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6

In Leibniz notation, even in the beginning, you weren't integrating functions: you were integrating differential forms. This is true even back in introductory calculus, even though you didn't have words to put to the concept. e.g. you didn't compute $\int f(x)$; you computed $\int f(x) \, \mathrm{d}x$. And even then is was important not to forget that $\...


6

Differential forms are not introduced to answer the question "How do I integrate functions of manifolds?" Instead, they are introduced to answer a different question, namely "What are the correct objects to integrate on manifolds?" The answer to that question is: differential forms! More precisely, the correct objects to integrate on an orientable $m$-...


5

By Stokes, our condition is necessary. The converse direction follows from Serre duality. Consider $$\omega_k=g(z)\,d\bar z\in\Gamma(\mathbb{P}_\mathbb{C}^1,\bar K\otimes \mathcal O(-k))$$ as well-defined forms with values in $\mathcal O(-k)$, by using the standard trivialization of $\mathcal O(-k)$ and the assumption that $g$ has compact support. By ...


4

Expanding my comment: \begin{align*}(g^*\omega)(x)(v_x^1,\dots,v_x^{2n+1}) & =\omega(g(x))(d_xg(v_x^1),\dots,d_xg(v_x^{2n+1})) \\ & =\omega(\color{red}{-x})(-v_x^1,\dots,-v_x^{2n+1})\\ & =\sum_{j=1}^{n+1} (-x_j) (-v_x^{n+j}) - (-y_j) (-v_x^j)\\ &=\sum_{j=1}^{n+1} x_jv_x^{n+j} - y_jv_x^j\\ &=\omega(x)(v_x^1,\dots,v_x^{2n+1}).\end{align*} ...


4

Write $\omega = \alpha_1 + \dots + \alpha_n$ where $\alpha_i = dx_{2i - 1} \wedge dx_{2i}$. The important observation is that if you expand $$ \omega^n = (\alpha_1 + \dots + \alpha_n) \wedge \dots \wedge (\alpha_1 + \dots + \alpha_n) = \sum_{i_1, \dots, i_n} \alpha_{i_1} \wedge \dots \wedge \alpha_{i_n} = \sum_{I} \alpha_I$$ then if $I$ contains a repeated ...


4

If $\beta$ and $\beta'$ are one-forms such that $d\beta = d\beta' = \alpha$, then $\beta' - \beta$ is closed, i.e. $d(\beta' - \beta) = 0$. It then follows from the Poincaré Lemma that $\beta' - \beta$ is exact, i.e. $\beta' - \beta = df$ for some function $f$. So we see that if $\beta$ and $\beta'$ satisfy $d\beta = d\beta' = \alpha$, then $\beta' = \beta + ...


3

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$First, $(dz + x\, dy)(-\dd_{z}) = -1$. Generally, working in $\Reals^{3}$ for concreteness, if $(x, y, z)$ denotes some coordinate system, $(\dd_{x}, \dd_{y}, \dd_{z})$ denotes the associated coordinate vector fields, and $(dx, dy, dz)$ denotes the coordinate $1$-forms,...


3

You are effectively asked to show that the $2$-forms $x_i\wedge x_j$ with $1\leq i<j\leq n$ form a basis for the vector space of all $2$-forms. It is instructive to verify that the set of all $2$-forms is indeed a (real) vector space. I'll first prove the fact without the hint, and later with the hint. Let $\omega^2$ be a $2$-form. Then $\omega^2=\...


3

Firstly observe that the equation is linear in $\theta$, so it suffices to consider the case $\theta = adx^k$ with some $a \in C^\infty(U)$. Thus, $$d\theta = \sum_{i=1}^{n}\frac{\delta a}{\delta x^i}dx^i\wedge dx^k.$$ Let us write the vector fields in terms of coordinates, i.e. $$X = \xi^1\frac{1}{\delta x^1} + \dots + \xi^n\frac{1}{\delta x^n},\ \ \ Y = \...


3

I pulled up my sleeves and did the computation for $M^2 \subseteq \Bbb R^4$. I'll post it here, but I'll leave the question open for a while in case someone sees a non-medieval way to do this. First of all, take $\{e_1,e_2\}$ an orthonormal positive basis of $T_xM$, and pick any orthonormal positive basis $\{n,\nu\}$ of $T_xM^\perp$ with $\{e_1,e_2,n,\nu\}$ ...


2

HINT: Generalized Laplace expansion by cofactors. The way I like to think of it is this. Reduce to the case of an oriented orthonormal basis $v_1,\dots,v_n$ for $\Bbb R^n$ (in your case, some will be tangent to $M$ and the rest normal to $M$). Consider the Hodge star operator on both forms and multivectors. Then it boils down to $$(dx^i\wedge dx^j)(v_1\wedge ...


2

Recall the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{|\alpha|} \alpha \wedge d\beta$. Hence, we have $$ d(\bar{z}_{\nu} d\bar{z}[\nu]) = d(\bar{z}_{\nu}) \wedge d\bar{z}[\nu] + \bar{z}_{\nu} \wedge d(d\bar{z}[\nu]). $$ Applying the product rule again, you see that $d(d\bar{z}[\nu])$ is a sum of a wedge of forms, each containing $d^...


2

That formula is often used as the definition! Since you're asking this question, I'll assume you're using the other common definition of $d$ for one-forms, which is the coordinate formula $d\omega_{ij} = \partial_i \omega_j - \partial_j \omega_i.$ Contracting with $X^i Y^j$ we get $$d \omega(X,Y) = Y^j X(\omega_j) - X^j Y(\omega_j).$$ From the Leibniz rule ...


2

No wonder you cannot prove that - your claim is false! Let $i : \Sigma \hookrightarrow T^*M$ be the natural embedding $i(x) = x$. Notice that $\theta = i^* \tilde \theta$. Let $\tilde \omega = \textrm d \tilde \theta .$ Notice that $$\omega = \textrm d \theta = \textrm d i^* \tilde \theta = i^* \textrm d \tilde \theta = i^* \tilde \omega ,$$ so $\omega$ ...


2

There is no need for orientations or bases to show that the two given (linear!) maps $$\nu_1:\ A\ \longmapsto\ \omega_A^1\qquad\text{ and }\qquad \nu_2:\ A\ \longmapsto\ \omega_A^2,$$ are isomorphisms. The spaces of $1$-forms and $2$-forms are of dimensions $\tbinom{3}{1}=3$ and $\tbinom{3}{2}=3$, respectively. So to show that $\nu_1$ and $\nu_2$ are ...


1

We use that $${\rm d}\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) -\alpha([X,Y]). \qquad (\ast)$$ Suppose that $\alpha \wedge{\rm d}\alpha = 0$ and let's check that $\xi$ is closed by the Lie bracket. Let $X,Y \in \xi$. Then $(\ast)$ becomes ${\rm d}\alpha(X,Y) = -\alpha([X,Y])$. Evaluating $\alpha \wedge {\rm d}\alpha$ at the triple $([X,Y],X,Y)$ and using ...


1

For a closed $k$-form defined in a star-shaped region relative to the origin, there is an algorithm for finding an antiderivative that I describe here. Applying this to $\alpha$: Step 1: substitute $x_k\to tx_k$ and $dx^k\to t\,dx^k+x_k\,dt$.$$\begin{align}x_2\,dx^1\land dx^2 &\to tx_2(t\,dx^1+x_1\,dt)\land(t\,dx^2+x_2\,dt)\\ &= tx_2(t^2\,dx^1\...


1

I think I see the issue. Let $\omega$ be an $m$-form on $M$, supported in $U \subset M$. Let $\tau \colon U \to \mathbb{R}^m$ be an oriented chart, with coordinates $(r_1, \ldots, r_m)$. On $U \subset M$, we can write $\omega = f\,dr_1 \wedge \cdots \wedge dr_m$. Let $\phi, \psi \colon U \to \mathbb{R}^m$ be oriented charts, also on $U \subset M$. It is ...


1

Note that $$d\omega = \sum_{i=1}^n(-1)^{i-1}dF_i\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n.$$ and $dF_i = \sum_{j=1}^n\frac{\partial F_i}{\partial x_j}dx_j$. As wedge product is skew-symmetric, if some $dx_k$ appears twice in a term, that term must be zero. So we see that \begin{align*} &\ (-1)^{i-1} dF_i\wedge dx_1\...


1

Let $x=(x_1,...,x_n)$ be (global) coordinates for $\mathbb{R}^n$. Then $p$ lives in the tangent space, $T\mathbb{R}^n|_x$, which has corresponding basis $\dfrac{\partial}{\partial x_1},...,\dfrac{\partial}{\partial x_n}$ The dual space to the tangent space, $T^{\star}\mathbb{R}^n$ is the cotangent space, which is the space of $1$-forms. It has the ...


1

Each cotangent space, i.e. the set of the 1-forms restricted to a point $x_0$ is a vector space. The set of the differential 1-forms over a manifold is more generally a module over the ring of the smooth functions: as you correctly pointed out, you can multiply differential forms by functions, not just by scalars. Nonetheless they form in particular also a ...


1

The contour integral $\int_{\partial D} f(z) \, \mathrm{d}z$ when written out becomes $$\int_{\partial D} (u+iv) \cdot (\mathrm{d}x + i \mathrm{d}y) = \Big(\int_{\partial D} u \, \mathrm{d}x - v \, \mathrm{d}y\Big) + i \Big(\int_{\partial D} v \, \mathrm{d}x + u \, \mathrm{d}y\Big).$$ This is not heuristic - complex integrals are defined to make this true. ...


1

No, you've got it backwards. :) Let $i : S^2 \hookrightarrow \Bbb R^3$ be the usual inclusion. If $\omega$ is a form on $\Bbb R^3$, then its restriction to $S^2$ is defined as $\eta = i^* \omega$ (the pull-back of $\omega$). Therefore, the fact that $\eta$ is closed does not mean anything relevant for $\omega$ because $0 = \Bbb d \eta = \Bbb d (i^* \omega) =...


1

Write $X=\sum_iX_i{\partial\over{\partial x_i}}$ $\theta(X)=\sum_i a_iX_i$, $Y.\theta(X)=\sum_ida_i(Y)X_i+a_idX_i(Y)$. This implies that $X.\theta(Y)-Y.\theta(X)-\theta([X,Y])=\sum_ida_i(X)Y_i+a_idY_i(X)-da_i(Y)X_i-a_idX_i(Y)-\sum_ia_i(dY_i(X)-dX_i(Y))$ $=\sum_ida_i(X)Y_i-da_i(Y)X_i.$


1

In polar coordinates you have $$ \omega=r\cos\phi\sin\phi \,d(r\cos\phi)+r(1+\sin^2\phi)\,d(r\sin\phi)=\\ r\cos^2\phi\sin\phi\, dr- r^2\cos\phi\sin^2\phi\, d\phi+r(\sin\phi+\sin^3\phi)\,dr+r^2(\cos\phi+\cos\phi\sin^2\phi)\,d\phi=\\ 2r\sin\phi\, dr +r^2\cos\phi\, d\phi=d(r^2\sin\phi). $$ Therefore, $$ \omega=d\left(y\sqrt{x^2+y^2}\right). $$


1

Let's start by making a quick "type list" of everything we're dealing with here. I'm going to rewrite some of your equations to help me keep everything together, because you're often conflating $V$ and $V'$. $x$ is a point in the space $\Omega \subset V$. $x'$ is a point in the space $\Omega' \subset V'$. $v_1, v_2$ are vectors in $V$. $v'_1, v'_2$ are ...



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