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

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

  • $\begingroup$ No, especially given that it doesn't make sense. What is the wedge product of two numbers? $\endgroup$ – user98602 Jan 3 '16 at 16:59
  • $\begingroup$ Good point. Thanks Mike. $\endgroup$ – kathleen Jan 3 '16 at 17:00
  • $\begingroup$ My question is related to this similar question: math.stackexchange.com/questions/608269/… $\endgroup$ – kathleen Jan 4 '16 at 1:51
  • $\begingroup$ I guess the wedge product of two numbers is just regular multiplication. $\endgroup$ – kathleen Jan 4 '16 at 1:54
  • 2
    $\begingroup$ That's correct, in which case the formula in your post is false - the definition of wedge product of 1-forms is $(a \wedge b)(v,w)=a(v)b(w)-a(w)b(v)$. $\endgroup$ – user98602 Jan 4 '16 at 1:58

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