# Can an oriented closed $n(\geq 2)$-dimensional manifold be smoothly embedded in $\mathbb{R}^{2n-1}$?

Can anyone provide me with an example of an orientable closed manifold $$M$$ of dimension $$n\geq 2$$, which cannot be smoothly embedded in $$\mathbb R^{2n-1}$$?

I know these cannot exist for $$n=1$$, i.e. $$S^1$$. If we ignore orientability, then if we take $$n=2^r$$, $$\mathbb RP^n$$ cannot be embedded in $$\mathbb R^{2n-1}$$.

While searching on the internet, I found some sharper results here.

So can we say anything when $$\dim(M)= 2^r$$ and $$M$$ is orientable?

It would be very helpful if someone could provide me with some more references. Thanks in advance.

A closed smooth $n$-manifold embeds into $\mathbb R^{2n-1}$ if and only if the normal $(n-1)$th Stiefel-Whitney class vanishes. This is due to Hirsch-Haefliger in dimensions $\neq 4$ and to Fang in dimension $4$. Massey showed that if the normal $(n-1)$th Stiefel-Whitney class is nonzero, then $M$ is non-orientable and $n$ is a power of $2$. Thus any smooth closed orientable $n$-manifold smoothly embeds into $\mathbb R^{2n-1}$. For references see here and here.

• I wish to add that Hirsch-Haefliger's result assumes $n>4$. The case $n=3$ has to be handled separately, and it was treated independently by Wall (see Rivin's answer below), or by Hirsch (quoted in the Hirsch-Haefliger's paper). – Igor Belegradek Jan 1 '16 at 23:43

Every closed orientable (smooth) $n$-manifold (smoothly) embeds in $\mathbb{R}^{2n-1}.$ This is true for $n> 4$ by Haefliger-Hirsch, $n=3$ by C.T.C. Wall (Wall's paper has the reference to Haefliger-Hirsch)

Wall, C. T. C.
All 3-manifolds imbed in 5-space.
Bull. Amer. Math. Soc. 71 1965 564–567.


For $n=4$ this is true in the topological category, and is open (as far as I know) in the smooth category [which means that no counterexample is known] - see Bruno Martelli's answer to this question.

For dim $$n=4$$ it has proven recently by Ghanwat and Pancholi (https://arxiv.org/pdf/2002.11299.pdf) that every closed oriented 4-manifolds smoothly embeds in $$\mathbb R^7$$.

The (beautiful) key idea of their proof is if we have a closed oriented smooth 4-manifold $$M$$ such that there exists two smoothly embedded 2-spheres $$S^2_a, S^2_b$$ that transversally intersect at one point and represent non-trivial element in $$H_2(M)/Tor$$, then any smooth closed oriented 4-manifold $$X$$ smoothly embed in $$M\times CP^1$$. This telling us in particular $$X$$ embeds smoothly in $$S^2\times S^2\times S^2= \partial (S^2\times S^2\times D^3)$$ which embeds in $$\mathbb R^7$$.

• Perhaps "reproven" instead of "proven" in the first sentence? (as the authors themselves say they are giving a "new proof" of this, as a corollary of their arguments to prove that every closed orientable smooth four-manifold embeds into $CP^3$) – Aleksandar Milivojevic Jun 30 '20 at 1:07