# Tag Info

0

Here's a hint: Can you see that the standard area 2-form on $S^2$ is $SO(3)$-invariant?

1

How many Euler angles are there? Can you compute new $\theta$ and $\phi$ in terms of them? Can you represent these transformations as $\theta\mapsto\theta+\epsilon$, $\phi\mapsto\phi+\delta$?

4

Recall that an $n$-form associates to each point $p\in M$ a map $\omega_p:(T_pM)^n\to \mathbb R$. Thus for any basis $x_1,\ldots,x_n$ of $T_pM$, we have an associated real number $\omega_p(x_1,\ldots,x_n)$, which is either positive or negative. If we let $x_1,\ldots,x_n$ vary continuously with $p$ such that $x_1,\ldots,x_n$ is always a basis for $T_pM$, then ...

4

First you have to have a good notion of orientation on a vector space $V$. Since the transition matrix from one basis to another is an invertible matrix it has non-zero determinant. Hence, it's either positive or negative. Two bases of $V$ determine the same orientation if the transition matrix has positive determinant. So, an orientation on $V$ is an ...

2

This is equivalent to showing that $$h^0(\Omega[n\cdot p]) > 0$$ for all n > 1. By Riemann-Roch, we have $$h^0(O_X [-n\cdot p]) - h^0(\Omega[n\cdot p]) = -n + 1 - g$$ Since the degree of the divisor is negative, $$h^0(O_X [-n\cdot p]) = 0$$ so, when rearranging, we get: $$h^0(\Omega[n\cdot p]) = n + g - 1$$ If n > 1, then n + g - 1 > 0 for all g.

3

If $X$ is an $n$-dimensional manifold with boundary then for any $n-1$-form $\mu$ with compact support, $$\int_Xd\mu=\int_{\partial X}\mu.$$ This is Stokes' theorem. If there is no boundary, i.e., if $\partial X=\emptyset$, then the boundary integral is zero, and hence so is the other integral. But otherwise, you cannot expect the integral to vanish.

3

You might also think about this: vector fields are in the dual space of forms. I'll exemplify this in $\mathbb{R}^3$. A vector field is a function over a region of three dimensional space that gives at each point a vector. For a point $P=(x,y,z)$ let the the vector field $V$ associates to $P$ the vector $$V(P) = ... 1 A different way to see that it's not possible is to read what it means for a form to vanish on a submanfold. When you have a submanifold W\subseteq M and an \omega\in \Omega^k(M), then \omega vanishes on W when the pullback by the embedding map \iota\colon W\to M is zero: \Omega^k(W)\ni\iota^*\omega=0. But pullback commutes with wedge product, ... 3 It's not possible. I assume you're just thinking about multilinear algebra since you say "subspace," but the reason I will give below translates easily to differential forms on manifolds. By the definition of the wedge product, we have that for 1-forms \omega and \phi on a vector space V,$$(\omega \wedge \phi)(u, v) = \frac{1}{2}(\omega(u)\phi(v) - ...

-1

The notation between the dollar sign is LaTeX (http://www.latex-project.org/) mathematics notation. It is meant to be very descriptive of mathematical expressions; it is not intended to "look" like standard mathematical notation. The closest equivalent in Mathematica itself is: Dt [Congruent] ([PartialD]/[PartialD]x)[Wedge]t

6

I don't think so! In fact you should set an axillary equation regarding to the OE as $$m^4-m^3=0$$ and so solve it: $$m=0,~~~\text{3 times}~~~m=1$$ and then write the proper solution as $$y=C_1+C_2x+C_3x^2+C_4\text{e}^x$$ Now put the initial conditions. I don't think you get another one but the Andrea's one.

3

There is no calculation needed. The family of solutions consists of linear combinations of $4$ functions. The initial conditions make all the constants $0$, so the solution is the identically $0$ function. Another way of thinking about it is that the solution of the initial value problem is unique. But identically $0$ works, so it is the only solution.

0

To begin with, what is $\theta$? The fact that $d\theta$ is the notation used for a particular $1$-form does not mean that we have such a thing as "$\theta$". After all, $d\theta$ could be just that, a convenient piece of notation. But yes, it's actually more than that... Suppose we have a family of functions $\theta_\alpha$, each of which is $C^\infty$ ...

1

Before discussing the boundary of a manifold, perhaps it would be best to look at the case with embedded hypersurfaces in general, and for this I will be following Lee's Intro to Smooth Manifolds. Let $M$ be a smooth, oriented, $n$-manifold and $S$ be an embedded ($n-1$) hypersurface. Let $\omega$ be an orientation form on $M$, so that $\omega$ is just a ...

1

You are correct that the question does not make sense.

1

First, you should tell us what $\omega$ is!! I am going to guess that it is your volume/area form. The point is to use the nowhere-vanishing property of this top-dimensional form. On your surface, a tangent vector $v_1$ to $z=y^2$ is only in the $z$ and $y$ directions. So we have $dx\wedge dy(v_0,v_1)=c\cdot dx(v_0)$ which needs to be nonzero. Thus $v_0$ ...

1

A "de-Rahm k-form" is an equivalence class of closed k-forms where two closed k-forms are in the same equivalence class if their difference is an exact k-form. But usually when dealing with equivalence classes you work in practice with a representative of that class, so a de-Rahm k-form will probably mean in your case some representative of such an ...

1

If you're in $\mathbb R^n$ (or on a Riemannian manifold) the $1$-form $\omega = \sum F_i dx_i$ naturally corresponds to the vector field $F =(F_1,\dots,F_n)$. But what I think you have in mind is to associate to $\omega(p)$ the hyperplane it annihilates, i.e., $V_p =\{v\in\mathbb R^n: \omega(p)(v)=0\}=\ker\omega(p)$. When $n=2$, this is a line field on the ...

0

I'm going to assume that you are working in Euclidean space $\mathbb{R}^n$. Given a vector space $V$ over $\mathbb{R}$, the dual space $V^*$ is defined to be the space of linear maps $V \to \mathbb{R}$ (check that $V^*$ is in fact a $\mathbb{R}$-vector space). The vectors at a given point $p \in \mathbb{R}^n$ form a vector field $V_p \cong \mathbb{R}^n$, ...

2

part 1) The differential forms approach is indeed very powerful, what Hestenes points out in his "From Clifford Algebra to Geometric Calculus" is that to give a complete treatment of differential geometry of manifolds you need various structures. In the book you will find an alternative. The starting point (as was pointed out above) is the notion of a ...

0

Note that $$\omega=\sum\limits_{k=1}^n x_{2k-1}\wedge x_{2k}$$ so $$\omega^n=\sum\limits_{k_1=1}^n\ldots\sum\limits_{k_n=1}^n x_{2k_1-1}\wedge x_{2k_1}\wedge\ldots\wedge x_{2k_n-1}\wedge x_{2k_n}$$ Summands here are non-zero iff $k_1,\ldots,k_n$ are all distinct numbers, so summands in bijective correspondence with permutations of numbers $1,\ldots, n$. ...

4

$\Omega^\bullet(M) = \bigoplus_{k=0}^{\dim M} \Omega^k(M)$ is the space of all differential forms, which is $\mathbb Z$ graded by degree.

0

Let $c$ be a one-dimensional submanifold of $\mathbb{R}^d$ with embedding $\iota: c \rightarrow \mathbb{R}^d$, $d \geq 2$. Let $\omega = f\mathrm{d}x + g\mathrm{d}y$ be a differential 1-form on $\mathbb{R}^d$, where $x,y$ denote the first two coordinates. Then $\int_c f\mathrm{d}x + g \mathrm{d}y := \int_c \iota^* \omega$, and the latter is defined via ...

1

By Stokes' theorem we have $\int_{\partial S} \omega = \int_S d \omega$, so the problem is whether a given volume form is exact, or equivalently whether it's zero in top de Rham cohomology. On a compact oriented Riemannian manifold, a volume form is never exact (in fact it generates top de Rham cohomology), but in the noncompact case this need not be true. ...

0

Does this in fact trivially follow from what I already showed? If we take the $v_i$ to all be $[dx_1,\dots,dx_n]^T$, then $\det[Δf_1(x),…,Δf_{n−k}(x),v_1,…,v_k]$ is certainly a form-field. The only remaining question is whether this always has a consistent sign.

0

Do you know about the (Hodge) star operator? This is a great way to get from an ($n-k$)-form to a $k$-form. This will accomplish what you've sketched.

1

It is true that your differential $$\omega=Pdx+Qdy:={y\over x^2+xy+y^2}\ dx-{x\over x^2+xy+y^2}\ dy$$ satisfies the integrability condition $Q_x-P_y\equiv0$. Therefore in simply connected subsets $\Omega\subset G$ there exist potential functions $f:\ \Omega\to{\mathbb R}$ with $df =\omega$, or $\nabla f=(P,Q)$. But there is no global $f:\ G\to{\mathbb R}$ ...

-1

You can write your $\omega$ $$\omega=\frac{y}{x^2+xy+y^2}dx-\frac{x}{x^2+xy+y^2}dy$$ For integral it has the form $$\int\bigg(Pdx+Qdy\bigg)$$ which you can test for beeing conservative such as $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$ where $$\frac{\partial P}{\partial y}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$ \frac{\partial ...

0

You may wish to consider the limit as $y\to 0$ of your function. This limit depends on whether $x>0, x=0,x<0$ and $y>0,y<0$. Is it possible to define a value anywhere along the line $y=0$ and keep continuity? (Remember that you can choose which arctan to take freely!) If so, how many places can you make a consistent definition? The question is: ...

2

You can prove this by proving the contrapositive. Suppose $\omega$ is not decomposable. That means that $\omega$ necessarily must be the sum of two (and only two) decomposable terms, like $\omega = x \wedge y + z \wedge w$. Consider what happens if you add a term like $y \wedge z$. You should realize that you can lump that into one of the two terms and ...

2

Well, the best way is to start by the definition. Let's begin with $1$-forms so that we can understand what's the reasoning behind it. Recall that $\omega \in \Omega^1(U)$ is simply a way to associate at each point a linear function of one single vector. In other words if $p \in U$ then $\omega(p) \in T^\ast_p U$. Our problem is: $\varphi$ transforms $U$ ...

2

One problem with your approach is that your basis is too generic, which is inconvenient for computations. Try some special basis: e.g. Combine $a_1e_1 \wedge e_2 + a_2 e_1 \wedge e_3 + a_3 e_1\wedge e_4$ as $e_1 \wedge (a_1e_2 + a_2e_3 + a_3e_4)$. Then let $e_2' = a_1e_2 + a_2e_3 + a_3e_4$ we can replace the sum of first three terms as $e_1 \wedge e_2'$. ...

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