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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 ...


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, ...


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 $S\in\wedge^{k}V^*$ ...


0

Because specific examples have already been given, I will make some general statements. I will assume (without proof) that an $n$ dimensional manifold $M$ is orientable if and only if there is a nonvanishing $n$ form on $M$. Let $M$ be a compact, orientable, $n$ dimensional manifold without boundary. Then, it is a simple application of Stokes' theorem to ...


0

An answer to the first question is Counterexample. Let $\mathbb T^2$ (the 2-torus) endowed with periodic coordinates $(\theta,\phi)$ and with the symplectic form $\Omega = \mbox{d}\theta \wedge \mbox{d}\phi$. Then a costant vector field on $\mathbb T^2$ is locally Hamiltonian but not Hamiltonian (Compare [1], chapter 5). This can be important, e.g, in ...


1

Parametrize in spherical coordinates!


1

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


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 ...


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.


0

If your form is defined on the whole $\mathbb{R}^3$, then there is always the linear homotopy which contracts every point to the origin.


1

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


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 $ ...


0

If $\omega$ is nowhere degenerate then at every point $p$ in $M$ the volume form must be non-zero; otherwise $\omega$ is degenerate at $p$. The second statement (the integral) relies on two unstated assumptions, namely continuity and the statement that $M$ is not only closed but also connected. With that, you show that for any two points $p_1$ and $p_2$ ...


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 ...


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 ...


0

You can visualize the exterior derivative in terms of a boundary operation. The fact that $d \circ d = 0$ can then be understood as following from the fact that the boundary of a boundary is empty. The idea is difficult to explain without pictures, so here's a short PDF that explains the idea and gives some nice diagrams to help you visualize differential ...


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 ...


0

This problem is basically just orthogonal diagonalization (which is always possible with a real symmetric matrix)...and the question specifies the method you need to use. The Schur complements method can be applied as follow (I'm using a theorem from the book Matrices and Linear Transformations from CG Cullen): You need to find a unit vector ...


2

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 ...



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