Let $M$, $N$ be compact, connected, oriented manifolds without boundary embedded in $\mathbb{R}^m$ of dimensions $p$ and $q$ respectively. I know that when $m=p+q+1$ we can define the linking between $M$ and $N$ in the sense given [here](https://math.stackexchange.com/questions/1653975/relation-between-alexander-duality-and-linking-numbers), but what if $M$ and $N$ do not satisfy this constraint? 


Example: Can we 'link' two surfaces in $\mathbb{R}^3$? What is the definition of linking in this case ?



Notion of linking between two general $p$ and $q$ manifolds embedded in a higher dimensional manifold
Sure, there are ways to make sense of linking beyond the constraints you mention. 


I suppose the most basic would be: two disjoint submanifolds $A$ and $B$ of a manifold $M$ are **unlinked** if you can find disjoint embedded $m$-dimensional discs $D\_1, D\_2 \subset M$ such that $A \subset D\_1$ and $B \subset D\_2$. Here $m=dim(M)$. 


The manifolds would be **linked** if they are not **unlinked**. 


A nice feature of this definition is you can talk about linking of three or more objects, so you can similarly talk about Brunnian properties, i.e. $k$ objects can be linked but any collection of $k-1$ sub-objects are unlinked.


For example, there is a classical theorem of Schubert that says if you have two (or more) disjoint non-trivial knot exteriors in $S^3$, then they are unlinked. By knot exterior I mean a compact 3-manifold diffeomorphic to the complement of an open tubular neighbourhood of a knot in $S^3$. As you likely know, if the knot exterior is trivial it would be diffeomorphic to $S^1 \times D^2$, and such objects can be linked in $S^3$. 


I believe this is a commonly-used notion of linking that generalizes the definition you mention. Using this definition, linking of surfaces in $\mathbb R^3$ or $S^3$ is a fairly easy definition to work with, once you are comfortable with the tools of 3-manifold theory. 


In your question you refer to "the linking between $M$ and $N$" but really you are talking about **linking numbers**. Interesting numerical linking invariants can take a fair bit of work to define, and John Klein's answer is directed towards that kind of question. But if all you care about is what the term "linking" means, that is much more elementary, and what my answer is directed towards. 


**edit:** here are more details to support my comment below. This argument is typical of Hassler Whitney's work. I'm not certain if he ever made this specific argument, but the general mode of argument very much is due to him. 


For convenience, let's say $M$ and $N$ are compact submanifolds of $\mathbb R^k$. We can get by without this assumption, but it makes the arguments less technical. 


By compactness, there is some vector $v \in \mathbb R^k$ so that $M+v$ is disjoint from $N$. Consider the function


$$F: [0,1] \times M \to [0,1]\times \mathbb R^k$$


given by $F(t,p) = (t, tv+p)$. 


Thus $F(0,M) = \{0\} \times M$ and $F(1,M) = \{1\} \times (v+M)$. 


So $F([0,1] \times M)$ is disjoint from $[0,1] \times N$ at the start and the end of the interval, but maybe not in the interior. 


We can perturb $F$ a little bit (not affecting its values at the endpoints) to make it transverse to $[0,1]\times N$. Provided $k+1 > (m+1)+(n+1)$ the transverse map would be disjoint from $[0,1]\times N$, i.e. $k \geq m+n+2$. 


Since the transversal perturbation can be made arbitrarily small in the $C^\infty$-topology, we can assume it represents an isotopy of $M$. Thus $M$ can be isotoped to be separable from $N$ by disjoint embedded discs. So by isotopy-extension, $M$ and $N$ can be separated by disjoint embedded discs. 


