I have two commuting vector fields $X,Y\in\mathfrak{X}(\mathbb{T}^2)$ tangent to a 2-torus, which means $[X,Y]=\mathcal{L}_X Y=0$. It is immediate to find a bi-vector field $\Pi$ which is invariant under the flow of $X$, i.e. so that $\mathcal{L}_X\Pi = 0$ (where $\mathcal{L}_X$ stands for the Lie derivative along $X$).
Indeed, we can set $\Pi := X\wedge Y$ and hence $\mathcal{L}_X\Pi = [X,X]\wedge Y + X\wedge [X,Y] =0$.
My question is: is there a natural differential 2-form $\mu$ which is again preserved by $X$, and can be obtained by the information introduced above?
It would be enough for me to have $\mu$ defined in local coordinates.
Posting from the comments: assuming $X$ and $Y$ are linearly independent everywhere on $\mathbb{T}^2$, there exist dual forms $\alpha, \beta$ (satisfying $\alpha(X) = 1, \alpha(Y)=0$ etc.) One easily shows $\alpha, \beta$ are invariant under the flows of $X,Y$. For example $$ 0 = \mathcal{L}_X(0)=\mathcal{L}_X(\alpha(Y)) = (\mathcal{L}_X\alpha)(Y)+ \alpha(\mathcal{L}_XY) = (\mathcal{L}_X\alpha)(Y) $$ and similarly $(\mathcal{L}_X\alpha)(X)=0$, implying that $\mathcal{L}_X\alpha = 0$.
If $x, y$ denote the angular coordinates on $\mathbb{T}^2$, then locally $$ \begin{pmatrix}X \\ Y \end{pmatrix} = \begin{pmatrix}X^1 & X^2\\ Y^1 & Y^2 \end{pmatrix}\begin{pmatrix}\partial_1 \\ \partial_2\end{pmatrix} \implies \begin{pmatrix}\alpha \\ \beta\end{pmatrix} = \frac{1}{X^1Y^2-X^2Y^1}\begin{pmatrix} Y^2 & -Y^1 \\ -X^2 & X^1 \end{pmatrix}\begin{pmatrix}dx \\ dy\end{pmatrix} $$ Then $\mu = \alpha\wedge\beta = \frac{1}{X^1Y^2-X^2Y^1} dx\wedge dy$ is invariant under the flows of $X$ and $Y$.