Discover Top Posts Tagged with #differential forms

Differential Forms

One of the best math series on YouTube I've ever seen.

#differential forms #exterior algebra #math #Michael Penn

the cotangent bundle (differential forms) is the feminine side of calculus-on-manifolds; the tangent bundle (vector-fields) is the masculine side.

Shing-Shen Chern, via Richard Montgomery

#Shing-Shen Chern #Richard Montgomery #Stephanie Frank Singer #differential forms #cotangent #bundles #vector bundle #tangent bundle #masculine #feminine #mathematics #differential geometry #physics #mechanics #maths #math #calculus #manifolds

When your inner math major comes out of hibernation

#It’s 3am send help #binormal vector #differential forms #fuck linear algbera #marry abstract algebra #kill geometry

Geometric calculus relationships to differential forms, and vector calculus identities.

[Click here for a PDF version of this post] Motivation. I was asked about the geometric algebra equivalents of some of the vector calculus identities from [1]. I’ll call the specific page of those calculus notes “the article”. The article includes identities like \begin{equation}\label{eqn:formAndCurl:20} \begin{aligned} \spacegrad (f g) &= f \spacegrad g + g \spacegrad f \\ \spacegrad \cross (f…

View On WordPress

#curl #differential forms #divergence #Geometric Algebra #Geometric Calculus #vector calculus #vector identities

Free download Tensors, Differential Forms, and Variational Principles (Dover Books on Mathematics) written by David Lovelock.

#Tensors #Differential Forms #and Variational Principles (Dover Books on Mathematics)

Natural?

I present here two proofs of the naturality of the exterior derivative. Given a (smooth) manifold, $M$, this is the result that $F^*d\omega = d F^* \omega$, where $\omega$ is any $n$-form and $F$ is a smooth map.

Proof 1

I will use the standard abuse of notation that is "multi-indices". We will write, for example

$$ \omega = \frac{1}{k!} \omega_{j_1...j_k} dy^{j_1} \wedge ... \wedge dy^{j_k} = \frac{1}{k!} \omega_J dy^J, $$

where $ J = ( j_1, ..., j_k ) $ is a "multi-index". Using this notation, the definition of pullbacks in coordinates looks like

$$ F^* \omega = \frac{1}{k!} \left( \omega_I \frac{\partial F^I}{\partial x^J} \right) dx^J $$

\begin{align*} d F^* \omega &= \frac{1}{k!} \frac{\partial}{\partial x^i} \left( \omega_I \frac{\partial F^I}{\partial x^J} \right) dx^i \wedge dx^J \\\\ &= \frac{1}{k!} \left( \frac{\partial \omega_I}{\partial x^i} \frac{\partial F^I}{\partial x^J} + \omega_I \frac{\partial}{\partial x^i}\frac{\partial F^I}{\partial x^J} \right) dx^i \wedge dx^J. \end{align*}

The second term,

$$ \omega_I \frac{\partial}{\partial x^i} \frac{\partial F^I}{ \partial x^J} dx^i \wedge dx^J, $$

is equal to zero. Why? The multi-index is a product of terms so, more explicitly, what we want to show is

$$ \omega_{i_1...i_k} \frac{\partial}{\partial x^i} \left( \frac{\partial F^{i_1}}{ \partial x^{j_1}}... \frac{\partial F^{i_k}}{\partial x^{j_k}} \right) dx^i \wedge dx^{j_1} \wedge ... \wedge dx^{j_k} = 0. $$

First note that if $i$ is equal to any of the $j$'s, the sum is zero because of the wedge products. Let's take an illustrative example, let $i=2$ and $j_1=4$ in the sum, then the above has a term

$$ \omega_{i_1...i_k} \frac{\partial F^{i_1}}{\partial x^2 \partial x^{4}}... \frac{\partial F^{i_k}}{\partial x^{j_k}} dx^2 \wedge dx^{4} \wedge ... \wedge dx^{j_k}. $$

There is another term where $i=4$ and $j_1=2$. This looks like

$$ \omega_{i_1...i_k} \frac{\partial F^{i_1}}{\partial x^4 \partial x^{2}}... \frac{\partial F^{i_k}}{\partial x^{j_k}} dx^4 \wedge dx^{2} \wedge ... \wedge dx^{j_k}, $$

which is the negative of the above term. Every term in the sum is either zero of has a negative in this way (switching all the wedges around works since we need $a$ swaps to take $dx^k$ to the position of $dx^i$ and $k-1$ to get $dx^i$ to $dx^k$'s old position, making a total odd number of swaps).

Hence, we have

$$ d F^* \omega = \frac{1}{k!} \frac{\partial \omega_I}{\partial x^i} \frac{\partial F^I}{\partial x^J} dx^i \wedge dx^J. $$

Doing things the other way round, we get

$$ d \omega = \frac{1}{k!} \frac{ \partial \omega_I}{\partial y^i} dy^i \wedge dy^I $$

and

$$ F^*d\omega = \frac{1}{k!} \frac{ \partial \omega_I}{\partial y^i} \frac{\partial F^i}{\partial x^j} \frac{ \partial F^I }{ \partial x^J} dx^j \wedge dx^J = \frac{1}{k!} \frac{ \partial \omega_I}{\partial x^i} \frac{ \partial F^I }{ \partial x^J} dx^i \wedge dx^J, $$

where the last step follows by the chain rule. Hence, $ d F^* \omega = F^* d\omega $, as required.

And now the alternative.

Proof 2

We first prove the result for differentials; let $d \phi \in \Lambda(M)$, for some $\phi \in C^{\infty}(M)$ and let $F: M \rightarrow N$. Let $V \in TM$, then

\begin{align*} F^*d\phi (V) &= d \phi (f_*V) \\\\ &= F_*V ( \phi) \\\\ &= V ( F^* \phi ) \\\\ &= d (F^* \phi) (V) \end{align*}

or $F^* d \phi = d F^* \phi$.

To extend this to $n$-forms, we note that an $n$-form, $\omega$, can be written

$$ \omega = g_{1...n} d \phi^1 \wedge ... \wedge d \phi^n $$

where $g_{1...n}, \phi^1, ..., \phi^n \in C^{\infty}(M)$.

Then,

\begin{align*} F^* d \omega &= F^* ( dg_{1...n} \wedge d \phi^1 \wedge ... \wedge d \phi^n ) \\\\ &= F^*dg_{1...n} \wedge F^*d \phi^1 \wedge ... \wedge F^*d \phi^n \\\\ &= d F^*g_{1...n} \wedge d F^* \phi^1 \wedge ... \wedge d F^* \phi^n \\\\ &= d ( F^*g_{1...n} \wedge d F^* \phi^1 \wedge ... \wedge d F^* \phi^n ) \\\\ &= d ( F^* \omega ). \end{align*}

#differential geometry #differential forms #geometry

Oriented?

Orientations of Vector Spaces

I wanted to write about orientations of manifolds. To do this, we first need to talk about orientations of vector spaces. When we talk about an orientation of, say, $\mathbb{R}^3$, we are asking, `given two basis vectors, does the third point up or down?’. Similarly, in $\mathbb{R}^2$, we are asking ‘given one basis vector, do I have to go clockwise or anticlockwise to get to the other?’.

This is captured by the transition matrix from the standard basis to your new basis. A transition matrix gives me my new basis in terms of the old; let $ ( E_i ) $ be one basis for a vector space, and $ ( F_i ) $ be a new one (we implicitly assume here that $i$ varies from $1$ to $n$, where $n$ is the dimension of the vector space), the transition matrix from $ ( E_i ) $ to $ ( F_i ) $ is the matrix, $A$, with components such that $ F_i = A_i^{\hspace{2mm} j} E_j $. The determinant measures a scale factor of enlargement $n$-volume when applying the matrix to $n$ vectors and, if this is negative, a `flip’ happens. This `flip’ is precisely a change in orientation of the vectors and we use this as our definition.

Definition (Orientation). Two bases are said to have the same orientation if the transition matrix from one to the other has a strictly positive determinant. This is an equivalence relation. A choice of one of the two equivalence classes is called and orientation of a vector space.

The basis, $ (u, v ) $ of $\mathbb{R}^2$ (blue), originally had an orientation. After we reflect the vectors in the line $y=-x$, a transformation whose matrix has negative determinant, the basis has the opposite orientation.

There is a fact about exterior algebras that lets us work with them instead of vector spaces. It is captured in the following proposition.

Proposition. Let $ \Lambda^k (V) $ denote the kth exterior power of an $n$-dimensional vector space, $V$, and $\mathrm{M}(n, V)$ the $ n \times n $ matrices on $V$. If $ \omega \in \Lambda^n (V) $ and $ A \in \mathrm{M}(n,V)$, then

$$ \omega ( A v_1, ..., A v_n ) = \det ( A ) \omega (v_1, ..., v_n). $$

Proof. Let $ (e_i ) $ be a basis for $ V $ and $ ( \varepsilon^i ) $ be the corresponding dual basis of $ V^* $. A member of $ \Lambda^n(V) $ must be proportional to $ \varepsilon^i \wedge ... \wedge \varepsilon^n $, so we only need to show the proposition for $ \omega = \varepsilon^i \wedge ... \wedge \varepsilon^n $. We note then that we only really need to show that

$$ \varepsilon^i \wedge ... \wedge \varepsilon^n ( A e_1, ..., A e_n ) = \det ( A ) \varepsilon^i \wedge ... \wedge \varepsilon^n ( e_1, ..., e_n ) $$

since the action of $ \varepsilon^1 \wedge ... \wedge \varepsilon^n $ on the other basis vectors is trivial. The right-hand side of this is just $ \det(A) $, since $ \varepsilon^1 \wedge ... \wedge \varepsilon^n ( e_1, ..., e_n ) = 1 $. So we are left with showing:

$$ \varepsilon^1 \wedge ... \wedge \varepsilon^n ( A e_1, ..., A e_n ) = \det ( A ). $$

Writing the components of $A$ in this basis as $A_i^{\hspace{2mm} j}$, we find

\begin{align*} \varepsilon^1 \wedge ... \wedge \varepsilon^n ( A e_1, ..., A e_n ) &= \varepsilon^1 \wedge ... \wedge \varepsilon^n ( A_1^{\hspace{2mm} i_1} e_{i_1}, ..., A_n^{\hspace{2mm} i_n} e_{i_n} ) \\\\ &= A_1^{\hspace{2mm} i_1}... A_1^{\hspace{2mm} i_n} \varepsilon^1 \wedge ... \wedge \varepsilon^n ( e_{i_1}, ..., e_{i_n} ) \\\\ &= A_1^{\hspace{2mm} i_1}... A_1^{\hspace{2mm} i_n} \varepsilon_{i_1...i_n}, \end{align*}

where the final epsilon is the Levi-Civita symbol. The above is precisely the Leibniz formula for determinants, so we have our result. $\square$

The nice thing about this is that we can just choose some non-zero $ \omega \in \Lambda^n (V) $ and take the bases that give a positive number when $\omega$ is applied to them to be our choice of equivalence class. This means we don’t have to worry about transition matrices; we can just specify an $\omega$ and we have implicitly oriented our vector space.

Orientations of Manifolds

We say an $n$-dimensional manifold is orientable if we can find a nowhere zero $n$-form on $M$. This $n$-form is called a volume form for $M$. We have cheated a bit here by hiding smoothness considerations in a differential form (remember forms are, by definition, smooth), but this is all we normally need (i.e. to do integration on manifolds).

If we want to think about this more, we can define an orientation of a manifold pointwise to be an orientation of each tangent space. We would then like to define a global frame to have the given orientation if it has the correct orientation in each tangent space. However, in general, we don’t have global frames, but only local ones. The way to deal with this is to demand that each point on the manifold is in the domain of a local frame that follows the pointwise orientation. It can be shown that this and the above definition are equivalent.

#differential geometry #orientation #differential forms #exterior product #geometry

On Differential Forms

On Differential Forms (WordPress Version)

Abstract. This article will give a very simple definition of -forms or differential forms. It just requires basic knowledge about matrices and determinants. Furthermore a very simple proof will be given for the proposition that the double outer differentiation of -forms vanishes.

MSC 2010: 58A10

(more…)