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I got confused on the operator of the wedge product on other 2 vectors. Please help.

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.

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  • $\begingroup$ This looks like an exercise so you must have some lecture notes. Where in your lecture notes is it mentioned that the wedge product of two vectors can be seen as a map? $\endgroup$ – Arnaud Mortier Feb 15 at 9:00
  • $\begingroup$ this notation indeed seems kind of off. Indeed one can define the exterior algebra over a universal property involving maps, but this does not make the wedge into functions, but as a universal space for alternating functions to facto over. Especially since there are 4 vectors appearing, but only 1 wedge $\endgroup$ – Enkidu Feb 15 at 9:51
  • $\begingroup$ See: math.stackexchange.com/questions/2842911/… $\endgroup$ – Sujit Bhattacharyya Feb 15 at 11:16
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If I understand the notation correctly, $$ (3e_1 \wedge 4e_3)((1,\alpha,0), (0,\beta,1)) = 12 (e_1 \wedge e_3)((1,\alpha,0), (0,\beta,1)) \\ = 12 (e_1 \otimes e_3)((1,\alpha,0), (0,\beta,1)) - 12 (e_1 \otimes e_3)((0,\beta,1), (1,\alpha,0)) \\ = 12\ e_1(1,\alpha,0)\ e_3(0,\beta,1) - 12\ e_1(0,\beta,1)\ e_3(1,\alpha,0) \\ = 12 ((1,0,0)\cdot(1,\alpha,0)) ((0,0,1)\cdot(0,\beta,1)) - 12 ((1,0,0)\cdot(0,\beta,1)) ((0,0,1)\cdot(1,\alpha,0)) \\ = 12 \cdot 1 \cdot 1 - 12 \cdot 0 \cdot 0 = 12 $$

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  • $\begingroup$ I think you forget the fraction $\frac{1}{2!}$ $\endgroup$ – Ngân Hoàng Nguyễn Feb 15 at 20:20
  • $\begingroup$ I was unsure if there should be such a factor. $\endgroup$ – md2perpe Feb 15 at 20:50

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