is a basis for the set of all k-fold tensor products, k(V ),
Solution:
We first state and prove a useful lemma.
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 | = ![]() | ||
= ∑
i=1m![]() | |||
= ∑
i=1ma
i![]() | |||
= ∑
i=1ma
i![]() | |||
= ∑
i=1ma
i![]() | |||
= ∑
i=1ma
i![]() | |||
= ∑
i=1m ∑
j=1na
ibj![]() |
then we’d have
0 | = ∑
i=1m ∑
j=1na
ibj![]() ![]() | ||
= ∑
i=1m ∑
j=1n(a
ibj - cij)![]() | |||
We can now prove is a basis for
k(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
is a basis for Tk-1(V ) ⊗ V * = Tk(V ). But this set is just , and thus
must be a basis for Tk-1(V ) ⊗ V * = Tk(V ), as
desired.
for any v1,…,vk ∈ V .
Furthermore, we have
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) | = ![]() | ||
= ![]() | |||
= -![]() | |||
= -Alt(T)(v1,…,vi,…,vj,…,vk) |
T(vσ(1),…,vσ(k)) | = T(v1,…,vk) | ∀σ ∈ Sk+ | |||||
T(vσ(1),…,vσ(k)) | = -T(v1,…,vk) | ∀σ ∈ Sk- |
Alt(T)(v1,…,vk) | = ![]() | ||
= ![]() ![]() | |||
= ![]() ![]() | |||
= ![]() ![]() | |||
= ![]() ![]() | |||
= T(v1,…,vk)![]() ![]() | |||
= T(v1,…,vk)![]() ![]() | |||
= T(v1,…,vk) |
where each fi is a real valued function on Rn. In other words
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) | |||
= ![]() | |||
= ![]() |
as desired.
(f*(dyi))(p) | = dyi(f(p)) | ||
= ∑
j=1n![]() |
as desired.
![]() | = f*![]() | ||
= f*(ω 1(p) + ω2(p)) | |||
= f*(ω 1(p)) + f*(ω 2(p)) | |||
= (f*ω 1)(p) + (f*ω 2)(p) | |||
= ![]() |
and let ω = -ydx + xdy. Thus we have
dx | = 2udu - 2vdv | ||
dy | = 2vdu + 2udv |
f*ω | = -(2uv)(2udu - 2vdv) + (u2 - v2)(2vdu + 2udv) | ||
= -4u2vdu + 4uv2dv + 2u2vdu + 2u3dv - 2v3du - 2uv2dv | |||
= (-2u2v - 2v3)du + (2uv2 + 2u3)dv |