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Maybe try this paper? Fast Matrix Multiplication: Limitations of the Laser Method It goes into enough detail for me to think I understand what's going on. I think the laser method is figuring out answers for N values with less rounding errors than what occurs using Strassen's method using some differential equation or numerical method. In recent years ...

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If $\frac{\partial^2}{\partial x \partial \lambda}$ is denoting an outer product of $\frac{\partial}{\partial x}$ with $\frac{\partial}{\partial \lambda}$ we have \begin{align} \partial_{x^m} \partial_{\lambda^n} (A^{ij} \lambda^i x^j + B^{ij} x^i \lambda^j) &= A^{ij} \delta^{in} \delta^{jm} + B^{ij} \delta^{im} \delta^{jn} \\ &=A^{nm} + B^{mn} ... 0 I think this is what you're after: If R is a ring, then the map f : R \times R \to R given by f(x, y) = x y is R-bilinear, so it induces a linear map \tilde{f}:R\otimes_R R \to R given by \tilde{f}(x \otimes y) = f(x, y) = x y. In general, tensor products of modules (or, more concretely, vector spaces) turn bilinear maps f : A \times B \to C ... 0 It is the first definition: Given a basis \mathcal F = (f_i) of a finite-dimensional vector space \Bbb V, the dual basis of the dual space, \Bbb V^*, is the unique basis (\hat f{}^i) of \Bbb V^* such that\hat f{}^i(f_j) = \left\{\begin{array}{rl}1,& i = j \\ 0,& i \neq j\end{array}\right.$$In particular, if v = ... 0 If you are having trouble with tensors, try this: Imagine that you want to locate a point in 3-dimensional Cartesian plane. You need (x,y,z) coordinates to locate the point. Now imagine that each of these points has a numerical value. Then this three dimensional plane with these numerical value will constitute a matrix. If the length of x, y and z axes, each ... 4 Tensors in general are multilinear coordinate-free objects that can be represented with respect to some basis by a multi-dimensional arrays indexed appropriately. Just like you can represent a bilinear form B \colon V \times V \rightarrow \mathbb{F} when V is an n-dimensional vector space (or a linear map T \colon V \rightarrow V) by a n \times n ... 1 I'ts a tensor. But so are normal numbers, vectors, and matrices which can be considered 1X1 tensor, 1Xn tensor, mXn Tensor. All data structures of this form (mXnX...)are what are called Tensors. 0 You can think of tensors as multidimensional arrays. 3 You can think of a rank three tensor as a three dimensional array. A matrix is a rank two tensor, or a two dimensional array. A vector then is a rank one tensor and scalar a rank zero. This is a simplification of the subject of tensors, but it's useful to think of them as a generalization of matrices. 0 In physics (and in mathematics, in differential geometry) you are working with so called manifolds M, which you can think of as smooth k- dimensional subsets of some \mathbb{R}^n. A vector (field) along M can be thought of as an assignment of a vector X(p) in each point p of M which is tangent to M. A differential form is an assignment of a ... 1 how exactly to represent specific linear transformations in tensor notation. Indeed every linear map A:V\rightarrow W has a representation as a tensor T\in V^{*}\otimes W such that \forall x\in V:Ax=T(x)\in W. In chosen bases,$$ T=\sum a_{ij}\;v_j^*\otimes w_{i} $$where a_{ij} is the matrix of the linear map and \{v_j^*\} is the basis in ... 1$$ u=\{\text{u1},\text{u2}\}, v=\{\text{v1},\text{v2}\}  \text{uv}=u\otimes v=\left( \begin{array}{cc} \text{u1} \text{v1} & \text{u1} \text{v2} \\ \text{u2} \text{v1} & \text{u2} \text{v2} \\ \end{array} \right)  \left| \text{uv}\right|=0 $$The determinant of the matrix that represents the tensor product of any 2 vectors will always be ... 0 Let E:=\mathbb{R}^n and E^* its dual. Here is a "concrete" way of seeing the equivalence of a linear transform L:E \rightarrow E with a linear application from L:E \otimes E^* \rightarrow \mathbb{R}: It is nothing more that the decomposition of matrix M=[m_{i,j}] of linear mapping L (with respect to canonical bases of E and E^*) under the ... 0 One can use the matrix for g within the construction of the reciprocal basis$$\partial^k=\sum_sg^{ks}\partial_s,$$or simply \partial^k=g^{ks}\partial_s. It happens that \partial_k\mapsto\partial^k is a change of basis in the tangent space. 1 Ok, lets formalise all that have been said in the comments: Let (v_1,\dots,v_n) and (v^1,\dots,v^n) be basis of V_p and V^*_p, respectively. Take a tensor \tau \in \mathcal{T}^r_s(V_p), and pick indexes k \leq r, l \leq s. Then we define the contraction C^k_l\tau \in \mathcal{T}^{r-1}_{s-1}(V_p) as ... 2 An element of the domain of T is of the form (f_1,\dots, f_k, v_1, \dots, v_l),\, f_1,\dots,f_k\in V^*, v_1,\dots,v_l \in V, so the first k elements of T should be able to take elements of V^* and the next l terms should be able to take elements of V, hence v_{\mu_1} \otimes \cdots \otimes v_{\mu_k}(f_1,\dots, f_k) first, and v^{{\nu_1}^*} ... 1 For finite dimensional vector spaces, we have an isomorphism V \cong V^{**}, so you are really using a basis of V^{**} 0 I'm going to write \varepsilon^i instead of e^i to make the notation clearer: we have \varepsilon^i(e_j) = \delta_{ij}. We want to write B as some linear combination of the \varepsilon^i \otimes \varepsilon^j. Note that (\varepsilon^i \otimes \varepsilon^j)(e_p \otimes e_q) = \delta_{ip} \delta_{jq} just as for the 1-dimensional case. This means ... 1 The second line is wrong. At first act with the transformation of the covariant derivative Y^{j'}_{,p}. Then you will get$$\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).$$So, you will get a term that will cancel the last term in last line in your solution. 3 The definition of the tensor product is (f \otimes g)(u,v) = f(u)g(v). You should be able to check that whenever f,g \in T^* (i.e. are linear), their product f \otimes g is bilinear; and the product f \otimes f is symmetric and weakly positive definite. Thus the action of dx \otimes dx on a pair u,v of tangent vectors is$$(dx \otimes dx)(u,v) = ...

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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 = ...

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