# Tag Info

0

Yes, @RealAnalysis, it's precisely that. Have you learned about pullback of differential forms? That's what either formula gives you.

1

You'd write it as \begin{align} \operatorname{Alt}(\phi_1\otimes\phi_2\otimes\phi_3) = \frac{1}{6}(&\phi_1\otimes\phi_2\otimes\phi_3 + \phi_2\otimes\phi_3\otimes\phi_1 + \phi_3\otimes\phi_1\otimes\phi_2\\ &- \phi_1\otimes\phi_3\otimes\phi_2 - \phi_2\otimes\phi_1\otimes\phi_3 - \phi_3\otimes\phi_2\otimes\phi_1) \end{align} More generally, for ...

2

It is $${\rm Alt}(\phi_1\otimes\phi_2\otimes\phi_3)= \phi_1\otimes\phi_2\otimes\phi_3-\phi_1\otimes\phi_3\otimes\phi_2+\phi_2\otimes\phi_3\otimes\phi_1-\phi_2\otimes\phi_1\otimes\phi_3+\phi_3\otimes\phi_1\otimes\phi_2-\phi_3\otimes\phi_2\otimes\phi_1$$ but some other will define $${\rm Alt}(\phi_1\otimes\phi_2\otimes\phi_3)=\frac{1}{6}( ... 3 d\omega=2x\,dx\,dy\,dz+2y\,dy\,dz\,dx+2z\,dz\,dx\,dy=2(x+y+z)\,dx\,dy\,dz\neq0 Hence \omega is not closed, therefore not exact. 1 If you restrict the inner product on \mathbb{R}^3 to the sphere \mathbb{S}^2, the result is a Riemannian metric. Every Riemannian metric comes with a canonical top-dimensional form, the volume form, which tells you which bases for a tangent space have unit volume. This 2-form is the volume form for the metric on the sphere. It's called an "area form" ... 4 The answer is yes, and no, and maybe. In sensitivity analysis, we can always run a model at different parameter values and use finite differencing methods to compute the directional derivatives in our parameter space. This is, as you might imagine, of limited use without any other external information. First, it requires many runs. Second, it only gives us ... 1 Any two-form gives a (skew-symmetric) bilinear form TM_p \times TM_p \rightarrow \mathbb{R}. The rank of the two-form at p is the rank of this bilinear form. What's the rank of a bilinear form? A bilinear map B: V \times V \rightarrow \mathbb{R} is equivalent to a linear map V \rightarrow V^* given by v \mapsto B(v,\cdot). The rank of this ... 3 I don't know where you're getting the (2k+1)-form. If you wedge \omega with itself k times, you get — as you yourself observed — a 2k-form. Let's try an example: With n=2 and \omega = dx_1\wedge dx_2+dx_3\wedge dx_4, we'll have \omega\wedge\omega = 2 dx_1\wedge dx_2\wedge dx_3\wedge dx_4. Since 2-forms commute with one another, you should be ... 0 It really depends on what you mean by the vanishing of \omega at a point p. Do you mean that \omega(p)(v,w) = 0 for all v,w\in\Bbb R^3, or do you wish to find a submanifold \alpha\colon M\hookrightarrow \Bbb R^3 with the property that \alpha^*\omega = 0 (i.e., \omega(p)(v,w)=0 for all p\in M and v,w\in T_pM)? For example, consider the ... 2 Comments: Your method seems to be a bit too complicated. It's possible that part of the reason you seem to be confused is that you're writing D^2 for two different things: One, the disc of radius 1 inside \mathbb{R}^2; this is the D^2 that appears in your statement of the problem and defintions etc. However, in the attempt you write up, you also ... 0 I believe this is just a consequence of Stokes' Theorem. First of all, all n-forms on an n-dimensional manifold M are closed, since H^{n+1}(M) = 0 In case this is not obvious: If \beta is an n+1 form then for each p \in M, \beta_p acts on n+1-elements of T_pM. Since \dim T_pM = n then necessarily any n+1 vectors are linearly dependent, ... 1 Starting with the vector \vec r given as$$\vec r = a (\sin \theta, \cos \theta) + (L−a\theta)(\cos \theta,− \sin \theta)$$we directly differentiate to obtain the differential d\vec r as$$d\vec r = a \frac{d}{d \theta}(\sin \theta, \cos \theta)d\theta + \frac{d}{d\theta}\left((L−a\theta)(\cos \theta,− \sin \theta)\right)d\theta$$Now, we use the ... 1 No, you cannot do that as the exterior derivative is not linear over functions. In general, if f is a smooth function and \omega is a k-form, d(f\omega) = df\wedge\omega + fd\omega which you'll note is not the same thing as fd\omega. 4 Logarithmic differentiation. Given y=f(x)>0, we can take \ln y = \ln f(x), and find that \dfrac{dy}{dx} = f(x) \dfrac{d}{dx} \ln f(x). However, if |y|=|f(x)|, then we obtain \ln |y| = \ln |f(x)|, and then \dfrac{d}{dx} \ln |y| = \dfrac{y}{|y|^2}y' =\dfrac{y'}{y}. So in fact whether f(x) is positive or negative is of no concern. This can ... 2$$\frac{d(u^v)}{dx} = \frac{d(e^{v \log u})}{dx} = e^{v \log u}\frac{d(v \log u)}{dx} = u^v (\frac{dv}{dx}\log u + v\frac{du/dx}{u})\frac{d(\log_u v)}{dx} = \frac{d}{dx}\left( \frac{\log v}{ \log u} \right)\arccos x = \frac{\pi}{2} - \arcsin x \therefore \frac{d}{dx}\arccos x = - \frac{d}{dx}\arcsin x \text{, and so on with $\arctan$ and ...

0

I think there is an importan point that has been overlooked in the above answers: The exterior derivative is the only linear natural operator in the list. This is explained with several variations the book by Kolar, Michor and Slovak cited in Yuri Viatkin's answer. The Lie derivative is also natural under general diffeomorphisms but only as a bilinear ...

6

Since the tangent bundle of a manifold $M$ is the disjoint union of tangent spaces $$TM=\bigcup_{p \in M}T_pM$$ usually a tangent vector is denoted by $(p,v) \in TM$, where $p \in M$ is a point and $v \in T_pM$ is a vector. This notation has the purpose to remember that the vector $v$ is tangent to $M$ in the point $p$. Now, it seems that the correct ...

2

A 2-form is a smooth choice of a skew-symmetric bilinear form on each of the tangent spaces of your manifold. A 2-form is non-degenerate if it is non-degenerate (as a bilinear form) when restricted to each tangent space. For a general introduction to differential forms and smooth manifolds see Lee's Introduction to Smooth Manifolds. For a general ...

4

Here's an attempt to make the "dinaturality" observation (more) precise. I will be leaving out many details that I haven't worked out, so I may go wrong somewhere; I hope the general outline makes sense though. Let $\mathcal N$ be the poset of natural numbers in the usual ordering (or $\mathcal N$ could be the universal chain complex, i.e. category with ...

0

In definition of dinatural transformation you have linked, there is the following: Let $F, G: C^{op} \times C \to D$ be functors. A dinatural transformation from F to G (...) Honestly I can't find such vast functors. But as for standard naturality first we need two parallel functors $(F,G:C\rightarrow D)$. In our case there is one firmly settled ...

1

Actually $D_1=C_1-C_2$ and $D_2=C_1-C_1=0$.

0

(1) If $\gamma(t)=(x(t),y(t))=r(t)(\cos\ s(t),\sin\ s(t))$ then on $\gamma$, $$\omega = \frac{xy' -yx' }{x^2+ y^2} dt = s'(t)dt$$ Consider a curve $\gamma$ with $$s(0)=0,\ 2\pi > s(t)>0\ (0<t<2\pi ),\ s(2\pi)=2\pi,$$ That is, this curve meets positive $x$-axis at $t=2\pi$ first (except starting), and maybe $r(0)\neq r(2\pi)$. Then  ...

2

Like GFR, I also have no interest in an extended discussion. This post is meant to be an extended comment for the sake of adding perspective. Personally, I like to define the exterior derivative $d$ in a coordinate-free, invariant way: Fact/Def: Let $M$ be a smooth manifold (possibly with boundary). Then there are unique operators $d \colon ... 1 I am not going to get engaged in a full rebuttal, for no other reason that it would take too much time/efforts, but a few points: The symbol$\nabla$is usually used for covariant derivatives, so$d= \partial\wedge\$ would be better. In fact I have seen this notation used somewhere, but cannot remember where. So yes, you could use this notation, personally ...

Top 50 recent answers are included