# Tag Info

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There is an identification taking place. Associated to the map $$T : \mathcal{T}(M)\times\mathcal{T}(M) \to \mathcal{T}(M),$$ we have the map $$\hat{T} : \mathcal{T}(M)\times\mathcal{T}(M)\times\mathcal{T}^*(M) \to \mathcal{C}^{\infty}(M)$$ given by $\hat{T}(X, Y, \omega) = \omega(T(X, Y))$. If $T$ is $\mathcal{C}^{\infty}(M)$-linear in both arguments, ...

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The comment as it is in your post is taken out of context: in your post, it seems to be a part of some considerations about vectors - in which case one naturally asks how could scalars be particular cases of vectors (exactly your question). In reality, this comment in its original context is a part of some considerations about tensors, and in this case ...

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Disclaimer: this is just a (graduate) educated guess based on this. I've never really worked with such objects. One way to think of finite dimensional tensors is as multilinear operators $$T:V^*\times\cdots\times V^*\times V\times\cdots\times V\rightarrow W$$ where $V,W$ are finite dimensional vector spaces and $V^*$ is the dual space of $V$. So, why ...

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A section of a bundle $E$ (with various properties) is a fancier way of referring to a "smoothly varying" choice of $s_p\in E_p$ (with the same properties) as $p$ varies over $M$. So (1) and (2) are identical. With regard to (2) and (3), we're just using the isomorphism (truly a definition) $\text{Hom}(E,\Bbb R) = E^*$ (where here $E=TM\otimes TM$). Notice ...

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Depends on what context did you study (old school, coordinate-based) differential geometry. Sometimes in physics courses, you have something called tensor calculus, which often just deals with doing tensor calculus in general coordinates in euclidean space. In that case, if you have a coordinate system $(u^1,...,u^n)$, and coordinate basis vectors $\mathbf{... 1$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Let$V$be a real vector space with basis$(\vec{\Basis}_{\mu})_{\mu=1}^{n}$, and assume$V$is equipped with an inner product$g$(non-degenerate symmetric bilinear pairing). The author of the notes defines two isomorphisms from$V$to the dual space$V^{*}$: On page 8,... 1 2. Which types of tensors admit a representation using geometric algebra? On p.4 of this document we can see that multivectors over a given Euclidean space$\mathbb{R}^n$do not have arbitrarily high grade/order; instead the highest order possible is$n$(the determinant/volume element). This is because of the two products available in geometric algebra, ... 1 I'll try to give an answer to this question that doesn't require knowing what a tensor is. Of course it will require some comfort with linear algebra, but I'll even try to keep the multilinear algebra to a minimum. Bye_World's answer in the comments is a good one, although there is slightly more that you can say: a matrix is naturally identified with a rank-... 1 Your list includes some subjects, some algebraic structures, and some objects within algebraic structures. I'll separate them to organize the list a little better. Anyone can feel free to contribute to/ edit this list as I'm certainly not an expert in all of this. Subjects: Linear Algebra: the study of vector spaces and the linear transformations ... 1 According to Wikipedia and some papers: Exterior algebra = Grassmann algebra (= differential forms, since they are a construction of the exterior algebra) (=derivations, since derivations are just one possible construction of the dual object to differential forms). EDIT: To make the claim that "exterior algebra=differential forms" precise, since as it ... 1 ($*$)$\to$($**$) is using the transformation law for tensors ($**$)$\to$($***$) is just changing the names of the summation labels ($rst$)$\to$($trs$). This is just like saying that$\sum_{i=1}^n i = \sum_{j=1}^n j$. ($***$)$\to$the final result follows by using$T^i_{jkl} = 3T^i_{ljk}$. The answer to this problem is really just a combination ... 1 How may we view$T=u_∗T_0$as a section of$T(M)$, when T is only a tensor above$π(u)$? We can't. Hovever by changing$u$in$L(M)$we can. In case of$G$structure wehere$G$is a group of transformations leaving$T_0$untached wewill get a tensor on$M$. Which leads to the answear to the second question. Now let$G<GL(n,R)\$ be the largest Lie ...

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