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Suppose we have a manifold $M$ of dimension $n$ and its cotangent bundle $T^*M$. The Liouville form $\lambda$ on $T^*M$ is defined as $\lambda_{\omega_p} = \pi^*(\omega_p)$ where $\pi$ is the standard projection map from $T^*M$ to $M$. The problem is to prove that $d\lambda$ is a nondegenerate 2-form on $T^*M$. I was able to solve this problem by simply expressing the form with the aid of local coordinates and deriving the result by straightforward computation. However, I also have a bogus proof where I can't identify the incorrect step. Exponents are repeated wedge products.

$d\lambda_{\omega_p} = d\pi^*(\omega_p)$

$(d\lambda_{\omega_p})^n = (d\pi^*(\omega_p))^n = \pi^*(d\omega_p)^n$

(Because pullbacks commute with exterior differentiation and wedge products).

However, $\pi^*(d\omega_p)^n = 0$ because it is the image of a $2n$ form on a $n$ dimensional manifold.

Where am I going wrong? I'm guessing I'm missing some hypothesis in order to use the commutative fact but I am not sure. Thank you.

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You can't differentiate a pointwise expression. In particular, $\lambda$ is not the pullback of a form on $M$.

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  • $\begingroup$ I think I understand what you're saying but just to make it clear I want to see if what I'm saying in my own words is correct. You can't differentiate a specific covector i.e. $d\omega_p$ makes no sense. However, if $\omega$ is a form on $M$ we do have $(d\omega)_p$. $\endgroup$
    – Memeozuki
    Aug 20, 2014 at 0:48
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    $\begingroup$ Right! Similarly, a calculus student must take the derivative and then evaluate, not vice-versa. $\endgroup$
    – Ted Shifrin
    Aug 20, 2014 at 1:59

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