# Evaluating a differential 2-form on test vector fields

This is hopefully a quick question stemming from misunderstanding of notation. I am aiming to understand the proof of Proposition 12.17 in Lee's "Introduction to smooth manifolds" which shows that for $\omega \in \Omega^1(M)$ a 1-form on a manifold $M$ and $X, Y \in \text{Vect}(M)$ vector fields, we have $$d \omega(X, Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y]).$$

We assume for a start that $\omega = u\ dv$ for smooth functions $u, v$, and calculate that $d \omega(X, Y) = (du \wedge dv)(X, Y)$. I would expect this to equal $(du)(X) (dv)(Y)$, however in Lee it is $$(du)(X) (dv)(Y) - (dv)(X) (du)(Y).$$

Where does this difference come from upon evaluation?

• To clarify: Are you asking why $(du \wedge dv)(X, Y) = du(X) dv(Y) - du(Y) dv(X)$ (the definition of the wedge product), and is not equal to $du(X) dv(Y)$ (the definition of the tensor product)? Jun 4, 2017 at 16:45
• @AndrewD.Hwang, yes, exactly. I think that I understand how the wedge product works on forms, but not sure how we evaluate the resultant form on vector fields Jun 4, 2017 at 16:46

## 2 Answers

In a basis we have $\omega = f_k \, \mathrm{d}x^k$ and $\mathrm{d}\omega = \partial_l f_k \, \mathrm{d}x^l \wedge \mathrm{d}x^k$.

We also have \begin{align} \omega(X) & = f_k \, \mathrm{d}x^k(X) \\ & = f_k \, \mathrm{d}x^k(X^l \partial_l) \\ & = f_k X^l \, \mathrm{d}x^k(\partial_l) \\ & = f_k X^l \delta^k_l \\ & = f_k X^k \end{align} and \begin{align} (XY)(\phi) & = X(Y(\phi)) \\ & = X^k \partial_k(Y^l \partial_l \phi) \\ & = X^k (\partial_k Y^l) \partial_l \phi + X^k Y^l \partial_k \partial_l \\ & = ((X^k \partial_k) Y^l) (\partial_l \phi) + X^k Y^l \partial_k \partial_l \phi \\ & = X(Y^l) (\partial_l \phi) + X^k Y^l \partial_k \partial_l \phi \end{align} so \begin{align} [X, Y](\phi) & = X(Y(\phi)) - Y(X(\phi)) \\ & = \left( X(Y^l) (\partial_l \phi) + X^k Y^l \partial_k \partial_l \phi \right) - \left( Y(X^l) (\partial_l \phi) + Y^k X^l \partial_k \partial_l \phi \right) \\ & = \left( X(Y^l) (\partial_l \phi) - Y(X^l) (\partial_l \phi) \right) + \left( X^k Y^l \partial_k \partial_l \phi - Y^k X^l \partial_k \partial_l \phi \right) \\ & = \left( X(Y^k) - Y(X^k) \right) \partial_k \phi \end{align} since \begin{align} X^k Y^l \partial_k \partial_l \phi - Y^k X^l \partial_k \partial_l \phi & = X^k Y^l \partial_k \partial_l \phi - Y^l X^l \partial_l \partial_k \phi \\ & = X^k Y^l \partial_k \partial_l \phi - X^l Y^l \partial_k \partial_l\phi \\ & = 0 \end{align}

Thus $$[X,Y]^k = X(Y^k) - Y(X^k).$$

This gives \begin{align} \mathrm{d}\omega(X,Y) & = \partial_l f_k (\mathrm{d}x^l \wedge \mathrm{d}x^k)(X,Y) \\ & = \partial_l f_k \, (\mathrm{d}x^l(X) \, \mathrm{d}x^k(Y) - \mathrm{d}x^k(X) \, \mathrm{d}x^l(Y)) \\ & = \partial_l f_k (X^l Y^k - X^k Y^l) \\ & = \partial_l f_k X^l Y^k - \partial_l f_k X^k Y^l \\ & = X^l (\partial_l f_k) Y^k - Y^l (\partial_l f_k) X^k \\ & = \left( X^l \partial_l (f_k Y^k) - f_k X^l \partial_l Y^k \right) - \left( Y^l \partial_l (f_k X^k) - f_k Y^l \partial_l X^k \right) \\ & = \left( X(f_k Y^k) - f_k X(Y^k) \right) - \left( Y(f_k X^k) - f_k Y(X^k) \right) \\ & = \left( X(\omega(Y)) - f_k X(Y^k) \right) - \left( Y(\omega(X)) - f_k Y(X^k) \right) \\ & = \left( X(\omega(Y)) - Y(\omega(X)) \right) - f_k \left( X(Y^k) - Y(X^k) \right) \\ & = \left( X(\omega(Y)) - Y(\omega(X)) \right) - f_k [X, Y]^k \\ & = X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y]) \end{align}

You can write the wedge product as the antisymmetrization of the tensor product $du\otimes dv$, i.e. $du\wedge dv = du\otimes dv - dv \otimes du$. This then leads to the answer by the standard relation between $V\otimes V$ and $V^*\times V^*$ where $V^*$ is the dual of $V$.

• Could you please elaborate on the antisymmetrization definition? I met the wedge product of two forms defined as an element of the exterior algebra on (in this case) $T^* M$, but I'm not sure how it follows from this definition. Jun 4, 2017 at 16:55
• The exterior algebra can be constructed as the antisymmetric part of the k-th tensor power of $V$. So for the exterior algebra of differential forms and $k=2$ this becomes the formula i gave. Jun 4, 2017 at 17:02
• See for example: nmt.edu/~iavramid/notes/diffforms.pdf or the main Wolfram page on exterior algebras: mathworld.wolfram.com/ExteriorAlgebra.html Jun 4, 2017 at 17:20
• This can be seen intuitively as follows. As differential forms they map vectors (vector fields) to $\mathbb{R}$. Which means they are a subspace of the linear forms on the space of vector fields. They are thus a subspace of $\bigotimes^k V^*$. As they are antisymmetric by definition they are a subspace of the antisymmetric part of this tensor power. Jun 4, 2017 at 17:30