Integral Separation of disjoint submanifolds of $\mathbb{R}^n$

Question

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.

Answer 1

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). Let $h_i$ be a smooth function supported in $U_i$ and equal to 1 on $M_i$; it's a theorem 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 $$\int_{M_i} \alpha_j = \lambda_{ij}.$$