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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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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*} ... 1 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 1 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}}$$makes it easier to see the symmetry. ... 3 Linear Independent Case A nice way of proving this is to use non-orthogonal basis for \mathbb{R}^3. Hence, consider the following definitions for the non-orthogonal basis$$\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 ...

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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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See my explanation of the difference between translations (functions) and pictures (arrows) of their effects

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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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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 ...

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I think I found the problem - while the article uses the notation $\textbf{v} = v^j \textbf{g}_j$ above, in the expression for the gradient it is implied that $\textbf{v} = v_j \hat{\textbf{e}}^j$ i.e. the covariant componenets are being taken with respect to the unit basis vectors. Using the Christoffel symbols from the article, I get ...

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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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"Geometric object" here is meant to convey the notion of basis independence. We identify the multilinear map $T$ with arguments in one basis with the map $T'$ with arguments in another basis if the give corresponding results with respect to the transformation between bases. This gives rise to the notion that the tensor is not tied to any particular basis at ...

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