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If $\underline{\underline{\underline y}}$ has $n$ independent components, then perhaps we can write $\underline{\underline{\underline y}}=\underline{\underline{\underline{\underline d}}}\cdot\underline{y'}$ for some $\underline{\underline{\underline{\underline d}}}$ and $n$ component vector $\underline{y'}$? Then note ...

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There is indeed such a formula. You would think it would be widely known, but in fact the only place I've seen it is in the Penrose Graphical Notation in Penrose's "Road to Reality". The formula for the determinant (in $n$ dimensions) is: $$\det(M)=\tfrac 1{n!}\varepsilon^{i_1i_2\dots i_n}\varepsilon_{j_1j_2\dots j_n}M^{j_1}_{\quad i_1}M^{j_2}_{\quad ... 2 I think the key here is to understand what is meant by the tensor product. For any vectors a, b, u, v, the tensor product a \otimes b means$$(a \otimes b)(u,v) = (a \cdot u)(b \cdot v)$$Or, perhaps, it might mean this instead:$$(a \otimes b)(v) = a (b \cdot v)$$The two notions are equivalent to each other, so mathematicians freely use one or the ... 2 Well, first you should recognize that we need some way of representing the B you have in your first definition in the second definition. Since this is the inertia tensor, we can consider B to be the angular velocity bivector. However the second definition is just a linear mapping without the thing that it maps. In fact it can just be thought of as a ... 1 Since no one has answered yet, I'm assuming your still having trouble with it. The trick for your question was to realise$$\begin{align} \implies [(\vec A \times \vec B) \times (\vec C \times \vec D)]_{i} &= [-(\vec B \times \vec A) \times (\vec C \times \vec D)]_{i} \\ &= [(\vec C \times \vec D) \times (\vec B \times \vec A)]_{i} \end{align}$$... 0 You have A(x+h) = A(x) + A(h), hence \|A(x+h)-A(x) - A(h) \| = 0, from which we conclude that DA(x)(h) = Ah, or, DA(x) = A. 2 You can't multiply by \epsilon^{ijk} because there's already a k in B_k and indices should not appear more than twice. 0 If you will browse papers on BSS, then you can notice that the authors make use of (numerical) linear algebra only. I think the best (shortest) way to learn rank, border rank, canonical polyadic decomposition, higher-order singular values decomposition, tensor trains, etc. will be to read the following review papers: 1 T.G. Kolda; Brett W. Bader. ... 1 For v_i\in V,w_i^*\in W^*, we can see \sum_i v_i\otimes w_i^* as an element of Hom(W,V), by setting :$$ \forall w\in W, \qquad (\sum_i v_i\otimes w_i^*)(w)= \sum_i w_i^*(w)v_i. \quad (A) $$Let (j_1^*,\cdots,j_n^*) be the dual base of B_W=(j_1,\cdots,j_n). (a) The matrix of \:e_i\otimes j_l^* in the bases you gave (B_W \: and \: (e_1, ... 1 An alternative approach, if A is invertible: First consider the case when A=I. Then the problem is to maximize u^Tx subject to x^T x\le r^2, where u is a fixed vector (e_\alpha in your case). By Cauchy-Schwarz, the solution is x=ru/|u|. For general invertible A, the problem is to maximize u^Tx subject to x^TA^TAx\le r^2. Let y=Ax. ... 1 Note that x^T A^T A x = (Ax)^T (Ax) = \lVert Ax \rVert^2, so this is expressible as a norm. We expect the derivative to be something like 2A^T Ax, which is what you have in the line$$ \frac{\partial L}{\partial x_l} = \delta_{\alpha l} + \lambda a_{jl}(a_{ji} x_i + a_{jk}x_k): $$changing the indices, this is$$ \frac{\partial L}{\partial x_l} = ...

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After some thinking I decided to add another answer to this question. The reason for doing this is that I want to keep my previous answer for the reference, while adding a comment or expanding the previous answer would make the texts unreadable. Let us begin with refining the statement of the original problem. Suppose we have an open set $U \subseteq ... 0 Note that $$0=[\partial_k,\partial_l] g_{mt} = R_{tmkl}+R_{mtkl}$$ Hence $$R_{t\ kl}^{\ s} + R_{\ tkl}^s =0$$ 2 It is useful to look at this situation from a more general setting. Suppose we have a hypersurface$S$in a$(n+1)$-dimensional Riemannian manifold$M$. (In your case$M = \mathbb{R}^3$with the standard (Euclidean) metric). By a hypersurface we mean a sumbanifold of codimension$1$, so$\dim S = n$. The unit normal vector$N$at each point$p \in S$is ... 0 (1) Levi-Civiata connection is defined as : $$Z_a(g(Z_b,Z_c))=g(\nabla_{Z_a} Z_b, Z_c) +g(\nabla_{Z_a} Z_c, Z_b)$$ and $$\nabla_{Z_a} Z_b= \nabla_{Z_b} Z_a$$ where$\{Z_a\}$is a coordinate vector field. (2) And define $$\Gamma_{ab}^c Z_c =\nabla_{Z_a}Z_b$$ (3) Christoffel symbol is uniquely well-defined ? $$(a)\ Z_a(g(Z_b,Z_c))=\Gamma_{ab}^k ... 0 First, I think it's helpful to think of a "function of a linear functional" and a "vector" as more or less equivalent, as you mentioned. A "function of a linear functional" is something that, when paired with a covector, produces a scalar. A vector fits this description exactly. Also, the dual space of a dual space is naturally isomorphic to the original ... 4 v\otimes w is either of the following: The rank 1 linear mapping V^*\to V given by \alpha\mapsto v.\langle \alpha,w\rangle where \langle\;,\;\rangle:V^*\times V\to \mathbb F is the duality. This is the easiest to visualize. The decomposable bilinear form V^*\times V^*\to \mathbb F given by (\alpha,\beta)\mapsto \langle\alpha,v\rangle ... 0 First notice that on account of the antisymmetry of F_{ij} (notice that F^{ij} will be antisymmetric whenever F_{ij} is and viceversa) the object written$$ T_{ijk}= \partial_i F_{jk} + \partial_{j}F_{ki} +\partial_{k}F_{ij} $$is totally antisymmetric under any index pair exchange (one can check this easily, such object is called the differential ... 2 Let V be a real, finite-dimensional vector space V and$$T: V^* \otimes V \to \mathbb{R}$$a (1, 1)-tensor. Now, given any \lambda \in V^*, we denote by$$T(\lambda, \,\cdot\,)$$the map V \to \mathbb{R} defined by$$X \mapsto T(\lambda, X).$$Since T is multilinear, the map T(\lambda, \,\cdot\,) is linear, but linear maps V \to \mathbb{R} ... 0 It seems to me this question presents an interesting and significant challenge, insofar as it concerns it self, as I read it, with the borderline between the intuitive and the formal views of a mathematical concept (in this case curvature), and asks us to investigate how the intuitive gives rise to the formal. If we accept the standard definition that a ... 1 This is defined on page 13 of the same book. A tensor field Y \in T^k_l(M) is a C^{\infty}(M)-linear map Y : (\Omega^1(M))^l\times (\mathfrak{X}(M))^k \to C^{\infty}(M). The trace of Y is the tensor field \operatorname{tr}Y \in T^{k-1}_{l-1}(M) satisfying$$\operatorname{tr}Y(\omega_1, \dots, \omega_{l-1}, V_1, \dots, V_{k-1}) = ... 1 It's very common to write symmetric products using juxtaposition with no product symbol. But you should never omit the wedge symbol in a wedge product. Maybe you're getting confused with the notation for integrals, where, for example$\int f\,dx^1\wedge dx^2$is defined to mean$\int f(x^1,x^2)\, dx^1dx^2$. The$dx^1dx^2$in the latter expression is not a ... 0 It seems to me that the question is not simply technical, but is about the concept of ''curvature'' in a Lorentz manifold. This concept is not so simple to define and the only intuition that I have is from my physical formation. So I start form a physical point of view about this concept, using the intuition proposed in the classical book ''Gravitation'' of ... 1 That equation is incorrect; I'm not used to this notation but I think it's actually $$e_{ijk} e_{lmk} = \delta_{il} \delta_{jm} + \delta_{im} \delta_{jl}.$$ In fact a tensor with this property does not exist in any dimension at least$2$. The reason is that it would imply the existence of a pair of linear maps$f : V \otimes V \to V$and$g : V \to V ...

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I don't think there is any (semi)closed-form solution, but you can simplify the problem quite a bit (when the dimension of the vector space is small). Presumably all the vectors here are real. Let $A=\sum_{u\in S_1}uu^T$ and $B=\sum_{v\in S_2}vv^T$. Since $$\|u\|^2-\langle u, s \rangle^2 = u^Tu - (s^Tu)(u^Ts) = \operatorname{trace}(uu^T)-s^Tuu^Ts,$$ it ...

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In chapter 2, when talking about Riemann Normal Coordinates, Carroll shows that you can cunningly pick coordinates such that at a given point the metric has its standard form $$\begin{pmatrix} -1 & & &\\ & 1& & \\ & &1 & \\ & & & 1\\ \end{pmatrix}$$ and so that the first derivatives all vanish $$\partial_\rho ... 0 Use metric compatibility on the left hand side. First, note that the traditional expression for metric compatibility is$$\nabla \sqrt{g} = 0$$Note also that this implies, for a unit pseudoscalar \hat \epsilon,$$\nabla [\sqrt{g} \hat \epsilon] = \nabla \epsilon = 0$$where \epsilon is the coordinate pseudoscalar (volume form). This is, perhaps, ... 1 The first thing to notice is in your last equation, the symmetry of g^{\mu\alpha} means that$$ g^{\mu\alpha} \partial_{\mu} g_{\nu \alpha} = g^{\mu\alpha} \partial_{\alpha} g_{\mu\nu}, $$so those two terms cancel and we are left with proving that$$ (\partial_{\nu} \log{g} =) \, \frac{\partial_{\nu} g}{g} = g^{\mu\alpha}\partial_{\nu} g_{\mu\alpha} ...

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When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Yes. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is ...

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$c$ means "act the element $R$, then do $\tau$". So claiming $c(v_0 \otimes v_1) = q^{-\lambda (\lambda-2)/2} v_1\otimes v_0$ is equivalent to $R (v_0\otimes v_1) = q^{-\lambda(\lambda-2)/2} v_0 \otimes v_1$. This follows from your description of $R$: you're right that $E^n \otimes F^n$ acts as zero on $v_0 \otimes v_1$ for $n \geq 1$, so only the action ...

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I found the answer on page 178 of Kassel's Quantum Groups. Given a braided bialgebra $(H,\mu,\eta,\Delta,\epsilon,R)$, with $V,W$ left $H$-modules, and a universal R-matrix $R = \sum_i s_i \otimes t_i \in H \otimes H$, we have that $c_{V,W}^R(v \otimes w) = \tau_{V,W}(R(v \otimes w)) = \sum_i t_i w \otimes s_i v$ for all $v \in V$, $w \in W$. That gets the ...

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For a given basis $\{e_a\}$, $g_{ab}=g(e_a,e_b)$ denotes the components of the metric $g$ with respect to the given basis. So (at each point of the manifold) $g$ is a bilinear form, and in a given basis it is represented by a matrix. While matrices generally do not commute, their components are just numbers and do commute. In your example $R_{bacd}$ is just ...

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Metric tensors are defined as symmetric bilinear forms, so we can write them as symmetric matrices. As general tensors, metric tensors are not commutative in general (try in dimension $2$ for example to construct two symmetric matrices that do not commute). Now, if $g^{ab}$ is defined as the inverse matrix of $g_{ab}$ , then $g^{ab}g_{ab}$ is the ...

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Yes. If you want to do it explicitly, just multiply both sides with the inverse metric to raise both indices.

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Consider the following matrix $$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ Then the $\det(\mathbf{A})$ can be written as \begin{align} \det(\mathbf{A}) &= (-1)^0a_{11}(a_{22}a_{33}-a_{23}a_{32})+(-1)^1a_{12}(a_{21}a_{33}-a_{23}a_{31})\\ ...

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Use $A=(A_{ij})$, for definition $$\det(\mathbf{A}) = \varepsilon_{ijk} A_{1i}A_{2j}A_{3k}$$ Or $$3!\det(\mathbf{A}) = \varepsilon_{lmn}\varepsilon_{ijk} A_{il}A_{jm}A_{kn}$$

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