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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.

Dadeslam
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    Is there any assumption that $X$ and $Y$ don't vanish at any point? If so, the answer is fairly straightforward: take dual 1-forms $\alpha, \beta$ (so satisfying $\alpha(X) = 1, \alpha(Y)=0$ etc.). Then it's not too difficult to show that $\alpha, \beta$ are invariant under the flows of $X$ and $Y$, and hence so is $\mu = \alpha\wedge \beta$. – user17945 Jul 20 '21 at 13:54
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    Thanks @user17945 , $X$ is assumed to be different from zero on any point of $\mathbb{T}^2$, and $X\wedge Y$ is assumed to be always different from 0 too. Thus what I expect, in local coordinates, is that something of the form $(X^1Y^2-X^2Y^1)dx\wedge dy$ is conserved, where $X = X^1\partial_x + X^2\partial_y$, $Y=Y^1\partial_x + Y^2\partial_y$, and $x,y\in S^1$ are the angular coordinates on $\mathbb{T}^2$. Do you think this is correct? – Dadeslam Jul 20 '21 at 13:59
  • Oh, I should have also mentioned that one requires $X, Y$ to be linearly independent (which is your condition that $X\wedge Y \ne 0$). You're close - in local coordinates, the conserved 2-form is actually $1/(X^1Y^2-X^2Y^1)dx\wedge dy$. You can prove this directly (using the coordinate form of $[X,Y]=0$), or on general grounds, using the method I suggested above. Specifically, suppose $e_i = \sum_ja_{ij} \partial_j$ are vector fields, and $\theta_i = \sum_j b_{ij}dx^j$ are the dual 1-forms. Then it's not too difficult to show that $b = (a^{-1})^T$ (cont.) – user17945 Jul 20 '21 at 14:39
  • In your case, $a = \begin{pmatrix} X^1 & X^2 \ Y^1 & Y^2\end{pmatrix}$ (implying $X\wedge Y = \det(a)(\partial_1\wedge \partial_2)$. Then $\alpha\wedge \beta = \det(b), dx\wedge dy = \frac{1}{\det(a)} dx\wedge dy$, as stated. – user17945 Jul 20 '21 at 14:41
  • Thank you @user17945 , I will workout the details and see if it gets more clear. If you copy these comments as an answer I will confirm it – Dadeslam Jul 20 '21 at 14:42

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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 $Y$.

user17945
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  • Thank you very much for the answer..one thing which still remains unclear to me is how you generate $\alpha,\beta$. Could you please send me a reference? Are they the interior product of $dx\wedge dy$ along respectively $X$ and $Y$? – Dadeslam Jul 20 '21 at 17:05
  • If you search for "dual basis", you will a lot of discussion of this. For example, see the question and answers here. – user17945 Jul 20 '21 at 17:11
  • Ok perfect, so now it is clear to me what you mean. Thank you again for the clear answer. – Dadeslam Jul 20 '21 at 17:12
  • You're welcome. – user17945 Jul 20 '21 at 17:13