# Tag Info

## New answers tagged differential-forms

0

In my opinion, a lot of these relationships are suggested by abusive notation, abuses that hide what's really going on. Don't get me wrong: some abuses of notation are harmless, or at the least, they help people get going on doing calculations. But they should still be understood to the fullest degree for those who wish to go beyond merely doing ...

3

Let me write $\alpha = (\alpha_1, \cdots, \alpha_k)$ and $I =(i_1, \cdots, i_k)$. Then using $$w = \sum_I w_I dx^{i_1}\wedge \cdots \wedge dx^{i_k} = \sum_\alpha w_\alpha dy^{\alpha_1}\wedge \cdots \wedge dy^{\alpha_k}$$ and $dy^{\alpha_j} = \frac{\partial y^{\alpha_j}}{\partial x^i} dx^i$, we have $$w_I = \sum_{\alpha} w_\alpha \frac{\partial ... 0 Daniel Fischer is right. To perform this calculation with differential forms, you need to use the "Hodge dual", or "star operator". In three dimensions, it transforms a 2-form \omega = X\,dy\wedge dz + Y\,dz\wedge dx + Z\,dx\wedge dy into the 1-form *\omega = X\,dx + Y\,dy + Z\,dz (and viceversa). In general: *(dx\wedge dy)= dz, and so on, ... 1 This can be answered easily with geometric calculus. The cross product is a duality operation:$$A \times B = -\epsilon (A \wedge B)$$The same goes for the curl:$$\nabla \times (A \times B) = -\epsilon \nabla \wedge [-\epsilon(A \wedge B)]$$The 3-vector \epsilon can be pulled out of the wedge product, at the cost of turning it into a contraction ... 1 Because people want to be able to do calculus without the presence of a metric. You know about \nabla from vector calculus. It's naturally written in terms of basis 1-forms. If the basis one-forms are e^1, e^2, \ldots, associated with the coordinates x^1, x^2, \ldots, then \nabla takes the form$$\nabla = e^1 \frac{\partial}{\partial x^1} + e^2 ...

0

There's an easy way to prove it using Cartesian Tensor Notation! Look at (2.10): http://phys.columbia.edu/~cheung/courses/MMSP2014/PS/s14_sol01.pdf I hope this is what you were looking for

1

Just follow the definitions. Write $\alpha$ as $adx+bdy+cdz$, then show that $b_x-a_y=C$ etc. On the way, you will need to use that $d\omega=0$, ie $A_x+B_y+C_z=0$.

1

Are you sure about your definition of the pullback of a one-form? My definition would be $$(\phi^*\alpha)_p(Z_p) = \alpha_{\phi(p)}(D \phi_p (Z_p)) = \langle X_{\phi(p)}, D \phi_p (Z_p) \rangle_{\phi(p)},$$ your expressions with $D\phi^{-1}$ don't make much sense (note that $D\phi^{-1}_{\phi(p)}$ maps into $T_pM$, so you can't take the scalar product at ...

3

GR can be a hard place to start for someone beginning their journey into differential geometry. Some concepts are easier to imagine in terms of manifolds embedded in ambient spaces, but GR is all very intrinsic--you're just plopping a metric on top of a space(time). To be honest, I'm not sure what you're saying with (1). Stuff about directional ...

0

Concerning differentials per se, the answer of xyz should clarify your question. Furthermore, the answer of user10676 settles the question concerning the line element. For those who do not know much about Riemannian geometry lets elaborate a bit on user10676's answer: Think a moment about lines. A possible way to view a (straight) line segment $l$ from $x$ ...

0

Yes it is mathematically rigorous. A line element is given by a Riemanian metric. Recall that a differential form on a vector space $V$ is a map ${\mathbb R}^2 \rightarrow V^*:={\rm Hom}(V,{\mathbb R})$, a Riemanian metric is a map $g:V \rightarrow {\rm Sym}(V^*)$ which is everywhere definite positive, and its length element $ds$ is the square root of $g$. ...

2

differentials (in a way of standard analysis) are not very rigorous in mathematics Standard or not has nothing to do with it. Operations on differential notation are fully rigorous in standard mathematics, as a well-defined representation of properties of differentiable functions. If you want the differentials themselves to be mathematical objects ...

2

The cochain complex of sheaves $$0 \to \mathbb{R} \to \Omega^0 \to \Omega^1 \to \cdots$$ is exact: this follows from the Poincaré lemma. (Any closed differential $(n+1)$-form on a sufficiently small open neighbourhood must be the exterior derivative of some differential $n$-form.) Thus, the cochain complex $$\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$$ ...

1

Good question! Here's a start. The ordinary derivative in one-variable calculus is a Lie derivative along a special vector field on $\mathbb{R}$; in particular, it is not a special case of the exterior derivative. The exterior derivative is instead some kind of "universal derivative": it records all of the information you would need to determine the ...

1

[This is best considered a comment, but a bit too long to post as one] I'm not an expert on this subject, but I can give you a reference that you might find helpful. The book I have in my collection is called "A Primer of Infinitesimal Analysis" by John L. Bell. It is an alternative approach to real analysis that rigorously develops an idea of an ...

2

Such a differential equation is called an Exact Differential Equation. The equation is solved by finding a multiplicative "integrating" factor $R$ such that the following form is exact: $$R(Qdx+Pdy) = QRdx+PRy=0.$$ In other words, $$QRdx + PRdy = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y} dx = 0.$$ for ...

1

That $d^2 = 0$ is probably something you were already taught in vector calculus. For instance, you probably remember that $\nabla \times \nabla \phi = 0$ for $\phi$ a scalar field, or that $\nabla \cdot (\nabla \times E) = 0$ for $E$ a vector field. It's a good exercise to show that both of these can be written as $d^2 f = 0$ and $d^2 E = 0$. Of course, ...

0

The interpretation is the following: given a differential 1-form $\omega=Pdx+Qdy$ in the plane, you are asked to find its integral curves, i.e. 1-dim submanifolds of $\mathbb R^2$ whose tangent line at each point is annihilated by the 1-form. For example, the integral curves of $xdx+ydy$ are the concentric circles around the origin. At a point where ...

4

Let $$\Omega^{1}(\Bbb{R}^{2}) \ni \omega = P(x,y)\,dx + Q(x,y)\,dy$$ be a 1-form on (an open subset of) $\Bbb{R}^{2}$, and think about the differential equation $$P(x,y) \, dx + Q(x, y) \, dy = 0. \tag{1}$$ This equations asks us to find the relationship between $x$ and $y$ that makes LHS vanish. Suppose we have find such a nice relation between $x$ ...

0

So in my other answer, we established that this corresponds to a particular line integral: if $\vec V(\vec r) = P(\vec r) \hat x + Q(\vec r) \hat y$, then the integral takes the form $$\oint_{\partial M} \vec V \cdot d\vec \ell = 0$$ over some closed path $\partial C$. That the integral is over come closed path is important; this effectively removes the ...

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