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Instead of specifying $A$ and then working with its inverse $A^{-1}$, do the opposite. Say calculate $B^{-1} = ..$ and the index with $B_{ij}$. I have done this often when I include the $\mbox{ }^{-1}$ in my matrix definition/calculation. Example: $$\Lambda^{-1} = s^\top T I^{-1}$$ $$\Lambda_{11} = m_1+m_2$$


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I have seen the following notation, for the $i,j$th entry of $A$ use $A_{i,j}$, then for the $i,j$th entry of $A^{-1}$ use $A^{i,j}$.


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There are two standard ways to index an element in a matrix. First. If you note the matrix with uppercase letters from the beginning of the english alphabet, then you can use the lowercase version of the letter while indexing. For example the matrix is $A$ and the element in the $i$-th row and $j$-th column is $a_{ij}$. Or for $B$ you use $b_{ij}$. Etc. ...


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Solution using box or scalar product.


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Using abstract index notation: There are two types of indices here: I will use $i,j,\ldots$ for indices relative to the domain (which appears to be $\mathbb{R}^2$) and $A,B,\ldots$ for indices relative to the co-domain (basically from the number of equations) which is $\mathbb{R}^N$. We can write $\mathbf{c}$ as the rank 4 tensor in index notation $$ ...


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$x^\prime$ is simply another coordinate on whatever manifold you're working on. I think that it's a typo, and should actually be $\partial_{\mu}^\prime=\dfrac{\partial}{\partial x^{\prime\mu}}$. If it is not a typo, then you can write $\partial^{\prime\mu}=g^{\mu\nu}\partial_{\nu}^\prime$.


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One thing that makes this problem feel unfamiliar is the presence of the metric tensor and the resulting need to distinguish upper/lower indices. We can make it look a lot more familiar if we swap the positions of the $\nu$ index in the first two terms: $$[-(k^2-m^2)g^{\mu\nu} + k^\mu k^\nu]D_{\nu \lambda}=[-(k^2-m^2)\delta^{\mu}_{\;\nu} + k^\mu ...


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You're absolutely right. You can only identify the dual space with the original Hilbert space when you can. (Tautology!) Which is almost never. Don't do that for the general proof. You need to think of the dual vectors as living somewhere else entirely and not at all related to the regular vectors except as you've said in your definition. Even in ...


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Your matrix can be written as $A(q) = S\cdot q + B$ where $S$ is a very sparse third order tensor. $B$ is also sparse, its only non-zero component is $B_{33} = 10$. The only non-zero components of $S$ are seen by inspection to be $$ \eqalign { S_{111} &= 1,\,\,\,S_{122} &= 1,\,\,\,S_{133} &= 1 \cr S_{211} &= 2,\,\,\,S_{222} &= ...


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The general definition is Frechet derivatie, or Gateaux derivative if a specific vector unit of differentiation is specified. If you applied the definition to the present question, the derivative is a tensor with elements $\Big[\frac{dA(q)}{dq}\Big]_{i,j,k}=\frac{[A(q)]_{i,j}}{dq_k}$.


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The answer to both of your questions is yes, and I think that it works completely analogously to what describe in the first part of your question. More specifically, The idea of the trace is the same one that you discuss in the first part of your question. The "$E$-Hessian" $\nabla^2 \varphi$ is a section of $E \otimes T^\ast M \otimes T^\ast M$. Use ...


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The simple answer is that the naive differentiation operator $\frac{\partial}{\partial x^\alpha}$ when acting on a tensor with $n$ covariant indices results, if the metric is that of flat space, in a tensor with $n+1$ covariant indices. Thus that differential operator is often written as $\partial_\alpha$. For example, if $\phi$ is a scalar-valued field, ...


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$\partial^\mu$ is often a shortcut for $\partial/\partial x_\mu$ and $\partial_\mu$ for $\partial/\partial x^\mu$ because of this. See here for example. $$ g^{\rho\mu} \frac{\partial}{\partial x^\rho} = g^{\rho\mu} \partial_\rho = \partial^\mu $$


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I realized how simple it was after a while. We have that: $$ \partial_i{\bf \bar \Theta}(\bar x(x))=\partial_i{\bf \Theta}(x) $$ And, knowing that: $$ \partial _i {\bf \Theta}(x)={\bf g}_i $$ Then, using the chain rule and the fact that $\chi= {\bf \bar x} = \mathbf{\bar \Theta}^{-1}\circ \mathbf{\Theta}$ $$ {\bf g}_i=\frac{\partial {\bf \Theta}}{\partial ...


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Note that $$\nabla \cdot [\mu (\nabla \mathbf{u}+\nabla \mathbf{u}^T)]=\nabla \mu \cdot(\nabla \mathbf{u}+\nabla \mathbf{u}^T)+ \mu \nabla^2\mathbf{u} + \mu \nabla(\nabla \cdot \mathbf{u}).$$ In terms of Cartesian components, the $\mathbf{e}_i$ component is (summing over $j =1,2$ in $2$D or $j =1,2,3$ in $3$D): $$\frac{\partial}{\partial x_j}\left[\mu ...


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No, you need the tensor product $\otimes$ to construct the most general tensors. The wedge product $\wedge$ produces skew-symmetric objects. Well, the space of tensors on a vector space V is constructed by taking tensor products of the form $V \otimes V \otimes \ldots V \otimes V^* \otimes V^* \ldots V^*$ where $V$ is the vector space and $V^*$ is its ...


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Any alternating $(r,s)$ tensor has a corresponding map that goes $\Lambda^r V \to \Lambda^s V$. Suppose $R \in \Lambda^r V$ and $\Sigma \in \Lambda^s V^*$. Then define $\underline T:\Lambda^r V \to \Lambda^s V$ such that $$T(R, \Sigma) = \Sigma[ \underline T(R)]$$ The uniqueness of $\underline T$ can be proved by taking a "gradient" with respect to the ...



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