Let p be an element of Rn

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All the vector spaces in this note are all real vector spaces.

The set of n-tuples of real numbers is denoted by Rn. Suppose that a is a real number and x = (x1, · · · , xn) and y = (y1, · · · , yn) are elements of Rn. We define the sum x +0y and the scalar product a ·0x by

x +0y = (x1+ y1, · · · , xn+ yn), a ·0x = (ax1, · · · , axn).

It is well known that the set Rntogether with the addition +0 and the scalar multiplication

·0 forms a vector space (Rn, +0, ·0). In fact, the set Rn can be equipped with other vector space structures. Let p be an element of Rn. For each element x of Rn, we represent x as a sum p +0v where v = x −0p. We denote p +0v by vp or (v)p. If x = (v)p and y = (w)p are elements of Rn and a ∈ R is a real number, we define x +py and a ·px by

x +py = (v + w)p, a ·px = (av)p.

Lemma 1.1. The set Rn together with the addition +p and the scalar multiplication ·p

forms a vector space.

Proof. This is left to the reader as an exercise. 

The vector space (Rn, +p, ·p) is denoted by TpRn and called the tangent space to Rn at p. We remark that T0Rn is the usual Rn we have learned before. One can also see that the vector space TpRnis isomorphic to Rnfor each p ∈ Rn. This can be proved by constructing a linear isomorphism from TpRn onto Rnby sending vp to v. If {e1, · · · , en} is the standard basis for Rn, {(e1)p, · · · , (en)p} forms a basis for TpRn.

We can also identify TpRn with the subset {(p, v) ∈ Rn× Rn : v ∈ Rn} of Rn× Rn by sending vp to (p, v). Instead of using the notation (Rn, +p, ·p) for TpRn, we may use the identification

TpRn= {(p, v) ∈ Rn× Rn: v ∈ Rn}.

On TpRn, we define the addition and the scalar multiplication by (p, v) +p(p, w) = (p, v + w), a ·p(p, v) = (p, av).

Furthermore, we introduce an inner product on TpRnby h(p, v), (p, w)iTpRn = hv, wiRn

where the right hand side of the equation is the standard Euclidean inner product of v, w on Rn.

Definition 1.1. Let U be an open subset of Rn. The set T U = S

p∈UTpRn is called the tangent bundle over U. A vector field on U is a function V : U → T U such that V (p) ∈ TpRn for any p ∈ U.

By definition, T U = U × Rn as a set.

Let U be an open subset of Rn and f : U → R be a C1-function. Let p ∈ U and vp ∈ TpRn. We define dfp(vp) to be the directional derivative of f at p along v, i.e.

dfp(vp) = d

dtf (p +0tv) t=0

. By chain rule, we know that

dfp(vp) = Df (p)(v).

1

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Here Df (p) : Rn → R is the derivative of f at p. For each vp, wp ∈ TpRn and a, b ∈ R, by linearity of Df (p), we obtain that

dfp(avp+ bwp) = dfp((av + bw)p)

= Df (p)(av + bw)

= aDf (p)(v) + bDf (p)(w)

= adfp(vp) + bdfp(wp).

This shows that dfp: TpRn→ R is a linear map.

Definition 1.2. Let V be a finite dimensional vector space. A linear map ψ : V → R is called a linear functional on V. The set of all linear functionals denoted by V is called the dual space to V.

By linear algebra, V forms a vector space. The dual space to TpRn is denoted by TpRn. For each 1 ≤ i ≤ n, let xi : Rn→ R be the function defined by

xi(p) = pi where p = (p1, · · · , pn).

Then {xi : 1 ≤ i ≤ n} are smooth functions on Rn and called the rectangular coordinate functions on Rn.

Example 1.1. Let p = (p1, p2, p3) be an element of R3 and {e1, e2, e3} be the standard basis for R3. Then {(e1)p, (e2)p, (e3)p} forms a basis for TpR3. Let vp= (v1, v2, v3)p be any vector in TpR3. For t ∈ R, p +0tv = (p1+ tv1, p2+ tv2, p3+ tv3) and hence

x1(p +0tv) = (p1+ tv1, p2+ tv2, p3+ tv3) = p1+ tv1. By definition,

(dx1)p(vp) = d

dtx1(p +0tv) t=0

= d

dt(p1+ tv1) t=0

= v1.

In general, one can show that (dxi)p(vp) = vi for 1 ≤ i ≤ 3. One can easily see that (dx1)p((e1)p) = 1, (dx1)p((e2)p) = 0, (dx1)p((e3)p) = 0,

(dx2)p((e1)p) = 0, (dx2)p((e2)p) = 1, (dx2)p((e3)p) = 0, (dx3)p((e1)p) = 0, (dx3)p((e2)p) = 0, (dx3)p((e3)p) = 1.

This example motivates the following definition:

Definition 1.3. Let V be an n-dimensional real vector space and {v1, · · · , vn} be a basis for V. A set of linear functionals {ϕ1, · · · , ϕn} on V is called the dual basis to {v1, · · · , vn} providing that

ϕi(vj) = δij, 1 ≤ i, j ≤ n.

Lemma 1.2. Let V be an n-dimensional real vector space and {v1, · · · , vn} be a basis for V. Suppose that {ϕ1, · · · , ϕn} is the dual basis to {v1, · · · , vn}. Then {ϕ1, · · · , ϕn} forms a basis for V.

Proof. Let us first prove that the set is linearly independent. Suppose a1ϕ1+· · ·+anϕn= 0.

By ϕi(vj) = δij, one has

0 = (a1ϕ1+ · · · + anϕn)(vj) = a1ϕ1(vj) + · · · + anϕn(vj) = aj. for 1 ≤ j ≤ n.

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Let ϕ : V → R be a linear functional on V. Define ψ = Pn

i=1ϕ(vjj ∈ V. Then ψ(vi) = ϕ(vi) for all 1 ≤ i ≤ n. Hence ψ(v) = ϕ(v) for any v ∈ V. We show that ψ = ϕ.

Thus ϕ is a linear combination of {ϕ1, · · · , ϕn}. We find that {ϕ1, · · · , ϕn} spans V.  The above lemma also implies that for any basis {v1, · · · , vn} for V and its dual basis {ϕ1, · · · , ϕn}, any linear functional ϕ : V → R on V has the following expression:

(1.1) ϕ =

n

X

i=1

ϕ(vii.

It follows from the definition that the set {(dx1)p, (dx2)p, (dx3)p} is the dual basis to {(e1)p, (e2)p, (e3)p} and hence is a basis for TpR3.

Let U be an open subset of Rn and f : U → R be a smooth function. Let p ∈ U. Since dfp: TpRn→ R is a linear functional, by equation (1.1) and Corollary 1.1, we have

dfp =

n

X

i=1

dfp((ei)p)(dxi)p. Let us now compute dfp((ei)p). By definition,

dfp((ei)p) = d

dtf (p +0tei) t=0

= ∂f

∂xi

(p).

From here, we obtain that

(dxi)p((ej)p) = ∂xi

∂xj

(p) = δij, 1 ≤ i, j ≤ n.

Lemma 1.2 implies the following results:

Corollary 1.1. The set {(dx1)p, · · · , (dxn)p} forms a basis to TpRn; it is the dual basis to {(e1)p, · · · , (en)p}.

Since {(dx1)p, · · · , (dxn)p} is the dual basis to {(e1)p, · · · , (en)p}, by Lemma 1.2 and its remark (1.1), we conclude that

(1.2) dfp =

n

X

i=1

∂f

∂xi

(p)(dxi)p.

Example 1.2. Let f : (a, b) → R be a smooth function and p ∈ (a, b). Then dfp = f0(p)dxp.

Theorem 1.1. (Riesz representation theorem) Let V be a finite dimensional inner product space. For any linear functional ψ : V → R, there exists a unique ξ ∈ V such that ϕ(v) = hv, ξi for all v ∈ V.

Proof. Let us prove the uniqueness first. Suppose ξ and η are two vectors in V such that ϕ(v) = hv, ξi = hv, ηi for all v ∈ V. Let α = ξ − η. Then hv, αi = 0 for all v ∈ V. We choose v = α. Then hα, αi = 0. By the property of inner product, α = 0 and hence ξ = η.

Let us prove the existence. Assume that dim V = n. Choose an orthonormal basis {e1, · · · , en} for V. We set ξ =Pn

i=1ϕ(ei)ei and define ψ : V → R by ψ(v) = hv, ξi. Then ψ is a linear functional on V. Furthermore, for 1 ≤ j ≤ n,

ψ(ej) = hej, ξi =

n

X

i=1

ϕ(ei)hej, eii = ϕ(ej).

This implies that ψ(v) = ϕ(v) for all v ∈ V and thus ψ = ϕ. 

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Remark. Let V be an n-dimensional inner product space and {ei : 1 ≤ i ≤ n} be an orthonormal basis for V. If ϕ : V → R is a linear functional on V, the unique vector ξ so that ϕ(v) = hv, ξi has the following representation:

ξ =

n

X

i=1

ϕ(ei)ei.

Now we are ready to introduce the notion of gradient of a smooth function at a point p.

Since dfp : TpRn→ R is a linear functional and TpRnis a finite dimensional inner product space, there is a unique vector in TpRn, denoted by ∇f (p), such that

dfp(v) = hv, ∇f (p)ip.

The vector ∇f (p) in TpRn is called the gradient vector of f at p.

Definition 1.4. If f : U ⊂ Rn→ R is smooth, ∇f defines a function

∇f : U → T U, p 7→ ∇f (p).

This vector field is called the gradient vector field of f on U.

By Theorem 1.1, and its remark, we see that

∇f (p) =

n

X

i=1

dfp((ei)p)(ei)p. Since dfp((ei)p) = fxi(p), we find that

∇f (p) =

n

X

i=1

∂f

∂xi(p)(ei)p =

n

X

i=1

∂f

∂xi(p)ei

!

p

Example 1.3. Let f : R2 → R be a smooth function and p ∈ R2. Then

∇f (p) = (fx(p), fy(p))p.

If f : U ⊂ Rn → R is smooth, dfp ∈ TpRn for any p ∈ U. We define the notion of cotangent bundle over an open subset U of Rn.

Definition 1.5. The cotangent bundle TU of an open subset U of Rn is the union S

p∈UTpRn.

An one-form on U is a function ω : U → TU. Let ω be an one-form on U. For each p ∈ U, ω(p) ∈ TpRn. By equation 1.1, if we set ai(p) = ω(p)((ei)p), then

ω(p) = a1(p)(dx1)p+ · · · + an(p)(dxn)p. We obtain functions a1, · · · , an: U → R. We write

ω = a1(x)dx1+ · · · + an(x)dxn.

We say that ω is a continuous/differentiable/Ck/smooth one-form if a1, · · · , an: U → R are continuous /differentiable/Ck/smooth functions on U. For example, let f : U ⊆ Rn → R be a Ck function. Then df defines a one-form df : U → TU by sending p to dfp. Since df = fx1dx1+ · · · + fxndxn and fxi are Ck−1 functions on U, df is a Ck−1 one-form.

Remark. The tangent bundle T U = U ×Rnover an open subset U of Rnand the cotangent bundle TU = U ×(Rn)over U are open subsets of T Rn= Rn×Rnand TRn= Rn×(Rn). We can define the continuities/differentiability/smoothness of a vector field V : U → T U over U or an one form ω : U → TU in the usual sense.

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Example 1.4. Let f (x, y) = excos y for (x, y) ∈ R2. Then

df = fxdx + fydy = excos ydx − exsin ydy.

Example 1.5. Let y = sin x for x ∈ R. Then

dy = f0(x)dx = cos xdx.

Example 1.6. Let f (x, y) = x3y + 2x2y for (x, y) ∈ R2. Then

df = fxdx + fydy = (3x2y + 2y2)dx + (x3+ 4xy)dy.

Example 1.7. Let U = R2\ {(0, 0)}. Then U is an open subset of R2. Define ω = −ydx + xdy

x2+ y2 for (x, y) ∈ R2. Then ω is a smooth one-form on U.

A one form on an open subset U of Rn is of the form ω = a1(x)dx1 + · · · + an(x)dxn. It is natural to ask when ω = df for some f ∈ C(U )? This is equivalent to the following family of differential equations

ai(x) = ∂f

∂xi

(x), 1 ≤ i ≤ n.

We will begin with the case when n = 2. Consider a smooth one-form on an open set U ⊂ R2 of the form

ω = M (x, y)dx + N (x, y)dy

and solve this problem when U is an open ball. If ω = df holds, then M = fx and N = fy. By smoothness of f,

My = fxy = fyx= Nx.

We find that if ω = df, then My = Nx. In fact, we have the following result:

Proposition 1.1. Let B be an open ball in R2and ω = M (x, y)dx + N (x, y)dy be a smooth one form on U. Then ω = df for some smooth function f : B → R if and only if My = Nx. Proof. We have proved one direction. Let us prove the reverse direction. We may assume that B the open unit ball B(0, 1) centered at 0 of radius 1. We define a function

f : B → R by f (x, y) = c + x Z 1

0

M (tx, ty)dt + y Z 1

0

N (tx, ty)dt.

Here c is any real number. By taking the partial derivative of f with respect to x, we obtain fx(x, y) =

Z 1 0

M (tx, ty)dt + x Z 1

0

Mx(tx, ty)tdt + y Z 1

0

Nx(tx, ty)tdt.

By the relation My = Nx, we see that Z 1

0

Nx(tx, ty)tdt = Z x

0

My(tx, ty)tdt.

Therefore

fx(x, y) = Z 1

0

M (tx, ty)dt + Z 1

0

t(Mx(tx, ty)x + My(tx, ty)y)dt.

By chain rule,

d

dtM (tx, ty) = Mx(tx, ty)x + My(tx, ty)y.

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Therefore fx can be rewritten as fx(x, y) =

Z 1 0

M (tx, ty)dt + Z 1

0

t d

dtM (tx, ty)

 dt

= Z 1

0

M (tx, ty)dt + tM (tx, ty)|1t=0− Z 1

0

M (tx, ty)dt (use the integration by parts)

= M (x, y).

Similarly, one can show that fy(x, y) = N (x, y). This proves that ω = df if My = Nx.  In calculus, we have learned the notion of antiderivative of a C1-function on a closed interval [a, b]. Let f : [a, b] → R be a continuous function. A C1-function F : [a, b] → R is said to be an antiderivative of f provided that F0(x) = f (x). If we denote ω = f (x)dx, then the condition F0(x) = f (x) is equivalent to the condition ω = dF. We now can define the notion of antiderivative of a smooth one form.

Definition 1.6. Let ω be a smooth one form on an open subset U of Rn. If there exists a smooth function f : U → R such that ω = df, then ω is said to have an antiderivative on U. In this case, f is said to be an antiderivative of f.

Proposition 1.1 says that a smooth one form ω = M (x, y)dx + N (x, y)dy on an open ball B in R2 has an antiderivative on B if and only if My = Nx. It is natural for us to ask when a smooth one form on an open set possess an antiderivative on that open set? What are the necessary and sufficient conditions for a smooth one form to have an antiderivative?

Example 1.8. Let ω = f (x)dx be a smooth one form on R. We define F (x) =

Z x 0

f (t)dt, x ∈ R.

By fundamental theorem of calculus, F0(x) = f (x) for any x ∈ R and hence ω = dF.

In other words, any smooth one form on R always possess an antiderivative on R by the fundamental Theorem of calculus.

Now we will express the condition My = Nx in terms of the “derivative of differential forms”. Let us introduce the notion of r-forms for any r ≥ 0.

2. Differential r-Forms

Suppose vp= (a, b)p and wp= (c, d)p are two tangent vectors to R2 at p. The area of the parallelogram spanned by {vp, wp} is the absolute value of the determinant

a c b d

.

Since dxp(vp) = a and dyp(vp) = b and dxp(wp) = c and dyp(wp) = d, the above determinant can be expressed as :

dxp(vp) dxp(wp) dyp(vp) dyp(wp)

. This motivates the following definition.

Definition 2.1. Let V be a finite dimensional vector space and ϕ, ψ be two linear func- tionals on V, i.e. ϕ, ψ ∈ V. We define the wedge product ϕ ∧ ψ of ϕ and ψ as follows. We define

ϕ ∧ ψ : V × V → R, (v, w) 7→

ϕ(v) ϕ(w) ψ(v) ψ(w) .

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Lemma 2.1. Let ϕ, ψ be linear functionals on a finite dimensional vector space V. Their wedge product ϕ ∧ ψ : V × V → R is a skew symmetric bilinear form on V, i.e.

(1) ϕ ∧ ψ : V × V → R is bilinear (2) (ϕ ∧ ψ)(w, v) = −(ϕ ∧ ψ)(v, w).

Especially, when ϕ = ψ, ϕ ∧ ϕ = 0.

Proof. The proof is left to the reader as an exercise. 

Definition 2.2. Let ψ1, · · · , ψr: V → R be linear functionals on a finite dimensional vector space V. We define the wedge product ψ1∧ · · · ∧ ψr: Vr → R by

1∧ · · · ∧ ψr)(v1, · · · , vr) = det[ψi(vj)]ri,j=1. Here (v1, · · · , vr) ∈ Vr.

Example 2.1. Let ψ1, ψ2, ψ3 : V → R be linear functions on V. For v1, v2, v3∈ V, one has (ψ1∧ ψ2∧ ψ3)(v1, v2, v3) =

ψ1(v1) ψ1(v2) ψ1(v3) ψ2(v1) ψ2(v2) ψ2(v3) ψ3(v1) ψ3(v2) ψ3(v3)

Lemma 2.2. Let ψ1, · · · , ψr : V → R and ψ1 ∧ · · · ∧ ψr : Vr → R be as above. Then ψ1∧ · · · ∧ ψr: Vr→ R is a alternating r-linear form on V.

Proof. This is left to the reader. 

Example 2.2. Let up= (a1, b1, c1)p and vp= (a2, b2, c2)p, and wp= (a3, b3, c3)p be vectors in TpR3 for p ∈ R3. In high school, we have learned that the volume of the parallelepiped spanned by up, vp, wp is the absolute value of the determinant

(dxp∧ dyp∧ dzp)(up, vp, wp) =

a1 b1 c1

a2 b2 c2 a3 b3 c3

Definition 2.3. Let V be a finite dimensional vector space. We set T0(V ) = R and T1(V) = V and Tr(V) be the space of all r-linear forms f on V, i.e. f : Vr → R is r-linear. Elements of Tr(V) are called r-tensors on V.

Let T1, T2, T : Vr → R be r-tensors and a ∈ R. We define the sum T1+ T2 and the scalar product aT by

(T1+ T2)(v1, · · · , vr) = T1(v1, · · · , vr) + T2(v1, · · · , vr), (aT )(v1, · · · , vr) = aT (v1, · · · , vr),

for (v1, · · · , vr) ∈ Vr.

Lemma 2.3. The set Tr(V) forms a vector space over R. (In fact, it is a vector subspace of the space of functions F (Vr, R) from Vr into R.)

For any linear functionals ψ1, · · · , ψr: V → R, we define the tensor product ψ1⊗· · ·⊗ψr: Vr → R by

1⊗ · · · ⊗ ψr)(v1, · · · , vr) = ψ1(v12(v2) · · · ψr(vr) for any (v1, · · · , vr) ∈ V. We leave it to the reader to verify that

ϕ ∧ ψ = ϕ ⊗ ψ − ψ ⊗ ϕ, see quiz 11.

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Proposition 2.1. Let β = {v1, · · · , vn} be a basis for a n dimensional vector space V and {ϕ1, · · · , ϕn} be the dual basis to β. Then {ϕi1 ⊗ · · · ⊗ ϕir : 1 ≤ i1, · · · , ir ≤ n} forms a basis for Tr(V); the dimension of Tr(V) is nr.

Proof. See quiz 11. 

A r-tensor T ∈ Tr(V) is alternating if

T (vσ1, · · · , vσr) = (sgn σ)T (v1, · · · , vr), (v1, · · · , vr) ∈ Vr for any permutation σ : {1, · · · , r} → {1, · · · , r},1 where

sgn σ = (

1 if σ is an even permutation;

−1 if σ is an odd permutation.

Lemma 2.4. Let Λr(V) the subset of Tr(V) consisting of all alternating r-tensors on V.

Then Λr(V) forms a vector subspace of Tr(V).

Proof. This is left to the reader as an exercise. 

Proposition 2.2. Let β = {v1, · · · , vn} be a basis for a n dimensional vector space V and {ϕ1, · · · , ϕn} be the dual basis to β. Then {ϕi1∧ · · · ∧ ϕir : 1 ≤ i1 < · · · < ir ≤ n} forms a basis for Λr(V). Hence the dimension of Λr(V) isn

r

 .

Proof. We will prove the case when r = 2. For general case, see quiz 11.  Definition 2.4. Let U be an open subset of Rn. We denote Λr(TU ) =S

p∈UΛr(TpRn).

A r-form on U is a function

η : U → Λr(TU ) such that η(p) ∈ Λr(TpRn) for any p ∈ U.

Since {(dxi)p: 1 ≤ i ≤ n} forms a basis of TpRn (dual to the standard basis {(ei)p: 1 ≤ i ≤ n} in TpRn,) the set {(dxi1)p∧ · · · ∧ (dxir)p : 1 ≤ i1 < · · · < ir ≤ n} forms a basis for the vector space Λr(TpRn) by Lemma 2.2. An element of Λr(TpRn) is of the form

X

1≤i1<···<ir≤n

ai1···ir(dxi1)p∧ · · · ∧ (dxir)p. If U is an open subset of Rn, a smooth r form on U is of the form

η = X

1≤i1<···<ir≤n

ηi1···ir(x)dxi1∧ · · · ∧ dxir, where ηi1···ir : U → R are smooth functions on U.

Example 2.3. A two form on an open subset U of R2 is of the form η = f (x, y)dx ∧ dy.

Here f : U → R is a function.

Example 2.4. A two form on an open subset U of R3 is of the form η = P (x, y, z)dx ∧ dy + Q(x, y, z)dx ∧ dz + R(x, y, z)dy ∧ dz;

a three form on U is of the form

α = g(x, y, z)dx ∧ dy ∧ dz.

Here P, Q, R, g : U → R are functions.

1A permutation on a nonempty set X is a bijection on X.

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Let Ir,n be the set of all r-tuple of nonnegative integers (i1, · · · , ir) such that 1 ≤ i1 <

· · · < ir≤ n. Let η be a r form on Rn. Then we denote η by

η = X

I∈Ir,n

ηI(x)dxI

where dxI = dxi1 ∧ · · · ∧ dxir for any I = (i1, · · · , ir) ∈ Ir,n. The set of all smooth r-forms on an open subset U of Rn is denoted by Ωr(U ).

Let η =P

IηIdxI and ω =P

IωIdxI be r-forms on U and f ∈ C(U ). We define η + ω and f η by

η + ω =X

I

I(x) + ηI(x))dxI, f η =X

I

f (x)ηI(x)dxI.

Proposition 2.3. The set Ωr(U ) forms a vector space over R such that for f, g ∈ C(U ) and ω, η ∈ Ωr(U ),

(1) f (η + ω) = f η + f ω;

(2) (f + g)η = f η + gη;

(3) (f g)η = f (gη);

(4) 1η = η

Proof. This is left to the reader as an exercise. 

Remark. In algebra, the above properties tell us that that Ωr(U ) is a C(U )-module.

Let η =P

IηI(x)dxIbe a r-form and ω =P

JωJdxJ be a s-form. We define their wedge product by

η ∧ ω =X

I,J

ηI(x)ωJ(x)dxI∧ dxJ. One can easily check that

η ∧ ω = (−1)rsω ∧ η.

Definition 2.5. Let η = P

IηI(x)dxI be a smooth r forms on U ⊆ Rn. We define the derivative drη of η by

drη =X

I

I∧ dxI. Here df denotes the differential of a smooth function f on U.

Example 2.5. Let ω = M (x, y)dx + N (x, y)dy be a smooth one form on an open set U ⊆ R2. The derivative of ω is given by

d1ω = (Nx− Mx)dx ∧ dy.

We see that the condition Ny = Mx is equivalent to the condition d1ω = 0.

Example 2.6. Let η = P dy ∧ dz − Qdx ∧ dz + Rdx ∧ dy be a smooth two form on an open subset of R3. Then

d2η = (Px+ Qy+ Rz)dx ∧ dy ∧ dz.

Definition 2.6. Let η be an r-form on an open subset U of Rn. If drη = 0, η is called a closed r-form. The set of all closed smooth r-forms on U is denoted by ZdRr (U ).

Lemma 2.5. The derivative dr of r-forms on an open subset of Rn defines a linear map dr: Ωr(U ) → Ωr+1(U ). Hence ZdRr (U ) = ker dr is a vector subspace of Ωr(U ).

Proof. The proof follows from the definition. 

(10)

We will set Ω0(U ) = C(U ); i.e. a smooth zero form on U is a smooth function on U.

We set d0 = d.

Example 2.7. Let f : U → R be a smooth function on an open subset U of R2. Then df = fxdx + fydy. The previous example tells us that

d1(d0f ) = (fyx− fxy)dx ∧ dy.

Since f is smooth, fyx = fxy on U. Hence d1(d0f ) = 0. Since f is arbitrary, we find d1◦ d0= 0.

In fact, we have a general result:

Lemma 2.6. The linear map dr : Ωr(U ) → Ωr+1(U ) satisfies the following properties:

(1) dr+1◦ dr= 0.

(2) (graded Leibniz rule) dr+s(η ∧ ω) = drη ∧ ω + (−1)rη ∧ dsω for any r form η and for any s form ω.

Let us go back to define the notion of antiderivatives of differential forms.

Definition 2.7. Let ω be a smooth r form on an open set U ⊆ Rn. We say that ω has an antiderivative on U if there exists a smooth r − 1 form η on U such that ω = dr−1η. If ω has an antiderivative on U, we say that ω is an exact r-form. The set of all exact r-forms on U is denoted by BrdR(U ).

The Proposition 1.1 can be reformulated as follows.

Proposition 2.4. Let B be an open ball in R2 and ω be a smooth one form on B. Then ω is an exact r-form on B if and only if it is a closed one form on U.

Since BdRr (U ) consists of r form of the form dr−1η for η ∈ Ωr−1(U ), by definition, BdRr (U ) = Im dr−1. Since dr−1is linear, BrdR(U ) is a vector subspace of Ωr(U ). Furthermore, since dr◦ dr−1= 0, dr(dr−1η) = 0 for any η ∈ Ωr−1(U ). This implies that Im dr−1⊆ ker dr, i.e. BrdR(U ) is a vector subspace of ZdRr (U ).

Now let us state the fundamental theorem of calculus for smooth differential forms on open sets in an Euclidean space.

Definition 2.8. Let U be an open subset of Rn. We say that the fundamental theorem of calculus holds for r forms on U if ZdRr (U ) = BdRr (U )

Proposition 2.7 can be reformulated as follows:

Proposition 2.5. The fundamental theorem of calculus for one forms holds on any open ball in R2.

To know whether the fundamental theorem of calculus holds or not, we need to study the “difference” between ZdRr (U ) and BdRr (U ). Thus we consider the quotient space

HdRr (U ) = ZdRr (U )/BrdR(U ).

The quotient space HdRr (U ) is called the r-th de Rham cohomology of U. Thus we can reformulate the fundamental theorem of calculus in terms of the quotient space HdRr (U ) : Proposition 2.6. Let r ≥ 1. The fundamental Theorem of calculus holds for smooth r-forms on U ⊆ Rn if and only if HdRr (U ) = {0}.

The Proposition 2.7 can be rewritten as:

Proposition 2.7. Let r ≥ 1. For any open ball B in R2, HdR1 (B) = {0}.

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