# wedge product (exterior algebra)

Let $$V=\mathbb R^3,e_1= (1,0,0),e_2= (0,1,0)$$, and $$e_3= (0,0,1)$$. Find: $$3e_1∧4e_3((1,α,0),(0,β,1))$$, where α,β are irrational numbers.
If I understand the notation correctly, $$(3e_1 \wedge 4e_3)((1,\alpha,0), (0,\beta,1)) = 12 (e_1 \wedge e_3)((1,\alpha,0), (0,\beta,1)) \\ = 12 (e_1 \otimes e_3)((1,\alpha,0), (0,\beta,1)) - 12 (e_1 \otimes e_3)((0,\beta,1), (1,\alpha,0)) \\ = 12\ e_1(1,\alpha,0)\ e_3(0,\beta,1) - 12\ e_1(0,\beta,1)\ e_3(1,\alpha,0) \\ = 12 ((1,0,0)\cdot(1,\alpha,0)) ((0,0,1)\cdot(0,\beta,1)) - 12 ((1,0,0)\cdot(0,\beta,1)) ((0,0,1)\cdot(1,\alpha,0)) \\ = 12 \cdot 1 \cdot 1 - 12 \cdot 0 \cdot 0 = 12$$
• I think you forget the fraction $\frac{1}{2!}$ – Ngân Hoàng Nguyễn Feb 15 at 20:20