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## Hot answers tagged tensors

7

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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(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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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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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 $\theta \in ... 3 There are three kinds of 2-tensors, of type$(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 ... 3 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 ... 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 ... 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$\nabla\cdot (f\,\nabla g)=\partial_k(f\,\partial_k g)$. 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 $${\overline T}_{ij} = T_{rs}\frac{\partial x^r}{\partial {\overline x}^i}\frac{\partial x^s}{\partial {\overline x}^j} \implies ... 1 I know of no algebra in which 3d arrays of numbers can be manipulated with the same ease as matrices. Part of that has to do with how any linear map from one space to another space can be represented with a matrix. You can chain such maps together sensibly (and really, only in one way up to the order of how you compose those operations together), whereas ... 1 Background: I work in the field of numerical relativity. I've read Carroll's book, but not recently. It's pretty common for physics students to reach this point in their education, not really knowing anything about what tensors are or how they're talked about in higher mathematics. That's not really the students' fault. If your education was anything ... 1 @mollyerin gives a nice answer. Furthermore, depending on the rigor that you are looking for I think that Roger Penrose's book Road to Reality gives a nice overview of the geometric interpretation of covariant derivatives. It should be noted that the book attempts to present topics such as the covariant derivative and other mathematical subjects using ... 1 Grassmann numbers can be embedded in a clifford algebra. Let g(a,b) be some scalar-valued, symmetric, bilinear function for a, b that are linear combinations of \theta_i--call a,b "vectors". Then the multiplication laws could be tweaked to read$$\theta_i \theta_j + \theta_j \theta_i = g(\theta_i, \theta_j)True grassmann numbers are the case ... 1 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 ... 1 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*} ... 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^* ... 1 We have the scalar $$\phi(x) =\Gamma_{\mu\nu}(x)a^{\mu}(x)a^{\nu}(x)$$ Going to a different coordinate systemx' = 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 ...

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