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When it comes to nonsymmetric tensors, the order of indices matter, even between covariant and contravariant indices. Let us take the difference between $T^a{}_b$ and $T_b{}^a$, multiplied by the metric $g_{ab}$ to raise and lower indices: $$g_{ac}(T^a{}_b-T_b{}^a)=g_{ac}T^a{}_b-g_{ac}T_b{}^a=T_{cb}-T_{bc},$$ which is zero only if $T_{bc}=T_{cb}$, i.e., if ...

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Gradient is covariant. Let's consider gradient of a scalar function. The reason is that such a gradient is the difference of the function per unit distance in the direction of the basis vector. We often treat gradient as usual vector because we often transform from one orthonormal basis into another orthonormal basis. And in this case matrix transpose and ...

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The components of $\text{Ric}$ transform during coordinate change $x^\mu\mapsto \tilde{x}^\mu$ as $\tilde{R}_{\mu\nu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\nu}R_{\sigma\rho}$. This is just the usual transformation rule for coordinate-components of tensors. Contracting over the two indices gives $$\tilde{... 2 Note that you can use the entanglement entropy to calculate the amount of entanglement in a bipartite pure state, but this is not a good measure for a general bipartite (mixed) state. In the general case there are several different entanglement measures currently used, which have certain desiderata: https://quantiki.org/wiki/axiomatic-approach. Invariance ... 12 The answer is no: whether or not the state can be written as a product state does not depend on the basis. And you are precisely correct: there is indeed a basis-independent invariant that characterizes the entanglement. It is called the "entanglement spectrum": the eigenvalue spectrum of the reduced density matrix produced by taking the partial trace over ... 4 No, the entanglement (yes/no) doesn't depend on the basis of the two subsystems, only on the way how the two subsystems are separated from one another. A non-entangled state is a state of the form |j\rangle \otimes |\alpha\rangle for some states |j\rangle,|\alpha\rangle of the two subsystems; all other states in the composite Hilbert space are entangled.... 0 We can do \small0=i in the single form equation$$\epsilon^{ijk}\partial_j F_{0k} + \epsilon^{ijk}\partial_0 F_{jk} + \epsilon^{ijk}\partial_i F_{jk} =0$$and because \epsilon^{ijk} permutes we can write$$\epsilon^{jik}\partial_j F_{ik} + \epsilon^{ijk}\partial_i F_{jk} + \epsilon^{kij}\partial_k F_{ij} =0 From this we can divide out $\epsilon^{ijk}$...

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$\Lambda_{\mu\nu} = {\Lambda_\mu}^\sigma\eta_{\sigma\nu}$. It doesn't "do" anything. $\delta_{\mu\nu}$ and $\delta^{\mu\nu}$ are not tensors, as I explain at length in this answer of mine. The matrix elements of the identity are $\delta_\mu^\nu$, which you could have determined by thinking about the fact that the identity must send vectors $v^\mu$ to other ...

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