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4

As a definition, you should take $\Omega^k (U)$ to be zero for all $k$ not in $\{0, 1, \dots, n\}$, where $n$ is the dimension of $U$. So, for example, "$d^{-1}$" is the zero map, and $\text{im} ~ d^{-1} = \{0\} \subset \Omega^0(U)$. Thus $H^0(U) = \text{ker} ~ d_0$ (which consists of locally constant functions). Similarly, $d_n = 0$, so $\text{ker} ~ d_n = ...


3

You can't differentiate a pointwise expression. In particular, $\lambda$ is not the pullback of a form on $M$.


2

Let $M$ be an n-manifold with boundary. Then $\partial M$ is the boundary of the manifold. The interior of $M$ is the set of points in $M$ such that they have a neighborhood which is homeomorphic to an open subset in $\mathbb{R}^n$. $M - IntM = \partial M$ If $M$ is a manifold without boundary, then $\partial M = \emptyset$ For example, consider the unit ...


2

Interior product is defined like this: if $\omega$ is a $k$-form and $X$ is a vector field, then $\iota_X \omega$ is a $(k-1)$-form defined by (remember, $\iota_X \omega$ "eats" $k-1$ vectors and returns a function): $$ [\iota_X \omega](V_1, V_2, \dots, V_{k-1}) := \omega(X,V_1, V_2, \dots, V_{k-1}) ,$$ i.e., we just put $X$ in the first argument of ...


2

Yes, your calculations are correct. Two things to remember: the wedge product is $C^\infty(M)$-bilinear rather than just $\Bbb R$-bilinear, and $du \wedge dv = -dv\wedge du$ (or, more generally, if $\omega$ is a k-form and $\mu$ an $l$-form, then $\omega \wedge \mu = (-1)^{kl}\mu \wedge \omega$). Using these, you should be able to simplify your last ...


2

So $dx$ is a $1$-form on $\Omega\subset \mathbb R^3$ for example where $(x,y,z)$ define the local coordinates. It means that $dx:\Omega\rightarrow \mathcal L(\mathbb R^3,\mathbb R)$ and $dx$ is simply the differential of the map $$\begin{array}{rccl}x:&\Omega&\rightarrow &\mathbb R \\ & p=(p_1,p_2,p_3)& \mapsto& p_1\end{array}$$ ...


2

Well, if those sets are pairwais disjoint, then everything splits as a direct sum: the $k$-forms $$\Omega^k(\cup_i U_i)\simeq\oplus_i\Omega^k(U_i)$$ is given by $\omega\mapsto (\omega|_{U_i})_i$ and the De Rham differential $d$ respects this splitting. So, both the kernel and image of $d$ respect this splitting as well.. Note, however, that for an infinite ...


1

If you consider $dt^1\wedge\cdots\wedge dt^k$ as a wedge product of $k$ 1-forms, the conclusion can't be more obvious. Recall that the wedge product of $k$ $1$-forms: $$ dt^1\wedge\cdots\wedge dt^k\triangleq\frac{k!}{1!\cdots 1!}\mathcal A(dt^1\otimes\cdots\otimes dt^k). $$ where $\mathcal A$ is the alternating operator: $$ \mathcal ...


1

It is the definition. Basically, $d$ is degree $1$ anti-derivation satisfying: 1.$df$ is the usual differential for $f$ a $0$-form 2.$d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta$ for $\alpha$ a $k$-form (antiderivation). 3.$d^2=0$ It can be proved that $d$ is uniquely determined by these axioms and the local coordinate version fits ...


1

Thank you for your answers and Phillip Andreae has the best answer so far. But I want to answer the question myself because I found an easier and more intuitive way to look at this problem. I just think of differential forms as alternating multilinear functions, for example the determinant ! Let's take a simpler example: $\omega = 3x dx\wedge dy$ $X = ...


1

The element, call it $g(a,b)$, of your Lie group can be viewed as the matrix $$ g(a,b)=\left(\begin{array}{rr}a&b\\0&1\end{array}\right) $$ acting on column vectors $(x,1)^T$. We see that the matrix of $1$-forms $$ ...


1

Another approach, ultimately equivalent to Jyrki's but from a different point of view, proceeds by first choosing arbitrarily a non-zero 2-form at the identity element (i.e., a vector in $\bigwedge^2T_e^*$) and then translating that vector to all other points by means of left-translations, so as to enforce left-invariance. (A different initial choice would ...


1

The first equation $\beta=\hat{Φ_1}^∗(x)$ is probably just a typo. I think you mean $\beta=\hat{Φ_1}^∗(\beta)$, which is the hypothesis. The second equation is also a typo. What you want is just use the fundamental theorem of calculus: $f(1) - f(\epsilon) = \int_{\epsilon}^1 \frac{\partial f}{\partial t} dt$ (apply it to $f(t)=\hat{Φ_t}^∗(\beta)$ and then ...


1

It is the case. One way to see it is dimension count, spaces of vectors and $n-1$ exterior forms have the same dimension, and $v\mapsto i_v\omega$ is linear and injective, hence surjective, on every tangent space. Inverses to a smooth family of invertible linear maps also form a smooth family.


1

Every differential form $\theta$ of degree n-1 has a corresponding 1 form $\eta$ with relation $\theta=i_\eta\omega$, it is called Hodge dual of $\theta$. To see this, assume $\theta\wedge\rho=\lambda\omega$, then $\theta$ can be seen as a linear functional $\theta:\Omega^1(M)\to\mathbb R$ by $\rho\mapsto \lambda$. By Rietz theorem, there exists a 1-form ...


1

Just write down the parametric form: $$ p=(x,y,z)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) $$ Now the two derivations (tangent vectors) can be computed by: $$ \begin{split} \partial/\partial\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta)\\ \partial/\partial\phi&=(-\sin\theta\sin\phi,\sin\theta\cos\phi,0) \end{split} $$ The only ...



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