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Home , Basis (linear algebra), Exterior algebra, Product (mathematics), Standard basis

Exterior → Differential Forms

Faraad M Armwood

North Dakota State

September 4, 2016

Faraad M Armwood (North Dakota State) → Differential Forms September 4, 2016 1 / 17

Let V , W be a f.d.v.s and let Hom(V , W ) = {f | f : V → W is linear}. V V Then the dual V to V is Hom(V , R). The elements of V are called 1-covectors or covectors on V . P k If we let e1, ..., en be a for V then if v ∈ V we have v = k v ek . i i i i 1 n Suppose α (ej ) = δj then α (v) = v where v = (v , ..., v ). Therefore we i i V have α : V → R is a and so α ∈ V . It follows that the {αi : i = 1, ..., n} are linearly independent and they span V V i.e dim(V ) = dim(V V ).

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 2 / 17 Example

2 1 2 2 Let V = R and x , x be the standard coordinates on R i.e if i T T p = (p1, p2) then x (p) = pi . Let e1 = (1, 0) , e2 = (0, 1) denote the i i then x (ej ) = δj . This example and the above demonstrate that for a f.d.v.s the is determined by the coordinates of a point in the standard basis.

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 3 / 17 Multilinear Functions

Let V k = V × · · · × V be the k- of a real vector space V . We say k f : V → R is k-linear if is linear in each of its k-arguments i.e;

f (...., av + bw, ...) = af (...., v, ...) + bf (...., w, ...) A k-linear on V is also called a k- on V . We will denote by Lk (V ) the of all k- on V . If f is a k-tensor, we also say that f has degree k.

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 4 / 17 Example

2 2 2 Let v, w ∈ TpR and let f = h·, ·i : TpR × TpR → R defined by; X hv, wi = v k w k k 2 2 then f is a 2-linear or bilinear function on R . Here TpR denoted the 2 on R i.e v, w are understood to be vectors.

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 5 / 17 Symmetric and Alternating Tensors

Let f ∈ Lk (V ) then we say f is symmetric if;

f (vσ(1), ..., vσ(k)) = f (v1, ..., vk ), ∀σ ∈ Sk We say f is alternating if;

f (vσ(1), ..., vσ(k)) = sgn(σ) · f (v1, ..., vk ), ∀σ ∈ Sk

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 6 / 17 Example

2 (1) The inner product define in the last example is symmetric on TpR n and even in the extension to TpR due to the commutability of R.

(2) A simple function f (x, y) = x + y is also symmetric

3 (3) The ~v × w~ on R is alternating

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 7 / 17 Symmetrizing and Alternatrizing Operators

Let f (v1, ..., vk ) ∈ Lk (V ) and σ ∈ Sk . Then we can define an action of a perm. on f by σf := f (vσ(1), ..., vσ(k)). We now demonstrate how to get an alternating and symmetric frunction from f ;

X Sf = σf (symmetric)

σ∈Sk

X Af = (sgnσ) σf (alternating)

σ∈Sk

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 8 / 17 Example

Let f ∈ L3(V ) and suppose v1, v2, v3 ∈ V then;

X Af (v1, v2, v3) = (sgnσ) σf

σ∈S3

Recall that S3 = {(1), (12), (13), (123), (132)} and to determine their sign we observe that (123) = (13)(12), (132) = (12)(13) and so;

Af = f (v1, v2, v3) − f (v2, v1, v3) − f (v3, v2, v1) + f (v3, v2, v1) + f (v2, v3, v1)

(12)Af = f (v2, v1, v3)−f (v1, v2, v3)−f (v3, v1, v2)+f (v3, v1, v2)+f (v1, v3, v2)

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 9 / 17

Let f ∈ Lk (V ) and g ∈ Ll (V ) then we defined their tensor product;

(f ⊗ g)(v1, ..., vk+l ) = f (v1, ..., vk )g(vk+1, ..., vk+l ) ∈ Lk+l (V )

The above is associative i.e (f ⊗ g) ⊗ h = f ⊗ (g ⊗ h).

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 10 / 17 Example

Let e1, ..., en be a basis for V and f = h·, ·i : V × V → R. Define P i P j hei , ej i = gij and v = i v ei , w = j w ej then by the previous remarks, αi (v) = v i , αj (w) = w j and so;

X j i X j i hv, wi = gij (α ⊗ α )(w, v) ⇒ h, i = gij α ⊗ α i,j i,j

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 11 / 17 The Wedge Product I

Let f ∈ Ak (V ) and g ∈ Al (V ) then we define their exterior product or wedge product to be; 1 f ∧ g = A(f ⊗ g) k!l! or explicitely;

1 X (f ∧ g)(v) = (sgnσ) f (v , ..., v )g v , ..., v  k!l! σ(1) σ(k) σ(k+1) σ(k+l) σ∈Sk+l

where v = (v1, ..., vk+l ) and dividing out by k!l! compensates for the repetition in the sum.

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 12 / 17 Discussion

(1) If σ(k + j) = k + j, ∀j = 1, ..., l then σf = sgn(σ) f and σg = g and so for each such σ you get fg in the amount of k! times. Similarly if τ(j) = j, ∀j = 1, ..., k then you get fg in the amount of l! times. Now convince yourself that there are no other repetitions in the sum.

(2) Let f ∈ A0(V ) and g ∈ Al (V ) then f is a constant function say c ∈ R and;

1 X c ∧ g(v , ..., v ) = (sgnσ)2 cg(v , ..., v ) = cg 1 l l! 1 l σ∈Sl

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 13 / 17 The Wedge Product II

Another way to compensate for the repeated terms in the sum for the tensor product is to arrange that σ(1) < ··· < σ(k) and σ(k + 1) < ··· < σ(k + l). If σ ∈ Sk+l is such a permutation, we say that σ is a (k, l) shuffle. Therefore we have;

X (f ∧ g)(v) = (sgnσ) f (vσ(1), ..., vσ(k))g(vσk+1, ..., vσk+l ) (k,l)−shuffles σ

The two definitions are not the same!

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 14 / 17 Example

To demonstrate that the two definitions differ, we take f ∈ A1(V ) and 3 g ∈ A2(V ) then the in discussion is S , but the only (1, 2) shuffle is the identity. Now compute f ∧ g by the original definition and see that they differ.

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 15 / 17 Properties of Wedge Product

kl (1) If f ∈ Ak (V ) and g ∈ Al (V ) then f ∧ g = (−1) g ∧ f .

(2) If f ∈ A2k+1(V ) then f ∧ f = 0

(3) (f ∧ g) ∧ h = f ∧ (g ∧ h), ∀g, h, f ∈ A∗(V ) i 1 k i (4) If α ∈ L1(V ) and vi ∈ V then; (α ∧ · · · ∧ α )(v1, ..., vk ) = det[α (vj )]

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 16 / 17 Basis for k-Covectors

1 n Lemma: Let e1, ..., en be a basis for a v.s V and α , ..., α be its dual basis V in V . If I = (1 ≤ i1 < ··· < ik ≤ n) and J = (1 ≤ j1 < ··· < jk ≤ n) are strictly ascending multi-indices of length k then;

I I α (eJ ) = δJ

I Proposition: The alternating k-linear function α , I = (i1 < ··· < ik ), form a basis for the space Ak (V ) of alternating k-linear functions on V i.e if f ∈ Ak (V ) then;

X I f = aI α I .

Corollary: If k > dim(V ) then Ak (V ) = 0.

Faraad M Armwood (North Dakota State) Exterior Algebra → Differential Forms September 4, 2016 17 / 17