# Tag Info

### Interpretation of "inverse" curl

$\newcommand\R{\mathbb R}$Yes, the expression $v\times\nabla$ makes sense. It is vector-like, and you have to decide what it's going to act on and how. If $f : \R^n \to \R$ is a scalar function then ...
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### The necessity of the Lie bracket of vector fields in the definition of the Riemann curvature tensor

The two definitions of the coefficients ${R^t}_{kij}$ coincide when $[X,Y]=0$ and this is the case for $X= \frac{\partial}{\partial x^i}$ and $Y= \frac{\partial}{\partial x^j}$. However the definition ...
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### The necessity of the Lie bracket of vector fields in the definition of the Riemann curvature tensor

Without the $\nabla_{[X,Y]}$ the expression for $R(X,Y)Z$ would not be a tensor because it would depend on derivatives of the $X,Y$ fields. A tensor $T(X,Y)$ depends only on the components of $X$, $Y$ ...
• 775
1 vote

### Integration by parts and divergence of curl

If boundary terms are $0$ then: $$\int (\nabla \times A_2(x))\cdot\nabla f(x) \ dx^3 =- \int \big[\nabla\cdot(\nabla \times A_2(x))\big] f(x) \ dx^3$$ by integration by parts which is what you are ...
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### Multivariable calculus: Can two trajectories of a vector field ever cross each other?

are you also in UIUC's NetMath for MVC? It sure sounds like it. I'm working on L6 right now; great to meet a fellow student outside my school. Are you at UIUC or also a high schooler? How did the ...
Accepted

### Why do these results not contradict Green's Theorem?

By definition, a conservative vector field is one which is the gradient of some scalar function. In $\mathbb{R}^2$, vector fields which have zero curl are only conservative on domains without holes (...

### How to approximate a vector-valued function?

Have you tried solving the following least-squares problem? $$\underset{{\bf X} \in \Bbb R^{2 \times 2}}{\text{minimize}} \quad \| {\bf X} {\bf A} - {\bf B} \|_{\text{F}}^2$$ If $\bf A$ has full row ...
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### Questions about the definitions of r

It seems to me that here $i$ and $j$ represent respectively the following unit vectors : $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. And $x,y$ are simply real ...
• 125
Accepted

### Questions about the definitions of r

These represent different things. Written in more common and clear mathematical notation, you have $$r = \sqrt{x^2 + y^2} \qquad \vec r = x \hat\imath + y \hat\jmath$$ respectively. The former ...
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### Left Invariant Vector field of matrix Lie group

Okay here is an attempt at a more full answer: You aren't pushing forward a matrix, you are pushing forward the map $L_A : G \to G; B \mapsto AB$. This is clearly a diffeomorphism of $G$ (by ...
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### What is the Jacobian matrix?

The Jacobian matrix finds multiple applications in robotics: a detailed explanation can be found in the book "Modern Robotics: Mechanics, Planning, and Control," by Kevin M. Lynch and Frank ...
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1 vote

### $N$-dimensional Anti-Curl Operator

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### Vector fields on a sphere: equivalence of two definitions

From $\Phi(x)$ we can construct a vector field in $\mathbb R^{n+1}$ (where we indentify $T_x\mathbb R^{n+1}$ with $\mathbb R^{n+1}$ ): $$X(x)=|x|\Phi\left(\frac{x}{|x|}\right).$$ Now i use the ...
• 417
1 vote

• 29.8k
1 vote
Accepted

Yes, that is correct. Equivalently, you may check directly that \begin{align}\det\left(\overrightarrow{PQ},\overrightarrow{PR},\overrightarrow{PS}\right) &=\begin{vmatrix}2&0&4\\3&1&... • 8,721 1 vote Accepted ### Lie bracket of vector fields - what have I done wrong? Informally, we have \frac{\partial~~}{\partial y} = \frac{\partial ~~}{\partial (2t)} = \frac{1}{2}\frac{\partial~~}{\partial t} = \frac{1}{2}\frac{\partial~~}{\partial x},  and \begin{align} v &...
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