The set of all linear functionals, denoted by V∗, forms a vector space

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JIA-MING (FRANK) LIOU

Let K be a field. All the vector spaces in this note are over K.

1. Some Linear Algebras

Let V be an n-dimensional vector space over K. A linear functional on V is a linear map ϕ : V → K. The set of all linear functionals, denoted by V, forms a vector space. The addition and scalar multiplication are given as follows. Let ϕ, ψ ∈ V, and a ∈ k. Define ϕ + ψ and aϕ by

(ϕ + ψ)(v) = ϕ(v) + ψ(v), (aϕ)(v) = aϕ(v).

Theorem 1.1. V has dimension n.

Proof. Let {v1, · · · , vn} be a basis for V. Let ψi : V → K such that ψi(vj) = δij. Claim that the set β = {ψi: 1 ≤ i ≤ n} forms a basis for V.

For any ϕ ∈ V, one can check that ϕ =

n

X

i=1

ϕ(eii. Then ϕ is a linear combinaton of elements of β.

Assume that Pn

i=1aiψi = 0 for some a1, · · · , an ∈ K. Evaluate the sum for vk, we see that ak= 0. This shows that β is linearly independent.

 Remark. The basis {ψi} for V in Theorem (1.1) is called the dual basis of {vi}.

Let ϕ1, ϕ2 ∈ V. We define a map ϕ1∧ ϕ2: V × V → K by (ϕ1∧ ϕ2)(w1, w2) = det [ϕi(wj)]2i,j=1.

We can check that ϕ1∧ ϕ2 : V × V → K is bilinear, and skew-symmetric, i.e. ϕ2∧ ϕ1 =

−ϕ1 ∧ ϕ2. We call ϕ1∧ ϕ2 the wedge product of ϕ1 and ϕ2. Let Λ2V be the set of all skew-symmetric bilinear maps from V × V to K.

Theorem 1.2. Λ2V is a vector space of dimensionn 2

 .

Proof. Let {vi} be a basis for V and {ψi} be its dual basis. Claim that {ψi∧ ψj : 1 ≤ i <

j ≤ n} forms a basis for V.

Suppose that

X

i<j

aijψi∧ ψj = 0.

Evaluate the equation for (vk, vl), we obtain akl= 0.

Let B be a skew-bilinear map. Denote bij = B(vi, vj). Then bij = −bji. For any x = Pn

i=1xivi and y =P

iyivi, we see that B(x, y) =X

i,j

bijxiyj 1

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=X

i<j

bijxiyj+X

i>j

bijxiyj

=X

i<j

bij(xiyj − xjyi).

Note that (ψi∧ ψj)(x, y) = (xiyj − xjyi). Hence we find B(x, y) = (X

i<j

bijψi∧ ψj)(x, y).

Hence we prove that B =P

i<jbijψi∧ ψj. 

Definition 1.1. Let p be a natural number and Sp be the symmetric group on p-letters.

A map f : ×pi=1V → K is alternating if

f (vσ1, · · · , vσp) = (sgn σ)f (v1, · · · , vp), for any v1, · · · , vp ∈ V. f is called p-linear if for each 1 ≤ k ≤ p,

f (v1, · · · , avk+ bwk, · · · , vp) = af (v1, · · · , vk, · · · , vp) + bf (v1, · · · , wk, · · · , vp), for any a, b ∈ K.

Let 1 ≤ p ≤ n. Given ϕ1, · · · , ϕp ∈ V, we define ϕ1∧ · · · ∧ ϕp : ×pi=1V → k by (ϕ1∧ · · · ∧ ϕp)(w1, · · · , wp) = det [ϕi(wj)]pi,j=1,

for all w1, · · · , wk ∈ V. Then we can check that ϕ1∧ · · · ∧ ϕp is p-linear and alternating.

Let ΛpV be the set of all p-linear alternating maps from ×pi=1V to k and denote Λ0V= V.

Theorem 1.3. For 0 ≤ p, ≤ n, the set ΛpV forms a vector space of dimensionn p

 . Proof. Let {vi} be a basis for V and {ψi} be a basis for V dual to V. We only need to show that the set {ψi1 ∧ · · · ∧ ψip : 1 ≤ i1< · · · < ip ≤ n} forms a basis for ΛpV.

 2. Tangent Vectors

Let p be a point in Rn. We denote (v)p = p + v, where + is the addition in Rn. Let TpRn be the set of all (v)p with v ∈ Rn. We define the addition and scalar multiplication on TpRn as follows. Let (v)p, (w)p ∈ TpRn, and a ∈ R. We define

(v)p+ (w)p = (v + w)p, a · (v)p= (av)p.

Note that the addition and scalar multiplication here are different from those of Rn. Proposition 2.1. The set TpRn forms an n-dimensional real vector space.

Proof. The proof is obvious. Let {ei : 1 ≤ i ≤ n} be the standard basis for Rn. Then {(ei)p: 1 ≤ i ≤ n} forms a basis for TpRn.

 Remark. We will use the notation vp for (v)p.

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We say that TpRn is the tangent space to Rn at p and elements of TpRn are tangent vectors at p. The dual space of TpRn is denoted by TRn and called the cotangent space of Rn at p.

Let U be an open set on Rn. The algebra of (real-valued) smooth functions on U is denoted by C(U ). Let p ∈ U and f ∈ C(U ), we define the directional derivative of f at p along a vector v ∈ Rn by

vp[f ] = d

dtf (p + tv) t=0

.

Let {ei : 1 ≤ i ≤ n} be the standard basis for Rn. Assume that v = Pn

i=1viei. From calculus, we know

vp[f ] =

n

X

i=1

vi

∂f

∂xi

(p).

Then we know

(1) vp : C(U ) → R is a linear functional, i.e. vp(af + bg) = avp(f ) + bvp(g) for all a, b ∈ R and

(2) vp(f g) = vp(f )g(p) + f (p)vp(g).

Note that if the tangent vector is given, the directional derivative only depends on the partial derivatives of the function at the point p. Suppose f, g are two smooth functions defined on some open sets, (not necessarily the same). If f and g agree on an open set containing p, then vp(f ) = vp(g). This leads to another definition of tangent vectors of Rn. Let V, W be open sets containing p in Rn. Let f be a smooth function on V and g smooth function on W. We say that the pair (f, V ) is equivalent to the pair (g, W ) if there exists an open set Z ⊂ V ∩ W contain p such that f = g on Z. The set of all equivalent classes [(f, V )]

is denoted by Cp. For simplicity, we will simply denote [(f, V )] by [f ]. An equivalent class [f ] is called a germ at p.

Proposition 2.2. The set Cpforms an algebra over R.

Let [f ] ∈ Cp. We define [f ](p) = f (p), where f is a representative of [f ]. Then [f ](p) is well-defined. Now, if f1, f2 belong to the same germ [f ] at p, we can check that vp(f1) = vp(f2). Hence we can define vp([f ]) = vp(f ) where f is a representative of [f ].

Definition 2.1. A point derivation δp at p is a linear functional on Cp such that δp([f ][g]) = δp([f ])[g](p) + [f ](p)δp([g]).

The set of all point derivations at p is denoted by Tp0Rn.

Remark. By definition, all tangent vectors are point derivation at p.

Given [f ] ∈ Cp, we define

∂xi

(p)[f ] = ∂f

∂xi

(p) for a representative f of [f ]. Then we know that ∂

∂xi(p) is a point derivation.

Theorem 2.1. The set

 ∂

∂xi(p) : 1 ≤ i ≤ p



forms a basis for Tp0Rn; Then Tp0Rn is an n-dimensional vector space.

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Proof. It is easy to verify that the set is linearly independent.

Let δp ∈ TpRn. Denote δp[xi] = vi. Let [f ] ∈ Cp. Choose a representative f of [f ].

Consider the Tayler expansion f (x) = f (p) +X

i

∂f

∂xi

(p)(xi− pi) +X

i,j

(xi− pi)(xj− pj) Z 1

0

(1 − t) ∂2f

∂xi∂xj

(p + t(x − p))dt.

Using the properties of tangent vectors, we find δp[f ] =X

i

vi

∂xi

(p)[f ].

In other words, we find δp =X

i

vi

∂xi

(p). 

We know that TpRnis a vector subspace of Tp0Rn. Since they both have the same dimen- sion, they must be equal. In fact, by elementary calculus,

(ei)p[f ] = ∂f

∂xi(p).

We find that (ei)p = ∂

∂xi

(p) by definition. Hence we conclude that the notion of tangent vectors are equivalent to the notion of point derivations.

Remark. Since we always choose a representative of a germ at p, from now on, we will simply use the notation f for [f ].

Let f be a germ at p. We define a linear functional dfp: TpRn→ R by dfp(vp) = vp(f )

Then dfp ∈ TpRn. Notice that for any vp = (v1, · · · , v)p ∈ TpRn, we have (dxi)p(vp) = vi. This implies that (dxi)p((ei)p) = δij for all i, j. Hence {(dxi)p : 1 ≤ i ≤ p} is the dual basis to {(ei)p: 1 ≤ i ≤ n}. We conclude that:

Theorem 2.2. The set {(dxi)p: 1 ≤ i ≤ n} forms a basis for TpRn. 3. Tangent Maps

Let U be an open subset of Rn. Suppose F : U → Rm is a smooth map, where F = (F1, · · · , Fm). For each p ∈ U, we can define a linear map, called the tangent map,

dFp : TpRn→ TF (p)Rm as follows.

Let f be a germ at F (p). Then f ◦ F is a germ at p. Given any vp∈ TpRn, we set (dFp(vp)) (f ) = vp(f ◦ F ).

Suppose F = F (y1, · · · , ym). By chain rule, ∂

∂xi(p)(f ◦ F ) =

m

X

j=1

∂f

∂yj(F (p))∂Fj

∂xi(p). By definition, dFp

 ∂

∂xi

(p)

 (f ) =

m

X

j=1

∂Fj

∂xi

(p)∂f

∂yj

(F (p)). Hence we see that

dFp

 ∂

∂xi(p)



=

m

X

j=1

∂Fj

∂xi(p) ∂

∂yj(F (p)).

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If we choose the standard basis for TpRnand TF (p)Rm, the tangent map dFp is represented by the Jacobi matrix of F at p.

Let U be an open set in Rn and F : U → Rm be a smooth map. Assume that V is an open set in Rm containing F (U ) and G : V → Rkis a smooth map. For any x ∈ U, we have (chain rule)

(3.1) d(G ◦ F )x: TxRn −−−−→ TdFx F (x)Rm

dGF (x)

−−−−→ TG(F (x)). 4. Differential Forms on Rn

Let U be an open set in Rn. We denote T U = [

p∈U

TpRn, TU = [

p∈U

TpRn.

Then we find that T U can be identify with U ×Rnand TU can be identified with U ×(Rn). We have a natural projection π : TU → U.

Let f be a smooth function on an open set U ⊂ Rn. The total derivative of f has the formal expression:

df =X

i

∂f

∂xidxi.

Since f is smooth, all the partial derivatives of f are all smooth. We shall use this idea to define the notion of one-form. A smooth one form ω on U is a formal expression

ω =X

i

ωidxi

with ωi ∈ C(U ). Now we can think of ω as a map from U to TU. In fact, for each p ∈ U, ω(p) =P

iωi(p)(dxi)p ∈ TpRn. Moreover, (π ◦ ω)(p) = p for all p ∈ U. Therefore a smooth one-form ω is a smooth map ω : U → TU such that π ◦ ω = 1U.

Let ΛkTU =S

p∈U ΛkTpRn. We can also consider the natural projection1 π : ΛkTU → U. We can check that ΛkTU can be identified with U × Λk(Rn).

Definition 4.1. A smooth k-form on U is a smooth map η : U → ΛkTU such that π ◦ η = 1U. The set of all smooth k forms on U is denoted by Ωk(U ). We denote Ω(U ) = L

k≥0k(U ).

By definition, a smooth k-form has the formal expression

η = X

1≤i1<···<ik≤n

ηi1···ikdxi1 ∧ · · · ∧ dxik with ηi1···ik ∈ C(U ).

Let I = (i1, · · · , ik) be a k-tuple. We write

dxI = dxi1 ∧ · · · ∧ dxik. A k-form η is also denoted by η =P

IηIdxI. The sum of two k-forms on U ω =P

IωIdxI and η =P

IηIdxI is defined to be

ω + η =X

I

(ω + η)dxI.

1Let us use the notation π for the projection but we have to know that this projection is different from the above.

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For any smooth function f on U, we define f ω =X

I

(f ωI)dxI. One can check that

Proposition 4.1. The set Ωk(U ) is a free C(U )-module of rankn k

 .

On Ω(U ), we define the exterior (wedge) product as follows. Given a s-form ω = P

IωIdxI and a k-form η =P

JηJdxJ, we define a s + k form by ω ∧ η =X

I,J

ωIηJdxI∧ dxJ.

The exterior product of forms on U has the following properties.

Theorem 4.1. Let ω, η, and θ be k, s, r-forms respectively. Then (1) (ω ∧ η) ∧ θ = ω ∧ (η ∧ θ),

(2) ω ∧ η = (−1)ksη ∧ ω,

(3) If r = s, then ω ∧ (η + θ) = ω ∧ η + ω ∧ θ.

Proof. Exercise. 

Let ω =P

IωIdxI. We define the exterior derivative dω of ω by dω =X

I

I∧ dxI.

Proposition 4.2. d : Ω(U ) → Ω(U ) is a linear map such that (1) d : Ωk(U ) → Ωk+1(U )

(2) d(ω ∧ η) = dω ∧ η + (−1)kω ∧ dη (3) d2 = 0.

Proof. Exercise. 

Let dk= d|k(U ). Then d2 = 0 implies that dk+1dk = 0. In other words, Im dk−1⊂ ker dk. Definition 4.2. The k-th de Rham cohomology of U is the quotient space defined by

Hk(U ) = ker dk/ Im dk−1.

Later, we will study more about the de Rham cohomology of a smooth manifold.

4.1. Pull back on Differential Forms. Let F : U ⊂ Rn→ Rm be a smooth map. Then F induces a linear map F from Ωk(V ) to Ωk(U ), where V is an open set containing F (U ).

Given a k-form ω on V, with k ≥ 1, we define a k-form Fω on U by

(Fω)(p)(v1, · · · , vk) = ω(F (p))(dFp(v1), · · · , dFp(vk)), ∀p ∈ U, where v1, · · · , vk ∈ TpRn. For k = 0, set Fg = g ◦ F.

Proposition 4.3. Let F : U ⊂ Rn→ Rm and g ∈ C(U ). Suppose ω, η are k-forms. Then (1) F(ω + η) = Fω + Fη,

(2) F(gω) = (Fg)(Fω),

(3) If ϕ1, · · · , ϕk are one-forms, F1∧ · · · ∧ ϕk) = Fϕ1∧ · · · ∧ Fϕk.

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Let us assume F = (F1, · · · , Fm), i.e. Fi = yi ◦ F. Suppose that ω = P

IωIdyI. Then Fω =P

IFωIFdyI. Using the properties, we have Fdyi1∧· · ·∧Fdyik = dFi1∧· · ·∧dFik. Then

Fω =X

I

ωI(F1, · · · , Fm)dFi1∧ · · · ∧ dFik. Using this identity, it is easy for us to prove the following corollary.

Corollary 4.1. Let ω, η be forms. Then F(ω ∧ η) = Fω ∧ Fη.

Proposition 4.4. Let F : U ⊂ Rn→ Rm and G : V ⊂ Rm → Rk, where V is an open set containing F (U ). Then

(G ◦ F )ω = F(Gω) for all k-forms ω.

Proposition 4.5. Let F : U → Rm be a smooth map and ω be a k-form. Then d(Fω) = Fdω.

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