# Is there a $4$-manifold which Immerses in $\mathbb{R}^6$ but doesn't Embed in $\mathbb{R}^7$?

I'm interested in both version of the question in the title, i.e. in the topological category and in the smooth category. By a topological immersion I mean a local embedding. I was asking in relation to this question:

https://math.stackexchange.com/questions/1801318/dimensions-of-immersions-vs-embeddings

In the very nice answer given in that thread, they work out almost all of the low-dimensional cases for smooth, compact manifolds and smooth immersions/embeddings. The only 'smooth, compact' case not covered by their answer is the one in the title of this question, so it would be interesting to know if there is a compact example, separately.

To be honest I'm more curious about the topological case, though. To what extent is there a topological version of the Cohen Immersion Theorem? Also, can someone clarify their answer wrt the constant in the Cohen theorem? They say it's $$2n - a(n) -1$$; is that an improvement for the compact case? If so then any smooth, compact manifold immerses in $$\mathbb{R}^6$$ so the Massey example suffices. But still curious about the non-compact case.

EDIT: A nice answer below completes the compact smooth case. For general $$4$$-manifolds the case of immersion in $$\mathbb{R}^5$$ is what remains:

Is every compact $$4$$-manifold which immerses in $$\mathbb{R}^5$$ smoothable?

Is there a $$4$$-manifold that immerses in $$\mathbb{R}^5$$ but doesn't embed in $$\mathbb{R}^6$$ (resp. $$\mathbb{R}^7$$)?

By results of Quinn, every open $$4$$-manifold is smoothable so it's sufficient to prove the smooth case for non-compact manifolds.

• Does $\mathbb{R}P^2\times \mathbb{R}P^2$ embed in $\mathbb{R}^7$? Also, Cohen states the general upper bound without the "$-1$" in the exponent. – Mark Grant Apr 22 at 21:13
• @MarkGrant That's my understanding of the Cohen result as well, but maybe it was improved at some point? I know that your manifold has non-vanishing Stiefel-Whitney classes in all dimensions; is that sufficient to imply that it doesn't embed in $\mathbb{R}^7$? That was my understanding; I hadn't realized it -obviously- immerses in $\mathbb{R}^6$ haha, oops! – John Samples Apr 22 at 22:11
• I'm still curious about the topological version, though. Also about the non-compact version for immersions into $\mathbb{R}^5$; does that imply embeddability in $\mathbb{R}^7$? – John Samples Apr 22 at 22:14
• My understanding is that arguments with Stiefel-Whitney classes can only give non-immersion results. I honestly don't know the answer to my question, but I'd be surprised if it's unkonwn. The Cohen bound is best possible already for surfaces (closed or compact with boundary). I'm afraid I don't know any results about immersions and embeddings for non-closed manifolds, or the topological case. – Mark Grant Apr 23 at 6:46
• Possibly relevant for the topological case is that $4$-manifolds are smoothable off a single point. But I'm not sure how to use that. – John Samples Apr 23 at 9:15

The manifold $$\mathbb{R}P^2\times \mathbb{R}P^2$$ smoothly immerses in $$\mathbb{R}^6$$, as a product of Boy's surfaces. However, the main result of
asserts that a closed smooth $$4$$-manifold $$M$$ embeds in $$\mathbb{R}^7$$ if and only if the normal Stiefel-Whitney class $$\bar{w}_3(M)$$ vanishes. A quick calculation shows that this is not the case for $$M=\mathbb{R}P^2\times \mathbb{R}P^2$$, which therefore doesn't embed in $$\mathbb{R}^7$$.
The cited article also gives necessary and sufficient conditions for topological embeddability of $$4$$-manifolds in $$\mathbb{R}^7$$. For example, a closed smooth non-orientable $$4$$-manifold $$M$$ admits a locally flat embedding in $$\mathbb{R}^7$$ if and only if $$\bar{w}_3(M)=0$$ and $$KS(M)=0$$, where $$KS$$ is the Kirby-Siebenmann invariant.
• Oh nice! Ok, then the compact smooth situation is resolved. I guess the only cases are what can happen with $4$-manifolds that immerse in $\mathbb{R}^5$, either the non-compact or non-smooth case. – John Samples Apr 23 at 8:56