# Tag Info

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 $$d(a_{i_1\cdots ... 2 In general, a k-form on a vector space \mathbb{V} over \mathbb{R} is an alternating map that takes k vectors and returns a scalar. So a 0-form is a map that takes no vectors at all and returns a scalar, but for many purposes we may as well just identify this map with the scalar itself. On an open set U \subset \mathbb{R}^n, a smooth k-form is ... 2 1) It's mainly a matter of notation. When wedging the 1-forms dx and dy, one obtains the 2-form dx\wedge dy. When integrating a real valued function on a domain in \mathbb{R}^2, it is customary to simply write dxdy. 2)Now we have a parametrization T:A\to S, where A\subset \mathbb{R}^2,S\subset\mathbb{R}^3, and consider for the moment the ... 2 So dx is a 1-form on \Omega\subset \mathbb R^3 for example where (x,y,z) define the local coordinates. It means that dx:\Omega\rightarrow \mathcal L(\mathbb R^3,\mathbb R) and dx is simply the differential of the map$$\begin{array}{rccl}x:&\Omega&\rightarrow &\mathbb R \\ & p=(p_1,p_2,p_3)& \mapsto& p_1\end{array}$$... 1 It is not a matter of notation; it is a matter of convention. The way a differential form integral is written can be understood as a shorthand. For instance, the integral$$\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$$where e_x and e_y are the tangent ... 1 If you consider dt^1\wedge\cdots\wedge dt^k as a wedge product of k 1-forms, the conclusion can't be more obvious. Recall that the wedge product of k 1-forms:$$ 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

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 ...

1

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: ...

1

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 $\wedge^{n}(TM^{*})$.

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

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 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 ...

1

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 ...

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