Definition of exterior product

From $r$ linear functionals $f_1,\cdots, f_r\in E^*$, we obtain the $r$-linear alternating form $f_1 \wedge \cdots \wedge f_r: E\times \cdots \times E \to \mathbb{R}$, defined by

$$(f_1 \wedge \cdots \wedge f_r)(v_1, \cdots, v_r) = \det(f_i(v_j)))$$

It is defined as a determinant. However, I have an exercise that asks me to calculate things like $e_1 \wedge e_2$. But where do I apply $e_1$ and $e_2$? In the definition the functionals $f_i$ are applied onto vectors $v_j$, but the exercise asks me to just take the product $e_1 \wedge e_2$.

$(e_1 \wedge e_2)(v_1, \cdots v_2)$?

• The correct symbol for the exterior product is not Λ (uppercase lambda) but \wedge. – md2perpe Oct 28 '17 at 8:26
• Can you say explicitly what the exercise is asking for? "Calculate $e_1 \wedge e_2$" doesn't really make sense. Could it be possibly, $e_1 = (a,b)$, $e_2=(c,d)$, so $e_1 \wedge e_2 = C (1,0) \wedge (0,1)$ for some constant $C$, calculate $C$? Then you would use bilinearity and alternating property: $e_1 \wedge e_2 = (a(1,0)+b(0,1))\wedge e_2 = a (1,0) \wedge e_2 + b (0,1) \wedge e_2$, then expand $e_2$, etc. – Zach Teitler Oct 31 '17 at 21:12

$$(e_1 \wedge e_2)(v_1, v_2) = e_1(v_1) e_2(v_2) - e_1(v_2) e_2(v_1) = \left| \begin{matrix} e_1(v_1) & e_1(v_2) \\ e_2(v_1) & e_2(v_2) \end{matrix} \right|$$
When I study smooth manifold,my teacher always let us keep in mind the crucial issue:$TM$ and $T^*M$ is a pair of dual spaces.In particular the restrict of them at 1 point,$TM_p,T^*M_p$ is also a pair of dual spaces. and dual and the wedge product or tensor product is exchangeable.for example $TM\wedge TM$ is the dual space of $T^*M\wedge T^*M$,$TM\otimes TM$ is the dual space of $T^*M\otimes T^*M$.
When you calculate the exterior product of two vector in $T^*M_p$,i.e. $e_1\wedge e_2$,in fact we are calculate the thing at every point $p$.and it is a functional on its dual space $TM\wedge TM$.so we just need to calculate$e_1\wedge e_2(f_1|_p,f_2|_p)$ for two function,i.e. member in $TM_p\wedge TM_p$.we need to observe $e_1(f)=df(e_1)=\lim_{h\to 0}\frac{f(p+he_1)-f(p)}{h}$.so in general we have: $(f_1\wedge f_2)(e_1, e_2)= (e_1\wedge e_2)(f_1, f_2)= det(f_i(e_j)))$.