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5

First, you need to assume that $p$ and $q$ are both positive; otherwise there are no nontrivial $(p-1,q-1)$-forms. You'll need to use both the ordinary Poincaré lemma and the $\partial$- and $\overline\partial$-Poincaré lemmas. First, if $\alpha$ is a $d$-closed $(p,q)$ form on $U$, then the ordinary Poincaré lemma implies that $\alpha=d\eta$ for some ...

3

They both may be right. If wedge product differs (Similar problem) and we set the definition of exterior derivative as $$d\omega=\sum_{I}d\omega_I\wedge dx^I,$$ then $d$ may differs as well (cause $\wedge$ appears). If we take axiomatic approach to exterior derivative, then one of axioms says ...

3

In complete generality, this is not possible. For example, consider the $1$-form $\omega = \sin x\,dx$ on $\mathbb R$. It's possible to find a countable open cover and partition of unity such that half of the terms $\int_{\phi_n(U_n)}(\phi_n^{-1})^*(\rho_n\omega)$ are equal to some constant $c$, and the other half are equal to $-c$. The limit of the sum ...

3

Let $M$ be a nice smooth manifold, $\alpha\in\Omega^k(M;\mathbb{R})$ be a differential $k$-form, and $X,X_1,\dots,X_{k+2}\in\mathfrak{X}(M;\mathbb{R})$ any collection of vector fields. A definition for the exterior derivative is ...

2

I think it is not difficult to get a proof from the "global" formula, which expresses the action of an exterior derivative on vector fields, i.e. $$d\phi (\xi_0,\dots,\xi_k)=\textstyle\sum_i(-1)^i\xi_i\cdot \phi(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)\\ ... 2 You need only the following three facts: H^1(X\sqcup Y) = H^1(X) \oplus H^1(Y), H^1 (X\times \mathbb R) = H^1(X), and H^1(\mathbb S^1) = \mathbb R. The first and the third facts can be proved directly, while the second one (sometimes referred as the Poincare lemma) might be a bit harder to show. Everything are well discussed in the book ... 2 We have$$\eta^{n-1} = (n-1)! \sum_{j=1}^n x_1 \wedge \dots \wedge x_{2n} \text{ (}x_{2j - 1} \wedge x_{2j} \text{ is missing).}$$Then$$x_{2j - 1} \wedge \eta^{n-1} = (n-1)! \sum_{j=1}^n x_1 \wedge \dots \wedge x_{2n} \text{ (}x_{2j} \text{ is missing)}$$and$$x_{2j} \wedge \eta^{n-1} = (n-1)! \sum_{j=1}^n x_1 \wedge \dots \wedge x_{2n} \text{ (}x_{2j-1}\text{ ...

2

In general, consider a one form $\beta = \sum_i \beta_i dx^i$ on $U\subset\mathbb R^m$. If $$\| \beta\|^2 = \beta_1^2 + \cdots + \beta_m^2 \neq 0$$ on $U$, then the $(n-1)$-form $$\alpha = \frac{1}{\|\beta\|^2} \sum_i (-1)^{i-1} \beta_i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots \wedge dx^m$$ on $U$ safisfies $$\beta \wedge \alpha = dx^1 ... 2 Generally a connection and its curvature are Lie algebra valued differential forms, a particular case of vector valued differential forms,see e.g. https://en.wikipedia.org/wiki/Vector-valued_differential_form . For the Lie group U(1) the Lie algebra is isomorphic to the real numbers so you get what they call an ordinary (real valued) differential form. 2 Note that the text assumes q is a quadratic differential, not an abelian one. I would like to explain the differences, motivate quadratic differentials a bit and show how they fit in the context of dynamical systems. In the meantime, I will answer your question, if you have not already seen the point of the square root. Let's start with abelian ... 2 Hint: The formula for the volume form of a hypersurface i:S^{n-1}\rightarrow \mathbb{R}^{n} is$$ds=i^{*}(\iota_\nu dV),$$where dV is a given volume form on the \mathbb{R}^{n} and \nu\in T\mathbb{R}^{n} is a smooth unit normal field to the surface; with interior product (contraction) and the pull-back (restriction) operations. Intuition: dV ... 1 Using that  \tilde S is the Hodge dual of S, we have$$*(S + \tilde S) = (S+ \tilde S),$$where * is the Hodge star operator (Acting on two forms). Then$$\partial_{\mu}(S^{\mu\nu}+\tilde{S}^{\mu\nu}) = 0$$if and only if d^* (S+ \tilde S) = 0, where d^* is the adjoint of d. But d^* = \pm *d*, so the equality is the same as$$0=d ...

1

The term to use here would be differential form. The operator $dx^i$ is a differential form. Specifically, it is a 1-form. It lives in the cotangent space, which is the dual space of the tangent space. The tangent space is where the tangent vectors live, and here it is equal to $Span\{x_1, ..., x_n\}$. The operator $dx^i$ acts on this tangent space via ...

1

$dx^i$ is a covector field. It eats vector fields and spits out functions. The way you integrate a covector field $\omega$ on $M$ along a parameterized path $\gamma: [0,1] \to M$ is by partitioning $[0,1]$ into subintervals $[x_i, x_{i+1}]$ and summing $\sum_1^n \omega(\gamma'(x_i)) (x_{i+1}-x_i)$, then taking a limit. This is what line integrals do.

1

Vector calculus is full of abbreviations and shortcuts that let you focus on what you're actually trying to compute, but when you examine those shortcuts with scrutiny, you might realize that there's a little more going on than you realized. This problem is a good example. First, what does it mean to take a line integral of a vector field? $$\int_C f ... 1 Since forms are antisymmetric, it will always be the case that \alpha\wedge\alpha = 0. In particular, dx_i\wedge dx_i = dx_i^2 = 0 for all i. So the answer to your first question is, "yes." To compute d\alpha, note first that the d operator obeys a Leibniz rule:$$ d(\omega\wedge\eta) = (d\omega)\wedge\eta + (-1)^\cdot \omega\wedge (d\eta) $$... 1 I found a self-contained answer that really requires no real computation. I'll reproduce a sketch of it below but it can be found in Natural Operations in Differential Geometry on pg. 66. I think the key is recognizing what generates the algebra  \Omega(M)  locally at least. Since d is a graded derivation of degree one,  d^2 = \frac12[d,d]  is a ... 1$$\begin{align}d_yf &\\ &=\sum_i \frac{\partial f}{\partial y^i}dy^i \\ &=\sum_i\left(\sum_j \frac{\partial f}{\partial x^j}\frac{\partial x^j}{\partial y^i} \right)\left(\sum_k\frac{\partial y^i}{\partial x^k}dx^k\right) \\ &= \sum_{i}\left(\sum_{j,k} \frac{\partial f}{\partial x^j}\frac{\partial x^j}{\partial y^i}\frac{\partial ...

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