Assume that $M\_1, M\_2, \ldots , M\_k$ are $k$ disjoint compact submanifolds of $\mathbb{R}^n$ of the same dimension $m$. Assume that $\lambda\_{ij}, \; 1\leq i,j\leq k$ are $k^2$ arbitrary real numbers.



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> Are there $k$ differential $m$- forms $\alpha\_1, \alpha\_2,\ldots, \alpha\_k$ with $\int\_{M\_{i}} \alpha\_{j}= \lambda{ij}$?
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The motivation for this post is that a positive answer to the above question implies that
the number of closed orbits of a planar vector field $X$ is less than the codimension of the range of the following linear operator:


$$L\_X:\Omega^1(\mathbb{R}^2)\to \Omega^1(\mathbb{R}^2) $$


Because if $\gamma$ is a closed orbit of $X$ then for every $\alpha \in \Omega^1(\mathbb{R}^2)$ we have $$\int\_{\gamma} L\_{X}\alpha =0 $$


Now if $\gamma\_1, \gamma\_2, \ldots, \gamma\_n$ are closed orbits of $X$ and we have $n$ elements $\alpha\_1, \alpha\_2,\ldots, \alpha\_n$ of $\Omega^1(\mathbb{R}^2)$ such that the matrix $(\int\_{\gamma\_{i}} \alpha\_{j})\_{i,j}$ is an invertible matrix, then no nontrivial linear combination $\sum c\_i \alpha\_{i}$ belongs to the image of the operator $L\_{X}$. This shows that the codimension of the range of $L\_X$ is more than the number of closed orbits.



Integral Separation of disjoint submanifolds of $\mathbb{R}^n$
In order for integration of differential forms to make sense, we need $M\_1,\dotsc,M\_k$ to be oriented. Let $\omega\_i\in\Omega^m(M\_i)$ be such that $\int\_{M\_i}\omega\_i = 1$.


Since $M\_1,\dotsc,M\_k$ are pairwise disjoint, compact subsets of $\mathbb R^n$, then they have open neighborhoods $U\_1\supset M\_1,\dotsc, U\_k\supset M\_k$ such that the $U\_i$s are pairwise disjoint (see [this Math.SE post](https://math.stackexchange.com/questions/575640)). Let $h\_i$ be a smooth function supported in $U\_i$ and equal to 1 on $M\_i$; it's [a theorem](https://en.wikipedia.org/wiki/Bump_function#Existence_of_bump_functions) that these functions exist.


The $\alpha\_j$ in question is
$$\alpha\_j := \sum\_{\ell=1}^k \lambda\_{\ell j} h\_\ell\omega\_\ell.$$


When restricted to $M\_i$, $h\_i = 1$ and $h\_j = 0$ if $j\ne i$, so $\alpha\_j|\_{M\_i} = \lambda\_{ij}\omega\_i$ and
$$\int\_{M\_i} \alpha\_j = \lambda\_{ij}.$$


