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Let $\omega := dx \wedge dy$ denote the standard area form on $\mathbb{R}^2$. As the question title suggests, does there exist a $1$-form $\alpha$ with $d\alpha = \omega$?

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$\newcommand{\dd}{\partial}$If $f$ and $g$ are real-valued functions of class $C^{1}$, and $\alpha = f\, dx + g\, dy$, then

$$

d\alpha = \left(\frac{\dd g}{dx} – \frac{\dd f}{dy}\right) dx \wedge dy.

$$

Consequently, $\omega = d\alpha$ if and only if

$$

\frac{\dd g}{dx} – \frac{\dd f}{dy} \equiv 1.

$$

The pleasure of finding as many $\alpha$ as necessary is left to you.

Well, $\mathrm d\omega = 0$ and $\omega$ is defined everywhere, so the answer to your question is yes.

You probably want to see such a 1-form, though. You can either make some educated guesses such as John Ma’s suggestion and try them out, or compute an antiderivative directly. I’ll do the latter here since it’s a simple illustration of the method.

*Step 1.* Substitute $x^i \to tx^i$ and $\mathrm dx^i\to x^i\,\mathrm dt + t\,\mathrm dx^i$; $$(x\,\mathrm dt + t\,\mathrm dx) \land (y\,\mathrm dt + t\,\mathrm dy) = tx\,\mathrm dt\land\mathrm dy + ty\,\mathrm dx\land\mathrm dt + t^2\,\mathrm dx\land\mathrm dy$$

*Step 2.* Discard all terms not involving $\mathrm dt$ and move it to the left in the remaining terms:$$tx\,\mathrm dt\land\mathrm dy – ty\,\mathrm dt\land\mathrm dx$$

*Step 3.* Treat the result as an ordinary integrand w/r $t$ and integrate from $0$ to $1$:$$\int_0^1 (tx\,\mathrm dy – ty\,\mathrm dx)\,\mathrm dt = \frac12(x\,\mathrm dy-y\,\mathrm dx). $$ This form might look familiar to you.

Note that, just as there’s an arbitrary constant of integration in elementary calculus, you can add $\mathrm df$, where $f$ is a real-valued function, to this and get another antiderivative of $\omega$. E.g., taking $f(x,y)=\frac12xy$ gives $x\,\mathrm dy$.

For the curious, here’s what’s going on above (asserted without proof).

Let the $k$-form $\omega$ ($k>0$) be defined in a star-shaped region $Q$ of $\mathbb R^n$. Define a function $\beta$ on $[0,1]\times Q$ that for each $p\in Q$ maps the interval $[0,1]$ to the line segment joining $p$ to the origin. For any function $g$ defined on $Q$, $(\beta^*g)(t;x^1,\dots,x^n)=g(tx^1,\dots,tx^n)$.

The pullback $\beta^*\omega$ is a sum of two types of terms: $$\tau_1(t)=A(t;x^1,\dots,x^n)\,\mathrm dx^{i_1}\land\cdots\land\mathrm dx^{i_k} \\ \tau_2(t)=B(t;x^1,\dots,x^n)\,\mathrm dt\land\mathrm dx^{i_1}\land\cdots\land\mathrm dx^{i_{k-1}},$$ i.e., terms that don’t involve $\mathrm dt$ and those that do. Define a linear operator $L$ as follows: $$\begin{align}

L\tau_1 &= 0 \\

L\tau_2 &=\left(\int_0^1 B\,\mathrm dt\right)\,\mathrm dx^{i_1}\land\cdots\land\mathrm dx^{i_{k-1}}

\end{align}$$ This operator has the property that $$\begin{align}

\mathrm dL\tau_1+L\mathrm d\tau_1 &= \tau_1(1)-\tau_1(0) \\

\mathrm dL\tau_2+L\mathrm d\tau_2 &= 0.

\end{align}$$ So, type $\tau_2$ terms can be ignored in $\mathrm dL(\beta^*\omega)+L\mathrm d(\beta^*\omega)$ and since $\tau_1(0)=0$ because of the factor of $t$ that accompanies each $\mathrm dx^i$, we have $$

\mathrm dL(\beta^*\omega)+L\mathrm d(\beta^*\omega) = \omega.

$$ If $\mathrm d\omega=0$, this becomes simply $\mathrm dL\beta^*\omega = \omega$.

The process at the top of the answer computes $L\beta^*\omega$. Step 1 is just forming the pullback; step 2 is some bookkeeping; and step 3 computes $L\tau_2$. The $\tau_1$ terms can be discarded at any point and the first two steps can of course be combined if you’re careful. This procedure can be extended to work for any region that’s the image of a star-shaped region under a smooth one-to-one mapping.

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