Dimensions of immersions vs embeddings

Question

Let's say that you have a manifold which you know can be immersed in $\mathbb{R}^n$. Is there a $k$ such that you can say, for sure, that the manifold is embedded in $\mathbb{R}^{n+k}$? I imagine that there is and that this is common knowmedge, but cursory googling did not throw up anything. Thank you in advance!!

Answer 1

This is nowhere near an answer. I'm just throwing some thoughts out there. Hopefully this inspires others to write down better results.

I assume the domain is closed. We can always take $k \leq n-2$. One path at attempting to get lower bounds at $k$ is to find manifolds that immerse into small-dimensional spaces but embed only in large-dimensional spaces. As an attempt at this, a parallelizable $n$-manifold immerses by Smale-Hirsch immersion theory into $\Bbb R^{n+1}$; how high of a dimension can you force for the embedding? It is conjectured that every parallelizable $n$-manifold embeds into $\Bbb R^{3n/2}$, so certainly no known examples can beat $k = (n-3)/2$ via this approach. I bet but have not made much effort to find parallelizable manifolds that achieve (roughly, at least) this bound.

Real projective spaces probably give nice bounds on $k$. Complex projective spaces do not.

Another approach is to use Cohen's immersion theorem, which says that every $n$-manifold immerses into $\Bbb R^{2n-\alpha(n)-1}$, where $\alpha(n)$ is the number of 1s in the binary expansion of $n$; this is not a particularly fruitful line of attack, because the best we can get from this is $k \geq \alpha(n)+1$. This is pitifully small, much smaller than the approach above.

Now for some small-dimensional examples. For $n=1, 2$ the answer is obviously $k=0$. For $n=3$ all we have are surfaces, which all embed in $\Bbb R^4$, so $k(3)=1$. For $n=4$ we have 3-manifolds, which famously all embed in $\Bbb R^5$. So $k(4) = 1$.

Suppose a 4-manifold immerses in $\Bbb R^5$. Then if $M$ is orientable, we see that its tangent bundle has trivial characteristic classes, including its Euler class and (most importantly!) Pontryagin class. (Normally one could only say its Pontryagin class is 2-torsion, but $H^4(M;\Bbb Z)$ does not have torsion.) This implies that its signature is zero, and Cappell and Shaneson proved that it embeds smoothly in $\Bbb R^6$. Danny Ruberman gives a proof here. For the non-orientable case, a simple Stiefel-Whitney class manipulation shows that (if $\nu$ is the normal bundle to the immersion in $\Bbb R^5$) that $w_1(\nu) = w_1(M)$ and $w_2(M) = w_1(M)^2$. This implies that the stable normal bundle of $M$ has $w_2(\nu) = 0$ by yet more algebraic manipulation. Now it is likely that such a manifold embeds in $\Bbb R^6$, but this does not appear to be known: see some partial progress in this paper of Fang. In particular we can safely conjecture that $k(5) = 1$.

It is not hard to find examples of 5-manifolds that immerse in $\Bbb R^6$ but only embed into $\Bbb R^8$. I find it likely that $k(6) = 3$, but have not put much effort into writing down or finding a proof.

Answer 2

Whitney's theorem state that: in $M$ has dimension $n$, then it can be embedded in $\mathbb{R}^{2n+1}$. A stronger version ensure that $M$ can be embedded in $\mathbb{R}^{2n}$.