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1

Substituting line two into the first line, we have $$a_{rm}\frac{dp^{m}}{dt}+\frac{1}{2}\left(\frac{ \partial a_{rm}}{\partial x^{n}}+\frac{ \partial a_{rn}}{\partial x^{m}}\right)p^{m}p^{n}-\frac{1}{2}\frac{\partial a_{mn}}{\partial x^{r}}p^{m}p^{n}=0$$ After a little bit of factoring and rewriting the first term we get: $$... 1 The way I know is the following: Let f(x_1,\ldots,x_n) be any homogeneous polynomial of the variables x_1,\ldots,x_n and of degree m \in \mathbb N. For simplicity I will assume that the coefficients of f are real. Each monomial in f is of the form$$c_{i_1,\ldots,i_m}x_{i_1}\cdot \ldots \cdot x_{i_m} \quad \text{ with } \quad c_{i_1,\ldots,i_m} ...

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This equality can be written in terms of matrices as $$x_i a_{ij} y_j=x^T A y=y_i a_{ij} x_j = y^T A x$$ where $x=(x_i),$ $y=(y_i)$, and $A=(a_{ij})$. If the vectors $x$ and $y$ are the same, then the above relation is identically true. As a side note, observe that $y^T A x= x^T A^T y$ and so the identity will hold for all $x,y$ if $A=A^T$ i.e. $A$ is ...

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If $x_i = y_i$ for all $i = 1, \dots, n$, then $a_{ij} x^i y^j = a_{ij} y^i y^j = a_{ij}y^i x^j = a_{ij} x^j y^i.$

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I will assume $x=(x_i)$ to be a $1\times N$ (row) vector, $y=(y_i)$ an $N \times 1$ (column) vector, and $A=(a_{ij})$ an $N\times N$ matrix. Then the terms on the RHS correspond to the matrix multiplications $x_i a_{ij} =(x A)_j$ and $a_{ij}y_j = (A y)_j$. Note that these are the components of a row vector and a column vector respectively. What about the ...

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Parsing the much-maligned notation yields $$\sum_{i,j} a_{ij} (x_i + y_j) \ne \sum_i a_{ij}x_i + \sum_j a_{ij}y_j$$ which is an evident inequality.

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No, actually $T^{i'}$ is a function in not primed coordinates. It's components show which value the primed coordinate will have depending of not primed one. Example: $$x=r \cos(\phi), y = r \sin(\phi)$$ Here primed coordinates are x and y. If you will take contravariant tensor of form $T=(f^1(\phi, r), f^2(\phi, r))^T$, you will get $T'=(\cos(\phi)f^1(\phi, ... 2 The gradient is not the real equivalent, even though many texts treat it as if it is. The derivative of a function$f:\mathbb{R}^n\to\mathbb{R}$at a point$p\in\mathbb{R}^n$is a$1$-tensor, which means that it eats a vector and spits out a number. More specifically, the thing is obtaining a function$I\to\mathbb{R}$, where$I$is an open segment in ... 1 I have written a somewhat eccentric course on this, which can be found here: http://ximera.osu.edu/course/kisonecat/m2o2c2/course/activity/week1/ A short answer to your question, though, is that if$f:\mathbb{R}^n \to \mathbb{R}$, then$D^{k+1} f$is a locally a$(k+1)\$-linear function with the property that $$D^{k}f\big|_{p+v_{k+1}}(v_1,v_2,...,v_k) ... 1 First, let's compute the mentioned example in detail: Example Find the Killing fields Z of the standard metric \bar{g} := dx^2 + dy^2 on the plane \mathbb{R}^2. If we write$$Z = f \partial_x + g \partial_y,$$for some functions f,g of (x, y), then$$g(Z, \cdot) = f \,dx + g \,dy$$and so the bilinear form (X, Y) \mapsto g(\nabla_X Z, Y) is ... 0 Your strategy is correct: you probably need to change notation a bit. I suppose you want to prove$$I:=\sum_{k,l=1}^n u_k A_{kl} u_l = 0~~\forall u\in\mathbb R^n \Rightarrow A_{kk}=0,~A_{kl}=-A_{lk},~~ k\neq l. $$To prove the first skewness relation for A we choose u=(0,\cdots,1,\cdots,0), where the 1 corresponds to the i-th coordinate, i.e. u_k ... 3 This is the polarization identity. Given that \omega(X,X)=0, we get that \omega(X+Y,X+Y)=0=\omega(X,X)+\omega(X,Y)+\omega(Y,X)+\omega(Y,Y)=2\omega(X,Y). Then assuming the characteristic of the ground field is not 2, we get that \omega(X,Y)=0, for all X, Y. 1 The determinante of \left(A_{ij}\right) is defined by Leibniz as$$\det \left(A_{ij}\right) := \sum_{\sigma}\text{sign}(\sigma)\Pi_i A_{i,\sigma(i)} $$With permutations \sigma and their sign \text{sign}(\sigma) (=+1 if Permutation is an even of 1,2,3,4,..., -1 if pertutation is odd, 0 else) This can be simplified with$$\det ...

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