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By the chain rule $$\frac{\partial x^c}{\partial x'^b}\frac{\partial^2x^d}{\partial x^c\partial x'^a}=\frac{\partial^2x^d}{\partial x'^b\partial x'^a}$$ and similarly the other term becomes $$\frac{\partial^2x^d}{\partial x'^a\partial x'^b}.$$ Then these terms cancel because, as Joe Hannon says in the comments, of "equality of mixed partial derivatives" ...


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I am using prior knowledge that this is an approach to eigenvalue decomposition to formulate my answer. Let $$R = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix},$$ where $q_1$ is an eigenvector of the matrix $A$. Then $$RAR^T = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix}^T A \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix} = \begin{bmatrix} q_1 \ \ ...


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In addition to the previous answer, there are two kinds of indexes in tensor equations: Free and Dummy. Although you can change the identifier for a given index, free or dummy, you need to be careful in how you do this to maintain consistency. For example, consider the following well known contravariant vector transformation equation: $$ \bar{x}^i = ...


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It needs to be written with all the indices the same $$a_i+b_i=c_i$$ This is so that when you have more than one index you know how to match them up. For example $$A_{ij}+B_{ij}$$ is different from $$A_{ij}+B_{ji}$$ where I've swapped $i$ and $j$ on the $B$. To work out the first one you would just add the matrices, whereas to calculate the second one you ...


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This doesn't use the symmetry of $S$, but you can use the following identity: $$\epsilon_{ijk}\epsilon_{mnp}=\delta_{im}\delta_{jn}\delta_{kp}+\delta_{in}\delta_{jp}\delta_{km}+\delta_{ip}\delta_{jm}\delta_{kn}-\delta_{im}\delta_{jp}\delta_{kn}-\delta_{ip}\delta_{jn}\delta_{km}-\delta_{in}\delta_{jm}\delta_{kp}$$ to get $$(A \times B)\cdot S\cdot(C \times ...


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I considered making a comment first, but i think there is only one way to resolve the vague way this question is put forth. I think a good notation here would be-- $$A_{ij}=(A^T)_{ji}$$ I do not think that It doesn't make sense to use the same notation for first-order tensors like $T_i$ and $c_i$, because their transpose changes them from a column vector to ...


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Your first equation is wrong: You left out the Levi-Civitta symbol coming from the middle cross product (and thus end up with an expression with no free indices, which you know can't be right). The correct starting point is $$ \epsilon_{krp} (\epsilon_{ijk}a_ib_j)\epsilon_{mnr}a_mc_n) $$ which simplifies as follows: $$ \begin{array}{c} \epsilon_{mnr} ...


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To clarify my remark in comments: Using any summation index more than twice renders the summation convention is insensible. This becomes obvious if you write in terms of $\Sigma's$: $$(\vec{a}\times \vec{b})^2=\left(\sum_{j,k=1}^3\epsilon_{ijk}a_j b_k\right)^2\neq \sum_{j,k=1}^3(\epsilon_{ijk}a_j b_k)^2.$$ What is correct is $$(\vec{a}\times ...


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Per my comment (which had a typo and should have said $b_jb_m$) \begin{align} \varepsilon_{ijk}a_ib_j\hat{e}_k\cdot\varepsilon_{lmn}a_lb_m\hat{e}_n &= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m(\hat{e}_k\cdot\hat{e}_n)\\ &= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m\delta_{kn}\\ &= \varepsilon_{ijk}\varepsilon_{lmk}a_ib_ja_lb_m\\ &= ...


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The main trick here is $$(x+y) \otimes (x+y) - x \otimes x - y \otimes y = x \otimes y + y \otimes x $$ I bet you're making things much harder by trying to do things in terms of a basis, rather than using tensor algebra. Also, you're probably contributing to your confusion by identifying elements of $V \otimes V$ with as bilinear forms on $V$: it would be ...


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This is a consequence of the symmetry of the metric tensor (which is what I'm assuming $g_{pq}$ is). Clearly, $$ \frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{\partial g_{pj}}{ \partial g_{kn}} g_{ql} + g_{pj} \frac{\partial g_{ql}}{\partial g_{kn}} $$ The partial derivative in this expression must give something symmetric. For a non-symmetric ...


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The answer for your first question is yes. The answer for your second is no. The next proposition is well known. It was proved here. Proposition: Let $V,W$ be vector spaces over $k$. Let $w=\sum_{i=1}^ra_i\otimes b_i=\sum_{i=1}^sv_i\otimes w_i\in V\otimes W$. If $\{a_1,\ldots, a_r\}$ is linear independent then $\text{span ...


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No, the expansion of last expression on the right does not have terms with $d$ in the third slot, whereas $T_{a[bcd]}$ does. One doesn't see nested brackets in practice (exactly because of the identity you're proving), but I'd interpret them to mean that one skews first over $bc$, and then skews over $bcd$. In other words, $$T_{a[[bc]d]} = ...


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(This will not be a complete result so much as a strategy, since I don't recall all the ins-and-outs of this calculation.) While one can integrate directly, the more 'physical' strategy is to take advantage of the symmetry and tensorial form of the integral. First, we note that the integral is invariant under rotations about the origin; for convenience, ...


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Up to a constant, there is only one such tensor, i.e. you know $$ \mu=C f^1\wedge f^2...\wedge f^n $$ and just need to determine that $C = 1$; evaluating on the basis does precisely this.


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HINT: Suppose not. Reduce to the case where the $f_i$ are linearly independent. Now choose $v_j$ so that $f_i(v_j)=\delta_{i,j}$. P.S. The standard notation would be $\text{Sym}(f_1\otimes \dots \otimes f_k)(v_1,\dots,v_k)$.


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Let $\omega$ be a 2 form. You need to write is as a linear combination of $e_i \wedge e_j$. What is your guess for, say, the coefficient of $e_1 \wedge e_2$?


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$\psi$ is antisymmetric iff $\psi(v_1,\ \cdots,\ v_k)=0$ where $v_i=v_j$ and $i<j$. Let $\{ e_i\}_{i=1}^n$ be a basis for $V$. Since $\psi$ is $k$-linear, $\psi( e_{i_1},\cdots ,e_{i_k})$ determines $\psi$. Note that $k> n$ so that there exists $j<l$ s.t. $e_{i_j}=e_{i_l}$ So $ \psi( e_{i_1},\cdots ,e_{i_k})=0$.


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Fix a basis of $V$. By multilinearity it is sufficient to prove every input consisting of basis vectors gets sent to zero. But inputting $k>n$ vectors from a basis of size $n$ results in a redundant entry, which by antisymmetry must map to zero.


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An answer to this question has been given as a comment: Because your operator $A$ is defined on all of $M$, so the bump function extends the use of the neighboorhood $\mathcal{U}$ to all of $M$ to allow us to use the operator $A$. – Chris Gerig Jan 24 at 8:13


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This question has been answered in comments: The implication $2\implies 1$ does not hold. Take for instance $\mathbb{Z}$-modules $A=\mathbb{Q}$, $B=\mathbb{Z}/2\mathbb{Z}$, $C=\mathbb{Z}/3\mathbb{Z}$. Then $A\otimes B=A\otimes C=0$, but $B$ and $C$ are not isomorphic $\mathbb{Z}$-modules. – adrido Jan 25 at 8:04 and Even easier: Take $A=0$ to see ...


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Comment turned into answer per request. $$\begin{align}u_{j,ij} &= \sum_{j=1}^3 \frac{\partial}{\partial x_j}u_{j,i} = \sum_{j=1}^3 \frac{\partial^2}{\partial x_j\partial x_i} u_j = \sum_{j=1}^3 \frac{\partial^2}{\partial x_i\partial x_j} u_j = \sum_{j=1}^3 \frac{\partial}{\partial x_i} u_{j,j}\\ &= \frac{\partial}{\partial x_i} \nabla\cdot u = ...


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I guess the issue in your example is that the index notation shows a more fine-grained picture of what's going on and thus is inevitably more complex and harder to parse. There are a couple of in-between notations you could use - abstract tensor expressions or multi-indices: $$ \langle F,G \rangle = (\text{tr}_g)^l (\text{tr}_{g^{-1}})^k (F \otimes G) = ...


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You probably know that $\partial_\mu v_\nu$ is not a tensor. The curl $\partial_\mu v_\nu-\partial_\nu v_\mu$ is however. The transformation law for the derivative of a covector is $$(\partial_\mu v_\nu)'=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial}{\partial x^\rho}\left(\frac{\partial x^\sigma}{\partial x'^\nu}v_\sigma\right)=\frac{\partial ...


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You can define a tensor agrees with the Levi-Civita symbol for orthogonal coordinate systems but that has the correct components for non-orthonormal systems. This, and other results, can be derived in the setting of clifford algebra. Clifford algebra deals with a "quotient" of the tensor algebra--an interesting subset, if you will, of tensors that ...


1

To answer the first part of your question: The first equation you have is incorrect as written for the following reason: the cross product $A \times B$ is a vector independent of any basis. On the right hand side, you have (in einstein summation convention) the components of this cross product in a cartesian basis. To set the equation right, you'll have to ...


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The (simple) bivector describing a plane in $\mathbb{R}^n$ belongs to a $n(n-1)/2$ dimensional space, which number equals 3 for n=3. In higher dimensions we cannot associate it with a unique n-dimensional normal vector.


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Here is a writeup by Lek-Heng Lim about generalising various properties of matrices to higher tensors.


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A metric tensor takes two tangent vectors and returns a number, their inner product. Under a coordinate transformation or a map between manifolds, tangent vectors $u$ are transformed (pushed-forward) by the differential of the map represented by the Jacobian matrix: $u\mapsto Ju$, and the Euclidean inner product $u^Tu\mapsto(Ju)^T(Ju)=u^TGu$, where $G=J^TJ$ ...


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A linear transformation is one that obeys the rule $f(\alpha x+\beta y)=\alpha f(x)+ \beta f(y)$. And we have clearly not $\sin(\alpha x+\beta y)=\alpha \sin(x)+ \beta \sin(y)$. But differential operators (like partial derivatives or differentials) are. The difference is that here the variables don't live on the reals or in a vector space, but that they are ...



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