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For a function $f$ and a form $\omega$, we have the product rule $d(f \omega) = df \wedge \omega + f d\omega$. Thus $$d (\phi \iota_v \rho ) = d\phi\wedge\iota_v \rho + \phi d(\iota_v \rho)=d \phi\wedge \iota_v \rho.$$ Now in general this is different to $d\phi(v) \rho$; but in 3 dimensions they are equal. To see this, let's choose coordinates so that at ...

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Differential forms are an appropriate generalisation of derivation and integration (i.e of calculus) for arbitary (within the scope of integration) manifolds and spaces. In order to arrive at such a generalisation (e.g like E. Cartan did) one starts with the basic definitions and operations of derivation and integration, how they affect the space they are ...

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Q1: Always true. Q2: True if you replace "surjective smooth map" with "submersion", otherwise only one direction of the implication holds. Remember the definition of the pullback: $$f^* \omega (X_1, X_2, \ldots) = \omega(Df (X_1), Df(X_2), \ldots).$$ One direction is obvious: if $\omega = 0$, then the RHS is always zero, and thus $f^* \omega(X_1,\ldots) = ... 2 From the first fundamental theorem of calculus: $$\int_{x_1}^{x_2} dF(x) = F(x_2)-F(x_1)$$ Your integral becomes: $$\int_{\beta(a)}^{\beta(b)} df(\beta)= f(\beta(b))-f(\beta(a))= f(q)-f(p)$$ 3 Chain rule is not needed. Using $$\beta^* df = d(f\circ \beta) = (f\circ \beta)'(t)dt,$$ we have $$\int_a^b \beta ^* df = \int_a^b (f\circ \beta)'(t) dt = (f\circ \beta) (b) - (f\circ \beta)(a) = f(q) - f(p).$$ 2$\omega$should be a smooth$1$-form as you want to integrate this along a curve$c$.$\omega$doesn't have to be of compact support, as$c$is compact anyway. Indeed, after the pullback$c:[a, b]\to X$, the calculations are done on$[a,b]$. Write$c^* \omega = g(x) dx$on$[a,b]$. Then $$(c\circ f)^* \omega = f^* c^* \omega = f^* (g(x) dx) = g(f(x)) ... 15 You have asked a good number of questions. I'll answer the one in the title. The point is that differential forms are "the things you can integrate on manifolds". Manifolds are more general objects than open subsets of \Bbb R^n, and in some sense one of the reasons one wants to introduce forms. Suppose you have a 1-form \alpha on a manifold M, and a ... 4 For a function f, the short answer is that the differential of f is set up to work that way. That is, the differential \operatorname{d}f is defined to be the best linear approximation to the change in the function’s value near a given point:$$\Delta_Pf(\mathbf h) = f(P+\mathbf h) - f(P) = \operatorname{d}f_P[\mathbf h]+\text{error},$$where the error ... 0 We use the polar transform \phi(x,y) =\big(r\cos(\theta), r\sin(\theta)\big) and find the pullback of d\omega under this mapping.$$\phi^*dx = d(r\cos(\theta)) = \frac{\partial r\cos(\theta)}{\partial r}dr + \frac{\partial r\cos(\theta)}{\partial \theta}d\theta = \cos(\theta)dr-r\sin(\theta)d\theta \\ \phi^*dy = d(r\sin(\theta)) = \sin(\theta)dr + ... 0 I've done it myself but the full proof is a little bit long so here is the pdf file: https://www.dropbox.com/s/d368qulnyzn82v4/Poisson.pdf?dl=0. The idea is that there exists a bracket on differential forms that is trivial on de Rham cohomology for any Poisson structure and that coincides with Schouten-Nijenhuis bracket if Poisson structure is ... 1 First of all, yes it is true. It's usual, when defining a quantity on a manifold to specify its representation in coordinates (or its local representation). For example, if$(x^1,\ldots,x^n)$is a coordinate system on a chart$U$of a manifold$M$, every$k$-form can be written $$\omega = a_i dx^i,$$ where$a_i \in C^{\infty}(U)are smooth functions. ... 3 Note that the two form is really $$\mathcal{K}=\frac{\sqrt{-1}}{2\pi} \sum_{j,k=1}^n g_{j\bar{k}}dz^j\wedge d\bar{z}^{\bar{k}},$$ so $$d\mathcal{K}=\frac{\sqrt{-1}}{2\pi} \left(\sum_{i,j,k=1}^n\partial_i g_{j\bar{k}} dz^i \wedge dz^j\wedge d\bar{z}^{\bar{k}} + \sum_{i,j,k=1}^n\partial_\bar{i} g_{j\bar{k}} d\bar z^\bar{i} \wedge dz^j\wedge ... 1 Assume M \times N is orientable. Fix a point p \in N and a basis \{v_1, \ldots, v_n\} of T_p N. Consider an orientation form \omega of M \times N and identify M with M \times \{p\} \subset M \times N. Now define \eta on M by \eta(e_1, \ldots, e_m) = \omega(e_1, \ldots, e_m, v_1, \ldots, v_n). (edited to make argument much simpler) 1 You need that M is finitely presented in addition to projective in order for duality to work nicely. The problem reduces to describing a natural nondegenerate pairing$$\wedge^k M \times \wedge^k M^{\ast} \to 1$$where 1 denotes the unit module R. Equipped with such a pairing, any isomorphism M \cong M^{\ast} gives an isomorphism \wedge^k M \cong ... 0 It might help to eschew the more complicated topic of differential geometry and look at vector calculus since everything locally amounts to doing vector calculus anyways. If f is a scalar function, then \mathrm{d}f is just the ordinary derivative \nabla f. And if, for example, \left[ \begin{matrix} x \\ y \end{matrix} \right] are the coordinates on ... 0 Let A be open in \Bbb{R}^{n}; let f: A \to \Bbb{R} be of class C^{1}; for every x \in A let df^{x}: v \mapsto \nabla f(x)\cdot v on \Bbb{R}^{n}. Then df, the differential of f, is a 1-form on A (provided that a 1-form is defined as an alternating 1-tensor). Each elementary 1-form dx_{i}, which is the differential of the ith ... 1 Possible recommendations - I'm not 100% sure if what you are looking for exists but these are all well-written and worth investigating. Edwards, Advanced calculus: a differential forms approach Bloch, A First Course in Geometric Topology and Differential Geometry Bachmann, A Geometric Approach to Differential Forms The last one is particularly nice, in ... 2 The set K cannot be assumed to be contained in U because it is a closure. In particular, if \omega is nowhere zero on U, then K=\overline{U}, which probably is not a subset of U. This is why the condition K \subseteq U is non-trivial; roughly speaking, it means \omega is zero near the boundary of U. On the other hand, it might have made ... 2 The Lie derivative is usually defined either to be equal to the Lie bracket L_XY=\left[ X,Y\right], or by using the flow of Y along X. You have said you already are familiar with the Lie bracket, so here is the second definition.$$(L_XY)_p=\lim_{t\to 0}\frac{\mathrm d\Phi_X^{-t}Y_{\Phi^t_X(p)}-Y_p}{t}$$Here p is a point in the manifold, \Phi^s_X ... 0 \begin{eqnarray*} L_{y\partial_x-x\partial_y}(x^2\partial_x) &=&[y\partial_x-x\partial_y\ ,\ x^2\partial_x]\\ &=&(y\partial_x-x\partial_y)(x^2\partial_x) -x^2\partial_x(y\partial_x-x\partial_y)\\ &=&y\partial_x(x^2\partial_x)-x\partial_y(x^2\partial_x)-x^2\partial_x(y\partial_x)+x^2\partial_x(x\partial_y)\\ ... 3 Let's start with a euclidean space for a moment, but impose upon this a general curvilinear coordinate system. The tangent vectors to the coordinate lines through a given point define the usual basis vectors, which are called various names. They constitute vector fields, at any rate, and for the purposes of this answer, I'll call them only the tangent ... 1 If y_1, \cdots y_l is another basis for V, then x_i = \sum_{j} A_{ij} y_j and$$\omega=\sum_{j=1}^\ell \omega_j x_j = \sum_{j,k} \omega_j A_{jk} y_k.$$As dA_{jk} = 0,$$\sum_{k} d\left(\sum_j\omega_j A_{jk}\right) y_k = \sum_{j,k} d \omega_j A_{jk} y_k = \sum_j d\omega_j x_j.$$Thus d is independent of the basis. The same argument shows ... 2 In \Bbb{R}^4, certainly, the volume form is dV = dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4. However, confined to the sphere S^3 in \Bbb{R}^4, you need a 3-form. (It is perhaps easier to see this for the sphere S^2 in \Bbb{R}^3, where what is wanted is for areas, not volumes.) 1 Hint: the form dV is defined on the Lie algebra su(2) of SU(2) you can represent the elements of su(2) by complex 2\times 2-matrices and use the trace and Lie bracket. su(2)=\pmatrix{ia & -c+id\cr c+id & -ia}a,b,c\in R. thus su(2) SU(2)=\pmatrix{a & -\bar b\cr b & \bar a}, a,b\in C \mid a\mid^2+\mid b\mid^2=1 thus it is a ... 3 Think back to what the Riemannian volume form is. It's a form that, when fed an oriented orthonormal frame, spits out 1. (Of course every manifold here is oriented.) What are the oriented orthonormal frames on \Sigma \subset M, where \Sigma is a (Riemannian) submanifold? Because it's a Riemannian submanifold, if (x_1, \dots, x_{d-1}) is an oriented ... 0 I found an answer to my question in Dupont's Fibre Bundles and Chern-Weil Theory [PDF]. Given a vector space valued form \omega: \bigwedge^k TM \to V, the exterior derivative commutes with post-composition by any linear map \varphi: V \to W. That is, d(\varphi \circ \omega) = \varphi \circ d \omega. So the symmetry of the map to my question was not ... 1 Okay, let's see. We'll start off with vector fields X and Y and the formula d\{Ad(k^{-1})\theta\}(X,Y)=X(Ad(k^{-1})\theta(Y))-Y(Ad(k^{-1})\theta(X)-Ad(k^{-1})(\theta([X,Y])), which is standard. All the action is in the first two terms, and by symmetry we only need to look at X(Ad(k^{-1})\theta(Y)). We begin by noting we have a product rule: ... 0 As Anthony noted in his comment, correcting the normalization coefficient by eliminating the denominator (k+\ell)! removes the problem and yields:$$\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$which allows for Cartan to be proved my way. Another way is this, which John Ma posted in his comment, but which is out ... 1 Given a vector valued function u:\>(\Omega\subset {\mathbb R}^n)\to{\mathbb R}^m and a point p\in\Omega the differential of u at p is a linear map$$du(p) :\quad T_p\to T_{u(p)}, \qquad X\mapsto du(p).X\ ,$$which is defined by$$u(p+X)-u(p)=du(p).X+o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{1}$$Since p will be fixed in the sequel I shall just ... 1 The crucial step towards this is differentiating the function f:M\to GL(\mathfrak g), which maps x to \text{Ad}(k(x))^{-1}. To do this, observe first that by definition T_xk\cdot\xi equals the value in k(x) of the left invariant vector field generated by X:=(k^*\omega_H)(\xi)(x). Using this, you see that T_xf\cdot\xi can be computed as the ... 0 Example. In \mathbb R^3, the form dx\wedge dy is clearly different from dy\wedge dz and from dy\wedge dx (which is the first one multiplied by -1). Indeed, the integral of the first form over the unit disk in the x,y plane defined by$$\{(x,y,z)\in\Bbb R^3 \mid x^2+y^2\le 1, z=0\}$$is the area of the disk; by contrast, the integral of the second ... 1 The tuple {x^1,...,x^k} has clearly determined order and elements. The tuple {}{x^{i_1},...,x^{i_p}} means that every element of the tuple can be any coordinate, the indices are numbered however to distinguish them, so that each indexed element can take on values independent of other indexed elements. If \omega is an n-form on \mathbb{R}^n (eg. it ... 1 Every element of SL_2(\mathbb{R}) is the product of elementary matrices$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \quad\text{and}\quad \begin{pmatrix}1&a\\0&1\end{pmatrix} $$so it suffices to check invariance under these. In terms of complex maps (az+b)/(cz+d), these are z\mapsto -1/z and z\mapsto z+a. The invariance under translation ... 0 If we let brackets denote unnormalized antisymmetrization to match your wedge convention, then in index notation we have (choosing a basis e_{i_k} with e_1 = X)$$\begin{align*} \iota_X (\alpha \wedge \beta)_{i_2 \ldots i_{k+l}} &= \alpha_{[i_1 \ldots i_{k}} \beta_{i_{k+1} \ldots i_{k+l}]}. \end{align*}$$Now let us expand just the i_1 in the ... 4 You have the integral$$\int_1^2 x\sqrt{x^2 +1}\; dx$$We can consider this to be the integral of the one-form f(x)\ dx = x\sqrt{x^2 +1}\; dx on the "1-dimensional manifold with boundary" M=[1,2]. The way that "pulling back" and "pushing forward" works in differential forms is summed up in this formula:$$\int_{N} \omega = \int_{M} \phi^{*}\omega\$ ...

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