# Tag Info

## New answers tagged differential-forms

0

I think you're onto something that is fully intended. edit: here's a quote I think could be relevant source on google books

1

Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S$where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity ...

0

You can imitate the argument that shows $\pi_1(S^n)=0$ if $n>1$ by removing a point! Try it!

3

I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular... P.S. Traying to write ...

-1

I have found the solution (with the help of Mathoverflow users). See here: http://mathoverflow.net/questions/239275/is-there-a-matrix-that-converts-the-gradient-of-every-possible-function-to-gradi

1

Looking at this a different way, $\omega=g^\ast\omega$ means that pointwise: $$(g^\ast\omega)_p=g^\ast(\omega_{g\cdot p})=\omega_p.$$ For a vector field $X$, $$(\omega(X))(p)=\omega_p(X_p)$$ On the other hand, $$(\omega(X))(p)=((g^\ast\omega)(X))(p)=(g^\ast\omega)_p(X_p)=(g^\ast(\omega_{g\cdot p}))(X_p)=\omega_{g\cdot p}(g_\ast X_p)=\omega_{g\cdot ... 3 \omega=g^\ast\omega means that for all p\in M, \omega_p=g^\ast(\omega_{g(p)}). Now, to consider your statement that \omega(X)=g^\ast\omega(X), we consider the statement pointwise. Suppose for now that the action is transitive and fix p\in M. Now consider (\omega(X))(q) for any q\in M. By transitivity, there is some g\in G such that ... 1 The exterior derivative operator d is real (i.e., real-valued on real-valued forms) and complex-linear (i.e., if \alpha and \beta are real p-forms, then d(\alpha + i\beta) = d\alpha + i\, d\beta, so yes, your calculation is correct. To deduce the Cauchy-Riemann equations from Green's theorem, it's convenient to differentiate the holomorphic ... 0 You take$$(f^{-1})^*i^*dx_j = (i\circ f^{-1})^*dx_j.$$Of course, i\circ f^{-1}:f^{-1}(S^2\cap\{x>0\})\to\mathbb{R}^{n+1} is given by$$(x_1,\ldots,x_n)\longmapsto\left(\sqrt{1-(x_1^2+\ldots+x_n^2)},x_1,\ldots,x_n\right).$$Now we have to compute the pullback of dx_j by this map (which we will call \phi from now on). To do this you can proceed as ... 0 Hint: Use the following invariant formula for the exterior derivative of a 1-form \theta:$$ d\theta(X,Y) = X(\theta(Y)) - Y(\theta(X)) - \theta([X,Y]). $$1 Let's start with equation (120). We have two assumptions, first that \omega is closed, d\omega = 0, and second that \omega \in Y_p, that is, in$$\omega = \sum_I f_I(x)\,dx_I$$all k-indices I only involve dx_i for i \leqslant p. We compute$$d\omega = \sum_I (d f_I)(x)\wedge dx_I = \sum_I \sum_{r = 1}^n (D_r f_I)(x)\, dx_r \wedge dx_I = ...

0

Hint: It is possible to find $f_i$ locally and show the identity. You may assume that on a contractible open subset $U$ of $M$, $d\theta_i=0$, the Poincare Lemma implies that on $U$, $\theta_i=df_i$, on $U$, $X_j(f_i)=\delta_{ij}$.

4

For $k < 2$ and $k > \dim V - 2$ it's always true that a $k$-form $\omega$ both is decomposable and satisfies $\omega \wedge \omega = 0$. For $k = 2$ (and provided the field underlying $V$ does not have characteristic $2$), the condition $\omega \wedge \omega = 0$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.) ...

1

In terms of the multilinear algebra, it looks like you're on the right track. We extend $\omega$ to a basis $\omega, v_2, \dots, v_n$ and then write $$\alpha = \sum a_I \omega \wedge v_{i_2} \wedge \dots \wedge v_{i_k} +$$ $$\sum b_J v_{j_1} \wedge \dots \wedge v_{j_k}$$ Then as you've noted, wedging with $\omega$ kills everything in the first term. ...

1

If $M$ is a manifold of dimension $n$, then for all $p > n$, $\Omega^p(M) = \{0\}$. In particular, as a point is a zero-dimensional manifold, $\Omega^p(\ast) = \{0\}$ for all $p > 0$.

1

Note that $$d\left(\frac{dz}{z}\right) = d\left(\frac{1}{z}dz\right) = \frac{\partial}{\partial z}\left(\frac{1}{z}\right)dz\wedge dz + \frac{\partial}{\partial\bar{z}}\left(\frac{1}{z}\right) d\bar{z}\wedge dz = 0$$ as $dz\wedge dz = 0$ and $\dfrac{1}{z}$ is holomorphic. Therefore $\dfrac{dz}{z}$ is closed. In general, a complex one-form on an open ...

1

I think this becomes easier by putting less emphasis on coordinates. View your forms a defining for each $x\in\mathbb R^n$ a linear map $\omega_x:\mathbb R^n\to\mathbb R^k$. By the assumption on linear independence, each of the maps $\omega_x$ is surjective. Now take a point $p$, choose a basis $\tilde X_1,\dots,\tilde X_{n-k}$ for $ker(\omega_p)$ and ...

0

You're completely correct with everything you said. To show what you asked for, use the map $\mathbb R^n -0 \to S^{n-1}$ which collapses the rays $\mathbb R_+x$. Now there are two ways; one elementary and one more algebraic (still being elementary if you're not purely a differential topologist). 1) Notice that the inclusion $S^{n-1} \to \mathbb R^n-0 \to ... 1 Cargo's answer doesn't seem to give any geometric intuition for the statements he/she claims; let me try and remedy that. Let's just talk about$\mathbb{CP}^n$. There are ways of calculating$H^{\bullet}(\mathbb{CP}^n;\mathbb{Z}) = \mathbb{Z}[x]/x^{n+1}$, where$x$has degree 2 (cellular homology + universal coefficients gives you the module structure; ... 2 Your De Rham cohomology algebra is$H^*(\mathbb P^3_\mathbb C)=\mathbb R[w]/(w^4)$, where$w=[\omega]$is the class of some closed$2$-form$\omega$. We then have$0\neq w^2=[\omega \wedge \omega]$, so that$\omega\wedge \omega$is not exact and a fortiori$\neq 0$. 0 I think that the biggest problem comes from the subtlety of the notation. First of all, there is a general formula for the derivative, and that is usually represented as$df\over dx$or$f^\prime(x)$. There is also the notion of a derivative at a point$x_0$, and that is represented as${df \over dx}|_{x=x_0}$, When you see the manipulation of ... 3 A differential is a$1$-form. At each point a$1$-form gives a linear functional in a tangent space, and the level sets of a linear functional are parallel affines subspaces. Now$df$in$\mathbb R^3$gives a sort of a ruling in each tangent. For a given$p$, the level sets of$df_p$are what we can see if we zoom in on the level sets of$f$in a ... 1 Hint: Use a local chart. Let$U\subset X$be a trivialization of$T^*X$,$T^*U=U\times R^n$with local coordinates$(x_1,...,x_n,y_1,..,y_n)$. You can write$d\alpha_X=\sum_{i=1}^{i=n}dx_i\wedge dy_j$. You have$\pi^*\mu= \sum u_{jk}dx_j\wedge dx_k$. This implies that$(d\alpha_X+\pi^*\mu)(\partial x_i,\partial y_j)=d\alpha_X(\partial x_i,\partial y_j)$. 3 Meditating on the p.d.e. for a minute, we can see that any function of the form$a(x, y) = b(xy)$satisfies$x a_x = y a_y$, and this is (I think) the general solution, but I don't see offhand how to prove this. Instead, here's another strategy for finding the glide-invariant$2$-forms that avoids solving a p.d.e.: Computing gives$\mathcal L_V (dx \wedge ...

0

I don't have Rudin at hand, but in the hope a picture is worth a thousand words: You presumably have a standard $2$-simplex $Q^{2}$, and are partitioning the unit square $I^{2}$ as shown: If $\omega$ is a $2$-form defined on the image $\Phi(I^{2})$, the integral breaks into two summands, one for each simplex, each an integral over the standard simplex. A ...

0

Let $r = x^2 + y^2$. Note that if the values are $a = \frac{x}{r}$ and $b = \frac{y}{r}$, then we will have: $w = \frac{x}{r}$ $dx$ + $\frac{y}{r}$ $dy$. Then, the following follows: $w(X) = \frac{1}{r} (-xy +xy)$ = $0$,$w(Y) = \frac{1}{r} (x^2+y^2)$ = $\frac{1}{r} r = 1.$ This contradicts the conditions given ($w(X) =1$, $w(Y) =0$). Thus, the solution in ...

0

Once you know the Frobenius Theorem, you know that the integral manifolds of $\omega = 0$ are (locally) given by level surfaces of some function $f$. This means that $df = \mu\omega$ for some nonzero function $\mu$.

1

We first note that $\omega(\gamma(t))$ (the $1$-form $\omega$ evaluated at $\gamma(t)$) is simply $\omega(\gamma(t))\gamma'(t)\,dt=\dfrac{x(t)dy-y(t)dx}{x(t)^2+y(t)^2}$. But $dx,dy$ map $\gamma'(t)=(x'(t),y'(t))$ to its components, so $\omega(\gamma(t))\gamma'(t)=\dfrac{x(t)y'(t)-y(t)x'(t)}{x(t)^2+y(t)^2}.$ Hence one regains the 'calc 3' notion of $(x,y)$ ...

3

When you have a binary operation $\newcommand{\bop}{\mathop{\scriptstyle\top}}\bop \colon S\times S \to S$, that induces a corresponding operation on the space of functions $D \to S$ by applying the operation pointwise, $$(f \bop g)(x) := f(x) \bop g(x).$$ The pointwise sum or product of real-valued functions are very familiar examples. The same ...

1

This is actually just calc 3, substitute $x=\cos t$ $y=\sin t$ and use $dx=-\sin t dt$ etc.It all simplifys to $\int dt$.

3

If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ... 2 The mistake is subtle. While - for sufficiently regular$g$- we have $$\int_{Q^k} g(u_1,u_2,u_3,\dotsc, u_k)\,d\mathbf{u} = \int_{Q^k} g(u_2, u_1,u_3,\dotsc,u_k)\,d\mathbf{u},\tag{\ast}$$ the$u_j$in the "denominator" of$\dfrac{\partial (\sigma_{i_1},\dotsc, \sigma_{i_k})}{\partial (u_1,\dotsc, u_k)}\$ are not arguments of the integrand, they just ...

0

Intuitively they are same and can be taken as simple. I state about how it appears to me. We consider small scalars and worry more about what is to be done among them in group operations rather than be bogged down what it deeply and definitively is.I say this as you are comfortable at advanced level and not so at the fundamental level. The Pythagoras ...

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