basis element is well defined (cf. wikipedia).


1

Let $\varphi: {\mathbb R}^n\to{\mathbb R}^n$ be an isometry with $\varphi(0)=0$. Then, for any $v\in {\mathbb R}^n$ and any $\lambda\in [0,1]$, we have $$(\dagger)\qquad \|\varphi(\lambda v)\| = \|\varphi(\lambda v) - \varphi(0)\|=\|\lambda v - 0\| = \lambda \| v\|=\lambda\|\varphi(v)\|\\(\ddagger)\qquad\|\varphi(v)-\varphi(\lambda v)\| = \|v - \lambda v\| = ...


1

The issue is that the normal coordinates $x^i$ depend on $p$ - I'll denote (some choice of) normal coordinates around $p$ by $x_p^i$ for clarity. It's certainly true that at each $p$ your formula holds, so you do have an equation of functions; but the function on the RHS has a lot of dependence on coordinates: if you write all the dependence out ...


1

I was finally able to answer the question. The goal is to give a Taylor expansion for the function $L$ above from mollyerin's answer: In order to fix notation, the Levi-Civita connection is denoted by $\nabla$. Pick geodesic normal coordinates centered at $p$. By using the Gauss-Lemma we find $$\begin{aligned}L(t) &=\int_0^t ...



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