# Tag Info

14

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

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: $$\int_{N} \omega = \int_{M} \phi^{*}\omega$$ ...

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 ...

4

In terms of differential forms, the standard integration you learned in calculus is actually about integrating $1$-forms: namely, the $1$-forms $f(x) \, dx$. In general, you integrate $n$-forms over $n$-dimensional manifolds. So a $0$-form would be integrated over a $0$-dimensional manifold: that is, a discrete set of points $P$, taken with orientation ...

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 ...

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 ...

3

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) = ... 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 $$d\mathcal{K}=\frac{\sqrt{-1}}{2\pi} \left(\sum_{i,j,k=1}^n\partial_i g_{j\bar{k}} dz^i \wedge dz^j\wedge d\bar{z}^{\bar{k}} + \sum_{i,j,k=1}^n\partial_\bar{i} g_{j\bar{k}} d\bar z^\bar{i} \wedge dz^j\wedge ... 3 Chain rule is not needed. Using$$\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).$$2 From the first fundamental theorem of calculus:$$\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)$$2 It is a typo. It is indeed true that d(xy)=ydx+xdy. In short, to calculate with differential forms you take total derivatives paired with the wedge product. The wedge product has dx_i \wedge dx_j= -dx_j\wedge dx_i hence dx_i \wedge dx_i=0 for all i. Then, you extend that rule naturally. In the same way, we could say complex numbers are just real ... 2 The Lie derivative is usually defined either to be equal to the Lie bracket L_XY=\left[ X,Y\right], or by using the flow of Y along X. You have said you already are familiar with the Lie bracket, so here is the second definition.$$(L_XY)_p=\lim_{t\to 0}\frac{\mathrm d\Phi_X^{-t}Y_{\Phi^t_X(p)}-Y_p}{t}$$Here p is a point in the manifold, \Phi^s_X ... 2 \omega should be a smooth 1-form as you want to integrate this along a curve c. \omega doesn't have to be of compact support, as c is compact anyway. Indeed, after the pullback c:[a, b]\to X, the calculations are done on [a,b]. Write c^* \omega = g(x) dx on [a,b]. Then$$(c\circ f)^* \omega = f^* c^* \omega = f^* (g(x) dx) = g(f(x)) ... 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.) 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 ... 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 ... 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 ... 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: ... 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 ... 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 ...

1

Hint. Let $f$ and $g$ be functions, $\phi$ and $\psi$ $1$-forms. Then \begin{align*} (1)&\;\mathrm{d}(fg)=\mathrm{d}f\;g+f\;\mathrm{d}g\\ (2)&\;\mathrm{d}(f\phi)=\mathrm{d}f\wedge\phi\\ (3)&\;\mathrm{d}(\phi\wedge\psi)=\mathrm{d}\phi\wedge\psi-\phi\wedge\mathrm{d\psi}. \end{align*} Notice that $\wedge$ denotes the wedge product.

1

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 ...

1

Your differentiation of $u$ is incorrect, leading to incorrect substitution. Also your limits do need to change when you change from $x$ to $u$. This does have the advantage that you don't need to substitute back in $x$, you can just apply the limits to $u$. Let $u=x^2+1$, hence $du=2xdx$. When $x=1$, $u=2$. When $x=2$, $u=5$. \begin{align} \int_1^2 ... 1 Your proof is correct. I don't think it can be done simpler, but perhaps the following is more "conceptual". The form \omega is the Hodge-* of the 1-form f_1\,dx+f_2\,dy+f_3\,dz. Hence, the volume element of its plane restriction is proportional to the dot product of (f_1,f_2,f_3) with the normal vector to the plane. No matter what (f_1,f_2,f_3) ... 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} $$so it suffices to check invariance under these. In terms of complex maps (az+b)/(cz+d), these are z\mapsto -1/z and z\mapsto z+a. The invariance under translation ... 1 If y_1, \cdots y_l is another basis for V, then x_i = \sum_{j} A_{ij} y_j and$$\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.$$Thus d is independent of the basis. The same argument shows ... 1 The tuple {x^1,...,x^k} has clearly determined order and elements. The tuple {}{x^{i_1},...,x^{i_p}} means that every element of the tuple can be any coordinate, the indices are numbered however to distinguish them, so that each indexed element can take on values independent of other indexed elements. If \omega is an n-form on \mathbb{R}^n (eg. it ... 1 First of all, yes it is true. It's usual, when defining a quantity on a manifold to specify its representation in coordinates (or its local representation). For example, if (x^1,\ldots,x^n) is a coordinate system on a chart U of a manifold M, every k-form can be written$$ \omega = a_i dx^i,  where $a_i \in C^{\infty}(U)$ are smooth functions. ...

1

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 ...

1

I think maybe you've misinterpreted what someone has said about integration. What's true is that the integral of a differential $k$-form over a $k$-manifold has a well-defined meaning, independent of any particular choice of parametrization. In other words, if you choose one parametrization and I choose another, we'll get the same answer for the integral. ...

Only top voted, non community-wiki answers of a minimum length are eligible