Building a two-form with wedge product of one-forms

Suppose $\alpha,\beta \in \Lambda ^1(X)$ where $X$ is a smooth manifold; and let $v,w \in TX$. Is the following an identity?

$(\alpha \wedge \beta)_x (v,w) = (\alpha_x(v)) \wedge (\beta_x(\omega))$

• No, especially given that it doesn't make sense. What is the wedge product of two numbers? – user98602 Jan 3 '16 at 16:59
• Good point. Thanks Mike. – kathleen Jan 3 '16 at 17:00
• My question is related to this similar question: math.stackexchange.com/questions/608269/… – kathleen Jan 4 '16 at 1:51
• I guess the wedge product of two numbers is just regular multiplication. – kathleen Jan 4 '16 at 1:54
• That's correct, in which case the formula in your post is false - the definition of wedge product of 1-forms is $(a \wedge b)(v,w)=a(v)b(w)-a(w)b(v)$. – user98602 Jan 4 '16 at 1:58