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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.

As a side-question, for topological $4$-manifolds the case of immersion in $\mathbb{R}^5$ is what remains to be fleshed out in the answer to that previous thread. Some relevant questions for that case:

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.

I cross-listed one of these on MSE expecting a quick counterexample, but still no takers. Someone gave a partial answer to the effect of "in high dimensions, codimension-$1$ locally flat immersion implies smoothability":

https://math.stackexchange.com/questions/4115222/is-every-compact-n-manifold-that-immerses-in-mathbbrn1-smoothable

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    $\begingroup$ 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. $\endgroup$
    – Mark Grant
    Apr 22, 2021 at 21:13
  • $\begingroup$ @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! $\endgroup$ Apr 22, 2021 at 22:11
  • $\begingroup$ 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 $\mathbb{R}^7$? $\endgroup$ Apr 22, 2021 at 22:14
  • $\begingroup$ 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. $\endgroup$
    – Mark Grant
    Apr 23, 2021 at 6:46
  • $\begingroup$ 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. $\endgroup$ Apr 23, 2021 at 9:15

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Edit: The answer below is incorrect. In fact, $\bar{w}_3(\mathbb{R}P^2\times\mathbb{R}P^2)=0$ (thanks to Rafal Walczak for pointing this out) so by the cited result $\mathbb{R}P^2\times\mathbb{R}P^2$ does embed in $\mathbb{R}^7$! So please strip me of my points and consider the question unanswered...

Edit 2: In fact what the below shows is that the answer to the question in the title is no in the smooth compact case. If $M^4$ immerses in $\mathbb{R}^6$ then $\bar{w}_3(M)=0$, which by the quoted result of Fang implies that $M$ embeds in $\mathbb{R}^7$.


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

Fang, Fuquan, Embedding four manifolds in (\mathbb{R}^ 7), Topology 33, No. 3, 447-454 (1994). ZBL0824.57014

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.

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    $\begingroup$ 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. $\endgroup$ Apr 23, 2021 at 8:56
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    $\begingroup$ RE the edit, well, dang! But that's ok, maybe it will draw more attention from the big guns on here! $\endgroup$ Jun 10, 2021 at 5:10

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