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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$1-forms, the conclusion can't be more obvious. Recall that the wedge product of$k1$-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 ...
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 ...