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Questions tagged [exterior-derivative]

For questions related to exterior derivative. The exterior derivative of a function $f$ is the one-form $df=\sum_i\frac{\partial f}{\partial x_i}dx_i$ written in a coordinate chart $(x_1,\dots,x_n)$.

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Why does calculating $d\omega(V_1...V_k)$ explicitly in local coordinates not give the same formula as in the literature?

Let $\omega$ be a $k$-form, and $V_i$ vector fields. To not have too many indices, let us suppose $k=1$ and we work in 3d. Now I cannot reconcile the following two ways of computing $d\omega(V,W)$. ...
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Two contradictory derivations of $\text{d}\omega$ for a $1$-form $\omega$ on a $2$-manifold.

I'm sure this is a silly question: there are two seemingly general equations for computing exterior derivatives that, when I apply them to a $1$-form on a $2$-manifold, yield different results. The ...
Sam's user avatar
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A theorem to describe of all connections on a given vector bundle.

In the answer to this question Any connection in the trivial line bundle is given by the exterior derivative $d: C^\infty(M) \to \Omega^1(M)$ Matsmir gave me two propositions that help to understand ...
some_math_guy's user avatar
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Any connection in the trivial line bundle is given by the exterior derivative $d: C^\infty(M) \to \Omega^1(M)$

I run into this statement: any connection in the trivial line bundle $\operatorname{pr}_M: M \times \mathbb{R} \to M$ is given by the exterior derivative $d: C^\infty(M) \to \Omega^1(M)$ with ...
some_math_guy's user avatar
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Counterexample of switching Pullback and exterior derivative

We have a theorem in class where for a $C^2$ function $F:U \to V$ and a differential $C^1$ $k$-form $\alpha \in \Omega(V)$, we have $d(F^*\alpha) = F^*(d\alpha)$. I would guess that the statement is ...
user007's user avatar
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Can we interchange the pullback and exterior derivative for a $C^1$ map?

In a recent lecture, the following statement was discussed: Suppose $U\subset \mathbb{R}^n$ is an open subset and $\varphi \in \Omega^k(U)$ a $C^1$ k-form, further suppose $V \subset \mathbb{R}^m$ is ...
user007's user avatar
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Identities involving exterior and interior derivatives

I am reading Arnold's Mathematical Methods of Classical Mechanics and get stuck in one of the practice problem. The problem is if $\tau$ is the 3-D volume element, we have $$i_{\nabla\times(\textbf{a}\...
Apocalypse's user avatar
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A Question Regarding Cartan’s Absorption Method

I want to ask a question from the book named “ Cartan for Beginners : Differential Geometry via Moving Frames and Exterior Differential Systems” as to how one can absorb an apparent torsion. Suppose ...
iliTheFallen's user avatar
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Is $\delta \Delta^{-1} d$ the identity operator?

Let $d, \delta, \Delta = (d\delta + \delta d)$ be the exterior derivative, codifferential and Laplace-de-Rham operator. Let $\omega$ be a closed $k$-form, one can then say $\Delta \omega = d \delta \...
Theo Diamantakis's user avatar
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Area 2-form and index 1-form on the Hyperbolic Semiplane

I'm having some trouble in defining/deducing a natural 2-form of area in the upper-half plane model of the hyperbolic space $\mathbb{H}^2$. Probably this is a very basic question. I set the context ...
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Is the exterior derivative of a form with values on a vector space well-defined?

I've recently asked a question on Physics SE about how one should interpret a particular object in the local expression for the spin covariant derivative. Namely, the object was $\mathrm{d}\psi(X)$, ...
Níckolas Alves's user avatar
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When is $h=d\omega$

Given $h\in \Lambda^{k+1}(M)$, where $M$ is some manifold, how do we know that $h=d\omega$ for some $\omega\in \Lambda^k(M)$? For example, consider $$h=h_1(x^1,x^2,x^3)\,dx^1\wedge dx^2+h_2(x^1,x^2,x^...
Irene's user avatar
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Writing the exterior derivative of a function in an arbitrary frame/coframe

Let $M$ be a smooth manifold of dimension $n$. Let $U \subseteq M$ be an open subset over which the tangent bundle $TM$ is trivial. Let $X_1,\ldots,X_n : U \to TM$ be a local frame for $TM$ and let $\...
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Does the Hessian correspond to the exterior derivative of the gradient 1-form? Or does its skew-symmetrization?

Question: Given a twice totally differentiable (not necessarily $C^2$) function $f: \mathbb{R}^m \to \mathbb{R}^n$, do its $n$ Hessian matrices correspond to the exterior derivatives of its $n$ ...
hasManyStupidQuestions's user avatar
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Kurzweil-Hensotck integral, an unclear point in a book for undergraduates by Fonda

I would like to understand how from $$d\omega_2(x)=(\frac{\partial g_{2,3}}{\partial x_1}-\frac{\partial g_{1,3}}{\partial x_2}+\frac{\partial g_{1,2}}{\partial x_3})dx_{1,2,3}$$ follows $$\...
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Infinitesimal Interpretation Of Exterior Derivative

I have a question regarding the conceptual understanding of the exterior derivative. I've read that one can view a $k$-form on an $n$-dimensional manifold as a collection of infinitesimal (oriented) ...
fweth's user avatar
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Is $\text{d}^{\dagger}\text{d}$ or $\text{d}\text{d}^{\dagger}$ by itself a valid operator?

If I simply consider just one of the combinations $\text{d}^{\dagger}\text{d}$ or $\text{d}\text{d}^{\dagger}$ both of them take from $\Omega^r(M)\to \Omega^r(M)$ for some manifold $M$. But do they ...
Dr. user44690's user avatar
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Exterior derivative of 1

In one of the examples of Loring W. Tu's Book "Introduction to Manifolds" (section 19.7 of 2nd ed.) the author takes the exterior derivative on both sides of the equation $x^2 + y^2 = 1$ ...
Andrea's user avatar
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1 vote
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Prove the coordinate-free formula for the exterior derivative

I can prove the following : There exist a unique map $d:\Omega^k(M)\to \Omega^{k+1}(M)$ such that $d$ is $\mathbb{R}$-linear $d^2=0$ For a $0$-differential form (a function) $d$ reduces to the usual ...
Laurent Claessens's user avatar
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2 answers
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Symplectic gradient is just giving me the gradient

I am thinking about the 2 dimensional vector calculus operation "grad perp" $\nabla^\perp := (-\partial_y , \partial_x)$. According to these notes, this can be defined using the Hodge star ...
Theo Diamantakis's user avatar
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Rotation in cylindrical coordinates and exterior derivative

Suppose a 1-form $A$ of $\mathbb{R}^3$ is represented as $A= A_r (r,\theta,z)dr + A_\theta (r,\theta,z)d\theta + A_z(r,\theta,z)dz$ using cylindrical coordinate system $(r,\theta, z)$. The external ...
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Application of properties of the exterior derivative to show submersion is preserved by translations by constant.

Let $p \in M$ be a smooth manifold and let $f \in C^{\infty}(M)$, and let it further be known that $df|_p$ is surjective for all $p \in M$ such that $f(p) = 0$. Now, am I correct in identifying the ...
Ben123's user avatar
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Exterior algebra question

I don't know how to approach this question from Flanders' Differential Forms. I see it was discussed here, but I don't believe that argument is correct and would apply to something like $2v_1 \wedge ...
Sam's user avatar
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1 answer
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Question regarding exterior derivative in local formulation.

Let $(U,\varphi)$ be a chart of a smooth manifold $M$, and let $\omega \in \Omega^k(M)$, where $\Omega^k(M)$ is the $C^{\infty}(M)$ module of differential k-forms on $M$. We write $\omega|_U = \sum_{I}...
Ben123's user avatar
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3 votes
1 answer
185 views

Why is $d(xdx) = 0$?

If the idea behind the exterior derivative $d$ is that it tells us how quickly a $k$-form changes along every possible direction, why is $d(xdx)=0$ even though $xdx$ varies with $x$? I understand the ...
Localth's user avatar
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4 votes
2 answers
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What kind of object is the (second) exterior derivative of a moving point $R$ in $\mathbb R^n$? What does $dR\wedge\omega \mathbf e$ mean?

TL;DR: Given $f:\mathbb R \to \mathbb R^n$ smooth enough, what are $df$ and $d^2f$ and how are they computed wrt possibly moving bases $\{e_i(t)\}$? I (believe that I) understand the answers in the &...
Al.G.'s user avatar
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Equivalence of differentials of equal coordinates

I've stumbled upon an ostensible inconsistency which probably has a simple resolution I'm overlooking. Consider for the time being a smooth function $f: \mathbb{R} \to \mathbb{R}$ giving rise to a ...
St3phenwalking's user avatar
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Confusion with covariant derivatives with vielbeins

I have some confusion regarding how the covariant derivative is defined for one forms on a manifold in the context of frames/vielbeins. I am a physics student and my reference is Sec 4.3 of the ...
newtothis's user avatar
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Could I see a drawing for $x^2 \mapsto 2x$ for this property: The derivative of a differentiable mapping is the linear map of the tangent spaces

I'm reading Vladimir Arnold's differential mechanics textbook and came across this definition and figure: Despite his drawing I still found it vague and hard to understand precisely what he means. ...
Nate's user avatar
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2 votes
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Computer algebra system for applying Cartan's test to systems of PDEs

It is my understanding that if one has a (possibly overdetermined) system of PDEs, one can check for compatibility by applying Cartan's test (see for example [1], Chapter 7). It involves first writing ...
Gateau au fromage's user avatar
7 votes
1 answer
233 views

Exterior Derivative and Lie Derivative on infinite dimensional manifolds

Lately I have been trying to understand the chapter in Abraham and Marsden's Foundations of Mechanics on infinite-dimensional Hamiltonian systems. Now that I've finally got a feeling for the canonical ...
whatever's user avatar
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Difficulty understanding a step in Weintraub's Differential Forms Textbook:

In Weintraub's Differential Forms, he gives this property: Let $f$ be a differentiable function defined on a region R of $\mathbb{R}^3$. Let $f_*$ be the derivative of $f$ with respect to some ...
Nate's user avatar
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Request of help for a problem on 2-forms

Let $y = f(x)$ and $z = g(x, y) $ be real functions of one variable $x$ and two variables $(x,y)$ respectively. Suppose $$dz \wedge dx =0.$$ What conclusion can be drawn from this statements? My ...
Mary_26's user avatar
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Closed $n$ form from Harvard Qualifying Exam

In the most recent Harvard Qualifying exam, one is asked to prove that if $M$ is a compact oriented manifold, and there are two nonvanishing orientation forms $\xi$ and $\eta$ who's integrals over $M$ ...
Chris's user avatar
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6 votes
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Why can the exterior derivative be interpreted as in this picture?

This paper provides a geometric intuition of differential forms. On page $5$ it reads: "Consider the case $xdy$. The number of horizontal lines is roughly proportional to $y$. In other words the ...
Sam's user avatar
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1 vote
1 answer
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Understanding Coulomb's law in the language of differential forms

I began studying differential forms to gain a deeper understanding of the Maxwell's equations. Let's say we have a $2$-form representing the electric flux density (the $D$-field). We can take the ...
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1 vote
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Intuition for the commutation of the pullback and the exterior derivative

I know that there are already several questions to this topic but I haven‘t seen a satisfactory answer. Deriving that the pullback and the exterior derivative commute is no problem but I want to know ...
Silas's user avatar
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1 answer
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Lie derivative of a differential form

I have a differential $1$-form $\omega = x\mathrm{d}x + x\mathrm{d}y$ and I need to find its Lie derivative along $X = (x+y)\partial_{x} - 2y\partial_{y}$. The first approach is by using Cartan ...
Matthew Willow's user avatar
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What does C.H. Edwards mean by "orientation" in the development of Stokes's theorem in n dimensions?

Edit: I am inclined to believe that the overall answer is going to be: it doesn't matter which sense we call positively oriented as long as we are consistent during our traversal. The following is ...
Steven Thomas Hatton's user avatar
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1 answer
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Generalization of Helmholtz theorem to differential forms?

In vector calculus, Helmholtz theorem says the divergence and curl of some vector field uniquely determines the vector field itself (with appropriate boundary conditions). Can this be generalized to ...
Aiden's user avatar
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Are $\phi$ and $\hat z$ 'dual' in $\mathbb R^3$?

Consider $\mathbb R^3$ and the following equation \begin{equation} \star d\omega=df \end{equation} where $f$ is a 0-form and $\omega$ is a 1-form. My question is whether this equation is satisfied by \...
dennis's user avatar
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2 votes
1 answer
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What is the name for this relationship between a 1-form and a vector? [duplicate]

all. I have a question about Visual Differential Geometry and Forms - A Mathematical Drama in Five Acts (by Tristan Needham). This book shows two relationships between Forms and vector. And I have a ...
yo-yos's user avatar
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1 answer
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Pullback of an exterior differential system

An exterior differential system (EDS) is a pair $(M,\mathcal{E})$ consisting of a smooth manifold $M$ and a homogeneous differentially closed ideal $\mathcal{E}$ of the graded algebra $\Omega^*(M)$ of ...
A. J. Pan-Collantes's user avatar
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Space of self-dual connections

Let $(M,g)$ be a 4-dimensional Riemannian manifold and $(P,M,\pi)$ a principal $G$-bundle over $M$. Denote by $ad (P)$ the adjoint bundle of $P$, that is, the vector bundle obtained by dividing $P\...
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How to compute the exterior derivative of $\omega := \frac{x^1dx^2 - x^2dx^1}{(x^1)^2+(x^2)^2}$?

Let $dx^I$ be local coordinates and let $$\omega := \frac{x^1dx^2 - x^2dx^1}{(x^1)^2+(x^2)^2}.$$ Compute $d\omega$ for $(x^1,x^2) \ne 0$. Remark: $d\omega$ means the exterior derivative, which we ...
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$k$-forms: What does $u^\times$ in $\omega_1 := \langle u,\cdot \rangle = \det(u^\times,\cdot)$ mean?

Consider the following $k$-forms $\omega_k$ in $U \subset \mathbb{R}^2$ (for $U$ open): \begin{align} &\omega_0 = f\\ &\omega_1 := \langle u,\cdot \rangle = \det(u^\times,\cdot)\\ &\...
3nondatur's user avatar
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Curvature defined as exterior covariant derivative of connection?

In Gauge Theory and Variational Principles by David Bleecker, the curvature of a $\mathfrak{g}$-valued 1-form connection is defined as $\Omega^\omega:=D^\omega\omega=(d\omega)^H$, where $\omega(d\...
user_not_found's user avatar
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Exterior derivative of a 2-form defined by composition of metric with an endomorphism

Im interested in calculating the exterior derivative of a 2-form defined by $\omega(x,y) = g(x,Ay)$ where $A \in \Gamma(End(T) )$ is skew and $g$ is some metric. I hope to reach some coordinate-...
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The exterior differential $d:\Omega^r(M)\to \Omega^r(M)$ defines a fibrewise linear mapping over the exterior bundle $\bigwedge^rT^\star M$.

Let $M$ be a finite-dimensional smooth manifold and $E=\bigwedge^rT^\star M$ be the vector bundle of $r$-forms in $M$. The smooth $r$-forms in $M$ constitutes the vector space smooth sections of $E$: $...
FUUNK1000's user avatar
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1 answer
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Intuition of curvature

The only physical intuition of curvature that I know: parallel transport along a closed loop doesn't close (e.g. parallel transport of a tangent vector on a sphere). The Ambrose-Singer theorem says &...
Alex's user avatar
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