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Linear Independent Case A nice way of proving this is to use non-orthogonal basis for $\mathbb{R}^3$. Hence, consider the following definitions for the non-orthogonal basis $$\matrix{ {{{\bf{g}}_1} = A} & {{{\bf{g}}_2} = B} & {{{\bf{g}}_3} = C} \cr } \tag{1}$$ and then the dual basis will be $$\matrix{ {V = \left( {{{\bf{g}}_1} \times ... 2 My old professor has a book that starts out going through this in detail. Look at pages 20 - 22 or so to start. Here's the draft of his book My Full Answer: If you ask "My question is: how to translate these rules into matrix equations?" I have to say that they already are matrix equations. Let me explain a bit. In an early course on Special relativity you ... 2 The metric specifies a canonical isomorphism between V and V^*--and therefore an invertible map g' : V \to V^*. Consider some T: V \times V^* \to \mathbb R. Now consider T': V \times V \to \mathbb R such that T'(A, B) = T(A, g'(B)). That is what we're doing when we raise or lower indices. We might have some multilinear function that takes ... 2 With tensor notation I assume you just mean Einstein's summations convention, i.e. the convention where repeated indices are summed over all the coordinates instead of having an explicity sum: a_ib_i \equiv \sum_{i=1}^n a_i b_i. The notation f({\bf x}) is just a shorthand for f(x_1,x_2,\ldots,x_n), i.e. to tell the reader that f takes points in ... 2 There are two relevant operations you want to be familiar with on tensors. First, given a (m,n) tensor T and an (k,l) tensor S, you can construct their tensor product T \otimes S which will be a (m + k, n + l) tensor. Invariantly,$$ (T \otimes S)(\varphi^1, \ldots, \varphi^m, \varphi^{m+1}, \ldots, \varphi^{m + k}, v_1, \ldots, v_n, v_{n+1}, ...

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Consider the properties of the $dx^i$ compared to the $x^i$. As linear functionals, the basis one-forms $dx^i$ are...well, linear on their arguments. Inner products are bilinear, but we are using two one-forms in each linearly independent term. Notational convention. My opinion? It's very misleading. I would almost always keep the tensor product explicit ...

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Given a vector valued function $u:\>(\Omega\subset {\mathbb R}^n)\to{\mathbb R}^m$ and a point $p\in\Omega$ the differential of $u$ at $p$ is a linear map $$du(p) :\quad T_p\to T_{u(p)}, \qquad X\mapsto du(p).X\ ,$$ which is defined by $$u(p+X)-u(p)=du(p).X+o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{1}$$ Since $p$ will be fixed in the sequel I shall just ...

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Defining geometric objects by their coordinate transformation laws is common in physics, particularly older physics texts, but it's not so common in math these days. To most modern geometers, the difference between positions and displacements is a difference in algebraic structure: the operations you can do with positions are different from the ones you can ...

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Let us rearrange your $$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma \Lambda^\beta{}_\delta T^{\gamma \delta},$$ into $$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma T^{\gamma \delta}\Lambda^\beta{}_\delta,$$ and one step more $$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma T^{\gamma \delta}(\Lambda^{\top})_\delta{}^\beta{}.$$ In this last equation anyone could ...

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It depends on how you define $S^2$ and $\Lambda^2$: the book may define $S^2(V) = \langle v\otimes w+w\otimes v:v,w\in V\rangle$ and $\Lambda^2= \langle v\otimes w-w\otimes v: v,w \in V\rangle$ in which case = is true. If you define them differently then perhaps = is not literally true, but it's very common to mildly abuse notation like this. Indeed it's ...

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The summation convention says that you sum any letter that appears twice. So $\epsilon_{ijk}\epsilon_{ljk}$, it really means $\sum_j\sum_k\epsilon_{ijk}\epsilon_{ljk}$ You are right that $\epsilon_{ijk}\epsilon_{ijk}=6$, but then $\epsilon_{ijk}\epsilon_{ijk}=\sum_i\sum_j\sum_k\epsilon_{ijk}\epsilon_{ijk}$ For the same reason, $\delta_{ii}=3,\delta_{11}=1$

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Rewriting: $${\delta _{nm}}{\varepsilon _{ijk}} - {\delta _{im}}{\varepsilon _{njk}} - {\delta _{jm}}{\varepsilon _{ink}} = {\varepsilon _{ijn}}{\delta _{km}}$$ as: $${\delta _{nm}}{\varepsilon _{ijk}} = {\delta _{im}}{\varepsilon _{njk}} + {\delta _{jm}}{\varepsilon _{ink}} + {\delta _{km}}{\varepsilon _{ijn}}$$ makes it easier to see the symmetry. ...

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The following argument just occured to me. (I probably half-remembered it from Cvitanovic's Birdtrack book.) We use the relation $$\varepsilon_{iab}\varepsilon_{ipq}=(\delta_{ap}\delta_{bq}-\delta_{aq}\delta_{bp})$$ to expand $$\varepsilon_{ani}\varepsilon_{abm}\varepsilon_{bjk}$$ in two different ways. First \begin{align*} ...

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This answer uses representation theory. Let $V=\mathbb{R}^3$ be the standard representation of $\mathrm{SO}(3)$, and suppose that $T:V\to V$ is a $\mathrm{SO}(3)$-equivariant map. Schur's lemma would say that $T$ must be a scalar multiple of the identity because $V$ is irreducible, but this is a real representation rather than a complex one so we can't say ...

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Let me answer first your second question (which is pure linear algebra) as I think it will clarify what goes on when you consider the second question (which is linear algebra applied at each point to the tangent space of a manifold). Given a real finite dimensional inner-product space $(V, g_{V})$, every vector $v \in V$ defines a linear functional ...

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If you feel uncomfortable with tensors, you can do everything in the setting of linear maps. Starting from a linear map $A:V\to L(V,V)$, you obtain a skew symmetric, bilinear map $\delta A:V\times V\to V$ via $\delta A(v,w)=A(v)(w)-A(w)(v)$. Now if you have a linear subspace $\mathfrak g\subset L(V,V)$, you can look at the restriction of $\delta$ to ...

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