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Your lists includes some subjects, some algebraic structures, and some more specific objects. I'll separate them to organize the list a little better. Anyone can feel free to contribute to this list as I'm certainly not an expert in all of this. Subjects: Linear Algebra: the study of finite dimensional vector spaces and the linear transformations ...

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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). Multilinear algebra contains differential forms as a special case, so exterior=Grassmann ...

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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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($*$) $\to$ ($**$) 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 ...

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Note that \eqalign{ \sum_k u_k^TXv_k &= \sum_k Xv_k:u_k \cr &= \sum_k X:u_kv_k^T \cr &= X:\sum_k u_kv_k^T \cr &= X:UV^T \cr } where the colon denotes the Frobenius Inner Product for matrices. For convenience, define a new variable \eqalign{ h &= (UV^T:X) - \sum_k b_k \cr dh &= UV^T:dX \cr\cr } Use this new variable to ...

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$$J_1 = \epsilon_0\sum_s \omega_{ps}^2\frac{i\omega E_1-\Omega_s E_2}{\omega^2-\Omega_s^2}=\sigma_{11}E_1+\sigma_{12}E_2.$$ Compare and find $\sigma_{11}$ and $\sigma_{12}$. Do the same for $J_2$.

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Easiest is to use indices: $$\left(\frac{dA}{d\alpha}\Delta \alpha\right)_{ij} = \sum_k \frac{\partial A_{ij}}{\partial \alpha_k} \Delta \alpha_k.$$ If you really want to store $\frac{dA}{d\alpha}$ separately, as you say you can do so in a three-dimensional array, where the $ijk$ entry is e.g. $\frac{\partial A_{ij}}{\partial k}$. Multiplying by vector $\... 0 Not a chance. Just consider the degrees of freedom. In dimension$n$the tensor$C$has$n^3$components and since you have only one free index on either side of your equation you could never hope to fix more than$n$of them. Maybe if$C$and$A$have a particular form then more might be possible. 4 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, ... 0 You get redundancy. With two cameras, each depth estimate depends on the intersection of a pair of lines. When those are near parallel, this estimate can be very bad -- a small error in the data can introduce a large error in the solution. Since your data are typically discretized (at the pixel level), such small errors can be quite common. With three ... 0 I have no idea how you are getting a rank-4 tensor from a rotation problem--perhaps you can explain more about that? Generally the term direction cosine refers to the fact that the rotation group on$\mathbb{R}^{n>1}$, e.g.$SO_n$, can always be written as an$n\times n$matrix representation where $$A_{ij} = \mathbf{b}_i\cdot\mathbf{r}_j,\ \ \ 1\leq i,... 1 How may we view T=u_∗T_0 as a section of T(M), when T is only a tensor above π(u)? We can't. Hovever by changing u in L(M) we can. In case of G structure wehere G is a group of transformations leaving T_0 untached wewill get a tensor on M. Which leads to the answear to the second question. Now let G<GL(n,R) be the largest Lie ... 2 The comment as it is in your post is taken out of context: in your post, it seems to be a part of some considerations about vectors - in which case one naturally asks how could scalars be particular cases of vectors (exactly your question). In reality, this comment in its original context is a part of some considerations about tensors, and in this case ... 0 Of course, writing "all calculations" is misleading. What one can say is the following: Given two vectors x, y\in V the sum z:=x+y is a well defined element of V which does not depend on the chosen basis. In the same vein: You obtain the the coordinates of z by adding the coordinates of x and of y, whatever the chosen basis is. Similarly, the ... 0 Yes, your derivation is correct. Incidentally, you can avoid expressing the Euclidean metric in cylindrical coordinates: Parametrize the hyperboloid by$$ \Phi(\rho, \varphi) = \left(\sqrt{R^{2} + \rho^{2}} \cos\varphi, \sqrt{R^{2} + \rho^{2}} \sin\varphi, \rho\right).$The partials are \begin{align*} \Phi_{\rho}(\rho, \varphi) &= \left(\frac{\... 1 Let me first remark that the tensors you consider here are of type(0,2)$and not of type$(2,2)$. Initially for such a tensor you have a bilinear form$b:V\times V\to\mathbb R$then the obvious way how a linear isomorphism$\phi:V\to V$acts on such a form is via$(v,w)\mapsto b(\phi(v),\phi(w))$. So the conditions you are referring to just express the ... 2 I've seen in the literature the notation$C$with some additional specifications for the contraction maps of all sorts, but the amount of decorations on the symbol$C\$ varied depending on the context. See, e.g., A.Gray, Tubes, p.56, where these maps are used in the case of somewhat special tensors, and therefore the notation is simpler. In general, there is ...

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