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,,vn be a basis for V and φ1,n be the dual basis for V * = T1(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, Tk(V ),

Solution:
We first state and prove a useful lemma.

Lemma 1. For any m-dimensional vector space W with basis ψ1,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,,am and b1,,bn such that α = a1ψ1 + ⋅⋅⋅ + amψm and β = b1φ1 + ⋅⋅⋅ + bnφn. Hence

α β = (       )
  m∑
     aiψi
  i=1β
= i=1m(aiψi ⊗ β)
= i=1ma i(ψi ⊗ β)
= i=1ma i(     (       ) )
        n∑
( ψi ⊗ (   bjφj) )
        j=1
= i=1ma i(  n          )
( ∑  (ψi ⊗ bjφj))
  j=1
= i=1ma i(             )
  ∑n
(    bj(ψi ⊗ φj))
  j=1
= i=1m j=1na ibj(ψi ⊗ φj)
implying that B spans W V *. Now if there existed reals cij 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=1m j=1na ibj(ψi ⊗ φj) - i=1m j=1nc ij    (ψi ⊗ φj)
= i=1m j=1n(a ibj - cij)(ψi ⊗ φj)
Since each ψi φj is nonzero, we must have that each aibj -cij = 0, implying that B is linearly independent. Hence B is a basis. Hence B is a basis. __

We can now prove A is a basis for Tk(V ) by induction. As a base case, we have that {φi} is a basis for T1(V ) since T1(V ) = V *. Assuming, now, that {φi1 ⋅⋅⋅φik-1} is a basis for Tk-1(V ), then lemma 1 informs us that the set

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

is a basis for Tk-1(V ) V * = Tk(V ). But this set is just A, and thus A must be a basis for Tk-1(V ) V * = Tk(V ), as desired.

2 Multilinear Algebra Problem 3


For a k-tensor T Tk(V ), define
                  1 ∑
Alt(T )(v1,...,vk) = k!   (- 1)σT (vσ(1),...,vσ(k))
                    σ∈Sk

for any v1,,vk V .

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


Let i,j be integers with 1 i < j k. For each σ Sk, define τσ = σ (i j) where (ij) is the element of Sk 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 Sk simply permutes the elements of Sk, we see that for v1,,vi,,vj,,vk V we have

Alt(T)(v1,,vj,,vi,,vk) = 1-
k! σSk(-1)σT(v σ(1),,vσ(j),,vσ(i),,vσ(k))
= 1-
k! σSk - (-1)τσ T(vτσ(1),,vτσ(i),,vτσ(j),,vτσ(k))
= -1-
k! τσSk(-1)τσ T(vτσ(1),,vτσ(i),,vτσ(j),,vτσ(k))
= -Alt(T)(v1,,vi,,vj,,vk)
as desired.

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


Denote the set of even and odd permutations of Sk by Sk+ and Sk-, respectively. Hence
T(vσ(1),,vσ(k)) = T(v1,,vk) σ Sk+
T(vσ(1),,vσ(k)) = -T(v1,,vk) σ Sk-
In particular, we note Sk+ and Sk- have the same cardinality, k!2, and that they partition Sk so that
Alt(T)(v1,,vk) = 1-
k! σSk(-1)σT(v σ(1),,vσ(k))
= 1
k!(                                                   )
  ∑                          ∑
(     (- 1)σT(vσ(1),...,vσ(k))+     (- 1)σT(vσ(1),...,vσ(k)))
  σ∈S+k                      σ∈S-k
= 1-
k!(                                          )
( ∑   T(vσ(1),...,vσ(k))-  ∑   T(vσ(1),...,vσ(k)))
  σ∈S+                 σ∈S-
     k                    k
= 1
k!(                                   )
  ∑                 ∑
(     T(v1,...,vk)-     - T(v1,...,vk))
  σ∈S+k             σ∈S-k
= 1-
k!( ∑                 ∑             )
(     T(v1,...,vk)+     T (v1,...,vk))
  σ∈S+             σ∈S-
     k                k
= T(v1,,vk) 1
k!(              )
   ∑      ∑
(     1 +     1)
  σ∈S+k    σ∈S-k
= T(v1,,vk)-1
k! (k!∕2+ k!∕2)
= T(v1,,vk)
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(Rpn,R), so that
df (p) = f1(p)dx1(p) + ⋅⋅⋅+ fn(p)dxn (p)

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

        1           n
df = f1dx + ⋅⋅⋅+ fndx

so we need only find these functions fi. However, because dx1(p),,dxn(p) is the dual basis to the basis (e1)p,,(en)p Rpn, then we know fi(p) = df(p)(ei)p for each i. Thus

fi(p) = df(p)(ei)p
= f(p)(ei)
= (                       )
  ∂f--  -∂f-      -∂f-
  ∂x1(p)∂x2 (p) ⋅⋅⋅∂xn (p)(ei)
= ∂f
∂xi(p)
Hence fi = -∂f
∂xi indicating
     ( ∂f)   1       ( ∂f )   n
df =  ∂xi  dx  +⋅⋅⋅+   ∂xi dx

as desired.

5 Differential Forms Problem 2


Let f : Rm Rn be a differentiable function.

(a) Show that f*(dyi) = j=1n(    )
  ∂fij
  ∂xdxj


For fixed p Rm we have
(f*(dyi))(p) = dyi(f(p))
= j=1n(∂f i)
 ∂xj-dxj(p)
so that
         n∑  (   i)
f*(dyi) =     ∂f-- dxj
         j=1  ∂xj

as desired.

(b) Show that f*(ω1 + ω2) = f*(ω1) + f*(ω2)


By the definition of f*ω and the linearity of f*, we have
  *
(f  (ω1 + ω2))(p) = f*((ω1 + ω2)(p))
= f*(ω 1(p) + ω2(p))
= f*(ω 1(p)) + f*(ω 2(p))
= (f*ω 1)(p) + (f*ω 2)(p)
=   *     *
(f ω1 + f ω2)(p)
for a fixed p.

(c) Show that f*() = (g f)f*(ω) = f*(ω)g*(ω)


(d) Show that f*(ω η) = f*(ω) f*(η)


6 Differential Forms Problem 3


Let f : R2 R2 be defined by
f(u,v) = (x(u,v),y(u,v)) = (u2 - v2,2uv)

and let ω = -ydx + xdy. Thus we have

dx = 2udu - 2vdv
dy = 2vdu + 2udv
so that
f*ω = -(2uv)(2udu - 2vdv) + (u2 - v2)(2vdu + 2udv)
= -4u2vdu + 4uv2dv + 2u2vdu + 2u3dv - 2v3du - 2uv2dv
= (-2u2v - 2v3)du + (2uv2 + 2u3)dv