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I've never had any experience with differential forms before, so I'm trying to work through a couple of examples to see if I understand what's going on. I think I understand what I've been doing so far, but I'd like a little bit of reassurance or correction before I go too much further.

First, consider the differential form $\alpha = x dy-\frac{1}{2}(x^2+y^2)dt$ in $\mathbb{R}^3=\{(x,y,t)\}$. Then $$ \begin{align*} d\alpha &= dx\wedge dy - \frac{1}{2}(2x dx+2y dy)dt \\&= dx\wedge dy- x dx\wedge dt- y dy\wedge dt \end{align*}$$ Then we also have that $$ dt\wedge d\alpha = dt\wedge dx\wedge dy = dx\wedge dy\wedge dt \qquad \text{by anticommutativity} $$ and $$ dx\wedge d\alpha = -y dx\wedge dy\wedge dt $$ Are these computations correct?

Next, I'd like to know if $\alpha$ is a differential of a globally defined smooth function on $\mathbb{R}^3$. I'm not entirely sure what this means, but I think the question is asking if there is some globally defined smooth function on $\mathbb{R}^3$ whose total derivative is the differential $\alpha$. A globally defined smooth function is one that has no discontinuities, is infinitely differentiable, and every derivative is a continuous function. It seems like it should be relatively easy to find such a function, but I haven't managed to do so. Is there some sort of criterion that makes this question easier?

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Your computations look sound to me (except for a missing $\wedge$ after the parentheses in the first expression for $d\alpha$).

As for the followup question of whether $\alpha = df$ for some $f$, it is highly worthwhile for you to convince yourself of the following identity: $$ d(d\beta) = 0\text{ for any differential form }\beta.$$ In particular, if $\alpha = df$ for some $f$, then we would have $d\alpha = d(df) = 0$. But you have calculated that $d\alpha\neq 0$, so there can be no such function $f$.

In fact, the smooth, globally defined differential forms with vanishing total derivative (the "closed forms") are exactly those which are themselves the differential of another form (the "exact forms"). If the forms are not defined on all of $\mathbb{R}^n$, this may no longer be true: the extent to which there are closed forms that are not exact is measured by de Rham cohomology.

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Thanks. Noting that $d^2\beta=0$ for any differential form $\beta$ is very helpful. Actually, that's incredibly helpful and exactly what I was looking for. Thank again. –  chris Oct 17 '12 at 4:21

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