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.

>Are there  $k$ differential $m$- forms $\alpha_1, \alpha_2,\ldots, \alpha_k$ with $\int_{M_{i}} \alpha_{j}= \lambda{ij}$?


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.