# Tag Info

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I speculate your problem is that you didn't take advantage of the fact that on spheres centered on the origin, we have an identity $$\sum_i x_i^2 + y_i^2 = r^2$$ and consequently $$\sum_i 2 x_i \mathrm{d} x_i + 2 y_i \mathrm{d} y_i = 0$$ is a linear dependence between the differential forms involved.

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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 ...

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 ...

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The form $d\theta$ is not defined at $0$ (and, indeed, can't be continued across $0$), so Stokes' theorem doesn't apply. For example, Stokes' theorem would give \begin{align*} \int_{\partial \Delta} d\theta = \int_{\Delta} d(d\theta) = 0,\;\;\;\;\;(!) \end{align*} although the first integral is clearly $2\pi$. The best you can do in this situation removing a ...

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The problem is the differential form $d\theta$. In Cartesian coordinates this so-called "form" is $$d\theta = \frac{xdy-ydx}{x^2+y^2}$$ This is not a well-defined form as it is not well-defined at $x=y=0$. Another way of seeing this is if $d\theta$ was a well-defined form then $\oint d\theta = 0$ but it is not! This is because $\theta$ is not a continuous ...

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 ...

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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 ...

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 $\omega^2=\... 0 So is this correct? First, let us assume that we talk about$\mathbb{R}^n$. By taking the standard basis$B={e_1,...,e_n}$, I then calculate the exterior product of$x_i, x_j$, the basic forms, element of the dual of$\mathbb{R}^n$on this basis. I get: (x_i \wedge x_j)(e_i,e_j)= \begin{vmatrix} x_i(e_i) & x_j(e_i) \\ x_i(e_j) & ... 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 ... 0 Let write the$1-form as $$\omega=A\:dx+B\:dy$$ with $$A=\frac{y}{2y+x},\quad B=\log (2y+x)+\frac{2y}{2y+x}.$$ One has $$\frac{\partial A}{\partial y}=-\frac{2 y}{(x+2 y)^2}+\frac1{x+2 y},\quad \frac{\partial B}{\partial x}=-\frac{2 y}{(x+2 y)^2}+\frac1{x+2 y}$$ giving \begin{align} d\omega=&\left(\frac{\partial A}{\partial x}dx+ \frac{\... 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 ,$$so \omega ... 0 I believe the kernel of the contraction \omega^\# at a point p\in \Sigma is dependent on the vector field X. For example, if we consider X such that X(p) \in (T_p\Sigma)^{\omega} for each p\in \Sigma then the one form is trivial. We can do this with a non-zero vector field anytime the submanifold is not symplectic so that we have a non trivial ... 1 Each cotangent space, i.e. the set of the 1-forms restricted to a point x_0 is a vector space. The set of the differential 1-forms over a manifold is more generally a module over the ring of the smooth functions: as you correctly pointed out, you can multiply differential forms by functions, not just by scalars. Nonetheless they form in particular also a ... 0 Agree with @user343900's answer. I would add that when one moves from \Bbb{C} to \Bbb{R}^2, the term \mathrm{i}u + v becomes the vector (v,u) and and \mathrm{d}x + \mathrm{i}\mathrm{d}y becomes the vector \mathrm{d}s = (\mathrm{d}x,\mathrm{d}y), i.e. becomes a line element (as is required by the integral being along the boundary of the region). ... 1 The contour integral \int_{\partial D} f(z) \, \mathrm{d}z when written out becomes$$\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) =... 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*} ... 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^...

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