A crash course on manifolds

Brynjulf Owren

July 14, 2015

Manifold. Let M be a set. A chart (U,ϕ) is a pair such that

· U ⊂M

· ϕ : U → ϕ(U) ⊂ Rn is a bijective map

· φ(m) = (x1, . . . , xn) are called coordinates of the point m,

Suppose (U,ϕ), (U ′, ϕ′) are overlapping, and set

V = ϕ(U ∩ U ′) ⊂ Rn, V ′ = ϕ′(U ∩ U ′) ⊂ Rn.

The charts (U,ϕ) and (U ′, ϕ′) are called compatible if

ϕ′ ϕ−1 : V → V ′ and ϕ (ϕ′)−1 : V ′ → V are C∞

Looking at the figure, the shaded regions at the bottom are both in Rn there-fore differentiability is well defined. M is a differentiable manifold of dimensionn if

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1. There is a collection of charts such that each m ∈ M is a member of atleast one chart

2. M is a union of compatible charts

Examples of manifolds are: Rn, the n-dimensional sphere, the set of invert-ible n× n-matrices etc.

In mechanics, the manifold of generalized positions is called the configurationmanifold

Tangent space. There are several equivalent definitions of tangent vectorsand tangent space, we shall just mention two.

1. Curves approach. We begin by using a fact that we shall not prove,namely that all manifolds can be embedded into some Euclidean spaceRm. This assumption is not necessary to define tangent vectors, but itmakes things easier for (some of) our purposes. Note that in this case onetypically describes M by introducing m− n constraints on the vectors ofRm. For example, we may embed the two-sphere into R3 by representingits elements as three-vectors x such that x2

1 + x22 + x2

3 = 1.

Consider a curve γ(t) ∈ M where γ(0) = m ∈ M . The velocity vectorv = γ(0) is a tangent vector at m, and we say that v ∈ TmM . Thecollection of all vectors obtained by differentiating differentiable curvesthrough M gives TmM .

Two curves will give the same tangent vector if they both pass through mwith the same “direction”. So a tangent vector is an equivalence class ofcurves, two curves γ(t) and ι(t) being equivalent if

γ(0) = ι(0) = m, γ(0) = ι(0)

2. Derivation approach. A tangent vector can be associated to a differ-ential operator acting on (germs of) functions on M . By this we mean:v ∈ TmM whenever

· v is a linear operator v : F(M) → R where F(M) is the ring ofsmooth functions on M .

· v[αf + βg] = αv[f ] + βv[g] (linearity)

· v[fg] = v[f ]g(m) + f(m)v[g] (derivation property)

It turns out that there is a one-to-one correspondence between such oper-ators and the tangent vectors defined with the curves approach. A simpleway to understand tangent vectors as derivation is to think of v[f ] as adirectional derivative of the function f in the direction v. More precisely,

v[f ] =d

dt

∣∣∣∣t=0

f(γ(t)), γ(0) = m, γ(0) = v.

In local coordinates, x1, . . . , xn on M , it takes the form

v1∂

∂x1+ · · ·+ vn

∂

∂xn

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Example 1 The 2-sphere can be represented as 3-vectors with unit Euclideannorm, i.e. x ∈ R3 : x2

1 +x22 +x2

3 = 1. So the components γi(t) of such a curvemust satisfy

3∑i=1

γi(t)2 = 1

Differentiating wrt. t, we obtain

3∑i=1

γi(t)γi(t) = 0

If at t = 0, we have γ(0) = r and γ(0) = v, we get the obvious result that thetangent space at x are precisely those vectors v which are orthogonal to theradius vector r.

Example 2 The Euclidean space Rn has TxRN ' Rn as its tangent space.There exists curves x+ tv for any v ∈ Rn.

Example 3 The set of all orthogonal n×n-matrices is a manifold which containsthe identity matrix, I. Consider a curve in this manifold, γ(t) through I, i.e.γ(0) = I. Assume γ(0) = v. We have for all t that γ(t)T γ(t) = I. Then

d

dtγ(t)T γ(t) = γ(t)T γ(t) + γ(t)T γ(t)

At t = 0 this reduces to vT + v = 0, the tangent space at I of the orthogonalmatrices are precisely the skew-symmetric ones.

Exercise 1 The space of all symmetric n × n-matrices is a manifold, what isits dimension, and what is the tangent space at an arbitrary point (symmetricmatrix) S?

Dual spaces and the cotangent space. In a linear space (vector space),V , there is the notion of a dual space, referring to the set of linear functions onV . Oftentimes this space is denoted V ∗. We shall here only be concerned withreal vector spaces, but in principle the definitions are valid for any field. Weshall also most of the time be working with linear spaces of finite dimension.

So the dual space consists of all forms f such that f(αu+βv) = αf(u)+βf(v)where α and β are real numbers. Note that if dimV = d, one can use a basise1, . . . , ed such that any element v ∈ V can be expressed in terms of the basis,

v = v1e1 + · · · vded, each vi ∈ R

We can then also find a basis for the dual space V ∗ by choosing linear functionsεi, 1 ≤ i ≤ d having the property that εi(v) = vi when v is expressed as above.This is equivalent to requiring εi(vj) = δij for all i, j.

The tangent space over a given point m ∈ M was denoted TmM . This isa linear space per se, and therefore has a dual space. This dual space is calledT ∗mM and consists of all linear forms on TmM , these forms we call cotangent

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vectors. We shall later see that whereas velocity vectors correspond to tangentvectors, the momentum vectors correspond to cotangent vectors.

Notation. For f ∈ V ∗, v ∈ V , we may write f(v) ∈ R, but often one also seesthe notation 〈f, v〉. Note that since V ∗ is a linear space in its own right, it alsohas a dual space (V ∗)∗. But as it turns out, (V ∗)∗ ' V . In fact, if w ∈ (V ∗)∗,we identify it with v ∈ V whenever 〈w, f〉 = 〈f, v〉 for all f ∈ V ∗. Then it isclear that under this identification the ordering of the arguments in a dualitypairing is irrelevant, i.e. 〈f, v〉 = 〈v, f〉.

Cotangent vectors naturally appear when we compute directional derivativesof functions. Recall the derivation approach to tangent vectors where v[f ] ∈ Ris the directional derivative of f in the direction of v. Since this operationbehaves linearly also in the tangent vector, we can interpret this operation as alinear function on the tangent vector v ∈ TmM , a common notation is to definedfm(v) = v[f ], so that dfm ∈ T ∗mM . The definition can be extended to a linearfunction on any of the tangent spaces, then we simply remove the superscript andwrite df . In Lagrangian mechanics, we work with the Lagrangian L(q, q) andwe need to compute directional derivatives with respect to q and q respectively.In this case, the notation dL would be inconclusive so we then write

∂L

∂q(q, q),

∂L

∂q(q, q),

and they will both be cotangent vectors.

Tangent and cotangent bundles. We have seen that for every m ∈ M wehave a tangent space TmM which is linear. For a differentiable manifold, thetangent space for all m ∈M can be glued together to form the tangent bundle

TM =⋃

m∈MTmM

This is in fact per se a manifold of twice the dimension as M , it is generally nota linear space. An element of TM is a tangent vector vx ∈ TxM at some pointx ∈ M . Local coordinates on TM are induced by coordinates on M . Supposea point x ∈ M has coordinates (x1, . . . , xn). By differentiating curves throughx, say γ(t), expressed in this coordinate system, we obtain vectors vx = γ(0) =vx,1, . . . , vx,n as explained before, and the 2n numbers (x1, . . . , xn, v1, . . . , vn)are taken as local coordinates for TM . So we may think that x determineswhich tangent space TxM we are considering and vx ∈ TxM is the tangentitself.

Maps and tangent maps. Suppose M and N are differentiable manifolds.Let Ψ : M → N be a smooth map. Such a map induces a derivative or tangentmap TΨm between tangent spaces TmM and TΨ(m)N . The curves approachis again useful for defining this. On the figure to the left, we have drawn thepoint m ∈ M on a curve γ(t), and this curve is defined such that γ(0) = mand γ(0) = v. Then the whole curve is mapped by Ψ over to N , where clearlyΨ(γ(0)) = Ψ(m). The tangent of the target curve defines the tangent map

d

dt

∣∣∣∣t=0

Ψ(γ(t)) =: w = TΨm(v)

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One meets different notations for TΨm in the literature, one is DΨm, anotheris Ψ′|m, or simply Ψm. Another curiosity is that if we choose the manifold N tobe R (yes it is also a manifold), then each tangent space of R can be identifiedwith R. So we simply get

TΨm(v) =d

dt

∣∣∣∣t=0

Ψ(γ(t)) = v[Ψ] = dΨm(v)

Later, we shall skip the reference to the base point of the map, i.e. we shalljust write TΨ, and it will act on tangent vectors at any tangent space. Thenwe shall also use the notation TΨ =: Ψ∗.

Exercise 2 Assume that local coordinates (x1, . . . , xm) have been introducedon M and similarly, (y1, . . . , yn) on N . This means that the map y = Ψ(x) canbe expressed as y1

...yn

=

Ψ1(x1, . . . , xm)...

Ψn(x1, . . . , xm)

Show, by using the definition of the tangent map that TΨ is just the Jacobianmatrix of Ψ, i.e.

TΨ =

∂Ψ1

∂x1· · · ∂Ψ1

∂xm

......

...∂Ψn

∂x1· · · ∂Ψn

∂xm

Vector fields and differential one-forms. As described earlier, the tangentand cotangent bundles of a manifold are again manifolds. An important objectis the natural projection π : TM → M such that π(q, vq) = q. We shalldefine a section of the (co)tangent bundle to be a map F : M → TM (respF : M → T ∗M) such that π F = IdM . This means that F assigns to m ∈M avector in the tangent space TmM (resp T ∗mM). Sections of the tangent bundleare called vector fields whereas sections of the cotangent bundle are differentialone-forms. We denote the set of smooth vector fields on the manifold by X (M)and the set of differential one-forms by Ω1(M).

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Example 4 An example of a vector field is the right hand side of a differentialequation. An example of a differential one-form is the differential (or directionalderivative operator) of a function.

By using the interpretation of tangent vectors as derivations, we can nowthink of vector fields as maps X : F(M) → F(M) where F(M) is the ringof smooth functions on M . From this viewpoint, the vector fields are linearderivations, i.e.

X[fg] = X[f ]g + fX[g]

Given coordinates (x1, . . . , xn) on the manifold. Then any vector field can bedescribed in terms of its coordinates X = (X1(x), X2(x), . . . , Xn(x)) and thecorresponding operator would be

X1(x)∂

∂x1+ · · ·+Xn(x)

∂

∂xn

We can also “pair” together smooth one-forms and vector fields to obtain func-tions, e.g. dH(X) ∈ F(M). We can write any one-form in terms of the differ-entials of coordinate functions as

θ = a1(x) dx1 + · · ·+ an(x) dxn (1)

If a tangent vector v is expressed in this system with coordinates (v1, . . . , vn)then dxi(v) = vi. A subspace of the one-forms are those which are the differen-tial of a function, their coordinate expression is

dH =∂H

∂x1dx1 + · · ·+ ∂H

∂xndxn

k-forms and the exterior product A k-form on a vector space V is a mapfrom V × · · · × V → R (k copies of V ), which is multilinear and completelyskew-symmetric, meaning that whenever two arguments are trading place, thesign of its value is negated, e.g. ω(ξ1, ξ2, . . . , ξn) = −ω(ξ2, ξ1, . . . , ξn). Thismeans in particular that if two arguments are identical, then the value is zero.

Exercise 3 Show that any k-form on Rn with n < k is identically zero.Hint. At least one of the arguments must be a linear combination of the other.

There is an important operation called the wedge-product, if ω and τ arek- and `-forms, then ω ∧ τ is a k + `-form. The interested reader can consultthe book by Tu[1] for a general definition of this product, but let us review theconstruction of a two-form ω from the wedge product of two one-forms. θ andτ , i.e. ω = θ ∧ τ

ω(ξ1, ξ2) = θ ∧ τ(ξ1, ξ2) = θ(ξ1)τ(ξ2)− θ(ξ2)τ(ξ1)

It is easy to check that this ω is bilinear and skew-symmetric. Moreover, weobserve that this wedge product itself is skew-symmetric θ ∧ τ = −τ ∧ θ. Givena basis ε1, . . . , εn for V ∗, we may construct a basis for the set of two-forms asfollows

εi ∧ εj , 1 ≤ i < j ≤ n,so the linear space of two-forms on a n-dimensional linear space is of dimension12n(n− 1).

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Differential k-forms and the exterior derivative. A differential k-formon M is a smoothly varying k-form on each tangent space TmM . We havealready seen in (1) how all such forms can be expressed in local coordinates.Functions on a manifold are also called 0-forms. So the operator d maps 0-formsto 1-forms. But d can be generalized to any positive integer k, such that it mapsk-forms to (k + 1)-forms. One has

1. df is the differential of f for smooth functions f

2. For any k-form α where k ≥ 0, one has d(dα) = 0

3. Derivation property. For any k-form α and `-form β

d(α ∧ β) = dα ∧ β + (−1)kα ∧ dβ

The last property can be used to find a formula for applying the exterior deriva-tive to any k-form expressed in local coordinates. For instance, if θ is given by(1) then

dθ = d(∑

aidxi

)=∑

d(ai ∧ dxi) =∑

dai ∧ dxi

since by 2., ddxi = 0.

Example 5 For symplecticity, the one-form τ =∑

i pidqi plays an importantrole. Its differential is

ω = dτ = d(∑i

pi ∧ dqi) =∑i

dpi ∧ dqi

This is the canonical symplectic two-form.

References

[1] L. W. Tu. An introduction to manifolds. Universitext. Springer, New York,second edition, 2011.

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