Given $k \leq n$, is there a general method that can help us determine whether a given $k-$ dimensional smooth manifold $M$ (let's assume $M$ has empty boundary) can be embedded in $\mathbb{R}^n$?
I know of some ways to know this in a couple of (very particular) cases:
$1)$ If $k = n$ and $M$ is compact, it's never possible (we would have $f(M) = \mathbb{R}^n $, since the embedding must be open in this case).
$2) $If $k = n-1$, then in order for this to be possible $M$ must be orientable (see the answer by Georges Elencwajg to this question).
But of course these are very limited (for example, using them I can't even refute the statement 'every $k-$dimensional manifold embeds into $\mathbb{R}^{k+2}$').
Are there other techniques that can help in more general cases? I'd appreciate any partial answers. Thanks in advance.
EDIT: I also know about Whitney's embedding theorem. The motivation for posting this question came from thinking about why a sharper bound can't be achieved.
Here is more details. Let $E \to X$ a vector bundle where $X$ is a manifold. The total Steifel-Whitney class of $E$ is an element $w(E) = 1 + w_1(E) + \dots \in H^*(X, \Bbb Z/2 \Bbb Z)$ where $w_i(E) \in H^i(X)$ such that :
Naturality (i.e $f^*w_i(E) = w_i(f^*(E)$) implies that the trivial bundle $V \times X$ has $w(V \times X) = 1$ as $V \times X = f^*V$ where $f : X \to pt$ is the constant map.
This observation and the fact that for a manifold $M \subset \Bbb R^n$ one has $TM \oplus NM \cong \Bbb R^n$ gives $w(M)w(NM) = 1$ by the second axiom, where $w(M) := w(TM)$ and $NM$ is the normal bundle of $M$. If $r$ is the last index where $w_r(TM) \neq 0$ and $s$ the last index where $w_s(NM) \neq 0$ this implies by the rank axiom that $n \geq r + s$.
As an example, one can use Euler exact sequence for compute $w(\Bbb RP^n) = (1+a)^{n+1}$ where $a$ is the generator of $H^1(\Bbb RP^n, \Bbb Z/2 \Bbb Z)$ (remember that coefficient are taken modulo $2$) for example $w(\Bbb RP^4) = (1+a)^5 =1 + a + a^4 $. Let's assume that $\Bbb RP^4$ is embedded, we have $w(\Bbb RP^4)w(N \Bbb RP^4) = 1 (\star)$.
We can solve $(\star)$ "degree by degree" : the first degree of this equation is $w_1(\Bbb RP^2) + w_1(N\Bbb RP^2) = 0$, i.e $w_1(N\Bbb RP^2) = a$. The second degree gives $w_2(N\Bbb RP^2) = w_1(N\Bbb RP^2)w_1(\Bbb RP^2) + w_2(\Bbb RP^2) = w_1(N\Bbb RP^2)w_1(\Bbb RP^2) = a^2$ and so one. This gives finally that $w(N \Bbb RP^2) = 1 + a + a^2 + a^3$, i.e that $\Bbb RP^4$ can only be embedded (in particular immersed) in $\Bbb R^7$. Indeed, for $k = 2^r$ the same conclusion holds ($\Bbb RP^{2^r}$ can only be immersed in $\Bbb RP^{2^{r+1} - 1}$ and this is the best bound by Whitney's theorem which says that any manifold of dimension $n$ can be immersed into $\Bbb R^{2n-1}$ ($n>1$ here).
Finally let me add you two references recommended by the author on this subject :