Smooth manifolds

. Note that df is a linear map ℝ → ℝ rather than a number.

This justifies the concept of derivatives as a fraction: Since dx is the identity map, we can compare df and dx, and thus recover the idea that f’ is the ratio of differentials df, dx.

Note that where . This expresses that dy is a kind of “polished” change of y. Of course this arises from the definition of the derivative, so the same property applies in the multivariate case in the form of .

Notation

Since dx(x, Δx) = Δx it is conventional use dx = Δx interchangeably. If y=f(x), we may write dy for df(x). So we get df = f’ dx.

y=ln x ⇒ x=e^y ⇒ dx/dy = e^y = x ⇒ dx = x d(ln x)

In , f is always a function of t so the fraction is a well-defined function that can be identified with a constant.

(Chain rule)

See

In the multivariable case, the total derivative is already a linear map and thus the (total) differential just “the same object”. Notation-wise as a linear functional, the differential should be written .

Note that this form is valid for any basis, but in use the x_i’s are generally instantiated as an actual basis.

This perspective that a differential is a linear-functional will be generalized.

Applications

Coordinates

There is an identity that expresses ∇ f in some coordinate system (here cylindrical) in terms of df: Keep in mind that . Note that f has two interpretations: as a mapping from , and the isomorphic mapping from the matrix representing the vector in .

Let the coordinate vector have coordinates . Let map coordinates to position: . So p is the isomorphism from coordinate matrices of in cylindrical coordinates to the space .

Then

(This following step requires basis vectors to be orthogonal to be well-defined)

So . Finally from directional derivative definition, .

The same process applies in general, giving us . Applying Taylor’s theorem also gives us this directly. Note-1

Experimental

Basic rules

[S11.pdf](Resources/Calculus of Differential Forms/S11.pdf)

For cylindrical coordinates,

Higher order

Higher order differentials are defined via so

Properties

Integration

sheds light on what moment of inertia means, but is useless for calculations. So we.introduce a parametrization I(r).Then . In the case of a thin disk of mass M, radius R, we have where dm = is the amount of mass in the ring dr by m(r)=f(r).

Exact differentials

A d

Holonomic constraints

is holonomic only if an integrating factor can be found turning it into an exact differential.

Scale factors

where

Intuition

https://juliadiff.org/ChainRulesCore.jl/stable/maths/propagators.html

https://sites.ualberta.ca/~vbouchar/MATH215/preface.html

[S11.pdf](Resources/Calculus of Differential Forms/S11.pdf)

What you should be integrating is not functions, but rather differential forms! We would like our theory to be reparametrization invariant: Be able to write . Where w is a “one-form”. We will need to be a parametric curve. However, we want to be defined intrinsically in terms of the geometry of the curve itself: we do not want the integral to depend on the choice of parametrization. We would like our theory to be oriented A differential form is most simply and a little inaccurately, a (multilinear) covector field (that is alternating on its inputs).

Forms in ℝ

0-forms correspond to vector fields.

Given a one-form , we can define an associated smooth vector field with component functions (f, g, h)

Functions of an attached vector that are linear at each point (where attached) are called a differential 1 form.

Exact 1 forms

An exact k-form is the differential of some (k-1)-form. This is the same concept as conservative for the vector field of that one form.

For now, just know we can take the differential of a smooth function f (0-form) to get the 1-form

If a 1-form is exact (), it is closed ().

Pullback of a 1-form

The pullback of dx is Let be a 1-form on , . Then

Integrals of one-forms over intervals are invariant under orientation-preserving reparametrizations

Let be a one-form on , and a C¹ function with . Then . Explicitly, by substitution rule.

Note that we need to be C¹ on an open set containing D, but this is implied by Tietze.

Line integrals

Let α: [a, b] parametrize a curve C. Let be a one-form over . Then is a one-form over Note-2

Via the associated vector field F of ω,

Let .

Let C be a curve parametrized by , be a one-form on , the line integral of ω along α is .

Reparametrization equivalence proof

In Physics, one may also parametrize by the position r. Considering , one can write a line integral as . Replacing [a, b] by C here makes sense since integral is invariant under reparametrizations. However, one should know that to evaluate the integral we need to compose the vector field with the parametrization first. One can also interpret together as a differential form directly, , recalling that dx_i form a basis for 1 forms.

Fundamental Theorem

Let .

k-forms

See. We state here some definitions A basic 1-form (1-blade) is dx. A 3-blade on ℝ⁴ is i.e. . In general, a k-blade . The antisymmetrized projection of x

Geometrically, the k-blade calculates the oriented area of the projection of k-object onto the hyperplane spanned by .

A k-form is a linear combination of basic-k-forms/k-blades.

Wedge products in ℝ³

We can associate 1, 2 forms with vector fields i.e.

Wedge of zero form is just scaling Wedge of two one forms, is the cross product of the associated vector fields. Wedge of one and two form is the dot product

follows directly from writing as

Exterior derivative

The exterior derivative of a 0-form f is Generalizing, the exterior derivative of a k-form where f_j are the component (of a certain basic k-form) of the k-form. Or .

In :

Div, Grad, Curl, are the exterior derivatives of the associated 0, 1, 2 forms.

Product rule:

By smoothness, The innermost term is 0 by smoothness.

Vector calc identities

Exact and closed

is generalized by . We say it is closed. Clearly, exact ⇒ closed.

Poincare: Let ω be a k-form defined on an open ball of ℝⁿ. ω is closed ⇔ exact.

De Rham cohomology: Consider the space of k-forms on a set U. Consider two k-forms equivalent (cohomologous) if they differ by an exact form. This quotient space is a vector space, H^k(U), the de Rham cohomology space. If U is an open ball in ℝⁿ, then all closed-forms are exact, and hence H^k(U)=0. Consider . We have .

Find the one form . so exists. We seek as WMA dz component is 0. (For any η, subtract dF where ) .

Top forms

See A n-form on is called a top-form. for .

Integrals

zero/one forms

Defining k-forms over k-dimensional spaces

define the orientation of a closed bounded region , and the notion of a canonical orientation. Define the induced orientation on
Define the integral (of a k-form) over D in terms of standard integrals from Calc_III
Define the notion of a parametric space, mapping a region D to . Show that the parametrization induces an orientation on S
To define the integral of a k-form on S, use the parametrization to pullback the k-form to D and integrate with standard calculus.
Show that the integral is invariant/sign-reversing on reparametrizations.
Study what happens for an exact k-form: Stokes.

Some concepts:

Oriented points
Integral of zero-form over oriented point
Orientation of interval, integral of 1-form over interval
(Step 3, 4, 5) Line integral
(Step 6) (FTOC) and

Orientation

Orientation of is determined by a choice of ordered basis on . The sign of the linear map between reorderings of the basis determine the two equivalence classes of orientations. The canonical orientation is given by the standard ordering of the standard basis (1,0 …), (0,1 …)…

The the orientation of a closed bounded region in is induced by the orientation of the ambient vector space. We may write D₊ or D for canonical, D_- the opposite. In the orientation is counterclockwise, in it is given by RHR.

Induced orientation of boundary: Let p be a boundary point, n the normal vector. Twirl n in following the canonical orientation until it intersects the tangent space. This defines a choice of tangent vector at all points on ∂D, and the (induced direction of motion?)/orientation on ∂D. In ℝ², if D has canonical orientation, the induced orientation of ∂D keeps the interior on the left.

Integrals in R²

By Fubini, we can do integrals over x or y-supported regions. Since closed bounded regions are the union of such regions Note-3 So we can do integrals over closed bounded regions D (Step 2). Explicitly, for .

Remark 5.3.5

Integrals are reparametrization equivalent:

Let D₁, D₂ be CBR, a bijective C¹ function. Assume ϕ is invertible on the interior of D. Say ϕ is orientation-preserving if det ϕ>0.

If ϕ is orientation preserving, by top forms. Apply substitution rule,

Consider the two-form ω = xy dx dy, and let D be the region bounded by the four curves y = 1/x, y = 4/x, y=x, y=4x, with canonical orientation. Evaluate the integral using . . . Since this is the correct orientation. It’s possible to also use .

Parametric surfaces

Let D be a simply connected CBR. A parametric surface is a C¹ function :

has full colspace on the interior of D.
α is injective on the interior.

A parametric surface is closed if it is the boundary of a solid in . If it is not closed, it has edges, the boundary consists of these edges. From the definition

Graphs satisfy all the conditions. Moreover, , the surface is just the continuous deformation of D in the z direction.

See

Grid curves

Consider the parametrization of a sphere The image of the rectangular lines are , the grid lines are found by fixing in the expression. In this case, latitude and longitude.

Tangent planes

Let α: D → ℝⁿ be a parametric surface with image S. For any point in the interior, we define the tangent plane at p as the n dimensional vector space spanned by the tangent vectors . It provides the best linear approximation of the surface at that point.

If we restrict to ℝ³ we can specify the tangent plane via a (normalized) normal vector .

Orientation of surfaces

We want to define a direction fo rotation like for ℝ² for surfaces in ℝ³, but this is not so simple as the surfaces can be deformed. Assign the normal vector as above, we say the surface is orientable if such a normal vector can be assigned in a continuous manner. This determines an “outside”. The mobius strip is non-orientable. From now on we assume our (parametric) surfaces are orientable.

Another way of thinking of it is that the the counterclockwise orientation on D induces the orientation given by the ordered basis on the tangent planes.

Induced orientation on the boundary of a parametric surface: See Orientation, everything still applies. In particular, for a surface in ℝ³, the orientation on the boundary is the direction induced by head pointing towards n and keeping surface on the left.

Orientation of reparametrizations: Let D₁, D₂ be simply-connected CBR, a bijective C¹ function. Assume ϕ is invertible. Then is another parametrization of S, with orientation determined by

We need only check tangent vectors are linearly independent.

Sphere

Area and volume of n-sphere?

General integration

https://en.wikipedia.org/wiki/Differential_form#Integration_over_chains

Over an open subset of , the integral of the smooth n-form is just the Lebesgue integral , (with the right sign). When is taken over a n-1 dim submanifold perpendicular to it, it is 0, why explicitly?

Pullback

Let M, N, A be smooth manifolds, ϕ : M → N, f : N → A smooth functions. Then the pullback of f by ϕ is the smooth function .

Examples

Pullback of bundles and sections

If E is a [fiber bundle](Topology_II.md#Fiber bundle) over N, a smooth map, the pullback bundle is a vector bundle over M whose fiber over x∈M is Precomposition defines a pullback of sections: Let s be a section of E, the pullback section is a section of over M.

Pullback of multilinear forms

Let ϕ: V → W be a linear map between vector spaces. We can form the pullback of a multilinear form F . Hence Φ^∗^ is a linear operator from multilinear forms on W to multilinear forms on V. In particular, Φ^*^ defines a linear map between dual spaces .

Extend?

Pullback of cotangent vectors and 1-forms

The differential of ϕ, is a vector bundle morphism . It has a transpose Now given a section α of T^*^N, apply the above bundle map pointwise to get the pullback of α by ϕ, which is a 1-form defined by

There is an associated linear map φ^*^ from the space of 1-forms on N to the space of 1-forms on M. More generally, any covariant tensor field may be pulled back to M via ϕ.

Properties

Axiomatic approach

. Note-4

Let be a smooth function, a (basic) k-form on U The pullback Note-5

Geometric approach

First we define pullbacks for linear maps . We get . Now for ϕ a smooth function, we define the pullback pointwise via its Jacobian .

Let . Calculate

Substitution rule

https://www.youtube.com/watch?v=aXL_0X1vMT0&list=PL22w63XsKjqzQZtDZO_9s2HEMRJnaOTX7&index=13&pp=iAQB

Pushforward

The differential/pushforward of φ at x is a linear map In total, it is a vector bundle morphism (over M)

The differential lets us push forward vectors, while the pullback lets us pull back k-forms.

Elemental differential?

Differential

Properties

Linearity: d(u+v)=du+dv Product rule: d(uv)=udv+vdu Chain rule:

PB1: PB2:

ROUNDOUT

How about coordinate change between 1-forms? I noticed Is there some sort of equivalence between dz, du dv, and dr dθ as coordinates?

ROUNDOUT
Exterior derivative vs Derivation vs Differential vs Pushforward

Differential k-forms

Definition

A differential k-form (on a manifold M) is a smooth section of the k^th exterior power of the cotangent bundle (of M). At each point p, a differential k-form defines a k-form Note-8

The set of all differential k-forms on a manifold M Ω^k(M) is a vector space. At each point, k-forms inherits most properties from the exterior algebra.

![ROUNDOUT] Can be understood as assigning each point an alternating k-linear functional: the functional is smooth. This allows us to do integrals: the functional tells us how much to accumulate over a tiny bit of k-dimensional space. The orientation allows us oriented integrals.

A k-form on is a family of real-valued functions that are k-linear alternating.

Given basis vectors for , a differential k-form can be written where each is a smooth function .

EXAMPLES

Motivation

Let U be an open set in ℝⁿ. A differential 0-form is defined to be a smooth function f on U. If v is any vector in Rⁿ, then f has a directional derivative ∂_vf(p). In particular, choosing a set of (orthogonal) coordinates, for . (Note that partial derivatives depend on the choice of coordinates). Now is a linear function of v, thus we may define as a linear functional on tangent vectors. This is the prototypical example of a 1 form. By linearity, df is determined by . Note that the coordinates/projections are functions and thus define 1-forms . It follows Note-6. More generally, is a differential form when are smooth functions.

Now given a 1-form α, when does ? This reduces to finding a function whose partial derivatives match those of the expansion above of α. By smoothness of f, α must have

A 1-form is then a linear-functional on the tangent-space of a manifold. It is coordinate-independent, but in order to work with it, we understand it in terms of coordinate charts i.e. w=3dx+dy

Intrinsic definition

Exact

A differential form is exact if it is the differential of some function .

ℝⁿ

Derivation

Given an algebra A over a ring K, a K-derivation is a K-linear map The collection of all K-derivations of A to itself is denoted by Der_K(A).

Graded algebra

Given a graded algebra A and a homogeneous linear map D of grade |D| on A, D is a homogeneous derivation if for all homogenous elements where ϵ=±1. When 1, just standard definition. When -1, D is called an anti-derivation.

Exterior derivative

Inner product

Hodge Dual

The Hodge dual can be simply defined so that

Vector calculus identities

Inner product

Given (for each k) an inner product for the space of k-forms, we can generalize the previous definition wrt to an inner product: is defined as the unique (n-k)-form:

Consider the matrix of an inner product on the one-forms . This inner product lifts(?) to the following inner product on k-forms, determined via the k-blades: where

COMPUTATION

Maxwell’s equations

Let . Note that . One finds that . Let . Setting to each of the four basic 1-forms we find .

Coordinates

See

The cartesian coordinate system consists of writing a position as the linear combination of 3 fixed orthonormal vectors. Note that in every coordinate system we should have where is the position, the coordinates. Coordinate systems in general differ from the Cartesian system by the fact that B is time-dependent. In a polar coordinate system, our basis vectors are also orthonormal, but vary with the position of the “special” particle p₀ (e.g. expression for position vector never involves ). In converting between different coordinate systems (e.g. polar), given scalars or the Cartesian coordinates of vector quantities of the system, we first want the polar coordinates of those vectors, then to determine the basis vectors, either in terms of Cartesian coordinates or in relations with other quantities (the coordinates of p₀). Together, they will be able to describe mechanics of the system in a useful way. It’s important to distinguish between functions whose inputs are vectors and whose inputs are coordinates. A change in the coordinate does not always correspond to a proportional change in the corresponding basis vector.

Archimedean spiral

Notation: Standardize?

Cylindrical

Definition/Converting quantities: We want to use these quantities in our coordinates in the cylindrical basis.

Converting basis vectors: Note that both basis vectors only depend on angle. Thus use this system when you wish to isolate the azimuthal angular component.

Time derivatives of basis vectors: Since our basis vectors rotate staying fixed relative to one another, for each basis vector, where is the angular velocity all the basis vectors/the spherical basis.

Position, velocity, acceleration:

. Applying change of basis matrix, ,

vs fictitious forces?

Gradient

In polar coordinates, i.e. for horizontal v,

r=(r₀e^βt, ω) has no radial acceleration for β=±ω.

Spherical

Definition:

Kugelkoord-lokale-Basis-s

Converting basis vectors: THETA AND PHI ARE REVERSED in the matrix

Time derivatives of basis vectors: Note-9

Position, velocity, acceleration:

Integrals

Let our curve be given by in cylindrical coordinates. The Euclidean distance of each infinitesimal segment of the arc can be given by . Then

Integral of acceleration in polar coords?

Differential equations

https://math.stackexchange.com/questions/2015872/what-is-the-differential-form-of-an-ode

https://math.stackexchange.com/questions/1577421/the-connection-between-differential-forms-and-ode

Coordinate change

Vector fields

See

As a differentiation operator

Vector fields determine a linear map given by differentiating in the direction of the vector field

The vector field describes a rotation . The calculation shows is invariant under rotation.

A standard notation is thus to write for the basis/coordinate vectors of the range. i.e. Every vector field can be written as .

Smooth vector fields form a a module over the ring of smooth functions.

Transformations

Scalar fields

Vector fields

Let V be a vector field. Let be two bases, with each a function of x. Then V is required to satisfy the (contravariant) transformation law

Covector fields

Stokes

Let M be an n-dimensional smooth manifold with boundary, and let ω be a (n−1)-form that is smooth on M. . If you have the derivative of a form, you can calculate its integral by doing a lower dimensional integral over the boundary. That is, the boundary operator is the formal adjoint of the differential operator.

ROUNDOUT
In practical terms, Faraday’s and Ampere’s law is how I conceptualize it, even if it is a simple case relating line integrals to surface integrals. Magnetostatically, your magnetic field tends towards closed loops, the strength of which relates to the flux of charge through the loop. Also the EMF around a closed loop like a piece of wire is going to be zero unless there are some changing magnetic fields passing through the loop; the basis of transformers and our entire power grid.

An m-cell is the image of a differential map . An m-chain is formal linear combination of m-cells . The boundary of a m-cell is the image . (Take formal difference of the slices corresponding to the 1-chains at 0, 1)

EXAMPLE

PROOF
We prove for a “simple” (m-1)-form over .
Next we can apply to by pullbacks. Next we can apply over m-cells by pullbacks.

![COMPUTATION]

Roadmap

Poincare duality for de Rham cohomologies, CW-complexes or even Atiyah-Signer theorem