0
votes
1answer
56 views

Why is the exterior derivative called exterior derivative

I am studying exterior calculus, and I think I have some grasp of what is the exterior derivative. However its name still eludes me - why is it called a derivative? Is it just because the operator $d$ ...
2
votes
1answer
94 views

Pullback of Differential-Form

Given a differential form $$x\,dy\wedge dz-y\,dx\wedge dz+z\,dx\wedge dy$$ I am supposed to prove that the it's pullback by a linear map of determinant one leaves it invariant. For example, if ...
1
vote
1answer
45 views

If $\omega\wedge\beta$ is exact for every closed form $\beta$, then $\omega$ is exact.

Let $\omega$ be a closed $k$-form. Then: If $\omega$ is exact, for every closed form $\beta$, the form $\omega\wedge\beta$ is exact. Proof: Let $\omega=d\alpha$. Now $d(\alpha\wedge\beta) = ...
6
votes
2answers
104 views

Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

What does an exterior (second) derivative such as in $d^2=0$ have to do with second derivatives as in single- or multi-variable calculus? Is this a correct start: Calculus derivatives are good for ...
0
votes
0answers
154 views

Hodge dual exterior derivative

The introduction of the Hodge dual to the structure of the cotangent space requires the reference to a specific basis or an inner product. I was wondering however, if the composition of hodge dual and ...
1
vote
2answers
68 views

Equality involving exterior product..

suppose you have a differential form $\omega$ writting in local coordinates as $$\omega=\sum_{i=1}^ndx_i\wedge dy_i.$$ Can anyone help me showing the following equality: $$\omega^n=n!(dx_1\wedge ...
2
votes
2answers
69 views

Why is $\theta \not \in C^{\infty}(S^1)$?

Why is $\theta \not \in C^{\infty}(S^1)$? I know that since $\int_{S^1} d\theta = 2\pi$ then $d\theta$ is not exact. Thus since $d(\theta)=d\theta$, $\theta$ must not be $C^{\infty}$ but it seems ...
5
votes
2answers
250 views

If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable.

I am trying to prove the following from a book I am reading through. Thm: If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable. Note ...
1
vote
1answer
53 views

Basis for the space of $p$-forms

According to the book I read, a general $p$-form can be written as: $$\omega=\omega_{a_1\ldots a_p} dx^{a_1}\wedge\ldots\wedge dx^{a_p},\hspace{0.5cm} a_1>a_2>\ldots>a_p$$ where I have used ...
1
vote
1answer
97 views

Use the Fundamental Theorem to deduce the formula for the area of an ellipse.

Use the Fundamental Theorem (Green's Theorem) to deduce the formula for the area of an ellipse. Hint: find a 1-form whose exterior derivative is $ dxdy $.
4
votes
1answer
225 views

Are these two definitions of exterior derivative equivalent?

I saw two definition of the exterior derivative of a $k$-form $\omega$. First definition: $$(d\omega)_p(v_0,\ldots,v_k)= \sum_{i=0}^k(-1)^id(\omega(v_0,\ldots,\hat{v_i},\ldots,v_k))_p(v_i)$$ Second ...
5
votes
2answers
187 views

$\alpha\wedge\beta = 0$ for all $\beta$ implies $\alpha = 0$ without using the Hodge dual

Let $\alpha$ be a differential $k$-form on an orientable smooth $n$-dimensional manifold. If $\alpha\wedge\beta = 0$ for every differential $(n - k)$-form $\beta$, then $\alpha = 0$ because we can ...
4
votes
2answers
231 views

algebraic manipulation of differential form

Suppose $\phi_1, \phi_2, \dots, \phi_k \in (\mathbb{R}^n)^*$, and $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ $(\mathbb{R}^n)^*$ stands for the space of all linear transformations that goes ...
2
votes
1answer
187 views

Wedge product with a non-degenerate form

Let $\alpha$ be a non-degenerate form in $\Lambda^k(V)$ for some vector space $V$, $\dim V = n$. (Here non-degenerate means that if $x\in V$ is nonzero, then $(y_1 , ... , y_{k-1}) \mapsto \alpha(x , ...
46
votes
5answers
4k views

Exterior Derivative vs. Covariant Derivative vs. Lie Derivative

In differential geometry, there are several notions of differentiation, namely: Exterior Derivative, $d$ Covariant Derivative/Connection, $\nabla$ Lie Derivative, $\mathcal{L}$. I have listed them ...
2
votes
1answer
252 views

Taking the exterior derivative of a 0-form

I'm attempting to show that $dg(\vec{x})=\alpha$, where $\alpha=\Sigma^n_{i=1}f_idx_i$ and $g(\vec x)={1\over{p+1}}\Sigma_{i=1}^nx_if_i(\vec x)$...and $d$ is the exterior derivative. The $f_i$ are ...
4
votes
3answers
148 views

elementary question regarding differential forms

Is it possible to give a high level explanation why changing the order of differentials will give rise to a minus sign ? I.e. why do we have $$ dx\,dt = - dt\,dx $$ (I am going to take a course on ...
2
votes
0answers
146 views

Calulation of pullback of form

If $M$ is $2n+1$ dimensional manifold, and $M'= M\times \mathbb R$ Let $x_1,y_1,... x_n, y_n,t', t$ be coordiante of $M'$. With $t$ for coordinate for $\mathbb R$. Let $$ \omega= \sum_{i=1}^n ...
6
votes
1answer
613 views

wedge product of differential form

If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$: $$\omega= d(r^2\alpha)$$ ...