# Tag Info

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.

0

If we denote inclusions $$I_{n_0 \dots n_p}^{n_i}:U_{n_0 \dots n_p}\hookrightarrow U_{n_0\dots \widehat{n_i} \dots n_p}\hspace{5pt}\text{and}\hspace{5pt}I_{n_0 \dots n_p}^{n_in_j}:U_{n_0 \dots n_p}\hookrightarrow U_{n_0 \dots \widehat{n_i} \dots \widehat{n_j} \dots n_p},$$ then $$\delta:C^p(\mathfrak{U},\Omega^k)\rightarrow C^{p+1}(\mathfrak{U},\Omega^k)$$ ...

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 ...

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 $R\colon U\to SO(n)$.

0

Of course, everything Michael Hardy says is right. Maybe the following will also help. If $f: \mathbb{C} \to \mathbb{C}$ is a smooth function, then we can talk about $df$. Here $df$ will take as input a real tangent vector to $\mathbb{C}$ and output a complex number. If you like, $df$ gives an element of $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C}, \mathbb{C})$ ...

0

We have that, $$V(W(f))(p)-W(V(f))(p)=[V,W](p)(f)=df(p)[V,W]=0$$ so, $$V(W(f))(p)=W(V(f))(p)$$

0

By definition of differential, for any vector field $W$, we have: $$W_p f=df_p W_p=0$$so your form is not only symmetric, it is identically null. My intuition is that a vector is a sort of partial derivative. But the total differential vanishes, so any partial derivative must also be zero.

0

Maybe a bit late, but how about this construction? By definition, $\Omega_{A/R} = I/I^2$, where $I= \ker[A\otimes_RA\rightarrow A, a\otimes b\mapsto ab]$. We have a direct sum decomposition $(A\otimes_RA)/I^2\cong A \oplus \Omega_{A/R}$ induced by $a\otimes b\mapsto (ab, a\cdot db)$ (where $d\colon A\rightarrow \Omega_{A/R}$, $a\mapsto (1\otimes a - a\otimes ... 2 Here's a fairly detailed sketch: Let$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 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 ... 0 Since$X_V = 0$, the equation is true at any point in$V$. Since$g = 0$outside$V$, the equation reduces to$X = (1-0)X$away from$V$, so it is true at any point not in$V$. 0 Since nobody answered me, I will try myself. I have to prove that if$\omega$is a volume form, then $$Vol(M):=\int_M 1=\int_M\omega>0$$ where the second equality follows by definition as well and$M$is taken positively oriented w.r.t$\omega$, i.e. for a local chart$\omega=\lambda dx_1\wedge\dots\wedge dx_n$and$\lambda(p)>0$for every$p$in the ... 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 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 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. 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 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 ... 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. 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] = ...

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. ...

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 ...

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 ... 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 ... 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 ... 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 $$\int_D (-x^2-y^2)dxdy = \int_D (-r^2) rdrd\theta = 2\pi \frac{-r^4}{4}\bigg|_0^1 = -\frac{\pi}{2}$$ 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$. 3 This is false. The contact condition is often known as being maximally non-integrable. Given an integrable distribution$\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 ... 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 ... 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 $$\Omega_2=\frac{1}{4\pi}(\xi^1 \mathrm ... 0 In response to a comment you posed:$$x_{j}=\sum_{i=1}^{4}a_{ji}y_{i},\;\;\;j=1,...,6$$Pulling back the given form I obtained$$B^{*}v=\sum_{k,i=1}^{4}(a_{1i}a_{4k}+a_{2i}a_{5k}+a_{3i}a_{6k})y_{i}\wedge y_{k}$$To make this vanish we respect the antisymmetric nature of the wedge product. All we need is for a_{1i}a_{4k}+a_{2i}a_{5k}+a_{3i}a_{6k} to be ... 0 Let me try to sum up the comments and understand at least the first half of the proof, i.e. how we get from a Liouville field to a contact form. First of all, we need two things. Firstly, \iota(X) is not a map applied to X, but rather the interior product, or interior derivative, of \omega and X. Given a field X, the interior derivative of a form ... 0 Let y_{k}=x_{k}=x_{k+3}, k=1,2,3$$B^{*}v=dy_{1}\wedge dy_{1}+dy_{2}\wedge dy_{2}+dy_{3}\wedge dy_{3}=0$$Now, there are a few remarks to be made. The image of \mathbb{R}^{3} under the map is not \mathbb{R}^{6} but this should be expected. The best we could hope for by such a linear map is that the image is a 3-dimensional subset of ... 3 I think you are encountering an issue with your calculation due to a misreading of the problem. The way I understand it, the question is, for any 2 form \omega, does there exist a basis \sigma_i such that$$\omega(u, v) = \sigma_1\wedge\sigma_2(u, v)+\dots +\sigma_{2r-1}\wedge\sigma_{2r}(u, v)$$In your example, it is fine to start with a basis ... 1 Yes, in spirit this is OK. You should be specifying, however, that U_i = \{x\in S^1: x_i\ne 0\}. And your "then v= ..." is not totally correct, as v could be any scalar multiple of this vector. An alternative approach (more amenable to generalization) is to observe that since S^1 is defined by the equation x_1^2+x_2^2=1, the 1-form$$\frac 12 ... 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 ... 0 Okay, figured it out. Normally we have two different arguments delta1 x and delta2 x. They are both vectors and represent the sides of a parallelogram extending from our base point. So if delta1 x = delta2 x then the sides are the same and so the parallelogram is "collapsed" and has no area at all. Then it is clear there should be no "flux" through such ... 0$\Delta x\wedge \Delta x =0\omega(\Delta x\wedge \Delta x)=\omega(0)=\omega(0 \Delta x)= 0\omega(\Delta x)=0$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 ...

0

The wedge product is define as: let $\omega\in \Omega^k(V)$ and $\tau\in \Omega^l(V)$; then $$\omega \wedge \tau=\frac{1}{k!l!}A(\omega\otimes\tau);$$ where $$A(\omega)(x_1,...,x_k)=\sum_{\sigma\in S_{k}}sgn(\sigma)\omega(x_{\sigma(1)},...x_{\sigma(k)})=\sum_{\sigma\in S_{k}}sgn(\sigma)\sigma\omega.$$  LEMMA: If $\omega\in\Omega^k(V)$, ...

1

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

0

Weird notation, usually people write $F^* \omega = \omega \circ F$... which is $\omega | _U$ in this case since $F|_U = id$... Your proof of the one direction doesn't make sense, how did the exterior differential come into play? Nevertheless, note that $F$ is the inclusion map, so we obviously have $F(p) = p$ since $p \in U$. Thus $$(F^* \omega)_p = \omega ... 3 If \omega=\alpha\wedge\beta, that is \omega is decomposable, then \omega\wedge\omega=0. For n\leq 3 the wedge product of any two form with itself vanishes, so we have chances to prove that any 2-form is decomposable. For n\geq 4 you can check that if \omega=e_1\wedge e_2+ e_3\wedge e_n you have \omega\wedge \omega\neq 0, hence there are forms ... 3 For \omega\in\wedge^k V, the minimal number s such that \omega has an expression$$ \omega=\omega^{(1)}+\dotsb+\omega^{(s)} $$where each \omega^{(i)} is of the form$$ \omega^{(i)}=\omega_1^{(i)}\wedge\dotsb\wedge\omega_k^{(s)} $$is called the rank of \omega. Now, if \{e_1,\dotsc,e_n\} is a basis for V, then every \omega\in\wedge^2 V has ... 2 Comments to the question (v2): One can define left and right exterior derivatives$$ d_L(\omega\wedge\eta)~=~(d_L\omega)\wedge\eta + (-1)^{|\omega|}\omega\wedge d_L\eta \tag{L},  d_R(\omega\wedge\eta)~=~(-1)^{|\eta|} (d_R\omega)\wedge\eta + \omega\wedge d_R\eta \tag{R},  d_R\omega~=~(-1)^{|\omega|}d_L \omega, \tag{C} where \$\omega, ...

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