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There are three kinds of 2-tensors, of type $(0,2)$, $(2,0)$ and $(1,1)$. Let's take the metric tensor $g$ of type $(0,2)$. If $p$ is a point in your manifold and $V:=T_p M$ the tangential space as a vectorspace, then the metric is a bilinear map $$g_p:V \times V \rightarrow \mathbb{R}$$ and "the matrix" of $g$ is a description of $g$ if you choose a ...

0

Consider the following matrix $$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}$$ By definition, the trace of $A$ is defined as $$\text{tr}(A) = a + e + i = a_{11} + a_{22} + a_{33}$$ So think about how we write tensor indices. If I have a ...

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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 ...

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To show that this quantity is a tensor we need to show that coordinate representation of the quantity is invariant under a change of coordinates. The question stipulates that, for the vectors with coordinates $\left(\begin{array}{c}x_1\\x_2\end{array}\right)$ and $\left(\begin{array}{c}y_1\\y_2\end{array}\right)$ in the cartesian system, the matrix ...

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I wrote a lot, due to boredom and the general hope that some of it is helpful. Much will probably be familiar to you. First: In order to work with tensors or to do calculus on manifolds at all, it's very important to start making the distinction between vectors and covectors (or "dual vectors" or whatever). Briefly, if $V$ is a (finite dimensional, real) ...

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I know of no algebra in which 3d arrays of numbers can be manipulated with the same ease as matrices. Part of that has to do with how any linear map from one space to another space can be represented with a matrix. You can chain such maps together sensibly (and really, only in one way up to the order of how you compose those operations together), whereas ...

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It's very common in tensor analysis to associate endomorphisms on a vector space with (1,1) tensors. Namely because there exists an isomorphism between the two sets. Define $E(V)$ to be the set of endomorphisms on $V$. Let $A\in E(V)$ and define the map $\Theta:E(V)\rightarrow T^1_1(V)$ by \begin{align*} (\Theta A)(\omega,X)&=\omega(AX). \end{align*} ...

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Let $T : V \mapsto W$. Then define $\tau: V \times W^* \mapsto K$ such that, for $a \in V$ and $\alpha \in W^*$, we have $$\tau(a, \alpha) = (\alpha \circ T)(a)$$ Note that it's typical to define tensor to mean a multilinear map that is a function of vectors only in the same vector space, or of covectors in the associated dual space, or some combination ...

1

To summarize as an answer what I wrote in various comments above: first beware that autors differ in their definition of tensor, even when using the same approach, i.e. using the tensor product in this case. For some authors a tensor is defined only as ... $$T\in \underbrace{V \otimes\dots\otimes V}_{n \text{ copies}} \otimes \underbrace{V^* ... 0 The rank is one provided that u\otimes v \neq 0. In particular, u, v \neq 0. Consider the matrix$$A = \left[\begin{array}{cccc} r_1s_1 & r_1s_2 & \dots &r_1s_n\\ r_2s_1 & r_2s_2 & \dots & r_2s_n\\ \vdots & \vdots & \ddots&\vdots\\ r_ms_1 & r_ms_2 & \dots & r_ms_n \end{array}\right].$$Note that ... 0 Let r be the column vector with entries r_i and let s be the column vector with entries s_j. Then if M=rs^T, we have M_{i,j}=r_is_j. Clearly M has rank one, since its column are all multiples of r. 1 Grassmann numbers can be embedded in a clifford algebra. Let g(a,b) be some scalar-valued, symmetric, bilinear function for a, b that are linear combinations of \theta_i--call a,b "vectors". Then the multiplication laws could be tweaked to read$$\theta_i \theta_j + \theta_j \theta_i = g(\theta_i, \theta_j)$$True grassmann numbers are the case ... 1 Background: I work in the field of numerical relativity. I've read Carroll's book, but not recently. It's pretty common for physics students to reach this point in their education, not really knowing anything about what tensors are or how they're talked about in higher mathematics. That's not really the students' fault. If your education was anything ... 0 By using chain rule I arrived at. \nabla\cdot (f\,\nabla g)=\partial_k(f\,\partial_k g) = \partial_k f \cdot \partial_k g + f \partial_k \partial_k g. 2 g and f are functions, so you don't have the indices on f or g, i.e. f\,\nabla g=f\,\partial_k g. So when you take divergence, we have \nabla\cdot (f\,\nabla g)=\partial_k(f\,\partial_k g). 0 f and g are scalar fields, so neither has an index. Try starting with f \nabla g = f \partial_j g and go from there. 0 A k-dimensional tensor can loosely be defined as a k-dimensional array of numbers (a_{i_1\cdots i_k})_{1\leq i_1,\ldots,i_k\leq n} which behaves "appropriately" under coordinate changes. The example from your question (A_{ij} \times B_{jk} = C_{ik}) is a so-called contraction of tensors, i.e. we sum over one index of each so that only the other ... 1 @mollyerin gives a nice answer. Furthermore, depending on the rigor that you are looking for I think that Roger Penrose's book Road to Reality gives a nice overview of the geometric interpretation of covariant derivatives. It should be noted that the book attempts to present topics such as the covariant derivative and other mathematical subjects using ... 0 Let me add that in order to find curves of minimal length between to positions in a surface for example, one sets an equation like$$\nabla_{C'}C'=0,$$which leads you to the classical expression \ddot{u}^k+\Gamma^k{}_{st}\dot{u}^s\dot{u}^t=0, where C'=\dot{u}^s\partial_s is the tangent field for C. This is pretty geometrical. 3 I'll say a few words about how I think about covariant derivatives, which is really just expanding on janmarqz's comment (hopefully others will contribute their own viewpoints as well): For me, the most important geometric idea behind a covariant derivative \nabla is that given a curve \gamma in a manifold M, \nabla gives you an isomorphism between ... 0 This is really something that you should learn from a textbook, but I'll try to answer the second part of your question. When you apply forces on any deformable body in equilibrium, it must develop internal resisting forces to keep every piece of it in equilibrium, by Newton's 2nd law. This is usually demonstrated by cutting the body into two by an ... 0 By a 'rank three tensor' I think you mean an element of \bigotimes^3V. This is not what 'rank' means in mathematics. An element of \bigotimes^kV is said to have order (or degree) k. A rank one tensor (or simple tensor) is a tensor of the form v_1\otimes\dots\otimes v_m. The rank of a tensor T is the minimal number of rank one tensors needed to ... 0 This is confusing because the usual way of writing integrals in differential forms leaves something implicit: the tangent k-vector of the manifold of integration, and the notation itself tends to make people confuse \mathrm dx---the cotangent basis vector associated with the coordinate x---with dx---the symbol that tells us we're integrating with ... 1 We have the scalar$$\phi(x) =\Gamma_{\mu\nu}(x)a^{\mu}(x)a^{\nu}(x)$$Going to a different coordinate system x' = x'(x) we have$$a^{\mu}(x') = \frac{dx^{\mu}}{dx'^\alpha} a^{\alpha}(x)$$Since \phi(x) transforms as a scalar we have \phi(x'(x)) = \phi(x) where$$\phi(x') = \Gamma_{\mu\nu}(x')a^{\mu}(x')a^{\nu}(x')$$Using the transformation law ... 2 You can indeed say that {\overline T}_{ik}{\overline T}^{kj}=\delta_i^j. This is because the definition of T^{ij} is that in each basis its components are such that it is the inverse of T_{jk}. Your line of working$${\overline T}_{ij} = T_{rs}\frac{\partial x^r}{\partial {\overline x}^i}\frac{\partial x^s}{\partial {\overline x}^j} \implies ...

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(i) Your understanding is correct. (ii) The significance of the ordering is mainly bookkeeping; in particular they provide an easy way to start from a basis $e_1,\dots,e_n$ for the vector space $V$ and extend it to a basis for the spaces of $p$-vectors and the exterior algebra. (iii) An algebra is a vector space with a product (that satisfies some rules). ...

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Non-orientablility is a global property. It can't be confirmed based on looking at one point or a small neighborhood of one point because any point has a neighborhood diffeomorphic to euclidean space, which is orientable. That is to say, locally any manifold is orientable. In terms of normal vectors, these are non-zero at any given point, but the question ...

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As much as I love Clifford / geometric algebra, I don't think it is of use here. CA allows you to deal in a basis-free way with multidimensional subspaces as algebraic objects, making various complicated derivations more elegant and transparent. In the case of neural nets, however, we really are dealing with a plain vector of parameters. If $\theta \in ... 1 This is a bit too long for a comment, so let me answer your first question. Let$(\cdot,\cdot)$denote the pairing of covector fields with vector fields, e.g.,$(\lambda,Y) := \lambda(Y)$for$\lambda$a covector field and$Y$a vector field. Think of the scalar field$(\lambda,Y)$as the product of the covector field$\lambda$with the vector field$Y\$ and ...

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