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3

A differential $2$-form can be understood in terms of integral operators, specifically what happens when you integrate them over (directed) 2-surfaces. Stoke's theorem, the higher-dimensional analog of the fundamental theorem of calculus, tells us that $$ \int_S d\omega = \int_{\partial S} \omega $$ where $\partial S$ is the boundary of $S$. (if $f$ is a ...


0

To develop some intuition, you might find it helpful to go back to the physics that inspired the definitions of grad, div and curl: electromagnetism. There are nice physical reasons why we would expect $d^2$ to be zero on both $0$-forms/scalar fields, $\nabla \times \nabla \phi = \vec{0}$, and $1$-forms/vector fields, $\nabla\cdot\nabla\vec{A} = 0$. ...


2

Henri Cartan, one of the greatest 20th century mathematicians wrote: ...utinam intelligere possim ratinationes pulcherrimae quae e propositione concisa DE QUADRATO NIHILO EXAEQUARI fluunt In the unlikely event that you don't speak Latin, o amice, this means: ...if I could only understand the beautiful consequences following from the concise proposition ...


1

The general statement has to do with $n$ forms on a compact $n$-dimensional manifold $M$ without boundary. It is a general fact that for such manifolds the $n$th de Rham cohomology group is $H^n_{dR}(M) = \mathbb R$. Now if $\alpha$ is an $n$-form that represents a non-zero class in $H^n_{dR}(M)$ (i.e. $\alpha$ is not exact), then every other $n$th degree ...


3

What I think you are asking is (relevant to the question) does the Jacobian change if we consider $$ \newcommand{\pwrt}[2]{\frac{\partial #1}{\partial #2}} J = \left|\det \pmatrix{ \pwrt xu & \pwrt xv\\ \pwrt yu & \pwrt yv}\right| $$ as opposed to $$ \newcommand{\pwrt}[2]{\frac{\partial #1}{\partial #2}} J = \left|\det \pmatrix{ \pwrt xv ...


0

It means not every form on $S^1$ is necessarily an exact form, meaning it being the differential of a function on $S^1$. In this case, $fd\theta=cd\theta+dg$, being a closed form (you should see that the OP is talking about closed form), is not exact due to the existence of the term $cd\theta$ (note that closed form $\omega$ means $d\omega=0$ and therefore ...


1

Try R.O. Wells' "Differential Analysis on Complex Manifolds" from 1973, Chapter 1 and 3.


2

Let $\omega_i = dx^{2i-1}\wedge dx^{2i}$, so $\omega = \omega_1 + \dots + \omega_n$. We have $$\omega^n = \underbrace{(\omega_1 + \dots + \omega_n)\wedge\dots\wedge(\omega_1 + \dots + \omega_n)}_{n\ \text{copies of}\ \omega}.$$ Every term in the expansion takes the form $\omega_{i_1}\wedge\dots\wedge\omega_{i_n}$ where $i_1, \dots, i_n \in \{1, \dots, ...


0

As you know, for every $i$, $dx^i\wedge dx^i=0$, and so after opening the brackets, the terms that survive are only those where each $dx^i$ appears exactly once. Computing the number of these terms is a not-too-hard exercise in combinatorics.


0

I've got a partial answer that in particular addresses Zhen Lin's objections. It requires relating integration in different dimensions in two different ways. We rely on two principles: External product: $\int_{X\times Y} \omega \boxtimes \eta = (\int_X \omega) \cdot (\int_Y \eta)$, whenever $\omega$ is a top-dimensional form on $X$ and $\eta$ is a ...


0

Let $\gamma:[a,b]\longrightarrow\Bbb R^3$ be a parametrized curve. Near a point $(t_0, \gamma(t_0))$, $$\gamma(t)\approx \gamma'(t_0)(t-t_0)+\gamma(t_0)$$ i.e. $$\|\gamma(t)-\gamma(t_0)\|\approx\|\gamma'(t_0)\||t-t_0|$$ or, $$\Delta(\text{length})\approx\|\gamma'(t)\|\Delta t.$$ What happens in the limit?


1

I don't know if this is good enough, but you could get some different expressions using the following facts: (1) One has $\Omega_{X \times_Z Y/Z} = \Omega_{X/Z} \boxplus \Omega_{Y/Z}$ (pull back to the product and take the direct sum). (2) In general, if $X \xrightarrow{f} Y \xrightarrow{g} Z$ are morphism with $f$ smooth (to get an injection on the left) ...


1

Have you thought of Green's theorem? It's quite simple to solve if you use Green's theorem.


1

If $b$ satisfies $db=\beta$, and $\alpha$ is a closed $k$-form, then $$d\omega=\alpha\wedge\beta$$ where $\omega=(-1)^k\alpha\wedge b$. The proof works for forms on any manifold.


0

Answering this: "...which is different from equation (1), because the $\alpha$ in equation (1) presumably means $\alpha_x$, not $\alpha_{\phi_t x}$." Actually, $\alpha$ in (1) does mean $\alpha_{\phi_t x}$. See also Ted's comment. You have indeed shown the equivalence of (1) and (2).


2

In general, let $V$ be a real or complex vector bundle of dimension $n$. What can we say about $V$ if it admits $k$ independent nonvanishing global sections? This is equivalent to admitting a splitting $V \cong W \oplus \mathbb{R}^k$ resp. $V \cong W \oplus \mathbb{C}^k$, and so it implies some conditions on characteristic classes. If $V$ is real, then ...


2

You can express $d\mathbf x_I$ in terms of the $n\times n$ determinant by putting standard basis vectors in all the remaining slots, yes, but it is not necessary to do so. For example, with vectors in $\Bbb R^4$, $$d\mathbf x_{12}(v_1,v_2) = D(v_1,v_2,e_3,e_4),$$ as you can check by expansion in cofactors. But the key fact is that for the different ...


1

This is false as stated. Let $A = \{(x,y,z): x^2+y^2=1, z=0\}$, and let $f\colon \Bbb R^3\to\Bbb R^3$ be given by $f(x,y,z) = (x,y,z+3)$. If $\omega = xz\,dy$, then $\int_A \omega = 0$ and $\int_{f(A)} \omega = 3\pi$. Upshot: You need $\omega$ to be a closed form (i.e., $d\omega=0$). EDIT: My apologies. I originally read the problem too quickly. Because ...


0

The motivation for the coefficient $\frac{1}{n!}$ is as follows : if $f : V\times...\times V\to\mathbb{K}$ is n-linear and alternate form, we define the alternator $\mathrm{Alt}$ so that $\mathrm{Alt}(f)=f$. That is $$f(x_1,...,x_n)=\frac{1}{n!}\sum_{\sigma\in S_n}\varepsilon(\sigma)f(x_{\sigma(1)},...x_{\sigma(n)})=\mathrm{Alt}(f)(x_1,...,x_n).$$


0

We can look at the equation $d\mathcal{Y}=f(C)dC$ as the definition of the exact differential. Then obviously $\mathcal{Y}$ depends only on $C$, so the solution is $\mathcal{Y}=g(C)$.


0

There is the theorem (Theorem 3 in Wedge Product) saying that $v_1\wedge v_2\wedge \cdots \wedge v_k=0$ if and only if $$v_j=\sum_{i=1,i\ne j} a_i v_i$$ for some $j$, $j\le 1\le k$. Applying it to (1) we get the most general solution in the form $$\mathcal{Y}=f dC,$$ with some arbitrary function $f$. The solution (2) is correct ONLY if we require ...


2

The statement in the article has to be taken with a huge grain of salt. Even for $1$-forms, there's no consistent way to interpret all wedge products of $1$-forms as intersections. If you restrict attention to nonvanishing decomposable forms (ones that can be written locally as wedge products of $1$-forms), and add a few extra restrictions, then there is ...


0

If $\mu_1 = \alpha\mu_2$ for some positive scalar function $\alpha$, then for any vector field $v$, $$ \operatorname{div}_1 v = \operatorname{div}_2v + v(\log\alpha). $$ Here's a proof. Note that $(\operatorname{div}_j v)\mu_j = d(i_v\mu_j)$ for $j=1,2$. Thus \begin{align*} (\operatorname{div}_1 v)\mu_1 &= d(i_v\mu_1)\\ &= d(i_v(\alpha\mu_2))\\ ...


0

I've actually worked out the answer to the question from the foillowing definition of the divergence : $\operatorname{div}(v)\mu = \operatorname{d}(\mathbf{i}_v \mu)$. Short-circuiting a few computation lines, the result is : $$ \alpha \operatorname{div}_1(v) = \operatorname{div}_2(\alpha v) $$


2

At a non-critical point $x$ of $f$, the kernel of the $1$-form $\text{d}f$ is the tangent space to the surface $X := f^{-1}(f(x))$ at $x$; similarly, for a non-critical point $x$ of $g$, $\text{d}g_x$ has the tangent space of $Y := g^{-1}(g(x))$ at $x$ as its kernel. Putting both together, at non-critical points $x$ for both $f$ and $g$ the radical of the ...


0

I'm writing an answer to my own question because it won't fit in comments and my original post would get too long if I edited it again. First of all, thanks to people for their answers. I'm just writing this one to check my own understanding, because I'm still a bit confused about differential forms. We have: $f^*{d\omega}=d(f^*\omega)$ On the other hand, ...


1

If you really want to go ahead in that way, then notice that, in very simple terms, you should have by definition $$ f^* \mathrm d y_i = \mathrm d y_i ( \mathrm d f) = (0,\ldots,1,\ldots,0) \left(\frac{\partial f_h}{\partial x_k}\right)_{h,k=1,\ldots,n} = \left(\frac{\partial f_i}{\partial x_1},\ldots,\frac{\partial f_i}{\partial x_n}\right) . $$ Then it is ...


1

Since the two sides of your equation are maps belonging at each point to a $1$-dimensional space, it suffices to check that they take the same value at a single tuple of tangent vectors, which we take to be $\left(\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}\right)$, simply because it's very easy to evaluate the right hand side on it, ...


5

No. For example, over $\mathbb{R}^4$, the $1$-form $\omega = x\; dy + z\; dw$ has $\omega\wedge \omega = 0$ (as any $1$-form has by (graded)-commutativity), but $d\omega = dx\wedge dy + dz \wedge dw$ has $$d\omega \wedge d\omega = 2\; dx\wedge dy\wedge dz\wedge dw\not =0.$$


0

On 1-dimensional manifolds, this is just the holonomy (for circles) or parallel transport (for intervals). One integrates the connection with the path-ordered exponential. It seems that there is a Stokes' theorem for higher gauge theory: There is a notion of 2-holonomy of surfaces for 2-connections, see for example An Invitation to Higher Gauge Theory or ...



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