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2

In the formula at pag. 19 after "Exterior differentiating this, we have" the author used the Cartan structural equation $$d\omega_k = \omega_{kj}\wedge \omega_j $$ which defines the $\omega_{kj}$'s, given the basis $\{\omega_k\}$ of the cotangent space $T^*_xM$, dual to the basis $\{e_k\}$ on $T_xM$, for all $x\in M$. In particular $$ d(a_{i_1\cdots ...


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In general, a $k$-form on a vector space $\mathbb{V}$ over $\mathbb{R}$ is an alternating map that takes $k$ vectors and returns a scalar. So a $0$-form is a map that takes no vectors at all and returns a scalar, but for many purposes we may as well just identify this map with the scalar itself. On an open set $U \subset \mathbb{R}^n$, a smooth $k$-form is ...


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1) It's mainly a matter of notation. When wedging the $1$-forms $dx$ and $dy$, one obtains the $2$-form $dx\wedge dy$. When integrating a real valued function on a domain in $\mathbb{R}^2$, it is customary to simply write $dxdy$. 2)Now we have a parametrization $T:A\to S$, where $A\subset \mathbb{R}^2,S\subset\mathbb{R}^3$, and consider for the moment the ...


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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 $$\begin{array}{rccl}x:&\Omega&\rightarrow &\mathbb R \\ & p=(p_1,p_2,p_3)& \mapsto& p_1\end{array}$$ ...


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It is not a matter of notation; it is a matter of convention. The way a differential form integral is written can be understood as a shorthand. For instance, the integral $$\int f \, \mathrm dx \wedge \mathrm dy$$ really means this: $$\int f(x,y) \, (\mathrm dx \wedge \mathrm dy)(e_x \wedge e_y) \, dx \, dy$$ where $e_x$ and $e_y$ are the tangent ...


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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 ...


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Think of the function $f_w(v) = v \cdot w$. I can write $f_w(v)$ for some vector $v$, or I can talk about the function $f_w$. Both are legitimate objects of study, just as $\sin(\pi)$ and the sine function (as a function) are of interest. The authors who write the second form are denoting a function that takes two vectors as arguments, but they're not ...


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To simplify, denote $$X=\sum_{i=1}^nx^i\frac{\partial}{\partial x^i}$$ and $\Omega=dx^1\wedge dx^2 \cdots \wedge dx^n$, then $\sigma=i(X)\Omega$, where $i$ is the Interior Product operator, that is, $\sigma(\cdots)=\Omega(X,\cdots)$. $\forall p\in \mathbb{R}^n$, let $q=r(p)$. Note that $X$ has the following property: ...


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A zero form is a smooth function defined on a manifold. Here smoothness means that the map $U\rightarrow \mathbb{R}$ is smooth by considering the smoothness of chart maps $\phi:\mathbb{R}^{n}\rightarrow U\rightarrow \mathbb{R}$. In general an $n$-form is a section of the anti-symmetric cotangent tensor bundle $\wedge^{n}(TM^{*})$.


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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 ...


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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 ...


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It is a matter of convention, and some authors do not follow it. If $Z$ is a $(1,0)$ vector field, then $IZ = iZ$. If $\alpha$ is a $(1,0)$-form, then one has $I^* \alpha = i \alpha$, where $I^*(\alpha)(-) = \alpha(I-)$ is the (pointwise) pullback of $\alpha$ by $I$. Ok, so one can, say, define $I$ as $I^*$. But then you have a rather annoying sign showing ...


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This might boil down to a question about defining differentials but in general, if $f$ is any function defined on your manifold, the value of the differential $df$ at a point $p$ and a tangent vector $X_p$ is $$df_p(X_p) = X_p f.$$ To be completely explicit, let's write $\partial_x|_p$ for the value of the vector field $\partial_x$ in a point $p$. In your ...



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