Tag Info

Hot answers tagged

6

One important thing to notice is that the assignment $M \mapsto \Gamma(M, TM)$ does not define a functor on the category of smooth manifolds and smooth maps, because tangent vector fields do not behave well under pushforwards or pullbacks by smooth maps. On the other hand, the assignment $M \mapsto \Gamma(M, \bigwedge^k T^*M)$ defines a contravariant functor ...


5

Nondegeneracy means precisely that $\underbrace{\omega\wedge\dots\wedge\omega}_{n\text{ times}}$ is nowhere $0$, so this is, by definition, a volume form. It follows in general that the integral is nonzero: The manifold must be orientable (use the nowhere-vanishing $2n$-form to decide whether a basis is "positive" or "negative") and then, assuming $M$ is ...


4

There are several things related, let me summarize a little bit. First of all, the one form $\cos \varphi d\theta$ is not defined on $\mathbb R^3\setminus\{0\}$, but only on $\mathbb R^3\setminus \{x=y=0\}$. That's because polar coordinate is generally not defined at $\{x = y =0\}$. You may, of course, have the same objection to $\omega = \sin\varphi ...


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 ...


4

In complex manifold theory, I think the most common convention is to use $\Omega^p(M)$ for the space of holomorphic $p$-forms, and some other notation like $\mathscr A^p(M)$ and $\mathscr A^{p,q}(M)$ (or $\mathscr E^p(M)$ and $\mathscr E^{p,q}(M)$) for smooth forms. But if you want to stick with $\Omega^p(M)$ and $\Omega^{p,q}(M)$ for smooth forms, one ...


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.


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}( ...


2

You have things sort of confused. For a $2$-form $\omega$, represented as $\omega = dt\wedge\omega_1 + \omega_2$ (I prefer to put the $dt$ first to spare sign problems), we define $$I(\omega) = I(dt\wedge\omega_1) = \sum_{i=1}^n\big(\int_0^1 a_i(x,t)dt\big)dx_i.$$ However, to prove your formula, we need to define the operator $I$, in general, on $p$-forms, ...


2

When $n>1$, $S^n$ is simply connected, and $H^1_{dR}(M) = 0$ for any simply connected manifold. (Hint: For any closed $1$-form $\omega$ and any closed curve $\gamma$, we have $\displaystyle\int_\gamma\omega = 0$.) Now, in order to deal with $\Bbb RP^n$, consider the $\Bbb Z/2\Bbb Z$ action given by the deck transformations. Show that any exact invariant ...


2

I think the point here is that an n-form which is non-zero everywhere defines a volume form and can be integrated to give a volume for the manifold which is positive. If you integrate an exact form you get zero, since there is no boundary. Hence the exact form cannot be a volume form.


2

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


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" ...


1

Parametrize in spherical coordinates!


1

If you look to $\omega$ as a vector rather than a 1-form, that is, $\omega(x,y) = (a(x,y), b(x,y)),$ then the following relation holds: $$ \int_\gamma \omega = \int_{t_0}^{t_1} \langle \omega(\gamma(t)),\gamma'(t)\rangle dt,$$ where $t_0$ and $t_1$ are the frontiers of the interval where $t$ lies. You can do this because there is an isomorphism between ...


1

Let $ (M,\mathcal{A}) $ be an $ n $-dimensional $ C^{\infty} $-manifold, meaning that $ M $ is a topological space and $ \mathcal{A} $ is a set of ordered pairs $ (U,\phi) $, where $ U $ is an open subset of $ M $ and $ \phi: U \to \Bbb{R}^{n} $ is a topological embedding, such that $ \displaystyle M = \bigcup_{(U,\phi) \in \mathcal{A}} U $ and for $ ...


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 ...


1

Yes, this is what you're asked to show.


1

Now use the product rule to calculate $\frac{d}{dt}\big\vert_{t=0}(\varphi_t^*\omega_1)\wedge(\varphi_t^*\omega_2)$.


1

First notice that there exists unique $\gamma:\wedge^{k+1}V^*\rightarrow \wedge^k V^*$ such that $$<\gamma(T),S>=<T,\xi\wedge S>.$$ This is due to the nondegeneracy of inner product $<,>.$ In fact, assume that there are two such $\gamma_1,\gamma_2.$ Fix arbitrary $T\in\wedge^{k+1}V^*.$ Now you get that for every $S\in\wedge^{k}V^*$ ...



Only top voted, non community-wiki answers of a minimum length are eligible