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This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and $\... 34 A second-order tensor can be represented by a matrix, just as a first-order tensor can be represented by an array. But there is more to the tensor than just its arrangement of components; we also need to include how the array transforms upon a change of basis. So tensor is an n-dimensional array satisfying a particular transformation law. So, yes, a third-... 33 The simplest way to explain the Christoffel symbol is to look at them in flat space. Normally, the laplacian of a scalar in three flat dimensions is: $$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$ But, that isn't the case if I switch from the$(x,y,z)$... 29 The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection$\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which$\nabla_{\mu}g_{\alpha \beta}$did not ... 27 You can decompose a rank two tensor$X_{ab}$into three parts: $$X_{ab} = X_{[ab]} + (1/n)\delta_{ab}\delta^{cd}X_{cd} + (X_{(ab)}-1/n \delta_{ab}\delta^{cd}X_{cd})$$ The first term is the antisymmetric part (the square brackets denote antisymmetrization). The second term is the trace, and the last term is the trace free symmetric part (the round brackets ... 27 Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an$x$-directed force pushing along an interface of constant$y$; this would be$\sigma_{xy}$. If we assemble all such combinations$\sigma_{ij}$, the collection of them is the stress tensor. Pressure is part of the stress tensor. The diagonal elements ... 21 Matrices are often first introduced to students to represent linear transformations taking vectors from$\mathbb{R}^n$and mapping them to vectors in$\mathbb{R}^m$. A given linear transformation may be represented by infinitely many different matrices depending on the basis vectors chosen for$\mathbb{R}^n$and$\mathbb{R}^m$, and a well-defined ... 18 For each surface on a unit cube (see below), the stress on that surface can point in each of the three directions. (source) Since it is not necessarily the case that$\sigma_{11}=\sigma_{31}=\sigma_{21}$(all pointing the in the same$\mathbf{e}_1$direction)--or any of the other$\sigma_{ij}$combinations, we need to have 9 components describing it, ... 15 The dual of a tensor you refer to is the Hodge dual, and has nothing to do with the dual of a vector. The word "dual" is used in too many different contexts, and in this case it is even used the same$*$symbol. One usually specifies "Hodge dual", or "Hodge star operator", to avoid confusion. Both these "duals" are isomorphisms between vector spaces endowed ... 15$\lvert A\rangle \langle B \rvert$is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space$H_1$(that's where the kets live) and of a dual of a Hilbert space$H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction ... 12 I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.] II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing $$\tag{1} V \otimes V^{*}~\stackrel{... 12 The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field \mathbb K (usually \mathbb R or \mathbb C). A formal definition is given below (there are many equivalent approaches). First, if V is a vector space, V^* denotes its algebraic dual space, namely the vector space of the linear ... 12 The answer is no: whether or not the state can be written as a product state does not depend on the basis. And you are precisely correct: there is indeed a basis-independent invariant that characterizes the entanglement. It is called the "entanglement spectrum": the eigenvalue spectrum of the reduced density matrix produced by taking the partial trace over ... 11 A (higher) n-rank tensor T^{\mu_1\ldots \mu_n} with n\geq 3 cannot always be decomposed into just a totally symmetric and a totally antisymmetric piece. In general, there will also be components of mixed symmetry. The symmetric group S_n acts on the indices$$(\mu_1,\ldots ,\mu_n)\quad \longrightarrow\quad (\mu_{\pi(1)},\ldots ,\mu_{\pi(n)})$$via ... 10 Physicists are always interested in what properties of a physical system are invariant under symmetries. If it's tricky to see the symmetry then they'll rearrange the system to make the symmetry more obvious. For example, consider a covariant rank two tensor like T^{ab}. In general the components of this tensor will change if the tensor is rotated in 3D. ... 9 This is true - in fact you could define \nabla^\sigma = g^{\sigma\rho} \nabla_\rho. I assume this meant to say$$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch g^{\sigma\rho} with \nabla_\nu, which you ... 9 We expect a vector to change in a certain way when we change the scale we use to measure distance. Consider the vector$$\vec{x}=(1, 0, 0)\,\mathrm{m}$$If we change scale and now measure in centimeters this vector becomes$$\vec{x}=(100, 0 ,0)\,\mathrm{cm}$$Now consider a vector representing a force:$$ \vec{F}=(1,0,0)\,\mathrm{J/m}$$where I've chosen ... 9 At the most basic level, you can just use the definition of the Christoffel symbols in terms of the metric: \Gamma^i_{jk} = g^{is} (\partial_j g_{sk} + \partial_k g_{sj} - \partial_s g_{jk}). Plugging this into the right-hand side of your expression will yield the left-hand side. However, one can obtain your expression directly from one of the ... 9 Slogan: Matrices are a tool to compute sums; tensors tell you which sums make sense. When you convert between rank-2 tensors and matrices, the decision as to which index of the tensor labels the rows and which one labels the columns is purely conventional. Matrix multiplication is no more than a convenient way to write products of the form$$K(i,k) = \sum\... 8 The covariant derivative is metric compatible, so$\nabla_{\alpha} g_{\beta \gamma} = 0$. This is the condition that the inner product is preserved under parallel transport. 8 Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange). The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis ... 8 A physicist would write your first equation$x^a = x^\mu e_\mu^a$. The notation$x^a$is invariant in your terminology. The$a$is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that$x$is a vector (i.e., rank 1,0 tensor.) Similarly for each$\mu$,$e^a_\mu$is ... 8 In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ... 8 I) The main point is that we usually only consider tensor products$V \otimes W$of vector spaces$V$,$W$(as opposed to general sets$V$,$W$). But groups (say$G$,$H$) are often not vector spaces. If we only consider tensor products of vector spaces, then the object$G \otimes H$is nonsense, mathematically speaking. With further assumptions on the ... 7 The notion of covariant derivative is equivalent to the notion of connection. More precisely, for every connection$\nabla$and vector field$X$, the operation$\nabla_X$is a covariant derivative. Connections$\nabla$on the tangent bundle$TM$of a manifold are usually induced by a metric, this is the so called Levi-Citiva connection. It is essentially ... 7 The (anti)symmetrization simply acts on all the enclosed indices (at the same "height" which are really enclosed between the brackets), regardless of their belonging to the same tensor or different tensors. For example, $$\delta^{[\alpha}{}_{[\gamma} R^{\beta]}{}_{\delta]} = \frac 12 \left(\delta^{[\alpha}{}_{\gamma} R^{\beta]}{}_{\delta} - \delta^{[\alpha}{... 7 I'll write this as an answer so that the math is more clear. So given an (p,q)-tensor T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}, this one transforms as:$$T'^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\frac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots \frac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\frac{\partial x^{\nu_1}}{\partial x^{\nu'_1}}\cdots\... 7 Let there be given a manifold$(M,\nabla)$equipped with a (not necessarily torsionfree) tangent bundle connection$\nabla$. I got the (possibly faulty) impression from reading the first lines in OP's question formulation (v18) that OP is asking: Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form$\...