Differential Forms

[trfrm]

Sanctus, Sanctus, Dominus Deus Sabaoth.

Markus I will get to your question in a mo, but for now notice that I said that $df=\frac{\partial f}{\partial x} dx+ \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$ is called the "total differential". This is wrong - it is in fact the gradient of our function $f(x,y,z)$ - one often writes $\nabla f$ for this guy True enough it is intimately related to the total differential, but they are not the same. Apologies.

Why the $df$ was regarded as $\nabla{f}$ ... ? Isn’t in vector calculus, $df=d\mathbf{r}\cdot\nabla{f}$, where $f$ depends on the position vector $\mathbf{r}$ ... ?

Consider the 3 real functions on the 3-manifold $R^3$ being $f,\,g,\,h$. Again I form the differential 1-form

$(\frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz)+ (\frac{\partial g}{\partial x}dx+ \frac{\partial g}{\partial y}dy + \frac{\partial g}{\partial z}dz )+(\frac{\partial h}{\partial x}dx+ \frac{\partial h}{\partial y}dy + \frac{\partial h}{\partial z}dz)$.

By an application of the exterior derivative operator $d$ I will have the 2-form

$= (\frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz) \wedge dx$$+ (\frac{\partial g}{\partial x}dx+ \frac{\partial g}{\partial y}dy + \frac{\partial g}{\partial z}dz) \wedge dy$$+(\frac{\partial h}{\partial x}dx+ \frac{\partial h}{\partial y}dy + \frac{\partial h}{\partial z})\wedge dz$

$=(\frac{\partial h}{\partial y} -\frac{\partial g}{\partial z})dy \wedge dz +(\frac{\partial f}{\partial z} -\frac{\partial h}{\partial x})dz \wedge dx+(\frac{\partial g}{\partial x} -\frac{\partial y}{\partial y})dx \wedge dy$ where I have the skew symmetry property of the exterior (wedge) product.

This is the analogue of the curl in our theory

For instance, I have an 1-form $\omega:=\mathbf{A}\cdot{d\mathbf{r}}=\sum_{j=1}^nA_jdx^j\in\Lambda^1((\mathbb{R}^n)^*)$ in Cartesian co-ordinate system ... .

As I have known, the exterior derivative of $\omega$ is

$\displaystyle\,d\omega=\sum_{j=1}^ndA_j\wedge{dx^j}$ ... .

Since $\displaystyle\,dA_j=\sum_{k=1}^n\frac{\partial{A_j}}{\partial{x^k}}dx^k$ and $dx^k\wedge{dx^j}=-dx^j\wedge{dx^k}$, then

$\displaystyle\,d\omega=\frac{1}{2}\sum_{j,k=1}^n\left(\frac{\partial{A_j}}{\partial{x^k}}-\frac{\partial{A_k}}{\partial{x^j}}\right)dx^k\wedge{dx^j}=(\nabla\times\mathbf{A})\cdot{d\mathbf{a}}$

where $\displaystyle\,d\mathbf{a}:=\frac{1}{2}\sum_{l,m=1}^n(dx^l\wedge{dx^m})(\hat{\mathbf{x}}_l\times\hat{\mathbf{x}}_m)$ is an element of area pseudo-vector ... .

Is it correct ... ? I don't know ... .

Haleluya.

[trfrm]

Salam damai Kristus.

But your calculus text will tell you that a vector field is defined as $df = \frac{\partial f}{\partial x}\vec{i} + \frac{\partial f}{\partial y}\vec{j} +\frac{\partial f}{\partial z} \vec{k}$, where the "vectorized" objects are unti vectors. I should not need to point out the parallel between 1-forms and fields - call them what you want! I may want to return to this in due course.

Excuse me ... .

Is the vector $\vec{i},\vec{j},\vec{k}$ in this quote correspond to $dx,dy,dz$ respectively ... , so $\nabla{f}$ corresponds to $df$ ... ? Isn’t in usual calculus, $df=d\mathbf{r}\cdot\nabla{f}$ ... ? Should not the usual calculus be consistent with the calculus of manifold ... ?

Here is an illustration ... .

The position (location) of a point $p$ of an $m$-dimensional manifold $M$, which be embedded in $\mathbb{R}^n$, where $m\leq{n}$, can be illustrated by the position vector $\mathbf{r}$ ... . The tangent space at $p$ is $T_pM$, which be like to a flat space $\mathbb{R}^m$ ... .

In terms of natural basis $\{\hat{\mathbf{x}}_1,\dots,\hat{\mathbf{x}}_n\}$ in Cartesian co-ordinate system,

$\displaystyle\mathbf{r}=\sum_{j=1}^nx^j\hat{\mathbf{x}}_j$,

where each $x^j$ depends on $m$ co-ordinates (as the parameters) on the manifold $M$, namely, $q^1,\dots,q^m$ ... .

The gradient of $\varphi$ is

$\displaystyle\nabla\varphi:=\sum_{j=1}^n\frac{\partial\varphi}{\partial{x^j}}\hat{\mathbf{x}}^j$ ... .

Here, $\hat{\mathbf{x}}^j\equiv\hat{\mathbf{x}}_j$ ... .

The contravariant basis in $T_pM$ is $\{\mathbf{e}_1,\dots,\mathbf{e}_m\}$, where

$\displaystyle\mathbf{e}_\mu:=\frac{\partial\mathbf{r}}{\partial{q^\mu}}=\sum_{j=1}^n\frac{\partial{x^j}}{\partial{q^\mu}}\hat{\mathbf{x}}_j$,

so the tangent vector $\mathbf{A}\in{T_pM}$ can be denoted as

$\displaystyle\mathbf{A}=\sum_{\mu=1}^mA^\mu\mathbf{e}_\mu$ ... .

Here, $A^\mu$ is contravariant component relative to the contravariant basis ... .

The covariant basis in $T_p^*M$ is $\{\mathbf{e}^1,\dots,\mathbf{e}^m\}$, where

$\displaystyle\mathbf{e}^\mu:=\nabla{q^\mu}=\sum_{j=1}^n\frac{\partial{q^\mu}}{\partial{x^j}}\hat{\mathbf{x}}^j$,

so the cotangent vector $\mathbf{A}\in{T_p^*M}$ can be denoted as

$\displaystyle\mathbf{A}=\sum_{\mu=1}^mA_\mu\mathbf{e}^\mu$ ... .

Here, $A_\mu$ is covariant component relative to the covariant basis ... .


Here is a relation that

$\displaystyle\mathbf{e}^\mu\cdot\mathbf{e}_\nu=\sum_{j,k=1}^n\frac{\partial{q^\mu}}{\partial{x^j}}\frac{\partial{x^k}}{\partial{q^\nu}}{\delta^j}_k=\sum_{j=1}^n\frac{\partial{q^\mu}}{\partial{x^j}}\frac{\partial{x^j}}{\partial{q^\nu}}=\frac{\partial{q^\mu}}{\partial{q^\nu}}={\delta^\mu}_\nu$ ... .


Therefore,
here is a relation between the vector $\mathbf{e}_\mu$ and the directional derivative $\partial/\partial{q^\mu}$, that $\mathbf{e}_\mu=\partial\mathbf{r}/\partial{q^\mu}$,
here is a relation between the co-vector $\mathbf{e}^\mu$ and the form $dq^\mu$, that $dq^\mu=d\mathbf{r}\cdot\nabla{q^\mu}\equiv{d}\mathbf{r}\cdot\mathbf{e}^\mu$ ... .


I’m sorry ... . Thank you and see you ... .

http://www.thescienceforum.com/mathematics/34909-differential-forms.html#post412052

Gloria in excelsis Deo.