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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 $X, Y$.


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 ...


2

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.


1

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 ...


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 ...


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 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 ...


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 $m \inĀ \mathbb N$. 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} ...


1

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


1

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. ...


1

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 $A^{i,j}$.


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) ...



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