Math 509: Advanced Analysis

Homework 11

Lawrence Tyler Rush

<me@tylerlogic.com>

April 28, 2015

http://coursework.tylerlogic.com/courses/upenn/math509/homework11

1 Multilinear Algebra Problem 2

Problem:
Let v₁,…,v_n be a basis for V and φ₁,…,φ_n be the dual basis for V ^* =

¹(V ). Show that the set

A = {φi ⊗ ⋅⋅⋅⊗ φi |1 ≤ i1,...,ik ≤ n} 1 k

is a basis for the set of all k-fold tensor products, ^k(V ),

Solution:
We first state and prove a useful lemma.

Lemma 1. For any m-dimensional vector space W with basis ψ₁,…,ψ_m, the set

B = {ψi ⊗ φj | 1 ≤ i ≤ m, 1 ≤ j ≤ n}

is a basis for W ⊗ V ^*.

Proof. Let α ⊗ β ∈ W ⊗ V ^*. Then there are reals a₁,…,a_m and b₁,…,b_n such that α = a₁ψ₁ + ⋅⋅⋅ + a_mψ_m and β = b₁φ₁ + + b_nφ_n. Hence

α ⊗ β	= ⊗ β
	= ∑ _i=1^m
	= ∑ _i=1^ma_i
	= ∑ _i=1^ma_i
	= ∑ _i=1^ma_i
	= ∑ _i=1^ma_i
	= ∑ _i=1^m ∑ _j=1ⁿa_ib_j

implying that

spans W ⊗ V ^*. Now if there existed reals c_ij for 1 ≤ i ≤ m and 1 ≤ j ≤ n such that

∑m ∑n α⊗ β = cij (ψi ⊗φj ) i=1j=1

then we’d have

0	= ∑ _i=1^m ∑ _j=1ⁿa_ib_j -∑ _i=1^m ∑ _j=1ⁿc_ij
	= ∑ _i=1^m ∑ _j=1ⁿ(a_ib_j - c_ij)

Since each ψ_i ⊗φ_j is nonzero, we must have that each a_ib_j -c_ij = 0, implying that

is linearly independent. Hence

is a basis. Hence

is a basis. __

We can now prove is a basis for ^k(V ) by induction. As a base case, we have that {φ_i} is a basis for T¹(V ) since T¹(V ) = V ^*. Assuming, now, that {φ_i₁ ⊗ ⋅⋅⋅ ⊗ φ_{i_k-1}} is a basis for T^k-1(V ), then lemma 1 informs us that the set

{ψ ⊗ φj | ψ ∈ {φi1 ⊗ φik-1}, 1 ≤ j ≤ n}

is a basis for T^k-1(V ) ⊗ V ^* = T^k(V ). But this set is just , and thus must be a basis for T^k-1(V ) ⊗ V ^* = T^k(V ), as desired.

2 Multilinear Algebra Problem 3

For a k-tensor T ∈

^k(V ), define

1 ∑ Alt(T )(v1,...,vk) = k! (- 1)σT (vσ(1),...,vσ(k)) σ∈Sk

for any v₁,…,v_k ∈ V .

(a) Show that Alt(T) is alternating.

Let i,j be integers with 1 ≤ i < j ≤ k. For each σ ∈ S_k, define τ_σ = σ ∘ (i j) where (ij) is the element of S_k that transpositions i and j. Hence we have

(- 1)τσ = - (- 1)σ

Furthermore, we have

( { σ(i) n = j τσ(n) = σ ∘(i j)(n) = ( σ(j) n = i σ(n) otherwise

for any integer n ∈{1,…,k}. Making use of these two equations and the fact that the action of (i j) on S_k simply permutes the elements of S_k, we see that for v₁,…,v_i,…,v_j,…,v_k ∈ V we have

Alt(T)(v₁,…,v_j,…,v_i,…,v_k)	= ∑ _{σ∈S_k}(-1)^σT(v_σ(1),…,v_σ(j),…,v_σ(i),…,v_σ(k))
	= ∑ _{σ∈S_k} - (-1)^τ_σT(v_{τ_σ(1)},…,v_{τ_σ(i)},…,v_{τ_σ(j)},…,v_{τ_σ(k)})
	= -∑ _{τ_σ∈S_k}(-1)^τ_σT(v_{τ_σ(1)},…,v_{τ_σ(i)},…,v_{τ_σ(j)},…,v_{τ_σ(k)})
	= -Alt(T)(v₁,…,v_i,…,v_j,…,v_k)

as desired.

(b) Show that Alt(T) = T whenever T is alternating

Denote the set of even and odd permutations of S_k by S_k⁺ and S_k^-, respectively. Hence

T(v_σ(1),…,v_σ(k))	= T(v₁,…,v_k)		∀σ ∈ S_k⁺
T(v_σ(1),…,v_σ(k))	= -T(v₁,…,v_k)		∀σ ∈ S_k^-

In particular, we note S_k⁺ and S_k^- have the same cardinality, k!∕2, and that they partition S_k so that

Alt(T)(v₁,…,v_k)	= ∑ _{σ∈S_k}(-1)^σT(v_σ(1),…,v_σ(k))
	=
	=
	=
	=
	= T(v₁,…,v_k)
	= T(v₁,…,v_k)
	= T(v₁,…,v_k)

as desired.

(c) Conclude that Alt(Alt(T)) = Alt(T)

Since Alt(T) is alternating for any T, then the previous part of this problem yields Alt(Alt(T)) = Alt(T)

3 Multilinear Algebra Problem 7

4 Differential Forms Problem 1

Fix an arbitrary point p ∈ Rⁿ. Then dx¹(p),…,dxⁿ(p) is a basis for the vector space Hom(R_pⁿ,R), so that

df (p) = f1(p)dx1(p) + ⋅⋅⋅+ fn(p)dxn (p)

where each f_i is a real valued function on Rⁿ. In other words

1 n df = f1dx + ⋅⋅⋅+ fndx

so we need only find these functions f_i. However, because dx¹(p),…,dxⁿ(p) is the dual basis to the basis (e₁)_p,…,(e_n)_p ∈ R_pⁿ, then we know f_i(p) = df(p)(e_i)_p for each i. Thus