- #1
brydustin
- 205
- 0
{(a_i)_j} is the dual basis to the basis {(e_i)_j}
I want to show that
((a_i)_1) \wedge (a_i)_2 \wedge... \wedge (a_i)_n ((e_i)_1,(e_i)_2,...,(e_i)_n) = 1
this is exercise 4.1(a) from Spivak. So my approach was:
\BigWedge_ L=1^k (a_i)_L ((e_i)_1,...,(e_i)_n) = k! Alt(\BigCross_L=1^k (a_i)_L)((e_i)_1,...,(e_i)_n)= k! Alt(T)((e_i)_1,...,(e_i)_n) = k!(1/k! Sum _ {permutations σ} sgn σ T ((e_i)_σ (1),...,(e_i)_σ (n))
where T = \BigCross_L=1^k (a_i)_L
So there is already a result on what T ((e_i)_1,...,(e_i)_n) is. 1 if all the sub-indices agree, and 0 otherwise. My question is... is T ((e_i)_σ (1),...,(e_i)_σ (n)) any different?
I'm assuming that in the one dimensional case we would say that T acts on one element in a linear fashion... but I'm kinda confused by the idea of having several arguments...
Otherwise,...is there an easier approach to the solution?
I want to show that
((a_i)_1) \wedge (a_i)_2 \wedge... \wedge (a_i)_n ((e_i)_1,(e_i)_2,...,(e_i)_n) = 1
this is exercise 4.1(a) from Spivak. So my approach was:
\BigWedge_ L=1^k (a_i)_L ((e_i)_1,...,(e_i)_n) = k! Alt(\BigCross_L=1^k (a_i)_L)((e_i)_1,...,(e_i)_n)= k! Alt(T)((e_i)_1,...,(e_i)_n) = k!(1/k! Sum _ {permutations σ} sgn σ T ((e_i)_σ (1),...,(e_i)_σ (n))
where T = \BigCross_L=1^k (a_i)_L
So there is already a result on what T ((e_i)_1,...,(e_i)_n) is. 1 if all the sub-indices agree, and 0 otherwise. My question is... is T ((e_i)_σ (1),...,(e_i)_σ (n)) any different?
I'm assuming that in the one dimensional case we would say that T acts on one element in a linear fashion... but I'm kinda confused by the idea of having several arguments...
Otherwise,...is there an easier approach to the solution?