# Tag Info

10

I don't know if this provides the type of intuition sought. And one might interpret this as just one more approach to proving the identity. But I thought that this might shed a bit of intuition as to what is going on here. To that end, we provide a way forward that exploits symmetry and inversion. First, let us write the function $f$ in terms of its ...

8

"Restriction of $\omega$ to $N$" is the same thing as the pullback form $i^*\omega$, where $i:N \to M$ is the inclusion. Then all you need to know is that for any smooth map $f$, $df^*\omega = f^*d\omega$. You can prove this in coordinates, for instance.

7

Here is the way I like to think of this. Whenever I try to gain intuition about integration, I always boil it down to step functions since the results extend nicely from there and step functions are oh-so-easy to work with. This is a good technique that can be used to visualize many of the standard identities from calculus and gain some intuition about them. ...

6

Suppose we have a nice map $g:\mathbb{R} \to \mathbb{R}$ that is surjective and $k$-to-$1$, both properties meant for generic $x$ and not necessarily all $x$. In the problem at hand, $k=2$. I claim that if the sum of the $k$ pre-images of $x$ is equal to $x+C$ for a constant $C$, then the function $G(x)$ validates the formula $\int f(x) = \int ... 4 Following Élie Cartan, you want to think of your differential system$dR - R\omega = 0$on$M\times SO(n)$. This$\mathfrak{so}(n)$-valued$1$-form is integrable, as you said, because of vanishing curvature. The integral manifolds of this differential system will locally be the graphs of functions$R\colon U\to SO(n)$. 3 Its self-wedge will be a 4-form, so we'd better take$n \geq 4$. Then $$\beta=dx_1 \wedge dx_2 + dx_3 \wedge dx_4$$ should do the trick. 3 This is not true. Take$\alpha$to be the following$2$-form on$\mathbb R^4$with coordinates$(w,x,y,z)$: $$\alpha = dz\wedge dw - x \,dy \wedge dw,$$ and let$X,Y$be the vector fields $$X = \frac{\partial}{\partial x}, \qquad Y = \frac{\partial}{\partial y} + x \frac{\partial}{\partial z}.$$ Then$\iota_X\alpha = \iota_Y\alpha =0$, but$[X,Y] = ...

3

This is false. The contact condition is often known as being maximally non-integrable. Given an integrable distribution $\xi$ defined by a 1-form $\alpha$, use that $$d\alpha(X,Y) = Y\alpha(X) - X\alpha(Y) - \alpha([X,Y]).$$ So if $X,Y \in \xi$, then $d\alpha(X,Y) = 0$. So in particular $\alpha \wedge d\alpha = 0$ everywhere. For 3-manifolds, the contact ...

3

I think you are encountering an issue with your calculation due to a misreading of the problem. The way I understand it, the question is, for any $2$ form $\omega$, does there exist a basis $\sigma_i$ such that $$\omega(u, v) = \sigma_1\wedge\sigma_2(u, v)+\dots +\sigma_{2r-1}\wedge\sigma_{2r}(u, v)$$ In your example, it is fine to start with a basis ...

3

If $f(x,y)=(x,y,x^2+y^2)$ then $$\int_M \omega = \int_{D} f^\ast \omega$$ where $D$ is a unit disk. Hence $$\int_D (-x^2-y^2)dxdy = \int_D (-r^2) rdrd\theta = 2\pi \frac{-r^4}{4}\bigg|_0^1 = -\frac{\pi}{2}$$

3

General idea: Take the set $U=U_{\alpha_0\ldots\alpha_p}$. By performing $\delta$ twice on some form on $U$, you will get the set $V=U_{\alpha_0\ldots\alpha_p\beta_0\beta_1}$ twice, and the two forms on $V$ will cancel each other out. This is due to the $(-1)^p$ in the definition of $\delta$. To get the right feel, you may calculate $\delta$ explicitly for ...

2

I'm not familiar with that particular book, but any good treatment of the subject should be very helpful to you. It will definitely answer your questions about the Jacobian. And it should also give you a good grounding in the general version of Stoke's theorem of which all the other "equivalents of the fundamental theorem of calculus" are special cases.

2

Volume is non-unique. Therefore, the smooth structure of a smooth, compact manifold is not enough for the manifold to have a canonical volume. What can happen is, that if your manifold has a richer structure, then you can associate a canonical volume to that richer structure. For example, if you have a Riemannian manifold $(M,g)$, then the differential form ...

2

Use Stokes theorem: $$\int_{M} d\omega = \int_{\partial M} \omega$$ Find $\omega_1$ such that $d \omega_1 = \omega$ and reduce integration to $\partial M = \{z=x^2+y^2 = 1\}$. EDIT: This could work if $\omega$ would be exact but it is not so: $d \omega = 3\cdot dx \wedge dy \wedge dz \neq 0$.

2

Well, I will be answering my question for the benefit of people who may search for this exact thing in the future. I appreciate PVAL's valuable comments. In the end, it seems I had to simplify the volume form a bit - after that it all worked out. First of all, I normalize the form $vol_{S^2}$ from my question $$\Omega_2=\frac{1}{4\pi}(\xi^1 \mathrm ... 2 Posting this to sum up all the stuff that came out in the huge comment discussion under Mike's answer, a discussion which needs 4 screenshots, 1 2 3 and 4, to fit in them. Will accept Mike's answer for the patience he must have used to go on with that discussion :). What emerged was the following. I forgot a "maximally" in my contact condition. "maximally ... 2 Here's a fairly detailed sketch: Let M be an n-manifold. Suppose, contrapositively, that M is orientable, and fix a maximal oriented atlas, i.e., a maximal atlas for which all the transition maps have Jacobian with positive determinant. Lemma 1: If (U, \phi) is a chart and U is connected, then (U, \phi) is either compatible with the ... 2 Symplectic geometry exists, so the answer is yes. Take the standard symplectic form on R^{2n} for example (for n=2 this is Micah's answer). https://en.wikipedia.org/wiki/Symplectic_manifold Wikipedia makes the same observation in its discussion of decomposable vectors. https://en.wikipedia.org/wiki/Exterior_algebra 2 You're almost right, but there's an index error in your last term. It should be$$ \dots =\sum_i\left(\dots +\frac{\partial V_i}{\partial u_j}\alpha_i\right). $$The 1-form you're trying to compute is called the Lie derivative of \alpha in the direction V. It should be denoted by L_V(\alpha) or L_V\alpha: Here, \alpha is meant to be the ... 2 If you regard \mathbb C as a real vector space, the identification z\mapsto (\Re z, \Im z) is well-behaved under "any" construction on the vector space \mathbb{R}^2; dualization \mathbb{R}\mapsto \hom_{\mathbb R}(\mathbb{R}^2, \mathbb R), as a functor \mathbf{Vect}^\text{op}\to \mathbf{Vect} is no exception. Indeed, by this very functoriality, the ... 2 Hint: A 1-form \omega in \mathbb{R}^3 looks something like \omega = F(x, y, z)dx + G(x, y, z)dy + H(x, y, z)dz where F, G, H are functions \mathbb{R}^3 \rightarrow \mathbb{R}. Taking one exterior derivative, we find d\omega = (F_y \ dy \wedge dx + F_z \ dz \wedge dx) + (G_x \ dx \wedge dy + G_z \ dz \wedge dy) + (H_x \ dx \wedge dz + H_y \ dy ... 2 As @TedShifrin pointed out, just assuming connectedness of the intersections is not enough. But there is a way to compute de Rham cohomology if you have an open cover with the property that all sets in the cover, and all intersections of finitely many sets in the cover, are contractible. Such a cover is called a good cover, and there always exists such a ... 1 Differential forms are things that live on manifolds. So, to learn about differential forms, you should really also learn about manifolds. To this end, the best recommendation I can give is Loring Tu's An Introduction to Manifolds. Tu develops the basic theory of manifolds and differential forms and closes with a exposition of de Rham cohomology, which ... 1 Mike Miller said you that \omega|_N is the same thing as i^*\omega. Why is it so? (I assume that \omega is a k-form) See first that \omega|_N, by definition of restriction, is just a function and for any x\in N$$\omega|_N(x)\in\bigwedge^kT_x^*\color{red}{M}$$So it is not what we expected to get. Hence by definition$$\omega|_N:=i^*\omega.$$In ... 1 Yes, in spirit this is OK. You should be specifying, however, that U_i = \{x\in S^1: x_i\ne 0\}. And your "then v= ..." is not totally correct, as v could be any scalar multiple of this vector. An alternative approach (more amenable to generalization) is to observe that since S^1 is defined by the equation x_1^2+x_2^2=1, the 1-form$$\frac 12 ...

1

1. question The totally antisymmetric symbol $\varepsilon_{i_1,i_2,\dots,i_n}$ ($=1$ if indexes not permuted) is not a tensor. Usually, it is just a symbol. The total antisymmetry means that it is antisymmetric in all indexes, meaning that it changes sign to $-1$ if the permutation of indexes is odd, and remains $+1$ if the permutation is even, and it ...

1

Following the OP's notations, we write $$g\left(r,\theta\right)=\left(r\cos\theta,r\sin\theta,r^2\right)$$ and then $$g_r\left(r,\theta\right)=\left(\cos\theta,\sin\theta,2r\right),\quad g_\theta\left(r,\theta\right)=\left(-r\sin\theta,r\cos\theta,0\right).$$ Now we compute ...

1

Let v,w be the inverse functions to $(x-1/x)$, $f$ an even function and $a>0$. The proposition follows if the integrals of corresponding intervals are equal : $$\int_{a}^{a+\epsilon} f(y)\, dy=?\int_{v(a)}^{v(a+\epsilon) } f\left(x-\frac{1}{x}\right)\, dx+\int_{w(a)}^{w(a+\epsilon) } f\left(x-\frac{1}{x}\right)\, dx$$ For small $\epsilon$ the LHS ...

1

Possible answer. Does it make sense? $$0=(\omega|_U)_p(v_1,\dots,v_q)=(F^*\omega)_p(v_1,\dots,v_q)=dF^*_p(\omega_{F(p)})(v_1,\dots,v_q)=dF^*_p(\omega_{p})(v_1,\dots,v_q)=\omega_p(dF_p(v_1,\dots,v_q))=\omega_p(v_1,\dots,v_q)$$

1

There's a reason $\dfrac{\omega^n}{n!}$ shows up all over complex geometry as the induced volume form. You are correct and there is an error in whatever you're reading.

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