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5

So, the vast majority of manifolds are not locally symmetric. The most obvious example that comes to mind is the standard torus in 3-space. This has positive curvature just to the left of the topmost circle, and negative curvature just to the right, and hence there cannot possibly be a geodesic symmetry at any of the points on that top circle. Similarly with ...


5

Here's the quick way to describe what's going on: In linear algebra, you learned primarily about linear transformations. In particular, $T:V \to W$ is a function that takes a single vector, and produces another vector in a linear way. In particular $T$ satisfies $T(ax + by) = aT(x) + bT(y)$. It turns out that linear transformations can naturally be ...


4

It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction. Given a connection \begin{align*} \nabla : \Gamma(TM) \times \Gamma(TM) &\to \Gamma(TM)\\ (X, Y) &\mapsto \nabla_XY \end{align*} on $TM$, there is an associated connection (which I will also ...


4

This seems like mainly a question about what the abstract index notation is asking you to do. Let's look at a single term from the right hand: $\nabla_X\nabla_YZ$. Recall that in abstract index notation, two tensors juxtaposed signifies their tensor product. For example, if $V^a$ and $W^b$ are vector fields, $V^aW^b$ is supposed to mean the tensor field $V ...


3

The start of the computation is OK, but the partial derivative does not make sense. I think the problems come from the fact that you misinterpret $\nabla_c\nabla_dZ^a$. Evaluating the second covariant derivative $\nabla^2Z$ on $X$ and $Y$, you do not get $\nabla_X\nabla_Y Z$. What you have to do is to differentiate $\nabla Z$ as a $\binom11$-tensor field. ...


3

Note that $$(D\theta)(X, Y) = (\nabla\theta)(X, Y) = (\nabla_X\theta)(Y) = X(\theta(Y)) - \theta(\nabla_XY).$$ So the skew-symmetric part of $(D\theta)(X, Y)$ is \begin{align*} \frac{1}{2}[(D\theta)(X, Y) - (D\theta)(Y, X)] &= \frac{1}{2}[X(\theta(Y)) - \theta(\nabla_XY) - Y(\theta(X)) + \theta(\nabla_YX)]\\ &= \frac{1}{2}[X(\theta(Y)) - ...


3

The definition of the tensor product is $(f \otimes g)(u,v) = f(u)g(v)$. You should be able to check that whenever $f,g \in T^*$ (i.e. are linear), their product $f \otimes g$ is bilinear; and the product $f \otimes f$ is symmetric and weakly positive definite. Thus the action of $dx \otimes dx$ on a pair $u,v$ of tangent vectors is $$(dx \otimes dx)(u,v) = ...


3

If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ...


2

A vector is probably not just a tuple of scalars unless your definition of "tuple" is very very broad (and you also probably need to assume some extras like AC). In general a vector space is quite a bit more abstract. A first easy example of a vector space where the vectors are not really "tuples" for most definition of what a "tuple" is would be functions ...


2

There are several misconceptions in the OP about both mathematicians' and physicists' use of the word "vector", and even about what scalars and tensors are. To keep this a concise overview I'll be linking to fuller explanations. Firstly, anything you've heard about magnitude and direction was just an attempt to help schoolchildren avoid certain fallacies ...


2

An element of the domain of $T$ is of the form $(f_1,\dots, f_k, v_1, \dots, v_l),\, f_1,\dots,f_k\in V^*, v_1,\dots,v_l \in V$, so the first $k$ elements of $T$ should be able to take elements of $V^*$ and the next $l$ terms should be able to take elements of $V$, hence $v_{\mu_1} \otimes \cdots \otimes v_{\mu_k}(f_1,\dots, f_k)$ first, and $v^{{\nu_1}^*} ...


1

Ok, lets formalise all that have been said in the comments: Let $(v_1,\dots,v_n)$ and $(v^1,\dots,v^n)$ be basis of $V_p$ and $V^*_p$, respectively. Take a tensor $\tau \in \mathcal{T}^r_s(V_p)$, and pick indexes $k \leq r, l \leq s$. Then we define the contraction $C^k_l\tau \in \mathcal{T}^{r-1}_{s-1}(V_p)$ as ...


1

If $\vec{g}=\left[\begin{array}{c}f_1\\\vdots\\f_m\end{array}\right]$ then the derivative of $\vec{g}$ is the matrix $$J\vec{g}=\left[\begin{array}{c}\nabla f_1\\\vdots\\\nabla f_m\end{array}\right],$$ which is an $m\times n$ - rectangular array. In components, you would see it as $$J\vec{g}=\left[\dfrac{\partial f_i}{\partial x_j}\right],$$ where $i$ is ...


1

We can add something to the already excellent answer given by Omnomnomnom. The wonderful Gravitation by Misner,Thorpe and Wheeler (affectionately called MTW) on page 75 describes the most general (m,n) tensor as:" a linear machine with n input slots for n 1-forms and m input slots for m vectors, given the requested input, it puts out a real number..." ...


1

First, it's pretty clear that the field $K$ does not matter in any way (if you know how to prove things for $\mathbb{R}$, then just check that you don't use any special property of this field). Then, be careful : the statement for tensor algebras is already false for finite-dimensional vector spaces : if $V$ has finite dimension, $T(V^*)$ has countable ...


1

Rank one tensors, on a vector space $V$ over the scalar field $\Bbb F$, are linear maps $$V\to\Bbb F$$ and $$V^*\to\Bbb F,$$ where $V^*$ is the dual space of $V$. Rank two tensors are bilinear maps $$V\times V\to\Bbb F,$$ $$V^*\times V\to\Bbb F,$$ $$V^*\times V^*\to\Bbb F.$$ Rank three tensors are trilinear maps $$V\times V\times V\to\Bbb F,$$ ...


1

Why do you think that this would be correct? You are differentiating completely different objects on the left and right hand side. $g$ is not the same as $R$ (I'm omitting the bars). Check this site (scroll down to 'Lie derivative' of tensor fields) to check how to calculate $${\frak{L}}_V (R(X, Z, Z, Y)) $$ (which results in $$({\frak{L}}_V R)(X, Z, Z, ...


1

A tensor is just a multilinear, scalar-valued function. If I write $V \multimap W$, for the collection of linear functions from a vector space $V$ to a vector space $W$, then, over the reals, a rank-$n$ tensor is just $V^{\otimes n}\multimap\mathbb R$ where $V^{\otimes n}$ means the $n$-fold tensor product, e.g. $V^{\otimes 2} = V\otimes V$. A tensor field ...


1

Given that I prefer index notation, that presents less ambiguities, I would write expression 1 as: $$ \mathbf{u}(\mathbf{u}\cdot\nabla)\rho+\rho(\mathbf{u}\cdot\nabla)\mathbf{u}+\rho\mathbf{u}(\nabla\cdot\mathbf{u}) $$ and expression 2 as $$ \nabla\cdot(\rho\mathbf{u}\otimes\mathbf{u}) $$


1

For finite dimensional vector spaces, we have an isomorphism $V \cong V^{**}$, so you are really using a basis of $V^{**}$


1

$\delta_{ij}=\begin{cases}1 & \text{if } i=j \\ 0 & \text{otherwise}\end{cases}$ https://en.m.wikipedia.org/wiki/Kronecker_delta $\epsilon_{ijk}=\begin{cases} sgn(ijk) & \text{as a permutation, if } i,j,k \text{ are different} \\ 0 & \text{otherwise}\end{cases}$ https://en.m.wikipedia.org/wiki/Levi-Civita_symbol And of course repeated ...


1

The second line is wrong. At first act with the transformation of the covariant derivative $Y^{j'}_{,p}$. Then you will get $$\frac{\partial x'^p}{\partial x^i}(Y^{j'})=\frac{\partial x'^p}{\partial x^i}(\frac{\partial x^{j'}}{\partial x^q} Y^p_q).$$ So, you will get a term that will cancel the last term in last line in your solution.


1

Since $\epsilon$ and $\nabla_b \epsilon$ (for fixed $b$) are both antisymmetric, the only non-zero terms in the sum $$\epsilon^{a_1 \cdots a_n} \nabla_b \epsilon_{a_1 \cdots a_n}$$ will be those where $a_1\cdots a_n$ is a permutation of $1 \cdots n$. Moreover, if we permute the indices back to this order in each term, the signs that each $\epsilon$ pick up ...



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